arrow-left

Only this pageAll pages
gitbookPowered by GitBook
1 of 80

CryptoBook

Loading...

Loading...

Loading...

Loading...

Loading...

Fundamentals

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Number Theory

Loading...

Loading...

Loading...

Loading...

Loading...

Abstract algebra

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Elliptic Curves

Loading...

Lattices

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Asymmetric Cryptography

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Loading...

Symmetric Cryptography

Loading...

Loading...

Loading...

Loading...

Loading...

Hashes

Loading...

Loading...

Isogeny Based Cryptography

Loading...

Loading...

Loading...

Appendices

Loading...

Loading...

Sample Page

A rough guideline to a page

hashtag
Introduction

Give a description of the topic, and what you hope the reader will get from this. For example, this page will cover addition of the natural numbers. Talk about how this relates to something in cryptography, either through a protocol, or an attack. This can be a single sentence, or verbose.

hashtag
Laws of Addition

For all integers, the addition operation is

  • Associative:

  • Commutative:

  • Distributive:

hashtag
Interesting Identity

hashtag
Sage Example

hashtag
Further Resources

  • Links to

  • Other interesting

  • Resources

hashtag

Modular Arithmetic

Authors: A~Z, perhaps someone else but not yet (or they've decided to remain hidden like a ninja)

hashtag
Introduction

Thinking not over the integers as a whole but modulo some integerinstead can prove quite useful in a number of situation. This chapter attempts to introduce to you the basic concepts of working in such a context.

Mathematical Notation

hashtag
Introduction

Throughout CryptoBook, discussions are made more concise by using various mathematical symbols. For some of you, all of these will feel familiar, while for others, it will feel new and confusing. This chapter is devoted to helping new readers gain insight into the notation used.

circle-info

CryptoBook

hashtag
About this project

CryptoBook is a community project, developed by members of to create a resource for people to learn cryptography. The focus of this project is to create a friendly resource for the mathematical fundamentals of cryptography, along with corresponding implementation.

Think of this as an alternative SageMath documentation source, where the focus is its application in solving cryptographic problems.

If you're interested in contributing, come chat to us in our

CryptoHackarrow-up-right
SageMatharrow-up-right
Discord Channelarrow-up-right

Contains an identity element: a+0=0+a=aa + 0 = 0 + a = aa+0=0+a=a

  • Has an inverse for every element: a+(−a)=(−a)+a=0a + (-a) = (-a) + a = 0a+(−a)=(−a)+a=0

  • Closed: ∀a,b∈Z,a+b∈Z\forall a, b \in \mathbb{Z}, a + b \in \mathbb{Z}∀a,b∈Z,a+b∈Z

  • a+(b+c)=(a+b)+ca + (b + c) = (a + b) + ca+(b+c)=(a+b)+c
    a+b=b+aa + b = b + aa+b=b+a
    a(b+c)=ab+aca(b + c) = ab + aca(b+c)=ab+ac
    (1+2+3+…+n)2=13+23+33+…+n3(1 + 2 + 3 + \ldots + n)^2 = 1^3 + 2^3 + 3^3 + \ldots + n^3(1+2+3+…+n)2=13+23+33+…+n3
    sage: 1 + (2 + 3) == (1 + 2) + 3
    True
    sage: 1 + 2 == 2 + 1
    True
    sage: 5*(7 + 11) == 5*7 + 5*11
    True
    sage: sum(i for i in range(1000))^2 == sum(i^3 for i in range(1000))
    True
    hashtag
    Congruences

    For the following chapter, we will assumennnis a natural integer, andaaaandbbbare two integers. We say thataaaandbbbare congruent modulonnnwhenn∣(b−a)n\mid (b-a)n∣(b−a), or equivalently when there is an integerkkksuch thata=b+kna=b+kna=b+kn. We denote this bya≡b [n]a\equiv b~ [n]a≡b [n]or a≡bmod  na \equiv b\mod na≡bmodn. I will use the first notation throughout this chapter.

    Remark: Whenb≠0b\neq0b=0, we havea≡r [b]a\equiv r~[b]a≡r [b], whererrris the remainder in the euclidean division ofaaaby

    This relation has a number of useful properties:

    • ∀c∈Z,a≡b [n]  ⟹  ac≡bc [n]\forall c\in \mathbb Z, a\equiv b~[n] \implies ac \equiv bc ~ [n]∀c∈Z,a≡b [n]⟹ac≡bc [n]

    • ∀c∈Z,a≡b [n]  ⟹  a+c≡b+c [n]\forall c \in \mathbb Z, a\equiv b~[n] \implies a+c\equiv b+c ~[n]∀c∈Z,a≡b [n]⟹a+c≡b+c [n]

    • ∀c∈Z,a≡b [n] and b≡c [n]  ⟹  a≡c [n]\forall c \in \mathbb Z, a \equiv b ~[n] \text{ and } b\equiv c~[n]\implies a\equiv c ~[n]∀c∈Z,a≡b [n] and b≡c [n]⟹a≡c [n]

    • The proofs are left as an exercise to the reader :p (Hint: go back to the definition)

    Seeing as addition and multiplication are well defined, the integers modulonnnform a ring, which we noteZ/nZ\mathbb Z/n\mathbb ZZ/nZ. In sage, you can construct such ring with either of the following

    Powering modulonnnis relatively fast, thanks to the double-and-squarearrow-up-right algorithm, so we needn't worry about it taking too much time when working with high powers

    As a side note, remember that if an equality holds over the integers, then it holds modulo any natural integernnn. This can be used to prove that a relation is never true by finding a suitable modulus, or to derive conditions on the potential solutions of the equation.

    Example: by choosing an appropriate modulus, show that not even god is able to find integersaaaandbbbsuch thata2=2+4ba^2 = 2 + 4ba2=2+4b

    hashtag
    Modular Inverse

    Since we can multiply, a question arises: can we divide? The answer is yes, under certain conditions. Dividing by an integercccis the same as multiplying by its inverse; that is we want to find another integerdddsuch thatcd≡1 [n]cd\equiv 1~[n]cd≡1 [n]. Sincecd≡1 [n]  ⟺  ∃k∈Z,cd=1+kncd\equiv 1~[n]\iff\exists k\in\mathbb Z, cd = 1 + kncd≡1 [n]⟺∃k∈Z,cd=1+kn, it is clear from Bézout's Identityarrow-up-right that such an inverse exists if and only ifgcd⁡(c,n)=1\gcd(c, n) = 1gcd(c,n)=1. Therefore, the units modulonnnare the integers coprime tonnn, lying in a set we call the unit group modulonnn: (Z/nZ)×\left(\mathbb Z/n\mathbb Z\right)^\times(Z/nZ)×

    Finding the modular inverse of a number is an easy task, thanks to the extended euclidean algorithmarrow-up-right (that outputs solutions indddandkkkto the equationcd−kn=1cd-kn=1cd−kn=1from above).

    nnn
    If you're reading a page and something is new to you, come here and add the symbol, someone else who understands it can explain its meaning

    hashtag
    Mathematical Objects

    hashtag
    Special Sets

    • C\mathbb{C}C: denotes the set of complex numbers

    • R\mathbb{R}R: denotes the set of real numbers

    • Z\mathbb{Z}Z: denotes the set of integers

    • : denotes the set of rational numbers

    • : denotes the set of natural numbers (non-negative integers)

    • : denotes the set of integers mod

    • We refer to unit groups by R×R^\timesR× or R∗R^*R∗. Example: (Z/nZ)×(\mathbb Z/n \mathbb Z)^\times(Z/nZ)×

    • We refer to finite fields with qqq elements by Fq\mathbb{F}_qFq​

    • We refer to a general field by kkk

    • We refer to the algebraic closure of this field by

    hashtag
    Relation operators

    • ∈\in∈means is an element of (belongs to)

    hashtag
    Logical Notation

    • ∀\forall∀means for all

    • ∃\exists∃means there exists. ∃!\exists!∃! means uniquely exists

    hashtag
    Operators

    • Pr(A)Pr(A)Pr(A) means the probability of an event AAAto happen. Sometimes denoted as Pr[A]Pr[A]Pr[A]or as P(A)P(A)P(A)

    Contributors

    Optional space to say that you've worked on the book

    hashtag
    Thank you!

    circle-check

    🥳 CryptoBook is the result of the hard work of the CryptoHack community. Thanks to all of our writers, whether it's been line edits or whole section creation. This only exists because of the generosity and passion of a group of cryptographers.

    hashtag
    Our Writers

    • ...

    • You?

    hashtag
    Join CryptoBook

    If you would like to join the team, come over to our and talk with us about your ideas

    Book Plan

    A summary we plan to cover

    hashtag
    Philosophy

    The aim of CryptoBook is to have a consolidated space for all of the mathematics required to properly learn and enjoy cryptography. The focus of any topic should be to introduce a reader to a subject in a way that is fun, engaging and with an attempt to frame it as an applied resource.

    The second focus should be to cleanly implement the various topics using SageMath, so that there is a clear resource for a new reader to gain insight on how SageMath might be used to create the objects needed.

    circle-check

    Write about what you love and this book will be a success.

    circle-info

    Descriptions of attacks against cryptosystems are strongly encouraged, however full SageMath implementations should not be included, as this has the potential for destroying CryptoHack challenges, or making all attacks known by so many people that CTFs become a total nightmare!!

    hashtag
    Proposed topics

    This list is not complete so please add to it as you see fit.

    hashtag
    Mathematical Background

    hashtag
    Fundamentals

    • Congruences

    • GCD, LCM

      • Bézout's Theorem

    hashtag
    Number Theory

    Mainly thinking things like

    • Prime decomposition and distribution

    • Primality testing

    • Euler's theorem

    hashtag
    Abstract Algebra

    Mainly thinking things like:

    • Groups, Rings, Fields, etc.

    • Abelian groups and their relationship to key-exchange

    • Lagrange's theorem and small subgroup attacks

    hashtag
    Basilar Cryptanalysis forms

    • Introduction to Cryptanalysis

    • A linear Approach to Cryptanalysis

    • Matsui's Best biases algorithm

    hashtag
    Elliptic Curves

    • Weierstrass

    • Montgomery

    • Edwards

    Generating Elliptic Curves

    • Generating curves of prime order

    • Generating supersingular curves

    hashtag
    Hyperelliptic curves

    • Generalization of elliptic curves

    • Recovering a group structure using the Jacobian

    • Example: genus one curves, jacobian is isomorphic to the set of points

    hashtag
    Security background

    • Basic Concepts

      • Confidentiality, Integrity etc

      • Encryption, Key generation

    hashtag
    Asymmetric Cryptography

    hashtag
    RSA

    • Textbook protocol

    • Padding

      • Bleichenbacher's Attack

    hashtag
    Paillier Cryptosystem

    • Textbook protocol

    hashtag
    ElGamal Encryption System

    • Textbook protocol

    • ElGamal Digital Signature Scheme

    hashtag
    Diffie-Hellman

    • Textbook protocol

    • Strong primes, and why

    hashtag
    Elliptic Curve Cryptography

    • ECDSA

    • EdDSA

    hashtag
    Symmetric Cryptography

    hashtag
    One Time Pad

    • XOR and its properties

    • XOR as One Time Pad

    • Generalized One Time Pad

    Block Ciphers

    • AES

    Stream Ciphers

    • Affine

    • RC4

    hashtag
    Hashes

    • Introduction

    • Trapdoor Functions

    • MD family

    hashtag
    Isogeny Based Cryptography

    • Isogenies

    • Isogeny graphs

    • Torsion poins

    hashtag
    Cryptographic Protocols

    hashtag
    Zero-knowledge proofs

    • Schnorr proof of knowledge for dlog

    • Core definitions

    • Proof of equality of dlog

    hashtag
    Formal Verification of Security Protocols

    • Definition of Formal Verification

    • Uses of Formal Verification

    • Handshake protocols, flawed protocols

    hashtag
    Usefull Resources ( Books, articles ..) // based on my material

    • Cryptanalytic Attacks on RSA (Yan, Springer, 2008)

    • Algorithmic Cryptanalysis (Antoine Joux, CRC Press, 2009)

    • Algebraic Cryptanalysis (Brad, Springer, 2009)

    Theorems of Wilson, Euler, and Fermat

    hashtag
    Wilson's Theorem

    A positive integer is a prime if and only if:

    hashtag

    Style Guide

    Work in progress

    hashtag
    Working together

    • If something doesn't make sense, make a comment using GitBook, or ask in the Discord.

    Zn = Zmod(5)
    Zn = Integers(5)
    Zn = IntegerModRing(5)
    # Ring of integers modulo 5
    Zn(7)
    # 2
    Zn(8) == Zn(13)
    # True
    pow(2, 564654533, 7) # Output result as member of Z/7Z
    # 4
    power_mod(987654321, 987654321, 7) # Output result as simple integer
    # 6
    Zmod(7)(84564685)^(2^100) # ^ stands for powering in sage. To get XOR, use ^^.
    # 5
    Zn = Zmod(10)
    Zn(7).is_unit()
    # True
    Zn(8).is_unit()
    # False
    3 == 1/Zn(7) == Zn(7)^(-1) == pow(7,-1,10) # member of Z/10Z
    # True
    inverse_mod(7, 10) # simple integer
    # 3
    Zn(3)/7
    # 9
    Zn(3)/8
    # ZeroDivisionError: inverse of Mod(8, 10) does not exist
    Zn.unit_group()
    # Multiplicative Abelian group isomorphic to C4 (C4 being the cyclic group of order 4)
    xgcd(7, 10) # find (gcd(a, b), u, v) in au + bv = gcd(a, b)
    # (1, 3, -2) <-- (gcd(7, 10), d, -k)
    """
    We can call each of these sets with Sage using the 
    following commands. Comments are the result of the
    input.
    """
    CC
    # Complex Field with 53 bits of precision
    RR
    # Real Field with 53 bits of precision
    ZZ
    # Integer Ring
    QQ
    # Rational Field
    NN
    # Non negative integer semiring
    Zmod(11) # or `Integers(11)` or `IntegerModRing(11)` 
    # Ring of integers modulo 11
    """
    Example of defining a field and then its 
    algebraic closure
    """
    GF(3)
    # Finite Field of size 3 , where GF stands for Galois Field 
    GF(3).algebraic_closure()
    # Algebraic closure of Finite Field of size 3
    """
    If you want to find which field an element belongs to you can use the 
    `.parent()` function
    """
    
    x = 7
    print(x.parent())
    # Integer Ring
    
    y = 3.5
    print(y.parent())
    # Real Field with 53 bits of precision
    """
    If you want to "lift" an element from a quotient ring R/I to the ring R
    use the `.lift()` function
    """
    R = ZZ
    RI = Zmod(11)
    x =  RI(5)
    
    print(x.parent())
    # Ring of integers modulo 11
    
    y = x.lift()
    print(y.parent())
    # Integer Ring
    
    print(y in R)
    # True
    ∀m∈N,a≡b [n]  ⟹  am≡bm [n]\forall m \in \mathbb N, a\equiv b~[n] \implies a^m\equiv b^m ~[n]∀m∈N,a≡b [n]⟹am≡bm [n]
    Q\mathbb{Q}Q
    N\mathbb{N}N
    Z/nZ\mathbb{Z}/n\mathbb ZZ/nZ
    nnn
    kˉ\bar{k}kˉ
    Discord Channelarrow-up-right

    Gauss' Lemma and its ten thousand corollaries

  • Euclid's algorithm

  • Modular Arithmetic

  • Morphisms et al.

  • Frobenius endomorphism

  • Factoring
  • Legendre / Jacobi symbol

  • A Differential Approach to Cryptanalysis
    Counting points (Schoof's algorithm)
  • Complex multiplication

    • Good reference, thanks Joachimarrow-up-right

  • Generating non-supersinular curves of low embedding degreearrow-up-right
  • Generating curves of arbitary order (hard)

    • Thesis on the topicarrow-up-right

    • Sage implementation ChiCube's scriptarrow-up-right

  • Mumford representation of divisors
  • Computing the order of the Jacobian

    • For characteristic 2^n: Example 56arrow-up-right

    • Hyper Metroid example

  • Attacker goals + Attack games

  • Defining Security - Perfect security, semantic security

  • Proofs of security + Security Reductions

  • OAEP

  • Coppersmith

    • Håstad's Attack

    • Franklin-Reiter Attack

  • Wiener's Attack

  • RSA's Integer fattorization Attacks

    • Fermat Factoring Attack

    • Quadratic Sieve Attack

    • Number Fielde Sieve Attack

  • RSA Digital Signature Scheme

  • Timing Attacks on RSA

  • RSA with Chinese Remainder Theorem (CRT)

    • Fault Attack on RSA-CRTarrow-up-right

    • Bellcore Attack (Low Voltage Attack)arrow-up-right

  • SHA family
  • BLAKE Hash family

  • // TODO: Insert Attacks

  • SIDH
  • SIKE

  • BIKE

  • Proof of knowledge of a group homomorphism preimage
    The external threat: Man-In-The-Middle attacks
  • Attacking the (flawed) Needham-Shroeder public key exchange protocol

  • RC4 stream Cipher and its variants (H. Rosen, CRC Press, 2013)
  • Formal Models and Techniques for Analyzing Security Protocols (Cortier, IOS Press, 2011)

  • Algebraic Shift Register Sequences (Goresky && Klapper, Cambridge Press, 2012)

  • The Modelling and Analysis of Security Protocols (Schneider, Pearson, 2000)

  • Secure Transaction Protocol Analysis (Zhang && Chen, Springer, 2008)

  • Generating Anomalous curvesarrow-up-right
    Wikipediaarrow-up-right

    Euler's Theorem in Detail

    circle-info

    Todo

    Polynomials With Shared Roots

    Algorithmic Number Theory

    • Polynomial GCD

      • Euclidean GCD

      • Half-GCD for speed when e=0x10001

      • demo application for that one RSA related message attack?

    • Resultant

      • eliminate multivariate polynomials at the expense of increasing polynomial degree

      • demo application for that one RSA Coppersmith short padding related message attack?

    • Groebner Basis

      • what if you did GCD and Resultants at the same time, like whoa

      • and what if it took forever to run!

    Quadratic Residues

    Sieves

    Pollard rho

    Euler's Theorem

    Let n∈Z+n \in \mathbb{Z}^{+}n∈Z+ and a∈Za \in \mathbb{Z}a∈Z s.t. gcd(a,n)=1gcd(a, n) = 1gcd(a,n)=1, then:

    hashtag
    Fermat's Little Theorem

    Let pppbe a prime and a∈Za \in \mathbb{Z}a∈Z, then:

    or equivalently:

    hashtag
    Reference

    1. Wilson's Theorem - Brilliantarrow-up-right

    2. Euler's Theorem - Brilliantarrow-up-right

    3. Fermat's Little Theorem - Brilliantarrow-up-right

    n>1n > 1n>1
    (n−1)!≡−1mod  n(n-1)! \equiv -1 \mod n (n−1)!≡−1modn
    aϕ(n)≡1mod  na^{\phi(n)} \equiv 1 \mod naϕ(n)≡1modn
    ap≡amod  pa^p \equiv a \mod pap≡amodp
    ap−1≡1mod  pa^{p-1} \equiv 1 \mod pap−1≡1modp

    Untitled

    If you are confident that something is wrong, just fix it. There's no need to ask.
  • If you think something doesn't have enough detail, expand on it, or leave a comment suggesting that.

  • If a page is getting too long, break it down into new pages. If you're unsure, then leave a comment or talk in the discord

  • If you want to write about something new and learn as you type, this is fine! But please leave a warning at the top that this is new to you and needs another pair of eyes.

  • If there's big subject you're working on, claim the page and save it, show us that that's what you're doing so we don't overlap too much

  • hashtag
    General Tips

    • Introduce new objects slowly, if many things need to be assumed, then try to plan for them to appear within the Book Plan somewhere.

    • It is better to cover less, and explain something well, than it is to quickly cover a lot. We're not racing

    • When explaining anything, imagine you are introducing it for a first time. Summaries exist elsewhere online, the goal of CryptoBook is education

    • Contribute as much or as little as you want, but try to only work on topics that

      • You are interested in

      • You have some experience of thinking about

    • External resources should be included at the end of the page. Ideally the book should be self-contained (within reason) but other resources are great as they offer other ways to learn

    circle-exclamation

    If anything on any page is unclear, then please leave a comment, or talk in the discord. We are all at different levels, and I want this to be useful for everyone. Let's work on this as a big team and create something beautiful.

    • Try and use the hints / tips blocks to break up dense text, for example:

    circle-info

    To use LaTeX\LaTeXLATE​X, you can wrap your text in $$maths here$$. If this is at the beginning of a paragraph, it makes it block, otherwise it is inline

    hashtag
    Page Structure

    • A page should have a clear educational goal: this should be explained in the introduction. References to prerequisites should be kept within the book and if the book doesnt have this yet, it should be placed into Book Plan.

    • The topic should be presented initially with theory, showing the mathematics and structures we will need. A discussion should be pointed towards how this appears within Cryptography

    circle-check

    Motivating a new reader is the biggest challenge of creating a resource. People will be coming here to understand cryptography and SageMath, so keep pointing back to the goal!

    • Within a discussion of a topic, a small snippet of code to give an example is encouraged

    • If you write code better than you write maths, then just include what you can and the page will form around that

    • An example page is given in Sample Page

    hashtag
    Formatting

    hashtag
    Mathematics notation

    There's no "right or wrong" but it's good to be consistent, I think?

    • All maths must be presented using LaTeX\LaTeXLATE​Xusing either both block and inline

    • We seem to be using mathbb for our fields / rings. So let's stick with that? Maybe someone has a good resource for notation we can work from?

    hashtag
    Code Blocks

    • Make sure all code blocks have the right language selected for syntax highlighting

    • Preference is to SageMath, then to Python, then others.

    • Code should be cope-pastable. So if you include print statement, include the result of the output as a comment

    hashtag
    Algorithms

    • Algorithms should be presented as??

    # Example
    a = 3
    b = 6
    print(a+b)
    # 9
    // todo

    Euclidean Algorithm

    hashtag
    Introduction

    Although we have functions that can compute our gcd⁡\gcdgcdeasily it's important enough that we need to give and study an algorithm for it: the euclidean algorithm.

    It's extended version will help us calculate modular inverses which we will define a bit later.

    hashtag
    Euclidean Algorithm

    Important Remark

    If and then . Therefore

    hashtag
    Algorithm

    We write the following:

    Or iteratively until we find a . Then we stop

    Now here's the trick:

    If then divides

    circle-check

    Pause and ponder. Make you you understand why that works.

    Example:

    Calculate

    Code

    Exercises:

    1. Pick 2 numbers and calculate their by hand.

    2. Implement the algorithm in Python / Sage and play with it. Do not copy paste the code

    hashtag
    Extended Euclidean Algorithm

    triangle-exclamation

    This section needs to be expanded a bit.

    Bezout's identity

    Let . Then there exists such that

    The extended euclidean algorithm aims to find given

    Continued Fractions

    hashtag
    Continued Fractions

    Continued fractions are a way of representing a number as a sum of an integer and a fraction.

    Mathematically, a continued fraction is a representation

    a0+b0a1+b1a2+b2⋱a_{0} + \frac{b_{0}}{ a_{1} + \frac{b_{1}}{ a_{2} + \frac{b_{2}}{ \ddots }}}a0​+a1​+a2​+⋱b2​​b1​​b0​​

    ai,bia_{i}, b_{i}ai​,bi​are complex numbers. The continued fraction with bi=1 ∀ib_{i} = 1\ \forall ibi​=1 ∀i is called a simple continued fraction and continued fractions with finite number of aia_{i}ai​ are called finite continued fractions.

    Consider example rational numbers,

    the continued fractions could be written as

    hashtag
    Notation

    A simple continued fraction is represented as a list of coefficients() i.e

    for the above example

    hashtag
    Computation of simple continued fractions

    Given a number , the coefficients() in its continued fraction representation can be calculated recursively using

    The above notation might not be obvious. Observing the structure of continued fraction with few coefficients will make them more evident:

    SageMath provides functions continued_fraction and continued_fraction_list to work with continued fractions. Below is presented a simple implementation of continued_fractions.

    hashtag
    Convergents of continued fraction

    The convergent of a continued fractionis the numerical value or approximation calculated using the firstcoefficients of the continued fraction. The firstconvergents are

    One of the immediate applications of the convergents is that they give rational approximations given the continued fraction of a number. This allows finding rational approximations to irrational numbers.

    Convergents of continued fractions can be calculated in sage

    Continued fractions have many other applications. One such applicable in cryptology is based on Legendre's theorem in diophantine approximations.

    Theorem: if, thenis a convergent of.

    Wiener's attack on the RSA cryptosystem works by proving that under certain conditions, an equation of the formcould be derived whereis entirely made up of public information andis made up of private information. Under assumed conditions, the inequalityis statisfied, and the value(private information) is calculated from convergents of(public information), consequently breaking the RSA cryptosystem.

    Division and Greatest common divisor

    Author: Zademn

    hashtag
    Introduction

    Two of the skills a cryptographer must master are:

    1. Knowing his way and being comfortable to work with numbers.

    2. Understanding and manipulating abstract objects.

    This chapter of fundamentals proposes to prepare you for understanding the basics of number theory and abstract algebra .We will start with the most basic concepts such as division and build up knowledge until you, future cryptographer, are able to follow and understand the proofs and intricacies of the cryptosystems that make our everyday life secure.

    We will provide examples and snippets of code and be sure to play with them. If math is not your strongest suit, we highly suggest to pause and ponder for each concept and take it slow.

    For the math-savy people we cover advanced topics in specific chapters on the subjects of number theory and group theory.

    So what are we waiting for? Let's jump right in!

    hashtag
    Division

    Let be the set denoting the integers.

    Definition - Divisibility

    For we say that divides if there is some such that

    Notation:

    Example

    For we have because we can find such that .

    Properties

    • and implies

      • Example: Let and

    Definition - Division with remainder

    Let with ,

    There exists unique such that and

    is called the quotient and the remainder

    Examples:

    • To find python offers us the divmod() function that takes as arguments

    • If we want to find only the quotient we can use the // operator

    • If we want to find the remainder we can use the modulo % operator

    Exercises:

    1. Now it's your turn! Play with the proprieties of the division in Python and see if they hold.

    hashtag
    Greatest common divisor

    Definition

    Let be 2 integers. The greatest common divisor is the largest integer such that and

    Notation:

    Examples:

    Remark:

    • for all other common divisors of we have

    hashtag
    Things to think about

    What can we say about numbers with ? How are their divisors?

    Fermat's Little Theorem in Detail

    Would you like to be an author?

    hashtag
    Introduction

    Since we can add, subtract, multiply, divide even... what would be missing? Powering! I'm not talking about some power fantasy here, but rather introduce some really really important theorems. Fermat little's theorem proves useful in a great deal of situation, and is along with Euler's theorem a piece of arithmetic you need to know. Arguably the most canonical example of using these is the RSA cryptosystem, whose decryption step is built around Euler's theorem.

    hashtag
    Fermat's Little Theorem

    Since we want to talk about powers, let's look at powers. And because I like 7, I made a table of all the powers of all the integers modulo 7.

    On the last row, there is a clear pattern emerging, what's going on??? Hm, let's try again modulo 5 this time.

    Huh, again?! Clearly, there is something going on... Sage confirms this!

    Claim (Fermat's Little Theorem): Leta prime.

    When, this is equivalent to what we observed:. There are several proofs of Fermat's Little Theorem, but perhaps the fastest is to see it as a consequence of the Euler's Theorem which generalizes it. Still, let's look a bit at some applications of this before moving on.

    A first funny thing is the following:. When, this means we have found a non-trivial integer that when multiplied toyields 1. That is, we have found the inverse of, wow. Since the inverse is unique modulo, we can always invert non-zero integers by doing this. From a human point of view, this is really easier than using the extended euclidean algorithm.

    Ideals

    hashtag
    Example: Ideals of the integers

    Definition - Ideal of Z\mathbb{Z}Z

    I⊆Z I \subseteq \mathbb{Z}I⊆Zis an ideal   ⟺  ∀ a,b∈I and,z ∈Z\iff \forall \ a, b \in I \text{ and} , z\ \in \mathbb{Z}⟺∀ a,b∈I and,z ∈Zwe have

    Example: - multiples of

    Remarks:

    1. we have

    2. is an ideal

    Example: Consider . This ideal contains

    Greatest common divisor

    Let be 2 integers. If

    Integer Factorization

    hashtag
    Overview

    Given a composite integer nnn, can it be decomposed as a product of smaller integers (hopefully as a unique product of prime factors)?

    As easy as it may sound, integer factorization in polynomial time on a classical computer stands one of the unsolved problems in computation for centuries!

    Lets start dumb, all we need to do is check all the numbers 1<p<n1 < p < n1<p<n such that p∣np|np∣nor programmatically n%p==0

    Seems like its an algorithm! whats all the deal about? By polynomial time, we mean polynomial time in when is a b-bit number, so what we looking at is actually a which is actually exponential (which everyone hates)

    Now taking a better look at it, one would realize that a factor of can't be bigger than Other observation would be, if we already checked a number (say 2) to not be a divisor, we dont need to check any multiple of that number since it would not be a factor.

    Groups

    Authors: Ariana, Zademn Reviewed by:

    hashtag
    Introduction

    Modern cryptography is based on the assumption that some problems are hard (unfeasable to solve). Since the we do not have infinite computational power and storage we usually work with finite messages, keys and ciphertexts and we say they lay in some finite sets M,K\mathcal{M}, \mathcal{K}M,K and C\mathcal{C}C.

    Furthermore, to get a ciphertext we usually perform some operations with the message and the key.

    circle-check

    For example in AES128 since the input, output and key spaces are 128 bits. We also have the encryption and decryption operations:

    The study of sets, and different types of operations on them is the target of abstract algebra. In this chapter we will learn the underlying building blocks of cryptosystems and some of the hard problems that the cryptosystems are based on.

    hashtag
    Definition

    A setpaired with a binary operation is a group if the following requirements hold:

    • Closure: For all - Applying the operation keeps the element in the set

    • Associativity: For all

    • Identity: There exists an elementsuch that

    For , means when and when . For , .

    If , then is commutative and the group is called abelian. We often denote the group operation by instead of and we typically use instead of .

    Remark

    • The identity element of a group is also denoted with or just if only one groups is present

    Examples of groups

    Integers modulo (remainders) under modular addition . Let's look if the group axioms are satisfied

    1. . Because of the modulo reduction

    2. Modular addition is associative

    3. is the identity element

    Therefore we can conclude that the integers mod with the modular addition form a group.

    Example of non-groups

    is not a group because we can find the element that doesn't have an inverse for the identity . is not a group because we can find elements that don't have an inverse for the identity

    Exercise

    Is a group? If yes why? If not, are there values for that make it a group?

    sɹosᴉʌᴉp uoɯɯoɔ puɐ sǝɯᴉɹd ʇnoqɐ ʞuᴉɥ┴ :ʇuᴉH

    hashtag
    Proprieties

    1. The identity of a group is unique

    2. The inverse of every element is unique

    3. . The inverse of the inverse of the element is the element itself

    hashtag
    Orders

    In abstract algebra we have two notions of order: Group order and element order

    Group order

    The order of a group is the number of the elements in that group. Notation:

    Element order

    The order of an element is the smallest integer such that . If such a number doesn't exist we say the element has order . Notation:

    circle-check

    We said our messages lay in some group . The order of this group is the number of possible messages that we can have. For we have possible messages.

    circle-check

    Let be some message. The order of means how many different messages we can generate by applying the group operation on

    hashtag
    Subgroups

    Definition

    Let be a group. We say is a subgroup of if is a subset of and forms a group. Notation:

    Proprieties

    1. The identity of is also in

    2. The inverses of the elements in are found in

    How to check ? Let's look at a 2 step test

    1. Closed under operation:

    2. Closed under inverses:

    hashtag
    Generators

    Let be a group,an element and . Consider the following set:

    This set paired the group operation form a subgroup of generated by an element .

    circle-check

    Why do we care about subgroups? We praise the fact that some problems are hard because the numbers we use are huge and exhaustive space searches are too hard in practice.

    Suppose we have a big secret values space and we use an element to generate them.

    If an elementwith a small order is used then it can generate only possible values and if

    Example

    For now, trust us that if given a prime , a value and we compute for a secret , finding is a hard problem. We will tell you why a bit later.

    hashtag
    Examples

    // subgroups, quotient groups

    // cyclic groups

    Fields

    A setwith two binary operations is a field if the following holds:

    • is a commutative group with identity

    • is a commutative group with identity.

    a+b∈I and az∈Ia + b \in I \text{ and } az \in Ia+b∈I and az∈I
    aZ={az : z∈Z}→2Z,3Z,4Z,…a\mathbb{Z} = \{az \ : \ z \in \mathbb{Z} \} \to 2\mathbb{Z}, 3\mathbb{Z}, 4\mathbb{Z}, \dotsaZ={az : z∈Z}→2Z,3Z,4Z,…
    aaa
    ∀a,b∈Z\forall a, b \in \mathbb{Z}∀a,b∈Z
    b∈aZ  ⟺  a∣bb \in a\mathbb{Z} \iff a | bb∈aZ⟺a∣b
    I1+I2={a1+a2 : a1∈I1,a2∈I2}I_1 + I_2 = \{a_1 + a_2 \ : \ a_1 \in I_1 , a_2 \in I_2\}I1​+I2​={a1​+a2​ : a1​∈I1​,a2​∈I2​}
    18Z+12Z18\mathbb{Z} + 12\mathbb{Z}18Z+12Z
    6=18⋅1+12⋅(−1)⇒18Z+12Z=6Z6 = 18 \cdot 1 + 12 \cdot (-1) \Rightarrow 18\mathbb{Z} + 12\mathbb{Z} = 6\mathbb{Z}6=18⋅1+12⋅(−1)⇒18Z+12Z=6Z
    a,b∈Za, b \in \mathbb{Z}a,b∈Z
    d=gcd⁡(a,b)⇒aZ+bZ=dZd = \gcd(a, b) \Rightarrow a\mathbb{Z} + b\mathbb{Z} = d\mathbb{Z}d=gcd(a,b)⇒aZ+bZ=dZ

    Distributivity: a(b+c)=ab+ac,(a+b)c=ac+bca(b+c)=ab+ac,(a+b)c=ac+bca(b+c)=ab+ac,(a+b)c=ac+bc

    // field extensions, algebraic elements

    FFF
    +,⋅:F×F→F+,\cdot:F\times F\to F+,⋅:F×F→F
    R,+R,+R,+
    000
    R−{0},⋅R-\{0\},\cdotR−{0},⋅
    111

    LLL reduction

    hashtag
    Introduction

    In this section, we hope to bring some intuitive understanding to the LLL algorithm and how it works. The LLL algorithm is a lattice reduction algorithm, meaning it takes in a basis for some lattice and hopefully returns another basis for the same lattice with shorter basis vectors. Before introducing LLL reduction, we'll introduce 2 key algorithms that LLL is built from, Gram-Schmidt orthogonalization and Gaussian Reduction. We give a brief overview on why these are used to build LLL.

    As the volume of a lattice is fixed, and is given by the determinant of the basis vectors, whenever our basis vectors gets shorter, they must, in some intuitive sense, become more orthogonal to each other in order for the determinant to remain the same. Hence, Gram-Schmidt orthogonalization is used as an approximation to the shortest basis vector. However, the vectors that we get are in general not in the lattice, hence we only use this as a rough idea of what the shortest vectors would be like.

    Lagrange's algorithm can be thought as the GCD algorithm for 2 numbers generalized to lattices. This iteratively reduces the length of each vector by subtracting some amount of one from another until we can't do it anymore. Such an algorithm actually gives the shortest possible vectors in 2 dimensions! Unfortunately, this algorithm may not terminate for higher dimensions, even in 3 dimensions. Hence, it needs to be modified a bit to allow the algorithm to halt.

    Polynomials

    // symmetric polynomials

    // discriminants

    // resultants

    Another take on groups

    // Visual

    // Symmetries

    // Permutations

    Applications

    We shall now provide a few instances where lattices are used in various algorithms. Most of these uses the LLL algorithm as it is quite fast.

    a=b⋅q+ra = b \cdot q + ra=b⋅q+r
    d=gcd⁡(a,b)d = \gcd(a, b)d=gcd(a,b)
    d∣rd | rd∣r
    gcd⁡(a,b)=gcd⁡(b,r)\gcd(a, b) = \gcd(b, r)gcd(a,b)=gcd(b,r)
    a=q0⋅b+r0b=q1⋅r0+r1r0=q2⋅r1+r2⋮rn−2=rn−1⋅qn−1+rnrn=0 a = q_0 \cdot b + r_0 \\ b = q_1 \cdot r_0 + r_1 \\ r_0 = q_2 \cdot r_1 + r_2 \\ \vdots \\ r_{n-2} = r_ {n-1} \cdot q_{n - 1} + r_n \\ r_n = 0a=q0​⋅b+r0​b=q1​⋅r0​+r1​r0​=q2​⋅r1​+r2​⋮rn−2​=rn−1​⋅qn−1​+rn​rn​=0
    rk−2=qk⋅rk−1+rkr_{k-2} = q_k \cdot r_{k-1} + r_krk−2​=qk​⋅rk−1​+rk​
    000
    gcd⁡(a,b)=gcd(b,r0)=gcd(r0,r1)=⋯=gcd⁡(rn−2,rn−1)=rn−1=d\gcd(a, b) = gcd(b, r_0) = gcd(r_0, r_1) = \dots = \gcd(r_{n-2}, r_{n-1}) = r_{n-1} = dgcd(a,b)=gcd(b,r0​)=gcd(r0​,r1​)=⋯=gcd(rn−2​,rn−1​)=rn−1​=d
    d=gcd⁡(a,b)d = \gcd(a, b)d=gcd(a,b)
    ddd
    r0,r1,...rn−1r_0, r_1, ... r_{n-1}r0​,r1​,...rn−1​
    gcd⁡(24,15)\gcd(24, 15)gcd(24,15)
    24=1⋅15+915=1⋅9+69=1⋅6+36=2⋅3+0⇒3=gcd⁡(24,15)24 = 1 \cdot 15 + 9 \\ 15 = 1 \cdot 9 + 6 \\ 9 = 1 \cdot 6 + 3 \\ 6 = 2 \cdot 3 + 0 \Rightarrow 3 = \gcd(24, 15) 24=1⋅15+915=1⋅9+69=1⋅6+36=2⋅3+0⇒3=gcd(24,15)
    gcd⁡\gcdgcd
    d=gcd⁡(a,b)d = \gcd(a, b)d=gcd(a,b)
    u,vu, vu,v
    au+bv=dau + bv = dau+bv=d
    d=gcd⁡(a,b), and u,vd = \gcd(a, b), \text{ and }u, vd=gcd(a,b), and u,v
    a,ba, ba,b
    1711=1+611116=1+5665=1+1551=5+0\frac{17}{11} = 1 + \frac{6}{11} \\[10pt] \frac{11}{6} = 1 + \frac{5}{6} \\[10pt] \frac{6}{5} = 1 + \frac{1}{5} \\[10pt] \frac{5}{1} = 5 + 01117​=1+116​611​=1+65​56​=1+51​15​=5+0
    51=565=1+15116=1+56=1+165=1+11+151711=1+611=1+1116=1+11+11+15\frac{5}{1} =5 \\[10pt] \frac{6}{5} = 1 + \frac{1}{5} \\[10pt] \frac{11}{6} = 1 + \frac{5}{6} = 1 + \frac{1}{\frac{6}{5}} = 1 + \frac{1}{1 + \frac{1}{5}} \\[10pt] \frac{17}{11} = 1 + \frac{6}{11} = 1 + \frac{1}{\frac{11}{6}} = 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{5}}}15​=556​=1+51​611​=1+65​=1+56​1​=1+1+51​1​1117​=1+116​=1+611​1​=1+1+1+51​1​1​
    a0+1a1+1a2+1⋱a_{0} + \frac{1}{ a_{1} + \frac{1}{ a_{2} + \frac{1}{ \ddots }}}a0​+a1​+a2​+⋱1​1​1​
    aia_{i}ai​
    x=[a0; a1, a2, a3, a4, a5, a6, …]x = [a_{0};\ a_{1},\ a_{2},\ a_{3},\ a_{4},\ a_{5},\ a_{6},\ \ldots]x=[a0​; a1​, a2​, a3​, a4​, a5​, a6​, …]
    1711=[1; 1, 1, 5]  ,116=[1; 1, 5]  ,65=[1;5]  ,51=[5;] \frac{17}{11} = [1;\ 1,\ 1,\ 5]\ \ ,\frac{11}{6} = [1;\ 1,\ 5]\ \ ,\frac{6}{5} = [1; 5]\ \ ,\frac{5}{1} = [5;]1117​=[1; 1, 1, 5]  ,611​=[1; 1, 5]  ,56​=[1;5]  ,15​=[5;]
    xxx
    aia_{i}ai​
    x0=xai=⌊xi⌋xi+1=1xi−aix_{0} = x \\[4pt] a_{i} = \lfloor x_{i} \rfloor \\[4pt] x_{i+1} = \frac{1}{x_{i} - a_{i}}x0​=xai​=⌊xi​⌋xi+1​=xi​−ai​1​
    x0=a0+1a1+1a2,   x1=a1+1a2,   x2=a2xi=ai+1xi+1xi+1=1xi−aix_{0} = a_{0} + \frac{1}{a_{1} + \frac{1}{a_{2}}},\ \ \ x_{1} = a_{1} + \frac{1}{a_{2}}, \ \ \ x_{2} = a_{2} \\[10pt] x_{i} = a_{i} + \frac{1}{x_{i+1}} \\[10pt] x_{i+1} = \frac{1}{x_{i} - a_{i}}x0​=a0​+a1​+a2​1​1​,   x1​=a1​+a2​1​,   x2​=a2​xi​=ai​+xi+1​1​xi+1​=xi​−ai​1​
    kthk^{th}kth
    x=[a0;a1, a2, a3, a4,…]x = [a_{0}; a_{1},\ a_{2},\ a_{3},\ a_{4},\ldots] x=[a0​;a1​, a2​, a3​, a4​,…]
    k−1k - 1k−1
    kkk
    a01,   a0+1a1,   a0+1a1+1a2, …, a0+1a1+⋱ak−2+1ak−1\frac{a_{0}}{1},\ \ \ a_{0} + \frac{1}{a_{1}}, \ \ \ a_{0} + \frac{1}{a_{1} + \frac{1}{a_{2}}}, \ \ldots,\ a_{0} + \frac{1}{a_{1} + \frac{\ddots} {a_{k-2} + \frac{1}{a_{k-1}}}}1a0​​,   a0​+a1​1​,   a0​+a1​+a2​1​1​, …, a0​+a1​+ak−2​+ak−1​1​⋱​1​
    ∣x−ab∣<1b2\mid x - \frac{a}{b} \mid < \frac{1}{b^{2}}∣x−ba​∣<b21​
    ab\frac{a}{b}ba​
    xxx
    ∣x−ab∣\mid x - \frac{a}{b} \mid∣x−ba​∣
    xxx
    ab\frac{a}{b}ba​
    ∣x−ab∣<1b2\mid x - \frac{a}{b} \mid < \frac{1}{b^{2}}∣x−ba​∣<b21​
    ab\frac{a}{b}ba​
    xxx

    3∣63 | 63∣6 and 3∣9⇒3∣(6⋅5+9⋅2)  ⟺  3∣483 | 9 \Rightarrow 3 | (6 \cdot 5 + 9 \cdot 2) \iff 3 | 483∣9⇒3∣(6⋅5+9⋅2)⟺3∣48 . We can find k=16k = 16k=16such that 48=3⋅1648 = 3 \cdot 1648=3⋅16

  • a∣ba | ba∣b and b∣c b | c b∣c implies a∣c a | ca∣c

  • if a∣ba|ba∣band b∣ab|ab∣a then a=±ba = \pm ba=±b

  • Z={…,−1,0,1,2,3… }\mathbb{Z} = \{\dots , -1, 0, 1, 2, 3 \dots \}Z={…,−1,0,1,2,3…}
    a,b,∈Za, b, \in \mathbb{Z} a,b,∈Z
    aaa
    bbb
    k∈Zk \in \mathbb{Z}k∈Z
    a⋅k=ba \cdot k = ba⋅k=b
    a∣ba | ba∣b
    a=2,b=6a = 2, b = 6a=2,b=6
    2∣62 | 62∣6
    k=3k = 3k=3
    6=2⋅36 = 2 \cdot 36=2⋅3
    a∣a, 1∣a and a∣0a | a, \ 1 | a \text{ and } a | 0a∣a, 1∣a and a∣0
    a∣ba | ba∣b
    a∣c a | c a∣c
    a∣(bu+cv) ∀u,v,∈Za | (bu + cv) \ \forall u, v, \in \mathbb{Z}a∣(bu+cv) ∀u,v,∈Z
    b=6,u=5b = 6, u = 5b=6,u=5
    c=9,v=2c = 9, v = 2 c=9,v=2
    a,b∈Za, b \in \mathbb{Z}a,b∈Z
    b≥1b≥1b≥1
    q,r∈Zq, r \in \mathbb{Z}q,r∈Z
    a=bq+r\boxed{a = bq + r}a=bq+r​
    0≤r<b0 \leq r < b0≤r<b
    qq q
    rrr
    q,rq, rq,r
    a,ba, ba,b
    qqq
    rrr
    a,b∈Za, b \in \mathbb{Z}a,b∈Z
    d∈Zd \in \mathbb{Z}d∈Z
    d∣ad | ad∣a
    d∣bd | bd∣b
    gcd⁡(a,b)=d\gcd(a, b) = dgcd(a,b)=d
    ccc
    a,ba, ba,b
    c∣dc | dc∣d
    a,b a, ba,b
    gcd⁡(a,b)=1\gcd(a, b) = 1gcd(a,b)=1

    6

    2

    0

    1

    4

    2

    2

    4

    1

    3

    0

    1

    1

    6

    1

    6

    6

    4

    0

    1

    2

    4

    4

    2

    1

    5

    0

    1

    4

    5

    2

    3

    6

    6

    0

    1

    1

    1

    1

    1

    1

    4

    4

    1

    3

    0

    1

    3

    2

    4

    4

    0

    1

    1

    1

    1

    Power

    0

    1

    2

    3

    4

    5

    6

    1

    0

    1

    2

    3

    4

    Power

    0

    1

    2

    3

    4

    1

    0

    1

    2

    3

    4

    2

    0

    ppp
    ∀a∈Z,ap≡a [p]\forall a\in\mathbb Z, a^p\equiv a~[p]∀a∈Z,ap≡a [p]
    a≠0a\neq 0a=0
    ap−1≡1 [n]a^{p-1}\equiv 1~[n]ap−1≡1 [n]
    ∀a∈Z,a⋅ap−2≡ap−1≡1 [p]\forall a\in\mathbb Z, a\cdot a^{p-2}\equiv a^{p-1}\equiv 1~[p]∀a∈Z,a⋅ap−2≡ap−1≡1 [p]
    p>2p>2p>2
    aaa
    aaa
    ppp

    5

    1

    O(n)O(n)O(n)
    bbb
    nnn
    O(2b)O(2^b)O(2b)
    nnn
    n\sqrt{n}n​
    def my_gcd(a, b):
        # If a < b swap them
        if a < b: 
            a, b = b, a
        # If we encounter 0 return a
        if b == 0: 
            return a
        else:
            r = a % b
            return my_gcd(b, r)
    
    print(my_gcd(24, 15))
    # 3
    # In sage we have the `xgcd` function
    a = 24
    b = 15
    g, u, v = xgcd(a, b)
    print(g, u, v)
    # 3 2 -3 
    
    print(u * a + v * b)
    # 3 -> because 24 * 2 - 15 * 3 = 48 - 45 = 3
    def continued_fraction_list(xi):
        ai = floor(xi)
        if xi == ai: # last coefficient
            return [ai]
        return [ai] + continued_fraction_list(1/(x - ai))
    sage: cf = continued_fraction(17/11)
    sage: convergents = cf.convergents()
    sage: cf
    [1; 1, 1, 5]
    sage: convergents
    [1, 2, 3/2, 17/11]
    q, r = divmod(6, 2)
    print(q, r)
    # 3 0 
    
    q, r = divmod(13, 5)
    print(q, r)
    # 2 3 
    # Note that 13 = 2 * 5 + 3
    q = 13 // 5
    print(q)
    # 2
    
    r = 13 % 5
    print(r)
    # 3
    # In python we can import math to get the GCD algo
    import math
    print(math.gcd(18, 12)) # -> 6
    # Sage has it already!
    print(gcd(18, 12)) # -> 6
    p, itworks = 1, True
    for _ in range(100):
        p = next_prime(p)
        Fp = GF(p) # Finite Field of size p
        itworks &= all(Fp(x)^(p-1) == 1 for x in range(1,p))
    
    print(itworks)
    # True
    def factors(n):
        divisors = []
        for p in range(1,n):
            if n%p==0:
                divisors.append(p)
        return divisors
    for all
  • Inverse: For all elements a∈Ga\in Ga∈G, there exists some b∈Gb\in Gb∈Gsuch that b⋅a=a⋅b=eb\cdot a=a\cdot b=eb⋅a=a⋅b=e. We usually denotebbbas a−1a^{-1}a−1

  • ✓\checkmark✓∀a∈Z/nZ\forall a \in \mathbb{Z}/ n\mathbb{Z} ∀a∈Z/nZwe take n−a mod nn - a \bmod nn−amodnto be the inverse of aaa. We check that

    a+n−a≡n≡0 mod na + n - a \equiv n \equiv 0 \bmod na+n−a≡n≡0modn

    n−a+a≡n≡0 mod nn - a + a \equiv n \equiv 0 \bmod nn−a+a≡n≡0modn

    ∀a,b∈G:\forall a, b \in G: ∀a,b∈G: (ab)−1=b−1a−1(ab)^{-1} = b^{-1}a^{-1}(ab)−1=b−1a−1

    Proof: (ab)(b−1a−1)=a(bb−1)a−1=aa−1=e.(ab)(b^{−1}a^{−1}) =a(bb^{−1})a^{−1}=aa^{−1}= e.(ab)(b−1a−1)=a(bb−1)a−1=aa−1=e.

    is small enough an attacker can do a brute force attack.
    K=M=C={0,1}128\mathcal{K} = \mathcal{M} = \mathcal{C} = \{0, 1\}^{128}K=M=C={0,1}128
    Enc:K×M→CDec:K×C→MEnc: \mathcal{K} \times \mathcal{M} \to \mathcal{C} \\ Dec: \mathcal{K} \times \mathcal{C} \to \mathcal{M}Enc:K×M→CDec:K×C→M
    GGG
    ⋅:G×G→G\cdot:G\times G\to G⋅:G×G→G
    a,b∈G: a, b \in G: \ a,b∈G: 
    a⋅b∈Ga\cdot b \in Ga⋅b∈G
    a,b,c∈G:a, b, c \in G: a,b,c∈G:
    (a⋅b)⋅c=a⋅(b⋅c)(a \cdot b) \cdot c=a\cdot (b\cdot c)(a⋅b)⋅c=a⋅(b⋅c)
    e∈Ge\in Ge∈G
    a⋅e=e⋅a=aa\cdot e=e\cdot a=aa⋅e=e⋅a=a
    n∈Zn\in\mathbb Zn∈Z
    ana^nan
    a⋅a…⋅a⏟n times\underbrace{a\cdot a\dots{}\cdot a}_{n\text{ times}}n timesa⋅a…⋅a​​
    n>0n>0n>0
    (a−n)−1\left(a^{-n}\right)^{-1}(a−n)−1
    n<0n<0n<0
    n=0n=0n=0
    an=ea^n=ean=e
    ab=baab=baab=ba
    ⋅\cdot⋅
    +++
    ⋅\cdot⋅
    nanana
    ana^nan
    GGG
    1G1_G1G​
    111
    nnn
    =(Z/nZ,+)= (\mathbb{Z} / n \mathbb{Z}, +)=(Z/nZ,+)
    Z/nZ={0,1,...,n−1}\mathbb{Z} / n \mathbb{Z} = \{0, 1, ..., n -1\}Z/nZ={0,1,...,n−1}
    ✓\checkmark✓
    ∀a,b∈Z/nZ let c≡a+b mod n\forall a, b \in \mathbb{Z}/ n\mathbb{Z} \text{ let } c \equiv a + b \bmod n∀a,b∈Z/nZ let c≡a+bmodn
    c<n⇒c∈Z/nZc < n \Rightarrow c \in \mathbb{Z}/ n\mathbb{Z} c<n⇒c∈Z/nZ
    ✓\checkmark✓
    ✓\checkmark ✓
    0+a≡a+0≡a mod n⇒00 + a \equiv a + 0 \equiv a \bmod n \Rightarrow 00+a≡a+0≡amodn⇒0
    nnn
    (Q,⋅)(\mathbb{Q}, \cdot)(Q,⋅)
    000
    111
    (Z,⋅)(\mathbb{Z}, \cdot)(Z,⋅)
    111
    (Z/nZ∖{0},⋅)(\mathbb{Z} / n \mathbb{Z} \smallsetminus \{0\}, \cdot)(Z/nZ∖{0},⋅)
    nnn
    ∀\forall∀
    a∈G :(a−1)−1=ga \in G \ : \left(a^{-1} \right) ^{-1} = ga∈G :(a−1)−1=g
    GGG
    ∣G∣|G|∣G∣
    a∈Ga \in Ga∈G
    nnn
    an=1Ga^n = 1_Gan=1G​
    nnn
    ∞\infty∞
    ∣a∣|a|∣a∣
    M\mathcal{M}M
    ∣M∣|\mathcal{M}|∣M∣
    M={0,1}128\mathcal{M} = \{0,1\}^{128}M={0,1}128
    ∣M∣=2128|\mathcal{M}| = 2^{128}∣M∣=2128
    m∈Mm \in \mathcal{M}m∈M
    mmm
    mmm
    (G,⋅)(G, \cdot)(G,⋅)
    HHH
    GGG
    HHH
    GGG
    (H,⋅)(H, \cdot)(H,⋅)
    H≤GH \leq GH≤G
    GGG
    H:H: H:
    1H=1G1_H = 1_G1H​=1G​
    HHH
    HHH
    H≤GH \leq GH≤G
    ∀a,b∈H→ab∈H\forall a, b \in H \to ab \in H∀a,b∈H→ab∈H
    ∀a∈H→a−1∈H\forall a \in H \to a^{-1} \in H∀a∈H→a−1∈H
    GGG
    g∈Gg \in Gg∈G
    ∣g∣=n|g| = n∣g∣=n
    {1,g,g2,...,gn−1}=denoted⟨g⟩.\{1, g, g^2, ..., g^{n-1}\} \overset{\text{denoted}}{=} \langle g\rangle.{1,g,g2,...,gn−1}=denoted⟨g⟩.
    GGG
    ggg
    GGG
    ggg
    g∈Gg \in Gg∈G
    nnn
    nnn
    nnn
    ppp
    g∈Z/pZg \in \mathbb{Z} / p \mathbb{Z}g∈Z/pZ
    y=gx mod py = g^x \bmod py=gxmodp
    xxx
    xxx
    a∈Ga\in Ga∈G

    Discrete Log Problem

    hashtag
    Discrete log problem

    Given any groupGGGand elementsa,ba,ba,bsuch that an=ba^n=ban=b, the problem of solving fornnnis known as the disctete log problem (DLP). In sage, this can be done for general groups by calling discrete_log

    hashtag
    Discrete log over

    Typically, one considers the discrete log problem in , i.e. the multiplicative group of integers. Explicitly, the problem asks forgiven . This can be done by calling b.log(a) in sage:

    This section is devoted to helping the reader understand which functions are called when for this specific instance of DLP.

    Whenis composite and not a prime power, discrete_log() will be used, which uses generic algorithms to solve DLP (e.g. Pohlig-Hellman and baby-step giant-step).

    When is a prime, Pari znlog will be used, which uses a linear sieve index calculus method, suitable for .

    When , SageMath will fall back on the generic implementation discrete_log()which can be slow. However, Pari znlog can handle this as well, again using the linear sieve index calculus method. To call this within SageMath we can use either of the following (the first option being a tiny bit faster than the second)

    hashtag
    Example

    Given a small prime, we can compare the Pari method with the Sage defaults

    We can also solve this problem with even larger primes in a very short time

    hashtag
    Discrete log over

    // elliptic curve discrete log functions

    Gram-Schmidt Orthogonalization

    hashtag
    Overview

    Gram-Schmidt orthogonalization is an algorithm that takes in a basis {bi}i=1n\left\{b_i\right\}_{i=1}^n{bi​}i=1n​ as an input and returns a basis {bi∗}i=1n\left\{b_i^*\right\}_{i=1}^n{bi∗​}i=1n​where all vectors are orthogonal, i.e. at right angles. This new basis is defined as

    bi∗=bi−∑j=1i−1μi,jbj∗μi,j=⟨bi,bj∗⟩⟨bj∗,bj∗⟩b_i^*=b_i-\sum_{j=1}^{i-1}\mu_{i,j}b_j^*\quad\mu_{i,j}=\frac{\langle b_i,b_j^*\rangle}{\langle b_j^*,b_j^*\rangle}bi∗​=bi​−j=1∑i−1​μi,j​bj∗​μi,j​=⟨bj∗​,bj∗​⟩⟨bi​,bj∗​⟩​

    where μi,j\mu_{i,j}μi,j​is the Gram-Schmidt coefficients.

    One can immediately check that this new basis is orthogonal, meaning

    Let be the matrix where the th row is given by andbe the matrix where the th row is given by , then the Gram-Schmidt orthogonalization gives us where and is the Gram-Schmidt coefficient. As an example, consider the basis of a subspace of :

    Instead of doing the Gram-Schmidt orthogonalization by hand, we can get sage to do it for us:

    This outputs two matrices, and :

    One can quickly verify that and that the rows of are orthogonal to each other.

    A useful result is that

    Intuitively, this tells us that the more orthogonal a set of basis for a lattice is, the shorter it is as the volume must be constant.

    hashtag
    Exercises

    1) Show that the basis is orthogonal.

    2) Verify that the output of sage is indeed correct.

    3) Show that and is a diagonal matrix whose entries are . Conclude that .

    4*) Given the Iwasawa decomposition where is a lower diagonal matrix with on its diagonal, is a diagonal matrix and an orthogonal matrix, meaning , show that and . Furthermore, prove that such a decomposition is unique.

    Introduction

    Lattices, also known as Minkowski's theory after Hermann Minkowski, or the geometry of numbers (deprecated!) allows the usage of geometrical tools (i.e. volumes) in number theory.

    The intuitive notion of a lattice (perhaps surprisingly) matches its mathematical definition. For example, lattices are formed by

    • points on an infinite checkerboard;

    • centers of a hexagonal tessellation;

    • integers on the real number line.

    circle-info

    The last example should hint at how we generalize this concept to arbitrary dimensions. In general, lattices consist of discrete points which appear at "regular intervals."

    hashtag
    Definitions

    A lattice is a subgroup of generated by , i.e.

    where are linearly independent vectors. Collectively, form a basis of.

    circle-info

    We say a set of vectors are linearly independent if the only solution to the equation is when all are zero.

    Taking a step back, this definition should resemble that of a vector space, with one exception: scalars are integers! The discrete nature of lattices comes from this restriction.

    Some more terminology from linear algebra will be useful. The dimension of a lattice, denoted, is . A lattice is complete if . Note that we can always choose a subspace of such that the lattice is complete, namely the subspace generated by .

    The region

    is known as the fundamental mesh.

    In the image above, we see the points of a lattice in . The red vectors are one set of basis vectors and the shaded region is the corresponding fundamental mesh. The green vectors also form another set of basis vectors with its corresponding fundamental mesh. We see here that the basis vectors and fundamental mesh is not unique to a lattice.

    Although the fundamental mesh is not unique, it turns out that the (dimensional) volume of the fundamental mesh is constant for any given lattice. Hence we can define the volume of a lattice as the volume of a fundamental mesh. However this definition can be hard to handle hence we provide an equivalent definition via determinants:

    Letbe amatrix whose rows are given by the basis vectors. Then the volume of a fundamental mesh is given by

    A subset of is known as centrally symmetric if implies . It is convex if for any , the line joining is contained in , i.e. . Finally we can introduce the most important theorem about lattices, the Minkowski's Lattice Point Theorem:

    Let be a complete lattice of dimension and be a centrally symmetric convex set. Suppose

    Then contains at least one nonzero point of . This result is primarily used to prove the existence of lattice vectors.

    Throughout this section, denotes the norm and denotes the inner product.

    hashtag
    Proof sketch of Minkowski's theorem

    This proof is by some sort of a pigeonhole argument on volumes. Consider the set

    We have , hence the inclusion cannot be injective, thus we can find some , . Hence is a nontrivial lattice point.

    hashtag
    Exercises

    1) Let be the lattice generated by (take the rows as basis vectors).

    • Compute the volume of this lattice

    • Show that generates the same lattice

    • Show that each row in is in the lattice butdoes not generate the lattice. This is one key difference from the case of linear algebra (over fields).

    2) Letbe matrices whose row vectors are basis for lattices . Both lattices are the same iff there exists some such that . Find for problem 1. Note that is the group of invertible matrices with integer coefficients, meaning and have integer coefficients.

    3) Show that the condition in Minkowski's lattice point theorem is strict, i.e. for any complete latticeof dimension , we can find some centrally symmetric convex setwithbut the only lattice point inis the origin.

    4*) Letbe the shortest nonzero vector for some lattice with dimension. Show that

    Rings

    A setRRRwith two binary operations +,⋅:R×R→R+,\cdot:R\times R\to R+,⋅:R×R→Ris a ring if the following holds:

    • R,+R,+R,+is a commutative group with identity 000

    • R,⋅R,\cdotR,⋅is a monoid (group without the inverse axiom) with identity111.

    • Distributivity:

    // ideals, diff types of domains

    LLL reduction

    hashtag
    Overview

    There are a few issues that one may encounter when attempting to generalize Lagrange's algorithm to higher dimensions. Most importantly, one needs to figure what is the proper way to swap the vectors around and when to terminate, ideally in in polynomial time. A rough sketch of how the algorithm should look like is

    There are two things we need to figure out, in what order should we reduce the basis elements by and how should we know when to swap. Ideally, we also want the basis to be ordered in a way such that the smallest basis vectors comes first. Intuitively, it would also be better to reduce a vector by the larger vectors first before reducing by the smaller vectors, a very vague analogy to filling up a jar with big stones first before putting in the sand. This leads us to the following size reduction algorithm:

    LLL reduced

    hashtag
    Definition

    Let . A basisis - LLL-reduced if it is size reduced and satisfy the Lovász condition, i.e.

    This notion of reduction is most useful to use for fast algorithms as such a basis can be found in polynomial time (see ).

    Minkowski reduced

    hashtag
    Definition

    The basis is Minkowski-reduced if has minimum length among all vectors in linearly independent from. Equivalently, has minimum length among all vectors such that can be extended to form a basis of. Such a notion is strongest among all lattice reduction notions and is generally extremely hard to compute. Another equivalent definition is

    HKZ reduced

    hashtag
    Definition

    Let as the projection to the orthogonal complement of .Then the basis is HKZ-reduced if it is size-reduced and . This definition gives us a relatively simple way to compute a HKZ-reduced basis by iteratively finding the shortest vector in orthogonal projections.

    Coppersmith algorithm

    This algorithm solves for small roots of polynomials modulo any integer, meaning given some polynomial of degree and any integer , then if , this algorithm can findwith time polynomial in and . The key idea behind this algorithm is to construct a polynomialsuch that in . As roots of polynomials over the reals can be found easily, this gives an easy way to find . We shall introduce the Coppersmith algorithm in a few iterations, with each iteration approaching the bound.

    hashtag
    Polynomials in lattices

    Lagrange's algorithm

    hashtag
    Overview

    Lagrange's algorithm, often incorrectly called Gaussian reduction, is the 2D analouge to the Euclidean algorithm and is used for lattice reduction. Intuitively, lattice reduction is the idea of finding a new basis that consists of shorter vectors. Before going into Lagrange's algorithm, we first recap the Euclidean algorithm:

    The algorithm primarily consists of two steps, a reduction step where the size of is brought down by a multiple of and a swapping step that ensures is always the largest number. We can adapt this idea for lattices:

    Short integer solutions (SIS)

    hashtag
    Introduction

    In this section we will study the short integer solution problem and a hashing algorithm that is based on this algorithm.

    hashtag

    Lattices of interest

    triangle-exclamation

    Needs review.

    hashtag
    Introduction

    In this chapter we will study some specific types of lattices that appear in cryptography. These will help us understand how certain problems we base our algorithms on reduce to other hard problems. They will also give insight about the geometry of lattices.

    Z5 = Zmod(5) # Technically it's a ring but we'll use the addition here
    print(Z5.list())
    # [0, 1, 2, 3, 4]
    
    print(Z5.addition_table(names = 'elements'))
    # +  0 1 2 3 4
    #  +----------
    # 0| 0 1 2 3 4
    # 1| 1 2 3 4 0
    # 2| 2 3 4 0 1
    # 3| 3 4 0 1 2
    # 4| 4 0 1 2 3
    
    a, b = Z5(14), Z5(3)
    print(a, b)
    # 4 3
    print(a + b)
    # 2
    print(a + 0)
    # 4
    print(a + (5 - a))
    # 0
    
    n = 11
    Zn = Zmod(n)
    a, b = Zn(5), Zn(7)
    print(n - (a + b))
    # 10
    print((n - a) + (n - b))
    # 10
    Z12 = Zmod(12) # Residues modulo 12
    print(Z12.order()) # The additive order 
    # 12
    a, b= Z12(6), Z12(3)
    print(a.order(), b.order())
    # 2 4
    print(a.order() * a)
    # 0
    
    print(ZZ.order()) # The integers under addition is a group of infinite order
    # +Infinity
    p = 101 # prime
    Zp = Zmod(p) 
    H_list = Zp.multiplicative_subgroups() # Sage can get the subgroup generators for us
    print(H_list)
    # ((2,), (4,), (16,), (32,), (14,), (95,), (10,), (100,), ())
    
    g1 = H_list[3][0] # Weak generator
    print(g1, g1.multiplicative_order())
    # 32 20
    
    g2 = Zp(3) # Strong generator
    print(g2, g2.multiplicative_order())
    # 3 100
    
    
    ## Consider the following functions
    def brute_force(g, p, secret_value):
        """
        Brute forces a secret value, returns number of attempts
        """
        for i in range(p-1):
            t = pow(g, i, p)
            if t == secret_value:
                break
        return i
        
    def mean_attempts(g, p, num_keys):
        """
        Tries `num_keys` times to brute force and 
        returns the mean of the number of attempts
        """
        total_attempts = 0
        for _ in range(num_keys):
            k = random.randint(1, p-1)
            sv = pow(g, k, p) # sv = secret value
            total_attempts += brute_force(g, p, sv)
        return 1. * total_attempts / num_keys
        
    ## Let's try with our generators
    print(mean_attempts(g1, p, 100)) # Weak generator
    # 9.850
    print(mean_attempts(g2, p, 100)) # Strong generator
    # 49.200
    
    sage: G = DihedralGroup(99)
    sage: g = G.random_element()
    sage: discrete_log(g^9,g) # note that if the order of g is less than 9 we would get 9 mod g.order()
    9
    a(b+c)=ab+ac,(a+b)c=ac+bca(b+c)=ab+ac,(a+b)c=ac+bca(b+c)=ab+ac,(a+b)c=ac+bc

    Ring-LWE

    Cryptographic lattice problems

    Learning with errors (LWE)

    hashtag
    Bounds
    δ∈(14,1)\delta\in\left(\frac14,1\right)δ∈(41​,1)
    {bi}i=1d\left\{b_i\right\}_{i=1}^d{bi​}i=1d​
    δ\deltaδ
    δ∥bi∗∥2≤∥bi+1∗+μi+1,ibi∗∥2\delta\left\lVert b_i^*\right\rVert^2\leq\left\lVert b_{i+1}^*+\mu_{i+1,i}b_i^*\right\rVert^2δ∥bi∗​∥2≤​bi+1∗​+μi+1,i​bi∗​​2
    LLL reduction
    ∥b1∥≤(44δ−1)d−14vol(L)1d∥bi∥≤(44δ−1)d−12λi(L)∏i=1d∥bi∥≤(44δ−1)d(d−1)4vol(L)\begin{align*} \left\lVert b_1\right\rVert&\leq\left(\frac4{4\delta-1}\right)^{\frac{d-1}4}\text{vol}(L)^\frac1d\\ \left\lVert b_i\right\rVert&\leq\left(\frac4{4\delta-1}\right)^{\frac{d-1}2}\lambda_i(L)\\ \prod_{i=1}^d\left\lVert b_i\right\rVert&\leq\left(\frac4{4\delta-1}\right)^{\frac{d(d-1)}4}\text{vol}(L) \end{align*}∥b1​∥∥bi​∥i=1∏d​∥bi​∥​≤(4δ−14​)4d−1​vol(L)d1​≤(4δ−14​)2d−1​λi​(L)≤(4δ−14​)4d(d−1)​vol(L)​
    hashtag
    Bounds

    The proof presented here is based off [Waerden 1956]. We proceed by induction. Letbib_ibi​be a Minkowski-reduced basis for some latticeLLL. The lower bound is immediate and for i=1i=1i=1, the upper bound is immediate as well.

    Let v1,v2…viv_1,v_2\dots v_iv1​,v2​…vi​be linearly independent vectors such that∥vj∥=λj(L)\left\lVert v_j\right\rVert=\lambda_j(L)∥vj​∥=λj​(L). Let Li−1L_{i-1}Li−1​be the sublattice generated by b1,b2,…bi−1b_1,b_2,\dots b_{i-1}b1​,b2​,…bi−1​. Evidently somekkkmust exist such thatvkv_kvk​is not in Li−1L_{i-1}Li−1​. Consider the new lattice L′=L∩span(b1,b2,…bi−1,vk)L'=L\cap\text{span}\left(b_1,b_2,\dots b_{i-1},v_k\right)L′=L∩span(b1​,b2​,…bi−1​,vk​). Letvk′v'_kvk′​be the shortest vector inL′−Li−1L'-L_{i-1}L′−Li−1​such thatb1,b2,…,bi−1,vk′b_1,b_2,\dots,b_{i-1},v'_kb1​,b2​,…,bi−1​,vk′​is a basis for L′L'L′and we have

    Ifn=1n=1n=1, then we are done sincevkv_kvk​can be extended to a basis of LLL, so ∥bi∥≤∥vk∥=λk(L)≤λi(L)\left\lVert b_i\right\rVert\leq\left\lVert v_k\right\rVert=\lambda_k(L)\leq\lambda_i(L)∥bi​∥≤∥vk​∥=λk​(L)≤λi​(L). Otherwise, we have n2≥4n^2\geq4n2≥4. Let vk′=p+qv_k'=p+qvk′​=p+qwherepppis the projection ofvk′v'_kvk′​inLi−1L_{i-1}Li−1​. Since by definition we have∥p∥2≤∥p±bi∥2\left\lVert p\right\rVert^2\leq\left\lVert p\pm b_i\right\rVert^2∥p∥2≤∥p±bi​∥2, we must have

    Furthermore, since

    we have ∥q∥2≤14λk2\left\lVert q\right\rVert^2\leq\frac14\lambda_k^2∥q∥2≤41​λk2​, hence we have

    but since λi(L)2≤∥bi∥2\lambda_i(L)^2\leq \left\lVert b_i\right\rVert^2λi​(L)2≤∥bi​∥2by definition, the case of i=2,3i=2,3i=2,3cannot occur here (hence n=1n=1n=1in these cases).

    hashtag
    Exercises

    1) Show that both definitions of Minkowski-reduced lattice are equivalent

    2) Consider the lattice L=(2000002000002000002011111)L=\begin{pmatrix}2&0&0&0&0\\0&2&0&0&0\\0&0&2&0&0\\0&0&0&2&0\\1&1&1&1&1\end{pmatrix}L=​20001​02001​00201​00021​00001​​. We have showed in a previous exercise that the successive minima are all222but no basisbib_ibi​can satisfy ∥bi∥=λi\left\lVert b_i\right\rVert=\lambda_i∥bi​∥=λi​, show that for any Minkowski reduced basis bib_ibi​, the basis must satisfy ∥bi∥2=max⁡(1,(54)i−4)λi(L)2\left\lVert b_i\right\rVert^2=\max\left(1,\left(\frac54\right)^{i-4}\right)\lambda_i(L)^2∥bi​∥2=max(1,(45​)i−4)λi​(L)2

    {bi}i=1d\left\{b_i\right\}_{i=1}^d{bi​}i=1d​
    bib_ibi​
    LLL
    {bj}j=1i−1\left\{b_j\right\}_{j=1}^{i-1}{bj​}j=1i−1​
    bib_ibi​
    vvv
    {b1,…,bi−1,v}\left\{b_1,\dots,b_{i-1},v\right\}{b1​,…,bi−1​,v}
    LLL
    ∥bi∥≤∥∑j=idcjbj∥gcd⁡(cj)=1\left\lVert b_i\right\rVert\leq\left\lVert\sum_{j=i}^dc_jb_j\right\rVert\quad\gcd\left(c_j\right)=1∥bi​∥≤​j=i∑d​cj​bj​​gcd(cj​)=1
    λi(L)2≤∥bi∥2≤max⁡(1,(54)i−4)λi(L)2\lambda_i(L)^2\leq\left\lVert b_i\right\rVert^2\leq\max\left(1,\left(\frac54\right)^{i-4}\right)\lambda_i(L)^2λi​(L)2≤∥bi​∥2≤max(1,(45​)i−4)λi​(L)2
    vk=a1b1+a2b2+⋯+ai−1bi−1+nvk′ai,n∈Z∥bi∥≤∥vk′∥v_k=a_1b_1+a_2b_2+\dots+a_{i-1}b_{i-1}+nv'_k\quad a_i,n\in\mathbb Z\\ \left\lVert b_i\right\rVert\leq\left\lVert v'_k\right\rVertvk​=a1​b1​+a2​b2​+⋯+ai−1​bi−1​+nvk′​ai​,n∈Z∥bi​∥≤∥vk′​∥
    ∥p∥2≤14∑j=1i−1∥bj∥2\left\lVert p\right\rVert^2\leq\frac14\sum_{j=1}^{i-1}\left\lVert b_j\right\rVert^2∥p∥2≤41​j=1∑i−1​∥bj​∥2
    λk2=∥vk∥2=∥a1b1+a2b2+…ai−1bi−1+p∥2+n2∥q∥2\lambda_k^2=\left\lVert v_k\right\rVert^2=\left\lVert a_1b_1+a_2b_2+\dots a_{i-1}b_{i-1}+p\right\rVert^2+n^2\left\lVert q\right\rVert^2λk2​=∥vk​∥2=∥a1​b1​+a2​b2​+…ai−1​bi−1​+p∥2+n2∥q∥2
    ∥bi∥≤14∑j=1i−1∥bj∥2+14λk2≤14∑j=1i−1max⁡(1,(54)i−4)λj(L)2+14λk(L)2≤14(1+∑j=1i−1max⁡(1,(54)i−4))λi(L)2{=max⁡(1,(54)i−4)λi(L)2i≥4<λi(L)2i=2,3\begin{align*} \left\lVert b_i\right\rVert&\leq\frac14\sum_{j=1}^{i-1}\left\lVert b_j\right\rVert^2+\frac14\lambda_k^2\\ &\leq\frac14\sum_{j=1}^{i-1}\max\left(1,\left(\frac54\right)^{i-4}\right)\lambda_j(L)^2+\frac14\lambda_k(L)^2\\ &\leq\frac14\left(1+\sum_{j=1}^{i-1}\max\left(1,\left(\frac54\right)^{i-4}\right)\right)\lambda_i(L)^2\\ &\begin{cases}=\max\left(1,\left(\frac54\right)^{i-4}\right)\lambda_i(L)^2&i\geq4\\<\lambda_i(L)^2&i=2,3\end{cases}\\ \end{align*}∥bi​∥​≤41​j=1∑i−1​∥bj​∥2+41​λk2​≤41​j=1∑i−1​max(1,(45​)i−4)λj​(L)2+41​λk​(L)2≤41​(1+j=1∑i−1​max(1,(45​)i−4))λi​(L)2{=max(1,(45​)i−4)λi​(L)2<λi​(L)2​i≥4i=2,3​​
    hashtag
    Bounds
    πi\pi_iπi​
    {bj}j=1i−1\left\{b_j\right\}_{j=1}^{i-1}{bj​}j=1i−1​
    ∣∣bi∗∣∣=λ1(πi(L))||b_i^*||=\lambda_1\left(\pi_i(L)\right)∣∣bi∗​∣∣=λ1​(πi​(L))
    4i+3≤(∣∣bi∣∣λi(L))2≤i+34\frac4{i+3}\leq\left(\frac{||b_i||}{\lambda_i(L)}\right)^2\leq\frac{i+3}4i+34​≤(λi​(L)∣∣bi​∣∣​)2≤4i+3​
    We first consider a criteria for a root of a polynomial moduloNNNto exists over the reals as well. Supposef(x)=∑i=0dfixif(x)=\sum_{i=0}^df_ix^if(x)=∑i=0d​fi​xiis a polynomial of degreeddd. Define the ℓ2\ell_2ℓ2​norm∥f(x)∥2\left\lVert f(x)\right\rVert_2∥f(x)∥2​ of a polynomial to be ∑i=0dfi2\sqrt{\sum_{i=0}^df_i^2}∑i=0d​fi2​​. Given ∣x0∣<B,f(x0)=0(modN)|x_0|<B,f(x_0)=0\pmod N∣x0​∣<B,f(x0​)=0(modN), if

    then f(x0)=0f(x_0)=0f(x0​)=0 in R\mathbb RR. The proof is a relatively straightforward chain of inequalities:

    and since f(x0)=0(modN)f(x_0)=0\pmod Nf(x0​)=0(modN)implies f(x0)=kNf(x_0)=kNf(x0​)=kNfor some k∈Zk\in\mathbb Zk∈Z, we know that kkkmust be000to satisfy the inequality above.

    With this, if we can find some polynomials fif_ifi​such that fi(x0)=0(modN)f_i(x_0)=0\pmod Nfi​(x0​)=0(modN), then if we can find some cic_ici​such that ∥∑icifi(x)∥2≤Nd+1\left\lVert\sum_ic_if_i(x)\right\rVert_2\leq\frac N{\sqrt{d+1}}∥∑i​ci​fi​(x)∥2​≤d+1​N​, then we can find x0x_0x0​easily. This gives a brief idea as to why lattices would be useful in such a problem.

    To use lattices, notice that we can encode the polynomial f(x)=∑i=0dfixif(x)=\sum_{i=0}^df_ix^if(x)=∑i=0d​fi​xias the vector with components fif_ifi​. In this way, adding polynomials and multiplying polynomials by numbers still makes sense. Lets suppose that f(x0)=0(modN),x0<Bf(x_0)=0\pmod N,x_0<Bf(x0​)=0(modN),x0​<Band fd=1f_d=1fd​=1(otherwise multiplyfffby fd−1(modN)f_d^{-1}\pmod Nfd−1​(modN). Consider the polynomials gi(x)=Nxig_i(x)=Nx^igi​(x)=Nxiand consider the latticeLLLgenerated by f(Bx)f(Bx)f(Bx)and gi(Bx)g_i(Bx)gi​(Bx), 0≤i≤d−10\leq i\leq d-10≤i≤d−1. As a matrix, the basis vectors are

    As every element in this lattice is some polynomial g(Bx)g(Bx)g(Bx), if f(x0)=0(modN)f(x_0)=0\pmod Nf(x0​)=0(modN), theng(x0)=0(modN)g(x_0)=0\pmod Ng(x0​)=0(modN). Furthermore, if ∣x0∣<B|x_0|<B∣x0​∣<Band a short vectorv(Bx)v(Bx)v(Bx)has length less than Nd+1\frac N{\sqrt{d+1}}d+1​N​, then we have v(x0)=0v(x_0)=0v(x0​)=0in R\mathbb RR.

    The volume of this lattice is NdBd(d+1)2N^dB^{\frac{d(d+1)}2}NdB2d(d+1)​and the lattice has dimension d+1d+1d+1. By using the LLL algorithm, we can find a vector v(Bx)v(Bx)v(Bx) with length at most

    As long as cδ,dNdd+1Bd2<Nd+1c_{\delta,d}N^{\frac d{d+1}}B^{\frac d2}<\frac N{\sqrt{d+1}}cδ,d​Nd+1d​B2d​<d+1​N​, then by the above criteria we know that this vector has x0x_0x0​has a root over R\mathbb RR. This tells us that

    While this isn't the N1dN^{\frac1d}Nd1​bound that we want, this gives us an idea of what we can do to achieve this bound, i.e. add more vectors such that the length of the shortest vector decreases.

    hashtag
    Achieving the N1dN^{\frac1d}Nd1​bound

    One important observation to make is that any coefficients in front ofNxN^xNxdoes not matter as we can simply brute force the top bits of our small root in O(1)O(1)O(1)time. Hence we only need to getB=kN1dB=kN^{\frac1d}B=kNd1​for some fixed constantkkk.

    In order to achieve this, notice that if f(x0)=0(modN)f(x_0)=0\pmod Nf(x0​)=0(modN), then f(x0)h=0(modNh)f(x_0)^h=0\pmod{N^h}f(x0​)h=0(modNh). This loosens the inequality required for a polynomial to have x0x_0x0​as a small root as our modulus is now larger. With this in mind, consider the polynomials

    where we will determinehhhlater. Here gi,j(x0)=0(modNh)g_{i,j}(x_0)=0\pmod{N^h}gi,j​(x0​)=0(modNh), hence we shall consider the lattice LLL generated by gi,j(Bx)g_{i,j}(Bx)gi,j​(Bx). As an example, if we have

    the basis vectors of our lattice would look like

    We have the following immediate computations of LLL:

    hence when using the LLL algorithm, the shortest LLL basis vectorv(Bx)v(Bx)v(Bx)has length

    and we need ∥v(Bx)∥2<Nhdh\left\lVert v(Bx)\right\rVert_2<\frac{N^h}{\sqrt{dh}}∥v(Bx)∥2​<dh​Nh​for v(x0)=0v(x_0)=0v(x0​)=0. Hence we have

    Since lim⁡h→∞h−1dh−1=1d\lim_{h\to\infty}\frac{h-1}{dh-1}=\frac1dlimh→∞​dh−1h−1​=d1​, this will achieve the N1dN^{\frac1d}Nd1​bound that we want. However as for big hhh, the LLL algorithm would take a very long time, we typically choose a suitably largehhhsuch that the algorithm is still polynomial in log⁡N,d\log N,dlogN,dand brute force the remaining bits.

    hashtag
    Exercises

    1) We often see h=⌈max⁡(d+dε−1d2ϵ,7d)⌉h=\left\lceil\max\left(\frac{d+d\varepsilon-1}{d^2\epsilon},\frac7d\right)\right\rceilh=⌈max(d2ϵd+dε−1​,d7​)⌉in literature. We shall now show where this mysterious7d\frac7dd7​comes from. The other term will appear in the next exercise. Typically, one sets δ=34\delta=\frac34δ=43​to simplify calculations involving the LLL algorithm as 44δ−1=2\frac4{4\delta-1}=24δ−14​=2. Suppose we want B>12Nh−1dh−1B>\frac12N^{\frac{h-1}{dh-1}}B>21​Ndh−1h−1​, show that this gives us dh≥7dh\geq7dh≥7.

    2) We show that we can indeed find small roots less thanN1dN^{\frac1d}Nd1​in polynomial time. In the worse case, the longest basis vector cannot exceed O(Bdh−1Nh)O\left(B^{dh-1}N^h\right)O(Bdh−1Nh). Hence the LLL algorithm will run in at mostO(d6h6(d+h)2log⁡2N)O(d^6h^6(d+h)^2\log^2N)O(d6h6(d+h)2log2N)time.

    Let

    and choose ε=1log⁡N\varepsilon=\frac1{\log N}ε=logN1​, then NεN^\varepsilonNεis a constant hence the number of bits needed to be brute forced is a constant. This gives us the approximate run time of O((d+1dlog⁡N)2log⁡8N)O((d+\frac1d\log N)^2\log^8N)O((d+d1​logN)2log8N).

    3) We shall show that this is the best bound we can hope for using lattice methods. Suppose there exists some algorithm that finds roots less than O(N1d+ϵ)O\left(N^{\frac1d+\epsilon}\right)O(Nd1​+ϵ)in polynomial time. Then consider the case whenN=p2N=p^2N=p2and f(x)=x2+pxf(x)=x^2+pxf(x)=x2+px. Show that this forces the lattice to have a size too big for the algorithm to run in polynomial time, assuming the algorithm finds all small roots.

    f(x)∈Z[x]f(x)\in\mathbb Z[x]f(x)∈Z[x]
    ddd
    NNN
    f(x0)=0(modN),∣x0∣<N1df(x_0)=0\pmod{N},|x_0|<N^{\frac1d}f(x0​)=0(modN),∣x0​∣<Nd1​
    x0x_0x0​
    log⁡N\log NlogN
    ddd
    g(x)g(x)g(x)
    g(x0)=0g(x_0)=0g(x0​)=0
    R\mathbb RR
    x0x_0x0​
    N1dN^{\frac1d}Nd1​
    ∥f(Bx)∥2=∑i=0d(fiBi)2≤Nd+1\left\lVert f(Bx)\right\rVert_2=\sqrt{\sum_{i=0}^d\left(f_iB^i\right)^2}\leq\frac N{\sqrt{d+1}}∥f(Bx)∥2​=i=0∑d​(fi​Bi)2​≤d+1​N​
    Nd+1≥∑i=0d(fiBi)2≥∑i=0d(fix0i)2≥1d+1∑i=0d∣fix0i∣≥1d+1∣∑i=0dfix0i∣\begin{align*} \frac N{\sqrt{d+1}}&\geq\sqrt{\sum_{i=0}^d\left(f_iB^i\right)^2}\\&\geq\sqrt{\sum_{i=0}^d\left(f_ix_0^i\right)^2}\\ &\geq\frac1{\sqrt{d+1}}\sum_{i=0}^d\left|f_ix_0^i\right|\\ &\geq\frac1{\sqrt{d+1}}\left|\sum_{i=0}^df_ix_0^i\right|\\ \end{align*}d+1​N​​≥i=0∑d​(fi​Bi)2​≥i=0∑d​(fi​x0i​)2​≥d+1​1​i=0∑d​​fi​x0i​​≥d+1​1​​i=0∑d​fi​x0i​​​
    B=(N00…000NB0…0000NB2…00⋮⋮⋮⋱⋮⋮000…NBd−10f0f1Bf2B2…fd−1Bd−1Bd)\mathcal B=\begin{pmatrix} N&0&0&\dots&0&0\\ 0&NB&0&\dots&0&0\\ 0&0&NB^2&\dots&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\dots&NB^{d-1}&0\\ f_0&f_1B&f_2B^2&\dots&f_{d-1}B^{d-1}&B^d\\ \end{pmatrix} B=​N00⋮0f0​​0NB0⋮0f1​B​00NB2⋮0f2​B2​………⋱……​000⋮NBd−1fd−1​Bd−1​000⋮0Bd​​
    ∥v(Bx)∥2=(44δ−1)d4⏟cδ,dvol(L)1d+1=cδ,dNdd+1Bd2\left\lVert v(Bx)\right\rVert_2=\underbrace{\left(\frac4{4\delta-1}\right)^{\frac d4}}_{c_{\delta,d}}\text{vol}(L)^\frac1{d+1}=c_{\delta,d}N^{\frac d{d+1}}B^{\frac d2}∥v(Bx)∥2​=cδ,d​(4δ−14​)4d​​​vol(L)d+11​=cδ,d​Nd+1d​B2d​
    B<N2d(d+1)(cδ,dd+1)−2dB<N^{\frac2{d(d+1)}}\left(c_{\delta,d}\sqrt{d+1}\right)^{-\frac 2d}B<Nd(d+1)2​(cδ,d​d+1​)−d2​
    gi,j(x)=Nh−jf(x)jxi0≤i<d,0≤j<hg_{i,j}(x)=N^{h-j}f(x)^jx^i\quad0\leq i<d,0\leq j<hgi,j​(x)=Nh−jf(x)jxi0≤i<d,0≤j<h
    f(x)=x3+2x2+3x+4h=3f(x)=x^3+2x^2+3x+4\quad h=3f(x)=x3+2x2+3x+4h=3
    (N3000000000BN3000000000B2N30000004N23BN22B2N2B3N20000004BN23B2N22B3N2B4N20000004B2N23B3N22B4N2B5N200016N24BN25B2N20B3N10B4N4B5NB6N00016BN24B2N25B3N20B4N10B5N4B6NB7N00016B2N24B3N25B4N20B5N10B6N4B7NB8N)\footnotesize{\left(\begin{array}{rrrrrrrrr} N^{3} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & B N^{3} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & B^{2} N^{3} & 0 & 0 & 0 & 0 & 0 & 0 \\ 4 \, N^{2} & 3 \, B N^{2} & 2 \, B^{2} N^{2} & B^{3} N^{2} & 0 & 0 & 0 & 0 & 0 \\ 0 & 4 \, B N^{2} & 3 \, B^{2} N^{2} & 2 \, B^{3} N^{2} & B^{4} N^{2} & 0 & 0 & 0 & 0 \\ 0 & 0 & 4 \, B^{2} N^{2} & 3 \, B^{3} N^{2} & 2 \, B^{4} N^{2} & B^{5} N^{2} & 0 & 0 & 0 \\ 16 \, N & 24 \, B N & 25 \, B^{2} N & 20 \, B^{3} N & 10 \, B^{4} N & 4 \, B^{5} N & B^{6} N & 0 & 0 \\ 0 & 16 \, B N & 24 \, B^{2} N & 25 \, B^{3} N & 20 \, B^{4} N & 10 \, B^{5} N & 4 \, B^{6} N & B^{7} N & 0 \\ 0 & 0 & 16 \, B^{2} N & 24 \, B^{3} N & 25 \, B^{4} N & 20 \, B^{5} N & 10 \, B^{6} N & 4 \, B^{7} N & B^{8} N \end{array}\right)}​N3004N20016N00​0BN303BN24BN2024BN16BN0​00B2N32B2N23B2N24B2N225B2N24B2N16B2N​000B3N22B3N23B3N220B3N25B3N24B3N​0000B4N22B4N210B4N20B4N25B4N​00000B5N24B5N10B5N20B5N​000000B6N4B6N10B6N​0000000B7N4B7N​00000000B8N​​
    dim⁡(L)=dhvol(L)=Ndh(h+1)2B(dh−1)dh2\dim(L)=dh\quad\text{vol}(L)=N^{\frac{dh(h+1)}2}B^{\frac{(dh-1)dh}2}dim(L)=dhvol(L)=N2dh(h+1)​B2(dh−1)dh​
    ∥v(Bx)∥2=(44δ−1)dim⁡(L)−14vol(L)1dim⁡(L)=(44δ−1)dh−14Nh+12Bdh−12\begin{align*} \left\lVert v(Bx)\right\rVert_2&=\left(\frac4{4\delta-1}\right)^{\frac{\dim(L)-1}4}\text{vol}(L)^{\frac1{\dim(L)}}\\ &=\left(\frac4{4\delta-1}\right)^{\frac{dh-1}4}N^{\frac{h+1}2}B^{\frac{dh-1}2}\\ \end{align*}∥v(Bx)∥2​​=(4δ−14​)4dim(L)−1​vol(L)dim(L)1​=(4δ−14​)4dh−1​N2h+1​B2dh−1​​
    B<4δ−14(1dh)1dh−1Nh−1dh−1B<\sqrt{\frac{4\delta-1}4}\left(\frac1{dh}\right)^{\frac1{dh-1}}N^{\frac{h-1}{dh-1}}B<44δ−1​​(dh1​)dh−11​Ndh−1h−1​
    ε=1d−h−1dh−1h=d+dε−1d2ϵ≈1dε\varepsilon=\frac1d-\frac{h-1}{dh-1}\quad h=\frac{d+d\varepsilon-1}{d^2\epsilon}\approx\frac1{d\varepsilon}ε=d1​−dh−1h−1​h=d2ϵd+dε−1​≈dε1​

    NTRU

    (Z/nZ)∗\left(\mathbb Z/n\mathbb Z\right)^*(Z/nZ)∗
    (Z/nZ)∗\left(\mathbb Z/n\mathbb Z\right)^*(Z/nZ)∗
    mod n\text{mod }nmod n
    xxx
    ax=b(modn)a^x=b\pmod nax=b(modn)
    nnn
    n=pn=pn=p
    p<1050∼2166p < 10^{50} \sim 2 ^{166}p<1050∼2166
    n=pkn = p^kn=pk
    E(k)E(k)E(k)
    ⟨bi∗,bj∗⟩={0i≠j∥bi∗∥2i=j\langle b_i^*,b_j^*\rangle=\begin{cases}0&i\neq j\\\left\lVert b_i^*\right\rVert^2&i=j\end{cases}⟨bi∗​,bj∗​⟩={0∥bi∗​∥2​i=ji=j​
    B\mathcal BB
    iii
    bib_ibi​
    B∗\mathcal B^*B∗
    iii
    bi∗b_i^*bi∗​
    B=μB∗\mathcal B=\mu\mathcal B^*B=μB∗
    μi,i=1,μj,i=0\mu_{i,i}=1,\mu_{j,i}=0μi,i​=1,μj,i​=0
    μi,j\mu_{i,j}μi,j​
    R4\mathbb R^4R4
    b1=(−1−231)b2=(−6−451)b3=(551−3)\begin{matrix} b_1 &= & (&-1&-2&3&1&)\\ b_2 &= & (&-6&-4&5&1&)\\ b_3 &= & (&5&5&1&-3&) \end{matrix}b1​b2​b3​​===​(((​−1−65​−2−45​351​11−3​)))​
    B∗\mathcal B^*B∗
    μ\muμ
    B=μB∗\mathcal B=\mu\mathcal B^*B=μB∗
    B∗\mathcal B^*B∗
    det⁡(BBT)=det⁡(B∗B∗T)=∏i∥bi∗∥\det\left(\mathcal B\mathcal B^T\right)=\det\left(\mathcal B^*\mathcal B^{*T}\right)=\prod_i\left\lVert b_i^*\right\rVertdet(BBT)=det(B∗B∗T)=i∏​∥bi∗​∥
    bi∗b_i^*bi∗​
    μμT=1\mu\mu^T=1μμT=1
    B∗B∗T\mathcal B^*\mathcal B^{*T}B∗B∗T
    ∥bi∗∥\left\lVert b_i^*\right\rVert∥bi∗​∥
    det⁡(BBT)=det⁡(B∗B∗T)=∏i∥bi∗∥\det\left(\mathcal B\mathcal B^T\right)=\det\left(\mathcal B^*\mathcal B^{*T}\right)=\prod_i\left\lVert b_i^*\right\rVertdet(BBT)=det(B∗B∗T)=∏i​∥bi∗​∥
    B=LDO\mathcal B=LDOB=LDO
    LLL
    111
    DDD
    OOO
    OOT=1OO^T=1OOT=1
    B∗=DO\mathcal B^*=DOB∗=DO
    μ=L\mu=Lμ=L
    circle-info

    We can further improve this by optimizing the Gram Schmidt computation as this algorithm does not modify B∗\mathcal B^*B∗at all. Furthermoreμ\muμchanges in a very predictable fasion and when vectors are swapped, one can write explicit formulas for howB∗\mathcal B^*B∗changes as well.

    Next, we need to figure a swapping condition. Naively, we want

    for all iii. However, such a condition does not guarantee termination in polynomial time. As short basis vectors should be almost orthogonal, we may also want to incorperate this notion. Concretely, we want ∣μi,j∣\left|\mu_{i,j}\right|∣μi,j​∣to be somewhat small for all pairs of i,ji,ji,j, i.e. we may want something like

    However, since μi,j=⟨bi,bj∗⟩⟨bj∗,bj∗⟩\mu_{i,j}=\frac{\langle b_i,b_j^*\rangle}{\langle b_j^*,b_j^*\rangle}μi,j​=⟨bj∗​,bj∗​⟩⟨bi​,bj∗​⟩​, this condition is easily satisfied for a sufficiently long bj∗b_j^*bj∗​, which is not what we want. The key idea is to merge these two in some way and was first noticed by Lovász - named the Lovász condition:

    It turns out that using this condition, the algorithm above terminates in polynomial time! More specifically, it has a time complexity of O(d5nlog⁡3B)O\left(d^5n\log^3B\right)O(d5nlog3B)where we havedddbasis vectors as a subset of Rn\mathbb R^nRnand BBBis a bound for the largest norm of bib_ibi​. 14<δ\frac14<\delta41​<δ ensures that the lattice vectors are ordered roughly by size and δ<1\delta<1δ<1ensures the algorithm terminates.

    hashtag
    Polynomial time proof

    This follows the proof provided by the authors of the LLL paper. We first prove that the algorithm terminates by showing it swaps the vectors finitely many times. Letdddbe the number of basis vectors as a subset of Rn\mathbb R^nRn. Let did_idi​be the volume of the lattice generated by {bj}j=1i\left\{b_j\right\}_{j=1}^i{bj​}j=1i​at each step of the algorithm. We have di=∏j=1i∥bj∗∥d_i=\prod_{j=1}^i\left\lVert b_j^*\right\rVertdi​=∏j=1i​​bj∗​​. Now consider the quantity

    This quantity only changes whenever some bi∗b_i^*bi∗​changes, i.e when swaps happen. Let's consider what happens when we swap bib_ibi​and bi+1b_{i+1}bi+1​. Recall the Gram-Schmidt algorithm:

    From this, see that when we swap bib_ibi​and bi+1b_{i+1}bi+1​, bi∗b_i^*bi∗​is replaced by bi+1∗+μi+1,ibi∗b_{i+1}^*+\mu_{i+1,i}b_i^*bi+1∗​+μi+1,i​bi∗​. Now using the Lovász condition, we see that we have∥bi+1∗+μi+1,ibi∗∥2<δ∥bi∗∥2\left\lVert b_{i+1}^*+\mu_{i+1,i}b_i^*\right\rVert^2<\delta\left\lVert b_i^*\right\rVert^2​bi+1∗​+μi+1,i​bi∗​​2<δ∥bi∗​∥2, hence the value of did_idi​must decrease by at least δ\deltaδ, i.e. the new did_idi​is less than diδ\frac{d_i}\deltaδdi​​. All other dj,j≠id_j,j\neq idj​,j=imust remain the same as the volume remains fixed when we swap basis vectors around. Hence at each swap, DDDdecreases by δ\deltaδ. This is why we need δ<1\delta<1δ<1.Now we are left with showing did_idi​is bounded from below then we are done.

    Let λ1(L)\lambda_1(L)λ1​(L)be the length of the shortest (nonzero) vector in the lattice. We can treat did_idi​as the volume of the lattice LiL_iLi​generated by{bj}j=1i\left\{b_j\right\}_{j=1}^i{bj​}j=1i​. Let xix_ixi​be the shortest vector in the lattice in LiL_iLi​. By using Minkowski's lattice point theorem, we have

    (Note that the value of CiC_iCi​isn't particularly important, one can use a easier value like i\sqrt ii​)

    Hence we see that did_idi​, and hence DDDhas a (loose) lower bound Dmin⁡=∏i=1ddi,min⁡D_{\min}=\prod_{i=1}^dd_{i,\min}Dmin​=∏i=1d​di,min​, meaning that there are at most log⁡Dlog⁡Dmin⁡δ\frac{\log D}{\log D_{\min}\delta}logDmin​δlogD​swaps. Since at each iteration,kkkeither increases by111when there is no swaps or decreases by at most111when there is swaps and kkkranges from222toddd, the number of time the loop runs must be at most 2log⁡Dlog⁡Dmin⁡δ+d2\frac{\log D}{\log D_{\min}\delta}+d2logDmin​δlogD​+d, hence the algorithm terminates.

    This proof also gives us a handle on the time complexity of the operation. LetBBBis the length of the longest input basis vector. Since we have di≤Bid_i\leq B^idi​≤Bi, D≤Bm2+m2D\leq B^{\frac{m^2+m}2}D≤B2m2+m​and the algorithm loops O(d2log⁡B)O\left(d^2\log B\right)O(d2logB)times. The Gram-Schmidt orthogonalization is the most expensive part in the entire process, taking up O(d2n)O\left(d^2n\right)O(d2n)arithmetic operations. By using classical algorithm for arithmetic operations, each takes O(nlog⁡B)O\left(n\log B\right)O(nlogB)time. From this, we deduce that the time complexity of the LLL algorithm is O(d5mlog⁡2B)O\left(d^5m\log^2B\right)O(d5mlog2B), a somewhat reasonable polynomial time algorithm.

    Let bib_ibi​be the output of the LLL algorithm, it turns out that we have the bound

    which requires δ>14\delta>\frac14δ>41​. Such bounds for the shortest vector will be elaborated in more detail in the section on reduced basis.

    hashtag
    Exercises

    1) Implement the LLL in sage and experimentally verify that DDDdoes indeed decrease byδ\deltaδeach time.

    2) Show that the time complexity analysis is correct, and indeed each loop takes at most O(d2n)O\left(d^2n\right)O(d2n)operations.

    ∥bi∥≤∥bi+1∥\left\lVert b_i\right\rVert\leq\left\lVert b_{i+1}\right\rVert∥bi​∥≤∥bi+1​∥
    ∣μi,j∣≤c|\mu_{i,j}|\leq c∣μi,j​∣≤c
    δ∥bi∗∥2≤∥bi+1∗+μi+1,ibi∗∥2δ∈(14,1)\delta\left\lVert b_i^*\right\rVert^2\leq\left\lVert b_{i+1}^*+\mu_{i+1,i}b_i^*\right\rVert^2\quad\delta\in\left(\frac14,1\right)δ∥bi∗​∥2≤​bi+1∗​+μi+1,i​bi∗​​2δ∈(41​,1)
    D=∏i=1ddiD=\prod_{i=1}^dd_iD=i=1∏d​di​
    bi∗=bi−∑j=1i−1μi,jbj∗μi,j=⟨bi,bj∗⟩⟨bj∗,bj∗⟩b_i^*=b_i-\sum_{j=1}^{i-1}\mu_{i,j}b_j^*\quad\mu_{i,j}=\frac{\langle b_i,b_j^*\rangle}{\langle b_j^*,b_j^*\rangle}bi∗​=bi​−j=1∑i−1​μi,j​bj∗​μi,j​=⟨bj∗​,bj∗​⟩⟨bi​,bj∗​⟩​
    λ1(L)≤xi≤2πΓ(i2+1)1i⏟Cidi1idi≥λ1(L)iCii=di,min⁡\begin{align*} \lambda_1(L)\leq x_i&\leq\underbrace{\frac2{\sqrt\pi}\Gamma\left(\frac i2+1\right)^{\frac1i}}_{C_i}d_i^\frac1i\\ d_i&\geq\frac{\lambda_1(L)^i}{C_i^i}=d_{i,\min} \end{align*}λ1​(L)≤xi​di​​≤Ci​π​2​Γ(2i​+1)i1​​​dii1​​≥Cii​λ1​(L)i​=di,min​​
    ∥b1∥≤(44δ−1)d−14vol(L)1d\left\lVert b_1\right\rVert\leq\left(\frac4{4\delta-1}\right)^{\frac{d-1}4}\text{vol}(L)^\frac1d∥b1​∥≤(4δ−14​)4d−1​vol(L)d1​
    sage: R = Integers(99)
    sage: a = R(4)
    sage: b = a^9
    sage: b.log(a)
    9
    x = int(pari(f"znlog({int(b)},Mod({int(a)},{int(n)}))"))
    x = gp.znlog(b, gp.Mod(a, n))
    p = getPrime(36)
    n = p^2
    K = Zmod(n)
    a = K.multiplicative_generator()
    b = a^123456789
    
    time int(pari(f"znlog({int(b)},Mod({int(a)},{int(n)}))")) 
    # CPU times: user 879 µs, sys: 22 µs, total: 901 µs
    # Wall time: 904 µs
    # 123456789
    
    time b.log(a)
    # CPU times: user 458 ms, sys: 17 ms, total: 475 ms
    # Wall time: 478 ms
    # 123456789
    
    time discrete_log(b,a)
    # CPU times: user 512 ms, sys: 24.5 ms, total: 537 ms
    # Wall time: 541 ms
    # 123456789
    p = getPrime(100)
    n = p^2
    K = Zmod(n)
    a = K.multiplicative_generator()
    b = a^123456789
    
    time int(pari(f"znlog({int(b)},Mod({int(a)},{int(n)}))")) 
    # CPU times: user 8.08 s, sys: 82.2 ms, total: 8.16 s
    # Wall time: 8.22 s
    # 123456789
    B = Matrix([
    [-1, -2, 3, 1],
    [-6, -4, 5, 1],
    [5, 5, 1, -3]])
    
    B.gram_schmidt()
    (
    [-1 -2  3  1]  [ 1  0  0]
    [-4  0 -1 -1]  [ 2  1  0]
    [ 0  3  3 -3], [-1 -1  1]
    )
    def LLL(B):
        d = B.nrows()
        i = 1
        while i<d:
            size_reduce(B)
            if swap_condition(B):
                i += 1
            else:
                B[i],B[i-1] = B[i-1],B[i]
                i = max(i-1,1)
        return B
    
    def size_reduce(B):
        d = B.nrows()
        i = 1
        while i<d:
            Bs,M = B.gram_schmidt()
            for j in reversed(range(i)):
                B[i] -= round(M[i,j])*B[j]
                Bs,M = B.gram_schmidt()
        return B
    Here μ\muμis actually the Gram-Schmidt coefficient μ2,1\mu_{2,1}μ2,1​and it turns out that this algorithm will always find the shortest possible basis! Using the basis

    the Lagrange reduction looks like

    and here we see it clearly gives the shortest vectors.

    hashtag
    Optimality proof

    Let LLLbe a lattice. The basis b1,b2b_1,b_2b1​,b2​is defined to be the shortest for any other basis b1′,b2′,∥b1′∥≤∥b2′∥b_1',b_2',\left\lVert b_1'\right\rVert\leq\left\lVert b_2'\right\rVertb1′​,b2′​,∥b1′​∥≤∥b2′​∥, we have ∥b1∥≤∥b1′∥\left\lVert b_1\right\rVert\leq\left\lVert b_1'\right\rVert∥b1​∥≤∥b1′​∥and ∥b2∥≤∥b2′∥\left\lVert b_2\right\rVert\leq\left\lVert b_2'\right\rVert∥b2​∥≤∥b2′​∥. Note that this generally cannot be generalized to other dimensions, however in dimension 2, this is possible and is given by Lagrange's algorithm. The proof is a somewhat messy sequence of inequalities that eventually lead to the conclusion we want.

    Let b1,b2b_1,b_2b1​,b2​be the output of the Lagrange reduction for some lattice LLL. To prove that Lagrange reduction gives the shortest basis, we first show that ∥b1∥\left\lVert b_1\right\rVert∥b1​∥is the shortest vector in LLL.

    We know that ∣⟨b1,b2⟩∣∥b1∥2≤12\frac{\left|\langle b_1,b_2\rangle\right|}{\left\lVert b_1\right\rVert^2}\le\frac12∥b1​∥2∣⟨b1​,b2​⟩∣​≤21​from the algorithm directly. Let v=mb1+nb2∈Lv=mb_1+nb_2\in Lv=mb1​+nb2​∈Lbe any element in LLL. We first show that ∥b1∥≤∥v∥\left\lVert b_1\right\rVert\leq\left\lVert v\right\rVert∥b1​∥≤∥v∥:

    Since m2−mn+n2=(m−n2)2+34n2m^2-mn+n^2=\left(m-\frac n2\right)^2+\frac34n^2m2−mn+n2=(m−2n​)2+43​n2, this quantity is only 000when m=n=0m=n=0m=n=0and is a positive integer for all other cases, hence ∥v∥≥∥b1∥\left\lVert v\right\rVert\geq\left\lVert b_1\right\rVert∥v∥≥∥b1​∥and ∥b1∥\left\lVert b_1\right\rVert∥b1​∥is a shortest vector of LLL. Note that we can have multiple vectors with the same norm as b1b_1b1​, for instance −b1-b_1−b1​. So this is not a unique shortest vector.

    Suppose there exists some basis b1′,b2′b'_1,b'_2b1′​,b2′​for LLLsuch that ∥b1′∥≤∥b2′∥\left\lVert b_1'\right\rVert\leq\left\lVert b_2'\right\rVert∥b1′​∥≤∥b2′​∥. We show that ∥b2∥≤∥b2′∥\left\lVert b_2\right\rVert\leq\left\lVert b_2'\right\rVert∥b2​∥≤∥b2′​∥. Let b2′=mb1+nb2b_2'=mb_1+nb_2b2′​=mb1​+nb2​.

    If n=0n=0n=0, then b2′=±b1b_2'=\pm b_1b2′​=±b1​as b1′,b2′b_1',b_2'b1′​,b2′​must form a basis. This means that ∥b1∥=∥b1′∥=∥b2′∥\left\lVert b_1\right\rVert=\left\lVert b_1'\right\rVert=\left\lVert b_2'\right\rVert∥b1​∥=∥b1′​∥=∥b2′​∥ and by the inequality above, we must have ±b1′=b2\pm b_1'=b_2±b1′​=b2​or ±b1′=b1+b2\pm b_1'=b_1+b_2±b1′​=b1​+b2​. The first case tells us that ∥b1′∥=∥b2∥\left\lVert b'_1\right\rVert=\left\lVert b_2\right\rVert∥b1′​∥=∥b2​∥. By squaring the second case, we get

    but since ∥b1∥\left\lVert b_1\right\rVert∥b1​∥is the shortest vector, ∥b1∥=∥b2∥\left\lVert b_1\right\rVert=\left\lVert b_2\right\rVert∥b1​∥=∥b2​∥.

    Otherwise, we have m,n≠0m,n\neq0m,n=0 and m2−mn+n2≥1m^2-mn+n^2\geq1m2−mn+n2≥1, so

    Hence proving Lagrange's algorithm indeed gives us the shortest basis vectors.

    hashtag
    Exercises

    1) Show that the output of Lagrange's algorithm generate the same lattice as the input.

    2) Find a case where ∥b1∥=∥b2∥=∥b1+b2∥\left\lVert b_1\right\rVert=\left\lVert b_2\right\rVert=\left\lVert b_1+b_2\right\rVert∥b1​∥=∥b2​∥=∥b1​+b2​∥. Notice that the vectors here is the equality case for the bound given in Exercise 4 of the introduction, this actually tells us that the optimal lattice circle packing in 2D is given by this precise lattice! It turns out that this is actually the optimal circle packing in 2D but the proof is significantly more involved. (See https://arxiv.org/abs/1009.4322arrow-up-right for the details)

    3*) Let μ2,1=⌊μ2,1⌉+ε=μ+ϵ\mu_{2,1}=\lfloor\mu_{2,1}\rceil+\varepsilon=\mu+\epsilonμ2,1​=⌊μ2,1​⌉+ε=μ+ϵ, show that

    and show that ∣μ∣≥2|\mu|\geq2∣μ∣≥2for all steps in the algorithm except the first and last, hence ∥b1∥∥b2∥\left\lVert b_1\right\rVert\left\lVert b_2\right\rVert∥b1​∥∥b2​∥decreases by at least 3\sqrt33​ at each loop and the algorithm runs in polynomial time.

    mmm
    nnn
    mmm
    b1=(−1.8,1.2)b2=(−3.6,2.3)\begin{matrix} b_1&=&(-1.8,1.2)\\ b_2&=&(-3.6,2.3) \end{matrix}b1​b2​​==​(−1.8,1.2)(−3.6,2.3)​
    ∥v∥2=∥mb1+nb2∥2=m2∥b1∥2+2mn⟨b1,b2⟩+n2∥b2∥2≥m2∥b1∥2−∣mn∣∥b1∥2+n2∥b1∥2=(m2−∣mn∣+n2)∥b1∥2\begin{align*} \left\lVert v\right\rVert^2&=\left\lVert mb_1+nb_2\right\rVert^2\\ &=m^2\left\lVert b_1\right\rVert^2+2mn\langle b_1,b_2\rangle+n^2\left\lVert b_2\right\rVert^2\\ &\geq m^2\left\lVert b_1\right\rVert^2-|mn|\left\lVert b_1\right\rVert^2+n^2\left\lVert b_1\right\rVert^2\\ &=\left(m^2-|mn|+n^2\right)\left\lVert b_1\right\rVert^2\\ \end{align*}∥v∥2​=∥mb1​+nb2​∥2=m2∥b1​∥2+2mn⟨b1​,b2​⟩+n2∥b2​∥2≥m2∥b1​∥2−∣mn∣∥b1​∥2+n2∥b1​∥2=(m2−∣mn∣+n2)∥b1​∥2​
    ∥b1′∥2=∥b1+b2∥2∥b1′∥2=∥b1∥2+2⟨b1,b2⟩+∥b2∥20=2⟨b1,b2⟩+∥b2∥2∥b1∥2≤∥b2∥2\begin{align*} \left\lVert b'_1\right\rVert^2&=\left\lVert b_1+b_2\right\rVert^2\\ \left\lVert b'_1\right\rVert^2&=\left\lVert b_1\right\rVert^2+2\langle b_1,b_2\rangle+\left\lVert b_2\right\rVert^2\\ 0&=2\langle b_1,b_2\rangle+\left\lVert b_2\right\rVert^2\\ \left\lVert b_1\right\rVert^2&\leq\left\lVert b_2\right\rVert^2\\ \end{align*}∥b1′​∥2∥b1′​∥20∥b1​∥2​=∥b1​+b2​∥2=∥b1​∥2+2⟨b1​,b2​⟩+∥b2​∥2=2⟨b1​,b2​⟩+∥b2​∥2≤∥b2​∥2​
    ∥b2′∥2=m2∥b1∥2+2mn⟨b1,b2⟩+n2∥b2∥2≥m2∥b1∥2−∣mn∣∥b1∥2+n2∥b2∥2=n2(∥b2∥2−∥b1∥2)+(m2−∣mn∣+n2)∥b1∥2≥(n2−1)(∥b2∥2−∥b1∥2)+∥b2∥2≥∥b2∥2\begin{align*} \left\lVert b'_2\right\rVert^2&=m^2\left\lVert b_1\right\rVert^2+2mn\langle b_1,b_2\rangle+n^2\left\lVert b_2\right\rVert^2\\ &\geq m^2\left\lVert b_1\right\rVert^2-|mn|\left\lVert b_1\right\rVert^2+n^2\left\lVert b_2\right\rVert^2\\ &=n^2\left(\left\lVert b_2\right\rVert^2-\left\lVert b_1\right\rVert^2\right)+\left(m^2-|mn|+n^2\right)\left\lVert b_1\right\rVert^2\\ &\geq\left(n^2-1\right)\left(\left\lVert b_2\right\rVert^2-\left\lVert b_1\right\rVert^2\right)+\left\lVert b_2\right\rVert^2\\ &\geq\left\lVert b_2\right\rVert^2 \end{align*}∥b2′​∥2​=m2∥b1​∥2+2mn⟨b1​,b2​⟩+n2∥b2​∥2≥m2∥b1​∥2−∣mn∣∥b1​∥2+n2∥b2​∥2=n2(∥b2​∥2−∥b1​∥2)+(m2−∣mn∣+n2)∥b1​∥2≥(n2−1)(∥b2​∥2−∥b1​∥2)+∥b2​∥2≥∥b2​∥2​
    ∥b2∥2≥((∣μ∣−12)2−ε2)∥b1∥2+∥b2−μb1∥\left\lVert b_2\right\rVert^2\geq\left(\left(|\mu|-\frac12\right)^2-\varepsilon^2\right)\left\lVert b_1\right\rVert^2+\left\lVert b_2-\mu b_1\right\rVert∥b2​∥2≥((∣μ∣−21​)2−ε2)∥b1​∥2+∥b2​−μb1​∥
    Short integer solution problem

    Definition

    Let SISn,m,q,βSIS_{n, m, q, \beta}SISn,m,q,β​ be a Short Integer Solution problem. We define it as such:

    Given mmm uniformly random vectors ai∈(Z/qZ)na_i∈(\mathbb{Z}/q\mathbb Z)^nai​∈(Z/qZ)n, forming the columns of a matrix A∈(Z/qZ)n×mA∈(\mathbb{Z}/q\mathbb Z)^{n×m}A∈(Z/qZ)n×m, find a nonzero integer vector z∈Zmz∈\mathbb{Z}^mz∈Zm of norm ‖z‖≤β‖z‖ ≤β‖z‖≤β (short) such that

    Without the constraint β\betaβ the solution would be as simple as Gaussian elimination. Also we want β<q\beta < qβ<q otherwise z=(q,0,...,0)∈Zmz = (q,0, ..., 0) \in \mathbb{Z}^mz=(q,0,...,0)∈Zm would be a fine solution.

    Notice that a solution zzz for AAA can be converted to a solution for the extension [A∣A′][A| A'][A∣A′] by appending 000s to zzz ⇒\Rightarrow⇒

    • big m⇒m \Rightarrowm⇒ easy (the more vectors we are given, the easier the problem becomes)

    • big n⇒n \Rightarrown⇒ hard (the more dimension we work in the harder the problem becomes)

    Solution existence is based on parameters set. One should think about them as follows:

    • nnn is the security parameter. The bigger it is the harder the problem becomes

    • mmm is set depending from application to application. Usually m≫nm \gg nm≫n

    • q=poly(n)q = \text{poly}(n)q=poly(n), think of it as q=O(n2)q = \mathcal{O}(n^2)q=O(n2)

    • the bound is set depending on application and

    hashtag
    SIS as a SVP problem

    // TODO

    hashtag
    Ajtai's hashing function

    • Parameters: m,n,q∈Zm, n, q \in \mathbb{Z}m,n,q∈Z, m>nlog⁡2qm > n \log_2 qm>nlog2​q

    • Key: A∈(Z/qZ)n×mA \in (\mathbb{Z}/q\mathbb Z)^{n \times m}A∈(Z/qZ)n×m

    • Input: x∈{0,1}m⇒x \in \{0, 1\}^m \Rightarrowx∈{0,1}m⇒ Short vector

    • Output: where

    hashtag
    Hash function properties:

    Compression

    We know x∈{0,1}m⇒∣X∣=2nx \in \{0, 1\}^m \Rightarrow |\mathcal{X}| = 2^nx∈{0,1}m⇒∣X∣=2n and Ax∈Y=(Z/qZ)n⇒∣(Z/qZ)n∣=qn=(2log⁡q)nAx \in \mathcal Y = (\mathbb{Z}/q\mathbb Z)^n \Rightarrow |(\mathbb{Z}/q\mathbb Z)^n| = q^n = (2^{\log q})^nAx∈Y=(Z/qZ)n⇒∣(Z/qZ)n∣=qn=(2logq)n. Since we chose m>nlog⁡q⇒∣X∣>∣Y∣m > n \log q \Rightarrow |\mathcal{X}| > |\mathcal{Y}| m>nlogq⇒∣X∣>∣Y∣.

    Collision resistance:

    triangle-exclamation

    halp here

    Sage example:

    hashtag
    Cryptanalysis

    Inverting the function:

    Given AAA and yyy find x∈{0,1}mx \in \{0, 1\}^mx∈{0,1}m such that Ax=y mod qAx = y \bmod qAx=ymodq

    Formulating as a lattice problem:

    Find arbitrary ttt such that At=y mod qAt = y \bmod qAt=ymodq

    • All solutions to Ax=yAx = yAx=y are of the form t+L⊥t + L^{\perp}t+L⊥ where L⊥(A)={x∈Zm:Ax=0∈(Z/qZ)n}{L}^\perp(A) = \{x \in \mathbb{Z}^m : Ax = 0 \in (\mathbb{Z}/q\mathbb Z)^n \}L⊥(A)={x∈Zm:Ax=0∈(Z/qZ)n}

    • So we need to find a short vector in t+L⊥(A)t + {L}^{\perp}(A)t+L⊥(A)

    • Equivalent, find v∈L⊥(A)v \in {L}^{\perp}(A)v∈L⊥(A) closest to ttt (CVP)

    hashtag
    Hermite normal form

    // TODO

    hashtag
    Security Reduction

    triangle-exclamation

    If somebody can explain the security bounds and reduction better, please do.

    hashtag
    Resources

    • https://simons.berkeley.edu/sites/default/files/docs/14967/sis.pdfarrow-up-right + https://www.youtube.com/watch?v=qZIjVX61NFc&list=PLgKuh-lKre10rqiTYqJi6P4UlBRMQtPn0&index=4arrow-up-right

    • https://crypto.stanford.edu/cs355/18sp/lec9.pdfarrow-up-right

    • https://eprint.iacr.org/2015/939.pdfarrow-up-right - page 18

    fA(z)=Az=∑iai⋅zi=0∈(Z/qZ)nz1a1⃗+z2a2⃗+...+zmam⃗=0f_A(z) = Az = \sum_i a_i \cdot z_i = 0 \in (\mathbb{Z}/q\mathbb Z)^n \\ z_1\vec{a_1} + z_2\vec{a_2} +...+ z_m\vec{a_m} = 0fA​(z)=Az=i∑​ai​⋅zi​=0∈(Z/qZ)nz1​a1​​+z2​a2​​+...+zm​am​​=0

    Intuitively, if we have a problem (1) in some lattice space we can reduce it to a hard problem (2) in another related lattice space. Then if we can prove that if solving problem (1) implies solving problem (2) then we can conclude that problem (1) is as hard as problem (2)

    Understanding this chapter will strengthen the intuition for the fututre when we will study what breaking a lattice problem means and how to link it to another hard lattice problem.

    hashtag
    Dual lattice

    Let L⊂RnL \subset \mathbb R^nL⊂Rnbe a lattice. We define the dual of a lattice as the set of all vectors y∈span(L)y \in span(L)y∈span(L) such that y⋅x∈Z y \cdot x \in \mathbb Z \ y⋅x∈Z for all vectors x∈Lx \in Lx∈L:

    circle-info

    Note that the vectors in the dual lattice L∨L^\veeL∨ are not necessarily in the initial lattice LLL. They are spanned by the basis vectors of the lattice LLL.

    Examples:

    1. (Zn)∨=Zn(\mathbb Z^n) ^ \vee = \mathbb Z^n(Zn)∨=Zn because the dot product of all vectors in Zn\mathbb Z^nZnstays in Zn\mathbb Z^nZn

    2. Scaling: (k⋅L)∨=1k⋅L(k \cdot L)^\vee = \dfrac 1 k \cdot L(k⋅L)∨=k1​⋅L Proof: If y∈(kL)∨⇒y⋅kx=k(x⋅y)∈Z ∀ x∈L⇒y∈1kL∨y \in (kL)^\vee \Rightarrow y \cdot kx = k(x \cdot y) \in \mathbb{Z} \ \forall \ x \in L \Rightarrow y \in \dfrac 1 k L^\veey∈(kL)∨⇒y⋅kx=k(x⋅y)∈Z ∀ x∈L⇒y∈k1​L∨ If y∈(1kL)∨⇒yv∈L∨⇒ky⋅x=k(x⋅y)=y⋅kx∈Z ∀ x ∈L⇒y∈(kL)∨y \in \left (\dfrac 1 kL\right )^\vee \Rightarrow yv \in L^\vee \Rightarrow ky\cdot x = k(x \cdot y) = y \cdot kx \in \mathbb{Z} \ \forall \ x \ \in L \Rightarrow y \in (kL)^\veey∈(k1​L)∨⇒yv∈L∨⇒ky⋅x=k(x⋅y)=y⋅kx∈Z ∀ x ∈L⇒y∈(kL)∨

    Plot: 2Z22\mathbb Z ^22Z2 - green, 12Z2\dfrac 1 2 \mathbb Z ^ 221​Z2 - red

    Intuition: We can think of the dual lattice L∨L^\veeL∨ as some kind of inverse of the initial lattice LLL

    hashtag
    Basis of the dual lattice

    We will now focus on the problem of finding the basis B∨B^\veeB∨ of the dual lattice L∨L^\veeL∨given the lattice LLL and its basis BBB.

    circle-check

    Reminder: We can think of the lattice LLL as a transformation given by its basis B∈GLn(R)B \in GL_n(\mathbb R)B∈GLn​(R)on Zn\mathbb Z^nZn.

    We have the following equivalences:

    Therefore L∨=(B−1)T⋅ZnL^\vee = (B^{-1})^T \cdot \mathbb Z^nL∨=(B−1)T⋅Znso we have found a base for our dual lattice:

    • https://en.wikipedia.org/wiki/Hermite_normal_formarrow-up-right

    Let's look at some plots. With green I will denote the original lattice and with red the dual. The scripts for the plots can be found in in the interactive fun section

    hashtag
    Properties

    1. L1⊆L2  ⟺  L2∨⊆L1∨{L}_1 \subseteq {L}_2 \iff {L}^\vee_2 \subseteq {L}^\vee_1L1​⊆L2​⟺L2∨​⊆L1∨​

    2. (L∨)∨=L=({L}^\vee)^\vee ={L} = (L∨)∨=L=The dual of the dual is the initial lattice (to prove think of the basis of L∨L^\veeL∨)

    3. det⁡(L∨)=det⁡(L)−1\det(L^\vee) = \det(L) ^{-1}det(L∨)=det(L)−1 (to prove think of the basis of L∨L^\veeL∨)

    4. For consider the vector dot product and addition - - has no geometric meaning, they are in different spaces

    hashtag
    Successive minima

    We've seen that we can find the basis of the dual lattice given the basis of the original lattice. Let's look at another interesting quantity: the successive minima of a lattice LLL and its dual L∨L^\veeL∨. Let's see what can we uncover about them.

    circle-info

    We recommend to try and think about the problem for a few minutes before reading the conclusions.

    What is λ1(2Z2)\lambda_1(2\mathbb Z^2)λ1​(2Z2)? What about λ1((2Z2)∨)\lambda_1((2\mathbb Z^2)^\vee)λ1​((2Z2)∨)? Can you see some patterns?

    circle-check

    Reminder: We defined the successive minima of a lattice LLLas such:

    λi(L)=min⁡(max⁡1≤j≤i(∥vj∥):vj∈L are linearly independent)\lambda_i(L)=\min\left(\max_{1\leq j\leq i}\left(\left\lVert v_j\right\rVert\right):v_j\in L\text{ are linearly independent}\right)λi​(L)=min(1≤j≤imax​(∥vj​∥):vj​∈L are linearly independent)

    Claim 1:

    Proof: By Minkowski's bound we know:

    λ1(L)≤n⋅det⁡(L)1/n\lambda_1(L) \leq \sqrt{n} \cdot \det(L)^{1 / n}λ1​(L)≤n​⋅det(L)1/n and λ1(L∨)≤n⋅det(L∨)1/n=ndet⁡(L)1/n\lambda_1(L^\vee) \leq \sqrt{n} \cdot det(L^\vee)^{1 / n} = \dfrac {\sqrt{n}} {\det(L)^{1/n}}λ1​(L∨)≤n​⋅det(L∨)1/n=det(L)1/nn​​. By multiplying them we get the desired result.

    From this result we can deduce that the minima of the LLL and L∨L^\veeL∨have an inverse proportional relationship (If one is big, the other is small).

    Claim 2:

    Proof:

    Let x∈Lx∈Lx∈L be such that ∥x∥=λ1(L)\|x\|=λ_1(L)∥x∥=λ1​(L). Then take any set (y1,...,yn)(y_1, . . . , y_n)(y1​,...,yn​) of nnn linearly independent vectors in L∨L^\veeL∨. Not all of them are orthogonal to xxx. Hence, there exists an iii such that yi⋅x≠0y_i \cdot x \neq 0yi​⋅x=0 . By the definition of the dual lattice, we have yi⋅x∈Zy_i \cdot x \in \mathbb Zyi​⋅x∈Z and hence 1≤yi⋅x≤∥yi∥⋅∥x∥≤λ1⋅λn∨1 \leq y_i \cdot x \leq \|y_i\| \cdot \|x\| \leq \lambda_1 \cdot \lambda_n^\vee1≤yi​⋅x≤∥yi​∥⋅∥x∥≤λ1​⋅λn∨​

    hashtag
    Geometry + Partitioning

    // TODO

    hashtag
    Q-ary lattices

    We've seen that in cryptography we don't like to work with infinite sets (like Z\mathbb ZZ) and we limit them to some finite set using the  mod \bmodmod operation (Z→Z/qZ\mathbb Z \to \mathbb Z/ q\mathbb{Z}Z→Z/qZ). We will apply the same principle to the lattices so let us define the concept of a q-ary lattice.

    Definition:

    For a number q∈Z, q≥3q \in \mathbb{Z},\ q \geq 3q∈Z, q≥3we call a lattice q-ary if

    Intuition:

    • qZn⊆Lq\mathbb{Z^n} \subseteq \mathcal{L}qZn⊆L is periodic  mod  q\bmod \ qmod q

    • We use arithmetic  mod  q\bmod \ qmod q

    We will now look at 2 more types of lattices that are q-ary. Let A∈(Z/qZ)n×mA \in (\mathbb{Z}/q\mathbb Z)^{n \times m}A∈(Z/qZ)n×m be a matrix with m>nm > nm>n. Consider the following lattices: Lq(A)={y∈Zm:y=ATx mod q∈ for some x∈Zn}⊂ZmL_q(A) = \{y \in \mathbb Z^m : y = A^Tx \bmod q \in \text{ for some } x \in \mathbb{Z}^n \} \subset \mathbb{Z^m}Lq​(A)={y∈Zm:y=ATxmodq∈ for some x∈Zn}⊂Zm Lq⊥(A)={y∈Zm:Ay=0 mod q}⊂ZmL^\perp_q(A) = \{y \in \mathbb Z^m : Ay = 0 \bmod q \} \subset \mathbb{Z^m}Lq⊥​(A)={y∈Zm:Ay=0modq}⊂Zm

    Intuition:

    • Think of Lq(A)L_q(A)Lq​(A) as the image of the matrix AAA, the matrix spanned by the rows of AAA

    • Think of Lq⊥(A)L_q^\perp(A)Lq⊥​(A) as the kernel of AAA modulo qqq. The set of solutions Ax=0Ax = 0Ax=0

    circle-info

    Remark: If the same matrix AAA is used (AAA is fixed ) then Lq(A)≠Lq⊥(A)L_q(A) \neq L_q^\perp(A)Lq​(A)=Lq⊥​(A)

    Claim:

    Lq(A)L_q(A)Lq​(A) and Lq⊥(A)L_q^\perp(A)Lq⊥​(A) are the dual of each other (up to scaling): Lq(A)=1qLq⊥(A)L_q(A) = \dfrac 1 q L_q^\perp(A)Lq​(A)=q1​Lq⊥​(A)

    Proof:

    Firstly we will show Lq⊥(A)⊆q(Lq(A))∨L_q^\perp(A) \subseteq q(L_q(A))^\veeLq⊥​(A)⊆q(Lq​(A))∨

    • Let y∈Lq⊥(A)⇒Ay≡0 mod q  ⟺  Ay=qzy \in L_q^\perp(A) \Rightarrow Ay \equiv 0 \bmod q \iff Ay = qzy∈Lq⊥​(A)⇒Ay≡0modq⟺Ay=qzfor some z∈Zmz \in \mathbb{Z}^mz∈Zm

    • Let y′∈Lq(A)⇒y′≡ATx mod q  ⟺  y′=ATx+qz′y' \in L_q(A)\Rightarrow y' \equiv A^Tx \bmod q \iff y' = A^Tx + qz'y′∈Lq​(A)⇒y′≡ATxmodq⟺y′=ATx+qz′ for some x∈Zn, z′∈Zmx \in \mathbb Z^n, \ z' \in \mathbb Z^mx∈Zn, z′∈Zm

    Then we have:y⋅y′=y⋅(ATx+qz′)=y⋅ATx+q(y⋅z′)=Ay⏟qz⋅x+q(y⋅z′)=qz⋅x+q(y⋅z′)y \cdot y' = y \cdot (A^Tx + qz') = y\cdot A^Tx + q (y \cdot z') = \underbrace{Ay}_{qz} \cdot x + q(y \cdot z') = qz \cdot x + q(y \cdot z')y⋅y′=y⋅(ATx+qz′)=y⋅ATx+q(y⋅z′)=qzAy​​⋅x+q(y⋅z′)=qz⋅x+q(y⋅z′)

    ⇒1qy⋅y′∈Z⇒1qy∈Lq(A)∨\Rightarrow \dfrac 1 q y \cdot y' \in \mathbb{Z} \Rightarrow \dfrac 1 q y\in L_q(A)^\vee⇒q1​y⋅y′∈Z⇒q1​y∈Lq​(A)∨

    The second part is left as an exercise to the reader :D. Show Lq⊥(A)⊇q(Lq(A))∨L_q^\perp(A) \supseteq q(L_q(A))^\veeLq⊥​(A)⊇q(Lq​(A))∨

    hashtag
    Resources

    • https://cims.nyu.edu/~regev/teaching/lattices_fall_2004/ln/DualLattice.pdfarrow-up-right

    • https://sp2.uni.lu/wp-content/uploads/sites/66/2019/06/DualLattice-Luca-Notarnicola.pdfarrow-up-right

    • https://simons.berkeley.edu/sites/default/files/docs/14953/intro.pdfarrow-up-right

    L∨={y∈span(L):y⋅x∈Z ∀ x∈L}L^\vee = \{y \in span(L) : y \cdot x \in \mathbb{Z} \ \forall \ x \in L\}L∨={y∈span(L):y⋅x∈Z ∀ x∈L}
    y∈L∨  ⟺  y⋅x∈Z ∀ x∈L  ⟺  BTy∈Zn  ⟺  y∈(B−1)T⋅Zn\begin{align*} y \in L^\vee & \iff y \cdot x \in \mathbb Z \ \forall\ x \in L \\ & \iff B^Ty \in \mathbb{Z}^n \\ & \iff y \in (B^{-1})^T \cdot \mathbb Z^n \end{align*}y∈L∨​⟺y⋅x∈Z ∀ x∈L⟺BTy∈Zn⟺y∈(B−1)T⋅Zn​
    B∨=(B−1)T∈GLn(R)B^\vee = (B^{-1})^T \in GL_n(\mathbb{R})B∨=(B−1)T∈GLn​(R)
    λ1(L)⋅λ1(L∨)≤n\lambda_1(L) \cdot \lambda_1(L^\vee) \leq nλ1​(L)⋅λ1​(L∨)≤n
    λ1(L)⋅λn(L∨)≥1\lambda_1(L) \cdot \lambda_n(L^\vee) \geq 1λ1​(L)⋅λn​(L∨)≥1
    qZn⊆L⊆Znq\mathbb{Z}^n \subseteq {L} \subseteq \mathbb{Z}^nqZn⊆L⊆Zn
    LLL
    Rn\mathbb{R}^nRn
    bib_ibi​
    L=∑i=1dZbi={∑i=1daibi∣ai∈Z}L=\sum_{i=1}^d\mathbb{Z} b_i = \left\{\left. \sum_{i=1}^d a_i b_i \right | a_i \in \mathbb{Z} \right\}L=i=1∑d​Zbi​={i=1∑d​ai​bi​​ai​∈Z}
    bib_ibi​
    {bi}i=1d\left\{b_i\right\}_{i=1}^d{bi​}i=1d​
    LLL
    viv_ivi​
    ∑iaibi=0\sum_{i} a_i b_i = 0∑i​ai​bi​=0
    aia_iai​
    dim⁡L\dim LdimL
    ddd
    d=nd=nd=n
    Rn\mathbb R^nRn
    bib_ibi​
    Φ={∑i=1dxibi∣0≤xi<1}\Phi=\left\{\left.\sum_{i=1}^dx_ib_i\right|0\leq x_i<1\right\}Φ={i=1∑d​xi​bi​​0≤xi​<1}
    R2\mathbb R^2R2
    mmm
    B\mathcal BB
    d×nd\times nd×n
    vol(L)=∣det⁡(BBT)∣\text{vol}(L)=\sqrt{\left|\det\left(\mathcal B\mathcal B^T\right)\right|}vol(L)=∣det(BBT)∣​
    XXX
    Rn\mathbb R^nRn
    x∈Xx\in Xx∈X
    −x∈X-x\in X−x∈X
    x,y∈Xx,y\in Xx,y∈X
    x,yx,yx,y
    XXX
    {tx+(1−t)y∣0≤t≤1}⊂X\left\{tx+(1-t)y|0\leq t\leq1\right\}\subset X{tx+(1−t)y∣0≤t≤1}⊂X
    LLL
    nnn
    XXX
    vol(X)>2nvol(L)\text{vol}(X)>2^n\text{vol}(L)vol(X)>2nvol(L)
    XXX
    LL L
    ∥v∥=∑ivi2\left\lVert v\right\rVert=\sqrt{\sum_iv_i^2}∥v∥=∑i​vi2​​
    ℓ2\ell_2ℓ2​
    ⟨a,b⟩=∑iaibi\langle a,b\rangle=\sum_ia_ib_i⟨a,b⟩=∑i​ai​bi​
    12X={12x∣x∈X}\frac12X=\left\{\frac12x|x\in X\right\}21​X={21​x∣x∈X}
    vol(12X)>vol(L)\text{vol}\left(\frac12 X\right)>\text{vol}(L)vol(21​X)>vol(L)
    12X→Rn/L\frac12X\to\mathbb R^n/L21​X→Rn/L
    x1=x2+ℓx_1=x_2+\ellx1​=x2​+ℓ
    x1,x2∈12X,ℓ∈L,x1≠x2x_1,x_2\in\frac12 X,\ell\in L,x_1\neq x_2x1​,x2​∈21​X,ℓ∈L,x1​=x2​
    x1−x2∈Lx_1-x_2\in Lx1​−x2​∈L
    LLL
    B=(−1981−8−7)\mathcal B=\begin{pmatrix}-1&9&8\\1&-8&-7\end{pmatrix}B=(−11​9−8​8−7​)
    B′=(101011)\mathcal B'=\begin{pmatrix}1&0&1\\0&1&1\end{pmatrix}B′=(10​01​11​)
    C=(101022)\mathcal C=\begin{pmatrix}1&0&1\\0&2&2\end{pmatrix}C=(10​02​12​)
    C\mathcal CC
    B,B′\mathcal B,\mathcal B'B,B′
    d×nd\times nd×n
    L,L′L,L'L,L′
    U∈GLd(Z)U\in\text{GL}_d(\mathbb Z)U∈GLd​(Z)
    B′=UB\mathcal B'=U\mathcal BB′=UB
    UUU
    GLd(Z)\text{GL}_d(\mathbb Z)GLd​(Z)
    UUU
    U−1U^{-1}U−1
    LLL
    nnn
    XXX
    vol(X)=2nvol(L)\text{vol}(X)=2^n\text{vol}(L)vol(X)=2nvol(L)
    XXX
    vvv
    LLL
    nnn
    ∥v∥≤2πΓ(n2+1)1nvol(L)1n\left\lVert v\right\rVert\leq\frac2{\sqrt\pi}\Gamma\left(\frac n2+1\right)^{\frac1n}\text{vol}(L)^\frac1n∥v∥≤π​2​Γ(2n​+1)n1​vol(L)n1​

    Lattice reduction

    hashtag
    Overview

    Having introduced the LLL reduction, we now provide a more general notions of a reduced basis for a lattice as well as provide bounds for the basis vectors. The key idea behind introducing these definitions is that once we know some basis vector is []-reduced, we can bound the sizes of the basis, which is important when algorithms require short vectors in a lattice. For fast algorithms, LLL-reduction is typically the most important notion as it can be computed quickly. Two main definitions appear often when discussing lattice reductions, which we will provide here.

    hashtag
    Definitions

    A basisis size-reduced if . Intuitively this captures the idea that a reduced basis being "almost orthogonal".

    Let be a lattice, , we define the successive minima as

    Intuitively, is the length of the " shortest lattice vector". This intuition is illustrated by the definition of :

    However this is not precise as if is the shortest lattice vector, then is also the shortest lattice vector.

    Unfortunately, a basisfor where for dimensions and above. This tells us that we can't actually define "the most reduced basis" in contrast to the 2D case (see ) and we would need some other definition to convey this intuition.

    An alternate definition ofthat will be helpful is the radius of the smallest ball centered at the origin such that the ball contains at leastlinearly independent vectors in.

    hashtag
    Exercises

    1) Show that both definitions of are equivalent

    2) Consider the lattice . Show that the successive minima are allbut no basiscan satisfy .

    Extensions of Coppersmith algorithm

    The Coppersmith algorithm can be made even more general. There are two main extensions, first to an unknown modulus, then to multivariate polynomials.

    hashtag
    Unknown modulus

    This extension of Coppersmith allows one to find small roots modulo an unknown some unknown factor of a number. More specifically, suppose that we have some unknown factorbbbofNNNsuch that b<Nβb<N^\betab<Nβand some monic polynomialfffof degreedddsuch that f(x0)=0(modb)f(x_0)=0\pmod bf(x0​)=0(modb)for some x0<Nβ2dx_0<N^{\frac{\beta^2}d}x0​<Ndβ2​. Then we can findx0x_0x0​in time polynomial in .

    One key reason why this is possible is because we dont need to explicitly know the modulus to determine if a small root exists, i.e. is sufficient for a root less thanto exist. The algorithm here is extremely similar to the Coppersmith algorithm, except we add more polynomials into the lattice. The polynomials that we will use are

    The latticegenerated by these polynomials would have

    As we require the shortest vector to have length at most , repeating the computations from the previous section, we obtain

    It turns out that the maxima of is as . One way to achieve this is by settingand we obtain

    and this indeed achieves the bound. Similar to the Coppersmith algorithm, one chooses a sufficiently big such that the remainding bits can be brute forced in constant time while the algorithm still remains in polynomial time.

    hashtag
    Exercises

    1) We show that the maximum ofis indeed . We can assume that . Since and , the maximum occurs when and , hence we have reduced this to maximizing which achieves its maximum of at .

    Hard lattice problems

    triangle-exclamation

    This section is not complete. Help is needed with relevance + examples in cryptography, algorithms + hardness, relations between problems.

    Also needs review from more experienced people.

    hashtag
    Introduction

    Now that we are comfortable with lattices we shall study why are they important to cryptography.

    Like we said, when we construct cryptosystems we usually are looking for hard problems to base them on. The lattice world provides us with such problems such as the shortest vector problem or the closest vector problem.

    What makes lattices even more special is that some cryptographic problems (which we will study in the next chapter) can be reduced to worst-case lattice problems which makes them crazy secure. Moreover, some problems are even secure against quantum computers.

    But enough talk, let's get right into it!

    hashtag
    Shortest vector problem + GapSVP

    Before we go into any problems we must first define the concept of distance in a lattice.

    Let:

    • Lattice

    • the basis of the lattice

    • the dimension of the lattice

    Distance function

    Given some distance function (Example: Euclidean norm) the distance from a vector to the lattice is the distance from the vector to the closest point in the in lattice.

    We will denote the length of the shortest vector with and the length of the next independent vectors in order with

    hashtag
    Shortest vector problem

    Given a lattice and an arbitrary basis for it our task is to find the shortest vector .

    Approximate SVP

    We relax the SVP problem a bit. Given an arbitrary basis find a shortest nonzero lattice vector such that . Here is some approximation factor.

    hashtag
    Decision SVP (GapSVP)

    Given a lattice with a basis we must distinguish if or

    hashtag
    Sage example

    hashtag
    Closest Vector problem + GapCVP

    Closest vector problem

    Given a lattice with an arbitrary basis and a vector find the closest lattice vector to

    Approximate CVP

    Given a lattice with an arbitrary basis and a vector find the closest lattice vector to

    hashtag
    Decision CVP (GapCVP)

    Given a lattice with a basis and a vector we must decide if

    • There exists s.t

    hashtag
    Sage example

    hashtag
    Bounded distance decoding

    Given a lattice with an arbitrary basis , a vector and a real number find a lattice vector s.t

    Remark

    • If we have the solution to the BDD problem is guaranteed to be unique.

    hashtag
    Shortest independent vectors (SIVP)

    Given a full rank lattice with an arbitrary basis find linearly independent lattice vectors of length at most or for the approximate version.

    hashtag
    Hardness of lattice problems

    hashtag
    Resources

    • Pictures taken from and "Cryptography made simple - Nigel Smart" and edited a bit

    • Or generated by me

    Interactive fun

    Inspired by:

    Lattice + LLL + Fundamental mesh plot

    hashtag
    Lattice + CVP

    Common Modulus Attack

    What to do when the same message is encrypted twice with the same modulus but a different public key?

    Imagine we have Alice and Bob. Alice sends the SAME message to Bob more than once using the same public key. The internet being the internet, a problem may happen; a bit is flipped, and the public key changed while the modulus stayed the same.

    hashtag
    What we know

    Let be the following:

    RSA application

    Tutorial for application with RSA. We are going to use openSSL, openSSH and pycryptodome for key generation, key extraction and some implementation with python

    hashtag
    Pycryptodome:

    Pycryptodome is a python library about cryptography, see the documentation below: There is an example of RSA key generation with pycryptodome:

    hashtag

    Resources and notations

    hashtag
    References/Resources

    1. Nguyen, P. Q., & Vallée, B. (Eds.). (2010). The LLL Algorithm. Information Security and Cryptography. doi:10.1007/978-3-642-02295-1

      Massive survey, lots of detail if you're extremely interested)

    Wiener's Attack

    Wiener's attack is an attack on RSA that uses continued fractions to find the private exponent when it's small (less than , where is the modulus). We know that when we pick the public exponent to be a small number and calcute its inverse

    hashtag
    Wiener's theorem

    Wiener's attack is based on the following theorem:

    def euclid(m,n):
        while n!=0:
            q = round(m/n)
            m -= q*n
            if abs(n) > abs(m):
                m, n = n, m
        return abs(m)
    def lagrange(b1,b2):
        mu = 1
        while mu != 0:
            mu = round((b1*b2) / (b1*b1))
            b2 -= mu*b1
            if b1*b1 > b2*b2:
                b1, b2 = b2, b1
        return b1, b2
    from Crypto.Util.number import long_to_bytes, bytes_to_long
    
    n, m, q = 20, 40, 1009
    set_random_seed(1337)
    A = random_matrix(Zmod(q),n, m)
    
    print(A.parent())
    # Full MatrixSpace of 20 by 40 dense matrices over Ring of integers modulo 1009
    print(A.lift().parent())
    # Full MatrixSpace of 20 by 40 dense matrices over Integer Ring
    
    msg = b'msg'
    x = vector(Zmod(q), [int(i) for i in bin(bytes_to_long(msg))[2:].zfill(m)]) # pad message
    print(len(x)
    # 40
    
    print(x.parent())
    # Vector space of dimension 40 over Ring of integers modulo 1009
    
    print(len(A * x))
    # 20
    n = 5 # lattice dimension
    
    B = sage.crypto.gen_lattice(m=n, q=11, seed=42)
    B_dual = sage.crypto.gen_lattice(m=n,  q=11, seed=42, dual=True)
    
    B_dual_ = (B.inverse().T * 11).change_ring(ZZ) # Scale up to integers
    B_dual_.hermite_form() == B_dual.hermite_form() # Reduce form to compare
    # True
    n = 5 # lattice dimension
    
    B = sage.crypto.gen_lattice(m=n, q=11, seed=42)
    B_dual = sage.crypto.gen_lattice(m = n,  q=11, seed=42, dual=True)
    
    l1 = IntegerLattice(B).shortest_vector().norm().n() 
    l2 = IntegerLattice(B_dual).shortest_vector().norm().n() / 11
    
    print(l1 * l2 < n)
    # True
    n = 5 # lattice dimension
    
    B = sage.crypto.gen_lattice(m=n, q=11, seed=42)
    B_dual = sage.crypto.gen_lattice(m = n,  q=11, seed=42, dual=True)
    
    l1 = IntegerLattice(B).shortest_vector().norm().n() 
    
    B_dual_lll = B_dual.LLL()
    lnd = 0
    for v in B_dual_lll:
        lv = v.norm()
        if lv > lnd:
            lnd = lv
    lnd = lnd.n() / 11
    
    print(lnd * l1 > 1) 
    # True
    log⁡N,d\log N,dlogN,d
    ∥f(Bx)∥2≤bd+1≤Nβd+1\left\lVert f(Bx)\right\rVert_2\leq\frac b{\sqrt{d+1}}\leq\frac{N^{\beta}}{\sqrt{d+1}}∥f(Bx)∥2​≤d+1​b​≤d+1​Nβ​
    BBB
    gi,j(x)=Nh−jf(x)jxi0≤i<d,0≤j<hgi,h(x)=f(x)hxi0≤i<t\begin{align*} g_{i,j}(x)&=N^{h-j}f(x)^jx^i&0\leq i<d,0\leq j<h\\ g_{i,h}(x)&=f(x)^hx^i&0\leq i<t \end{align*}gi,j​(x)gi,h​(x)​=Nh−jf(x)jxi=f(x)hxi​0≤i<d,0≤j<h0≤i<t​
    LLL
    dim⁡(L)=dh+t=nvol(L)=Ndh(h+1)2B(n−1)n2\dim(L)=dh+t=n\quad\text{vol}(L)=N^{\frac{dh(h+1)}{2}}B^{\frac{(n-1)n}2}dim(L)=dh+t=nvol(L)=N2dh(h+1)​B2(n−1)n​
    Nβhn\frac{N^{\beta h}}{\sqrt{n}}n​Nβh​
    B<4δ−14(1n)1n−1N2βhn−1−dh(h+1)n(n−1)B<\sqrt{\frac{4\delta-1}4}\left(\frac1n\right)^{\frac1{n-1}}N^{\frac{2\beta h}{n-1}-\frac{dh(h+1)}{n(n-1)}}B<44δ−1​​(n1​)n−11​Nn−12βh​−n(n−1)dh(h+1)​
    2βhn−1−dh(h+1)n(n−1)\frac{2\beta h}{n-1}-\frac{dh(h+1)}{n(n-1)}n−12βh​−n(n−1)dh(h+1)​
    β2d\frac{\beta^2}ddβ2​
    n,h→∞n,h\to\inftyn,h→∞
    n=dβhn=\frac d\beta hn=βd​h
    B<4δ−14(1n)1n−1N(h−1)β2dh−βB<\sqrt{\frac{4\delta-1}4}\left(\frac1n\right)^{\frac1{n-1}}N^{\frac{(h-1)\beta^2}{dh-\beta}}B<44δ−1​​(n1​)n−11​Ndh−β(h−1)β2​
    O(Nβ2d)O\left(N^{\frac{\beta^2}d}\right)O(Ndβ2​)
    h,nh,nh,n
    2βhn−1−dh(h+1)n(n−1)\frac{2\beta h}{n-1}-\frac{dh(h+1)}{n(n-1)}n−12βh​−n(n−1)dh(h+1)​
    β2d\frac{\beta^2}ddβ2​
    d≥2,h≥1,0<β≤1,n≥2d\geq2,h\geq1,0<\beta\leq1,n\geq2d≥2,h≥1,0<β≤1,n≥2
    hn−1(2β−dh+1n)\frac h{n-1}\left(2\beta-d\frac{h+1}{n}\right)n−1h​(2β−dnh+1​)
    hn−1<h+1n\frac h{n-1}<\frac{h+1}nn−1h​<nh+1​
    h,n→∞h,n\to\inftyh,n→∞
    hn−1,h+1n→x\frac h{n-1},\frac{h+1}n\to xn−1h​,nh+1​→x
    2βx−dx22\beta x-dx^22βx−dx2
    β2d\frac{\beta^2}ddβ2​
    x=βdx=\frac\beta dx=dβ​
    {bi}i=1d\left\{b_i\right\}_{i=1}^d{bi​}i=1d​
    ∣μi,j∣≤12\left|\mu_{i,j}\right|\leq\frac12∣μi,j​∣≤21​
    LLL
    1≤i≤dim⁡L=d1\leq i\leq\dim L=d1≤i≤dimL=d
    ithi^\text{th}ith
    λi(L)\lambda_i(L)λi​(L)
    λi(L)=min⁡(max⁡1≤j≤i(∥vj∥):vj∈L are linearly independent)\lambda_i(L)=\min\left(\max_{1\leq j\leq i}\left(\left\lVert v_j\right\rVert\right):v_j\in L\text{ are linearly independent}\right)λi​(L)=min(1≤j≤imax​(∥vj​∥):vj​∈L are linearly independent)
    λi(L)\lambda_i(L)λi​(L)
    ithi^\text{th}ith
    λ1\lambda_1λ1​
    λ1(L)=min⁡(∥v∥:v∈L)\lambda_1(L)=\min\left(\left\lVert v\right\rVert:v\in L\right)λ1​(L)=min(∥v∥:v∈L)
    vvv
    −v-v−v
    bib_ibi​
    LLL
    λi(L)=∥bi∥\lambda_i(L)=\left\lVert b_i\right\rVertλi​(L)=∥bi​∥
    555
    λi(L)\lambda_i(L)λi​(L)
    iii
    LLL
    λi\lambda_iλi​
    L=(2000002000002000002011111)L=\begin{pmatrix}2&0&0&0&0\\0&2&0&0&0\\0&0&2&0&0\\0&0&0&2&0\\1&1&1&1&1\end{pmatrix}L=​20001​02001​00201​00021​00001​​
    222
    bib_ibi​
    ∥bi∥=λi\left\lVert b_i\right\rVert=\lambda_i∥bi​∥=λi​
    Lagrange's algorithm

    MITM

    Explanation of the MITM (Man In The Middle) with the Diffie-Hellmann key exchange

    Elliptic Curve Cryptography

    Low Private Component Attacks

    m the message in plaintext

  • e1 the public key of the first ciphertext

  • c1 the first ciphertext

  • e2 the public key of the second ciphertext

  • c2 the second ciphertext

  • n the modulus that is common to both ciphertexts

  • All of these but m are essentially given to us.

    hashtag
    Conditions of the attack

    Because we are going to need to calculate inverses for this attack, we must first make sure that these inverses exist in the first place:

    hashtag
    The math behind the attack

    We know that RSA goes as follows:

    From the conditions above we also know that e1e1 e1 and e2e2e2 are co-prime. Thus using Bezout's Theorem we can get:

    Using this, we can derive the original message mm m :

    NB: all the calculations are done mod nn n

    In general, Bezout's Theorem gives a pair of positive and negative numbers. We just need to adapt this equation a little to make it work for us. In this case, let's assume yyy is the negative number:

    Now to truly recover the plaintext, we are actually doing:

    gcd(e1,e2)=1gcd(c2,n)=1gcd(e_1, e_2) = 1 \newline gcd(c_2, n) = 1 gcd(e1​,e2​)=1gcd(c2​,n)=1
    c=me mod nc = m^e\ mod\ nc=me mod n
    xe1+ye2=gcd(e1,e2)=1xe_1 +ye_2 = gcd(e_1, e_2) = 1xe1​+ye2​=gcd(e1​,e2​)=1
    C1x∗C2y=(me1)x∗(me2)y=me1x+e2y=m1=mC_1^x * C_2^y = (m^{e_1})^x*(m^{e_2})^y \newline = m^{e_1x+e_2y} \newline = m^1 = m C1x​∗C2y​=(me1​)x∗(me2​)y=me1​x+e2​y=m1=m
    Let y=−aC2y=C2−a=(C2−1)a=(C2−1)−yLet\ y = -a \newline C_2^y = C_2^{-a} \newline = (C_2^{-1})^a \newline = (C_2^{-1})^{-y}Let y=−aC2y​=C2−a​=(C2−1​)a=(C2−1​)−y
    C1x×(C2−1)−y mod nC_1^x \times (C_2^{-1})^{-y}\ mod\ n C1x​×(C2−1​)−y mod n
    β=\beta = β=
    β≪q\beta \ll qβ≪q
    fA(x)=Ax mod q\boxed {f_A(x) = Ax \bmod q}fA​(x)=Axmodq​
    fA:{0,1}m→(Z/qZ)nf_A : \{0, 1\}^m \to (\mathbb{Z}/q\mathbb Z)^nfA​:{0,1}m→(Z/qZ)n
    x∈L,y∈L∨x \in {L}, y \in {L}^\veex∈L,y∈L∨
    x⋅y∈Zx \cdot y \in \mathbb{Z}x⋅y∈Z
    x+yx + yx+y
    https://cseweb.ucsd.edu/~daniele/papers/FOSAD11.pdfarrow-up-right

    Sets and Functions

    OpenSSL:

    OpenSSL is a robust, commercial-grade, and full-featured toolkit for the Transport Layer Security (TLS) and Secure Sockets Layer (SSL) protocols. It is also a general-purpose cryptography library

    hashtag
    OpenSSH:

    https://www.pycryptodome.org/en/latest/arrow-up-right
    from Crypto.Util.number import getPrime, bytes_to_long
    
    
    def generate_keys():
        e = 0x10001    #public exponent e, we generally use this one by default
        while True:
            p = getPrime(512)
            q = getPrime(512)
            phi = (p - 1) * (q - 1)    #Euler's totient 
            d = pow(e, -1, phi)    #Private exponent d
            if d != -1:
                break
    
        n = p * q
        public_key = (n, e)
        private_key = (n, d)
        return public_key, private_key
    
    
    def encrypt(plaintext: int, public_key) -> int:
        n, e = public_key
        return pow(plaintext, e, n)    #plaintext ** e mod n
    
    
    def decrypt(ciphertext: int, private_key) -> int:
        n, d = private_key
        return pow(ciphertext, d, n)   #ciphertext ** d mod n
    
    
    message = bytes_to_long(b"super_secret_message")
    public_key, private_key = generate_keys()
    ciphertext = encrypt(message, public_key)
    plaintext = decrypt(ciphertext, private_key)
    Q-ary plots
    @interact
    def draw_lattice(v1x = input_box(label = "v1 x =", default = 1),
                     v1y = input_box(label = "v1 y =", default = 0),
                     v2x = input_box(label = "v2 x =", default = 0),
                     v2y = input_box(label = "v2 y =", default = 1),
                     box = 5, search = 10,
                     plot_LLL = True,
                     plot_C = True,
                     plot_F = True):
    
        v1 = vector((v1x, v1y))
        v2 = vector((v2x, v2y))
        vecs = []
        # Generate vectors
        for i in range(-search,search):
            for j in range(-search,search):
                vecs.append(i*v1 + j*v2)
        # Plot stuff
        G = Graphics()
        for p1 in vecs:
            x1, y1 = p1
            if x1 > -box and x1 < box and y1 > -box and y1 < box:
                G += point(p1, color = 'green', size = 30)
                G += line([p1, p1 + v2], linestyle = '--', alpha = .20)
                G += line([p1, p1 + v1], linestyle = '--', alpha = .20)
        G+= arrow((0, 0), v1, color = 'red', arrowsize = 2)
        G+= arrow((0, 0), v2, color = 'red', arrowsize = 2)
        G+= text('v1', v1 + .2 * v1, color = 'red')
        G+= text('v2', v2 + .2 * v2, color = 'red')
        
        # LLL
        if plot_LLL:
            v1_, v2_ = matrix([v1, v2]).LLL()
            G+= arrow((0, 0), v1_, color = 'purple', arrowsize = 2)
            G+= arrow((0, 0), v2_, color = 'purple', arrowsize = 2)
            if plot_C:
                G += circle(center = (0, 0), radius = norm(v1_) if norm(v1_) > norm(v2_) else norm(v2_), alpha = .5, color = 'purple')
        # Fundamental mesh
        if plot_F:
            F = polygon([[0, 0], v1, v1 + v2, v2], color='red', alpha = .1)
            G += F
        G.show(axes = False, figsize = (7, 7))
    
    
    https://crypto.katestange.net/lattice-tools/arrow-up-right

    May, A. (2003). New RSA Vulnerabilities Using Lattice Reduction Methods. Universität Paderborn.

    Excellent exposition to LLL and coppersmith as well as showing some RSA attacks via LLL

  • Lenstra, A. K., Lenstra, H. W., & Lovász, L. (1982). Factoring polynomials with rational coefficients. Mathematische Annalen, 261(4), 515–534. doi:10.1007/bf01457454

    The original LLL paper, quite a nice read overall + proof that LLL works

  • Coppersmith, D. (1996). Finding a Small Root of a Univariate Modular Equation. Lecture Notes in Computer Science, 155–165. doi:10.1007/3-540-68339-9_14

  • Coppersmith, D. (1996). Finding a Small Root of a Bivariate Integer Equation; Factoring with High Bits Known. Lecture Notes in Computer Science, 178–189. doi:10.1007/3-540-68339-9_16

    Both of these paper introduces the coppersmith algorithm as well as provide some examples

  • Waerden, B. L. (1956). Die Reduktionstheorie Der Positiven Quadratischen Formen. Acta Mathematica, 96(0), 265–309. doi:10.1007/bf02392364

  • hashtag
    Notation

    • LLL lattice

      • dim⁡(L)\dim(L)dim(L)dimension of lattice

      • vol(L)\text{vol}(L)vol(L)volume of lattice

    • a chosen basis for

      • matrix whose th row vectors is

    • Gram-Schmidt orthogonalization of (without normalization)

      • matrix whose th row vectors is

    • Gram-Schmidt coefficients

    • the th successive minima of

    Let
    , with
    . Let
    . Given
    and
    with
    , the attacker can efficiently recover
    .

    hashtag
    Some observations on RSA

    In order to better understand Wiener's Attack, it may be useful to take note of certain properties of RSA:

    We may start by noting that the congruence ed≡1mod  ϕ(n)ed \equiv 1 \mod \phi(n)ed≡1modϕ(n) can be written as the equality ed=kϕ(n)+1ed = k\phi(n)+1ed=kϕ(n)+1 for some value kkk, we may additionally note that ϕ(n)=(p−1)(q−1)=pq−p−q+1\phi(n) = (p-1)(q-1) = pq - p - q + 1ϕ(n)=(p−1)(q−1)=pq−p−q+1, since both ppp and qqq are much shorter than pq=npq = npq=n, we can say that ϕ(n)≈n\phi(n) \approx nϕ(n)≈n.

    Dividing the former equation by dϕ(n)d\phi(n)dϕ(n) gives us eϕ(n)=k+1d\frac{e}{\phi(n)} = \frac{k+1}{d}ϕ(n)e​=dk+1​, and using the latter approximation, we can write this as en≈kd\frac{e}{n} \approx \frac{k}{d}ne​≈dk​. Notice how the left-hand side of this equation is composed entirely of public information, this will become important later.

    It is possible to quickly factor nnn by knowing nnn and ϕ(n)\phi(n)ϕ(n). Consider the quadratic polynomial (x−q)(x−p)(x-q)(x-p)(x−q)(x−p), this polynomial will have the roots ppp and qqq. Expanding it gives us x2−(p+q)x+pqx^2 - (p + q)x + pqx2−(p+q)x+pq, and substituting for the variables we know we can write this as x2−(n−ϕ(n)+1)x+nx^2 - (n - \phi(n) + 1)x + nx2−(n−ϕ(n)+1)x+n. Applying the quadratic formula gives us ppp and qqq: p∧q=−b±b2−4ac2ap \wedge q = \frac{-b \pm \sqrt{b^2-4ac}}{2a}p∧q=2a−b±b2−4ac​​, where a=1a = 1a=1, b=n−ϕ(n)+1b = n - \phi(n) + 1b=n−ϕ(n)+1, and c=nc = nc=n.

    Wiener's attack works by expanding en\frac{e}{n}ne​ to a continued fraction and iterating through the terms to check various approximations of kd\frac{k}{d}dk​. In order to make this checking process more efficient, we can make a few observations (this part is optional):

    • Since ϕ(n)\phi(n)ϕ(n) is even, and eee and ddd are both by definition coprime to ϕ(n)\phi(n)ϕ(n), we know that ddd is odd.

    • Given the above equations and the values of eee, nnn, ddd, and kkk, we can solve for ϕ(n)\phi(n)ϕ(n) with the equation ϕ(n)=ed−1k\phi(n) = \frac{ed-1}{k}ϕ(n)=ked−1​, thus we know that ed−1ed-1ed−1 has to be divisible by kkk.

    • If our is correct, the polynomial will have roots and , which we can verify by checking if .

    hashtag
    The Attack

    Suppose we have the public key (n,e)(n, e)(n,e), this attack will determine ddd

    1. Convert the fraction en\frac{e}{n}ne​ into a continued fraction [a0;a1,a2,…,ak−2,ak−1,ak][a_0;a_1,a_2, \ldots , a_{k-2},a_{k-1}, a_k][a0​;a1​,a2​,…,ak−2​,ak−1​,ak​]

    2. Iterate over each convergent in the continued fraction: a01,a0+1a1,a0+1a1+1a2, …,a0+1a1+⋱ak−2+1ak−1,\frac{a_{0}}{1},a_{0} + \frac{1}{a_{1}},a_{0} + \frac{1}{a_{1} + \frac{1}{a_{2}}}, \ \ldots, a_{0} + \frac{1}{a_{1} + \frac{\ddots} {a_{k-2} + \frac{1}{a_{k-1}}}},1a0​​,a0​+a1​1​,a0​+a1​+a2​1​1​, …,a0​+a1​+ak−2​+ak−1​1​⋱​1​,

    3. Check if the convergent is kd\frac{k}{d}dk​ by doing the following:

      • Set the numerator to be and denominator to be

      • Check if is odd, if not, move on to the next convergent

      • Check if , if not, move on to the next convergent

      • Set and find the roots of the polynomial

      • If the roots of the polynomial are integers, then we've found . (If not, move on to the next convergent)

    4. If all convergents have been tried, and none of them work, then the given RSA parameters are not vulnerable to Wiener's attack.

    Here's a sage implementation to play around with:

    //TODO: Proof of Wiener's theorem

    hashtag
    Automation

    The Python module owiener simplifies the scripting process of Wiener's attack:

    Here is a Wiener's attack template:

    ddd
    13n4\frac{1}{3}\sqrt[4]{n}31​4n​
    nnn
    eee
    d≡e−1mod  ϕ(n)d \equiv e^{-1} \mod \phi(n)d≡e−1modϕ(n)
    n=pqn = pqn=pq
    q<p<2qq < p < 2qq<p<2q
    d<13n4d < \frac{1}{3}\sqrt[4]{n}d<31​4n​
    nnn
    eee
    ed≡1mod  ϕ(n)ed \equiv 1 \mod \phi(n)ed≡1modϕ(n)
    ddd
    L=L= L=
    B=\mathcal B = B=
    n=n = n=
    ttt
    LLL
    μ(t,L)=min⁡v∈L∥t−v∥\mu(t, L) = \underset{v \in \mathcal{L}}{\min}{\|t-v\|}μ(t,L)=v∈Lmin​∥t−v∥
    ∥v∥=λ1(L)\|v\| = \lambda_1(L)∥v∥=λ1​(L)
    λi(L)⇒λ1(L)≤λ2(L)≤...≤λn(L)\lambda_i(L) \Rightarrow\lambda_1({L}) \leq \lambda_2({L}) \leq ... \leq \lambda_n({L})λi​(L)⇒λ1​(L)≤λ2​(L)≤...≤λn​(L)
    LLL
    B\mathcal{B}B
    v∈Lv \in Lv∈L
    B\mathcal{B}B
    v∈Lv \in Lv∈L
    v<γ(n)⋅λ1(L)v < \gamma(n)\cdot \lambda_1(L)v<γ(n)⋅λ1​(L)
    γ(n)>1\gamma(n) > 1γ(n)>1
    LLL
    B\mathcal BB
    λ1(L)≤1\lambda_1(L) \leq 1λ1​(L)≤1
    λ>γ(n)\lambda > \gamma(n)λ>γ(n)
    LLL
    B\mathcal BB
    w∈Rnw \in \mathbb{R}^nw∈Rn
    www
    v∈L,∥v−w∥≤μv \in {L}, \|v-w\| \leq \muv∈L,∥v−w∥≤μ
    LLL
    B\mathcal BB
    w∈Rnw \in \mathbb{R}^nw∈Rn
    www
    v∈L,∥v−w∥<γ(n)⋅μv \in {L}, \|v-w\| < \gamma(n) \cdot \muv∈L,∥v−w∥<γ(n)⋅μ
    LLL
    B\mathcal BB
    www
    v∈Lv \in Lv∈L
    ∥v−w∥≤1\| v - w\| \leq 1∥v−w∥≤1
    ∀v∈L:∥v−w∥>γ(n)\forall v \in L: \|v - w\| > \gamma(n)∀v∈L:∥v−w∥>γ(n)
    LLL
    BBB
    w∈Rnw \in \mathbb{R}^nw∈Rn
    d∈Rd \in \mathbb{R}d∈R
    v∈Lv \in {L}v∈L
    ∥w−v∥<d⋅λ1(L)\|w-v\| < d \cdot \lambda_1({L})∥w−v∥<d⋅λ1​(L)
    d<12d < \dfrac 12d<21​
    LLL
    B\mathcal BB
    nnn
    λn(L)⇒max⁡i∥vi∥≤λn(L)\lambda_n(L) \Rightarrow \max_i\|v_i\| \leq \lambda_n(L)λn​(L)⇒maxi​∥vi​∥≤λn​(L)
    max⁡i∣vi∣≤γ(n)λn(L)\max_i|v_i| \leq \gamma(n) \lambda_n(L)maxi​∣vi​∣≤γ(n)λn​(L)
    https://simons.berkeley.edu/sites/default/files/docs/14953/intro.pdfarrow-up-right

    Proof of correctness

    We now consider cd=med=mc^d = m^{ed} = mcd=med=mnecessary for the successful description of an RSA ciphertext. The core of this result is due to Euler's theoremarrow-up-right which states

    aϕ(n)≡1mod  na^{\phi(n)} \equiv 1 \mod naϕ(n)≡1modn

    for all coprime integers (a,n)(a,n)(a,n) and ϕ(n)\phi(n)ϕ(n) is Euler's totient functionarrow-up-right.

    circle-info

    As a reminder, we say two integers are coprime if they share no non-trivial factors. This is the same statement that gcd⁡(a,n)=1\gcd(a,n)=1gcd(a,n)=1.

    From the definition of the protocol, we have that

    for some . Combining this with Euler's theorem, we see that we recover from the ciphertext

    When the requirement does not hold, we can instead look at the equivalences modulo and respectively. Clearly, when we have that and our correctness still holds. Now, consider the case where we have that and Since we have already excluded the case that we can conclude that as is prime. This means that and by the multiplicative properties of the function, we determine that We conclude by invoking the Chinese Remainder theorem with

    that The case for follows in a parallel manner.

    RSA

    Will be introduced in this page the fundamentals of RSA, mathematical requirement and also some application with python and openSSL.

    circle-exclamation

    This page is pretty long, probably could be split up

    Edit: I haved deleted the last part, application with RSA, and i made a special part for this. Maybe we can do the same with the second part: Arithmetic for RSA.

    hashtag
    I- Introduction:

    RSA is a that is widely used in the world today to provide a secure transmission system to millions of communications, is one of the oldest such systems in existence. The RSA comes from the surnames of , , and , who publicly described the algorithm in 1977. An equivalent system was developed secretly, in 1973 at (the British agency), by the English mathematician . That system was in 1997.

    All public-key systems are based on the concept of , functions that are simple to compute in one direction but computationally hard to reverse without knowledge of some special information called the trapdoor. In RSA, the trapdoor function is based on the . The function involves the use of a public keyto encrypt data, which is (supposed to be) encrypted in such a way that the function cannot be reversed without knowledge of the prime factorisation of, something that should be kept private. Except in certain cases, there exists no efficient algorithm for factoring huge integers.

    Further reading:

    triangle-exclamation

    Formalize the introduction and include a discussion of the security based on the hardness of factoring integers.

    hashtag
    II- Arithmetic for RSA

    Before starting to introducing you RSA, a few arithmetic notions need to be introduce to understand perfectly other steps.

    hashtag
    III- Key generation

    • We pick two primes and

    • Using and , we calculate modulus and its Euler's totient

    • Now, choose the public exponent

    hashtag
    IV- Signature

    A digital signature is a proof of the authenticity of a message, i.e. a proof that the message has not been tampered with. RSA allows us to sign messages by "encrypting" them using the private key, the result being a signature that anyone can verify by "decrypting" it with the public key and comparing it to the associated message. Any attempt to tamper with the message will result in it no longer matching the signature, and vice-versa. Futhermore, a signature can only be generated using the private key, making it a secure and efficient method of confirming the authenticity of messages.

    Say Alice wants to send a message to Bob, but does not want Mallory, who has established herself as a middleman, to make changes to the message or swap it out entirely. Fortunately, Bob knows Alice's public key, and since and are inverses such that , Alice can sign her message by "encrypting" it with the private key such that , where is the signature verifying that the message came from Alice. Alice can now send and to Bob, who is now able to check the authenticity of the message by checking if . If Mallory tries to change , this congruence no longer holds, and since she does not have the private key, she is also unable to provide a maching for her tampered message.

    hashtag
    V- Format

    Recovering the Modulus

    When you want to recover N given some (plaintext, ciphertext) pairings

    hashtag
    Scenario

    Consider the case that that you know a set of (plaintext, ciphertext) pairings - this may be that you are provided them, or that you have access to some functionality that returns ciphertexts for provided plaintexts. If you do not know the modulus, but know the exponent used (note: this may be prone to a brute-force regardless), then given these pairings you can recover the modulus used.

    hashtag
    What we know

    Let the following be known:

    • plaintext && ciphertext pairings:

    • public exponent e (e.g. e = 65537 = 0x10001)

    hashtag
    Process

    The idea behind this attack is effectively finding common factors between pairings. Recall that, under general RSA encryption, we have:

    and recall what modular arithmetic tells us about the relation between these terms, namely that:

    This, rearranged, tells us that

    What this means for our known pairings is that, given we know and , we can form the relationship:

    Thus we can calculate for the value , though don't know either value individually - we want to somehow derive .

    Observe that any two pairings will equate to such a value, both with as a factor. We can take the gcd of these two values, and it is probable that the resulting value will be our value, such that:

    However, this is only true for the case that

    i.e., both and are coprime. In the case that they are not, i.e. , we have that

    In such a case, we don't have sufficient information to completely recover the modulus, and require more plaintext-ciphertext pairs to be successful. In general, the more pairings you have, the more confident you can be the value you calculate is . More specifically:

    Thus:

    hashtag
    Practical Notes

    • In reality, you're likely to only need two or three (plaintext, ciphertext) pairings (in the context of ctf challenges and exercises), and as such computations can be manual if needed, but shouldn't be too complex

    • As it's likely you'll be dealing with large numbers, overflows and precision errors may arise in code - using libraries like gmpy provide support for integers of (theoretically) infinite size, and some nice accompanying features too (like in-built gcd and efficient modular exponentiation)

    hashtag
    Code Example

    Boneh-Durfee Attack

    hashtag
    What is Boneh-Durfee Attack

    Boneh-Durfee attack is an extension of Wiener's attack. That is, it also attacks on low private component dddwith a further relaxed condition. If dddsatisfies:

    d<N0.292d < N^{0.292} d<N0.292

    Then we can use Boneh-Durfee attack to retrive ddd

    this, using a graphical directed point of view, can be seen as:

    {E,n}→d<N0.292P{d}\{E, n\} \xrightarrow[d < N^{0.292}]{P} \{d\} {E,n}Pd<N0.292​{d}

    Consider for first, see that

    As stated above, the RSA's totient function can be espressed as:

    continuing with the equation, we see that

    and if we decide to consider and , we will have:

    At this point, finding is equivalent to find the 2 small solutions and to the congruence

    now let and this will preserve the scomposed subtraction

    consider (with any ), we deduct that must be really closed to because is in the same order of the length of (so ), we will get

    hashtag
    Sage Implementation

    Diffie-Hellman

    hashtag
    Overview

    triangle-exclamation

    We need to make some changes: separate the explanation from the code, add a subpart about the MITM and maybe to develop more the instructions

    Let's say Alice and Bob want to exchange a secret over an insecure channel. In other words, anyone can read the messages they send, but the goal is to ensure that only Alice and Bob can calculate the secret key.

    Diffie-Hellman key exchange provides a solution to this seemingly impossible task. Since code may be easier to understand than a detailed explanation, I'll provide it first:

    Here's a brief explanation of the code:

    • We choose a prime and a generator

    • Alice picks a private key

    • Bob picks a private key

    circle-info

    So anybody observing the messages sent between Alice and Bob would see , but they wouldn't be able to calculate the shared key .

    This is because given and , it should be infeasible to calculate . If this sounds familiar, that's because it's the .

    The original paper can be found . It uses the group of integers modulo a prime to perform the key exchange. In practice however, any group with a hard discrete log problem can be used.

    Rijndael Finite Field

    circle-info

    A first time reader might skip this section and go directly to the description of the round transformations, then come back later (it is mostly useful to understand the construction of the operation MC and SB).

    Each byte in AES is viewed as an element of a binary finite field of 256 elements, where it can always be represented as a polynomial of degree at most 7 with coefficients inF2\mathbf{F}_2F2​. The construction of the finite field is made as the quotient ringF2[x]/f(x)\mathbf{F}_2[x]/f(x)F2​[x]/f(x), wherefffis an irreducible polynomial of degree 8 inF2[x]\mathbf F_2[x]F2​[x]so the ring becomes a field.

    In AES, the choice foris

    We can check with SageMath that it is irreducible:

    hashtag
    Matching Bytes as Finite Field Elements

    A byte is composed of 8 bits and is matched to a polynomial as

    For instance, take the byte 3a whose binary decomposition is and becomes the polynomial

    hashtag
    Polynomial Reduction

    Polynomials of degree 8 or more can always be reduced, using the fact that in the finite field, we have , so we have the relation

    Why not ? In fact, that's also true, but the coefficient are in so the additive inverse of is itself.

    In SageMath, this reduction can be produced in one of the following methods.

    Method 1: Remainder of an Euclidean division by

    Method 2: Image in the quotient ring

    Method 3: Using the Finite Field class of SageMath directly.

    On this page we use this last method. Also, this helper converts an element of the finite field to the hexadecimal representation of a byte, and could be useful in the examples:

    hashtag
    Addition

    The addition of two polynomials is done by adding the coefficients corresponding of each monomial:

    And as the addition of the coefficients is in, it corresponds to the bitwise xor operation on the byte.

    hashtag
    Multiplication

    Multiplication of two polynomials is more complex (one example would be the , more efficient than the naive algorithm). For an implementation of AES, it is possible to only use the multiplication by , whose byte representation is 02.

    Letan element and we consider the multiplication by:

    All coefficients are shifted to a monomial one degree higher. Then, there are two cases:

    • Ifis, then we have a polynomial of degree at most 7 and we are done;

    • Ifis, we can replacebyduring the reduction phase:

    This can be used to implement a very efficient multiplication bywith the byte representation:

    1. A bitwise shiftleft operation: (b << 1) & 0xff;

    2. Followed by a conditional addition with 1b if the top bit of is .

    Here an example in SageMath (we use the finite field construction of method 3):

    Round Transformations

    This page gives a description of the four operations that compose a round of AES. Each has been designed to satisfy criterias, one of them is that all must be invertible.

    hashtag
    Add Round Key

    circle-check

    This is the only operation that involves a key. It is obvious that omitting it would mean no encryption.

    Round keys are derived from the master key (see the Key Schedule section) and are all composed of 16 bytes. We simply xor byte by byte the state by the bytes of the round key in the according position.

    Its inverse is itself: if we xor again, we get back the original value of the state.

    hashtag
    Mix Columns

    circle-check

    A major goal of MC is the diffusion property: each byte has an impact on the whole column, so a modification propagates to other positions.

    This operation mixes each column of the state independently from each other: each byte of the column is replaced by a (slightly different) combination of the four bytes of the column.

    Let the quadruplet of elements of a column, the operation MC is done by multiplying with a matrix .

    The calculations are performed in the finite field. If you are not familiar enough with the notions, you can skip to the next part and retain that this operation is also invertible using another matrix.

    This matrix is circulant: each row is the same as the one above but is shifted by one column. So we can construct it in one line of SageMath (recall that the bytes 02 and 03 correspond respectively to and in the field):

    hashtag
    Inverse of Mix Column

    It is needed to reverse MC for the decryption process. Fortunately, there exists a matrix such that , the identity matrix.

    circle-info

    We remark that the coefficient of the matrix are less friendly for the inverse operation as it involves polynomial of higher degree. This means that the decryption is a bit slower.

    hashtag
    Shift Rows

    circle-check

    The goal of shifting rows is reinforce the diffusion property: the bytes of a column are shifted so they are all positioned on different columns. Combined with MC, then a one byte modification will have an effect to the whole state after several rounds: this is the .

    This operation shifts the rows of the state in the following manner:

    The rows are shifted from top to bottom respectively by an offset of 0, 1, 2 or 3 columns to the left. Bytes that overflow on the left are put back to the right of the row.

    The inverse is almost the same: the rows are shifted to the right instead by the same offsets and exceeding bytes are put back to the left.

    circle-info

    Small exercise: to what extent would be the impact on the security of AES if no shifting were present?

    hashtag
    Substitution Box

    circle-check

    Last, but not least, the SB design criterias is to bring the confusion property to make correlation between a plaintext and a ciphertext as small as possible.

    The precedent operations shuffle the plaintext in such a way that any modification of a byte has an impact to the whole state after several rounds. Though, this is not enough as those operations are linear: it means that the ciphertext could be expressed as linear equations from the plaintext and the master key. From a known couple plaintext/ciphertext it would be easy to solve the system and find the key.

    The substitution box is the operation that breaks the linearity: each byte of the state is replaced by another following the table below.

    circle-info

    If 3b is the input, its output is on row 30 and column 0b and is the byte e2.

    Let sbox the name of this table, then for any bytes b1 and b2, we have

    which is the desired property.

    Though, this is not enough and the design of the sbox is made to include a sufficient algebraic complexity. The construction of the sbox table is done in two steps:

    1. An elementis replaced by ( has no inverse and is mapped to );

    2. An affine transformation on the coefficients of :

    The table we gave above has been construted with SageMath and applying the two steps:

    The inverse of the table is simple to produce: we only need to reverse the match between an input and its output.

    circle-info

    Exercise: write the inverse of the substitution box in SageMath using the inverse of the affine transformation.

    Encryption

    Author: Chuck_bartwoski

    hashtag
    Introduction

    A typical application of cryptography is secure communication. Informally a secure communication channel is one that provides both confidentiality and Integrity of the messages. In this section we investigate confidentiality, therefore we may assume that integrity is already guaranteed by some other means. (see section on integrity...#TODO)

    We assume that two parties that want to communicate share a secret key kkk. Prior to sending a message, the sender encrypts the message with the secret key, this produces a ciphertext that is then sent. The receiver uses the same key to decrypt the message and recover the message.

    circle-info

    Intuitively: A secure encryption scheme will prevent an eavesdropper to learn the content of the message since the ciphertext is unintelligible. The security requirement will be formalized later.

    hashtag
    Formal definition

    We introduce some notation first: We will use to denote the set of al possible message, is the set of all possible keys and is the set of ciphertexts.

    A symmetric encryption scheme is a tuple of efficiently computable functions .:

    • Selects a key at random from the key space.

    • . Encrypts the message with the key into a ciphertext . Sometimes written as

    • . Decrypts the ciphertexts with the key

    circle-exclamation

    For to be useful we need that . This is also called correctness.

    After all what's the point of confidentially sending a nice Christmas card to your grand children if they wont be able to read its content

    TODO: security notions and examples

    AES

    Advanced Encryption Standard

    hashtag
    Introduction

    The Advanced Encryption Standard most known as AES is one of the most used ciphers nowadays. Created by Vinent Rijmen and Joan Daemen under the name Rijndael, it won the NIST competition that resulted in its standardization in 2001 to replace older algorithms such as DES (and its variant 3DES). In fact, it is six times faster than 3DES.

    AES encrypts a block of 16 bytes only at a time, though ciphertexts tend to be much longer. To accomodate this, cipherexts are cut in blocks of 16 bytes using an operating mode [see future section on mode]. We only focus on the encryption of a single block.

    The array of 16 bytesare arranged from up to bottom, column by column inmatrix. During the encryption, the state of this matrix changes and results in a 16-bytes ciphertextwhose output can be read following the same ordering:

    A key is involved and three sizes are possible: 128, 192, or 256 bits. Depending of the size, there are a few differences which will be explained later. For now, it is sufficient to know that round keys are derived from this master key.

    Our interest is to look at what goes inside the transformation between the plaintext and the ciphertext. Basically, there are four operations on the state matrix, each important for the security of AES:

    • AK: add round key;

    • SR: shift row;

    • SB: substitution box;

    All these operations are executed a several number of times in what are called rounds to mix the plaintext enough. A look on the flow of an encryption is given in the figure below.

    Two particular cases can be noticed:

    • the first round is preceded by an additional AK;

    • last round is missing MC.

    The number of rounds NR is different depending on the master key length:

    Isogenies

    hashtag
    Motivation

    circle-info

    Prerequisites: in this section we assume the reader is somewhat familiar with elliptic curves and begin by considering morphisms (maps) between elliptic curves.

    Introduction to Isogeny Cryptography

    Making this section as a motivation to make sure this is part of the book. Something to work towards.

    hashtag
    Page Plan

    • Describe that there is a drive towards post-quantum algorithms

    The Birthday paradox / attack

    Authors: Zademn, ireland Reviewed by:

    Prerequisites

    • Probability theory (for the main idea)

    • Hashes (an application)

    Introduction / overview

    Authors: Zademn Reviewed by:

    hashtag
    Introduction

    Another desired propriety of our cryptographic protocols is data / message integrity. This propriety assures that during a data transfer the data has not been modified.

    Suppose Alice has a new favourite game and wants to send it to Bob. How can Bob be sure that the file he receives is the same as the one Alice intended to send? One would say to run the game and see. But what if the game is a malware? What if there are changes that are undetectable to the human eye?

    The One Time Pad

    Author: chuck_bartowski

    hashtag
    Introduction

    The One Time Pad (OTP) is a well known example of encryption schemes that provide "perfect secrecy". Informally, this means that observing a ciphertext does no give any information to the eavesdropper. A proof of this fact will be provided later. Crucially we will assume that the sender and the receiver have both access to a common source of random bits.

    @interact
    def draw_cvp(v1x = input_box(label = "v1 x =", default = 1),
                 v1y = input_box(label = "v1 y =", default = 0),
                 v2x = input_box(label = "v2 x =", default = 0),
                 v2y = input_box(label = "v2 y =", default = 1),
                 wx = input_box(label = "w x =", default = 1.8),
                 wy = input_box(label = "w y =", default = 1.7),
                 box = 5, search = 10):
    
        v1 = vector((v1x, v1y))
        v2 = vector((v2x, v2y))
        print(v1, v2)
        vecs = []
        # Generate vectors
        for i in range(-search,search):
            for j in range(-search,search):
                vecs.append(i*v1 + j*v2)
        # Plot stuff
        G = Graphics()
        for p1 in vecs:
            x1, y1 = p1
            if x1 > -box and x1 < box and y1 > -box and y1 < box:
                G += point(p1, color = 'green', size = 30)
                G += line([p1, p1 + v2], linestyle = '--', alpha = .20)
                G += line([p1, p1 + v1], linestyle = '--', alpha = .20)
        G+= arrow((0, 0), v1, color = 'red', arrowsize = 2)
        G+= arrow((0, 0), v2, color = 'red', arrowsize = 2)
        G+= text('v1', v1 + .2 * v1, color = 'red')
        G+= text('v2', v2 + .2 * v2, color = 'red')         
        
        # Cvp
        L = IntegerLattice(matrix([v1, v2]))
        w = vector((wx, wy))
        t = L.closest_vector(w)
        G += point(w, color = 'purple', size = 30)
        G+= text('w', w + .2 * v1, color = 'purple')         
    
        G += point(t, color = 'red', size = 30)
        G+= text('t', t + .2 * v1, color = 'red')         
        G+= circle(center = w, radius=norm(t - w), color = 'purple', alpha = .5)
        G.show(axes = False, figsize = (7, 7))
    @interact
    def draw_dual(v1x = input_box(label = "v1 x =", default = 1),
                     v1y = input_box(label = "v1 y =", default = 0),
                     v2x = input_box(label = "v2 x =", default = 0),
                     v2y = input_box(label = "v2 y =", default = 1),
                     box = 3, search = 6,
                     plot_lattice_points = True,
                     plot_lattice_lines = True,
                     plot_dual_points = True,
                     plot_dual_lines = True):
    
        v1 = vector((v1x, v1y))
        v2 = vector((v2x, v2y))
        vecs = []
        v1_, v2_ = matrix([v1,v2]).inverse().T
        vecs_ = []
        # Generate vectors
        for i in range(-search,search):
            for j in range(-search,search):
                vecs.append(i*v1 + j*v2)
        for i in range(-search,search):
            for j in range(-search,search):
                vecs_.append(i*v1_ + j*v2_)
        # Plot stuff
        G = Graphics()
        for p1 in vecs:
            x1, y1 = p1
            if x1 > -box and x1 < box and y1 > -box and y1 < box:
                if plot_lattice_points:
                    G += point(p1, color = 'green', size = 70)
                if plot_lattice_lines:
                    G += line([p1, p1 + v2], linestyle = '--', alpha = .20, color = 'green')
                    G += line([p1, p1 + v1], linestyle = '--', alpha = .20, color = 'green')
        if plot_lattice_lines:
            G+= arrow((0, 0), v1, color = 'green', arrowsize = 2)
            G+= arrow((0, 0), v2, color = 'green', arrowsize = 2)
            G+= text('v1', v1 + .2 * v1, color = 'green')
            G+= text('v2', v2 + .2 * v2, color = 'green')
        
        # Dual
        for p1 in vecs_:
            x1, y1 = p1
            if x1 > -box and x1 < box and y1 > -box and y1 < box:
                if plot_dual_points:
                    G += point(p1, color = 'red', size = 25)
                if plot_dual_lines:
                    G += line([p1, p1 + v2_], linestyle = '--', alpha = .20, color = 'red')
                    G += line([p1, p1 + v1_], linestyle = '--', alpha = .20, color = 'red')
        if plot_dual_lines:
            G+= arrow((0, 0), v1_, color = 'red', arrowsize = 2)
            G+= arrow((0, 0), v2_, color = 'red', arrowsize = 2)
            G+= text('v1\'', v1_ + .2 * v1, color = 'red')
            G+= text('v2\'', v2_ + .2 * v2, color = 'red')
        
        G.show(axes = False, figsize = (7, 7))
    # I'm not sure this does what is supposed to do
    # but the plots are nice
    
    from sage.modules.free_module_integer import IntegerLattice
    @interact
    def draw_qary(v1x = input_box(label = "v1 x =", default = 1),
                     v1y = input_box(label = "v1 y =", default = 0),
                     v2x = input_box(label = "v2 x =", default = 0),
                     v2y = input_box(label = "v2 y =", default = 1),
                     q = input_box(label = "q =", default = 5),
                     box = 7, search = 6,
                     plot_lattice_points = True,
                     plot_lattice_lines = True,
                     plot_qary_points = True,
                     plot_qary_lines = True):
    
        v1 = vector((v1x, v1y))
        v2 = vector((v2x, v2y))
        L = IntegerLattice(matrix([v1, v2]))
        vecs = []
        v1_ = v1.change_ring(Zmod(q))
        v2_ = v2.change_ring(Zmod(q))
        vecs_ = []
        # Generate vectors
        for i in range(-search,search):
            for j in range(-search,search):
                v = i*v1 + j*v2
                vecs.append(v)
                v = v.change_ring(Zmod(q)).change_ring(ZZ)
                if v not in vecs_:
                    vecs_.append(v)
                    
        # Lattice
        G = Graphics()
        for p1 in vecs:
            x1, y1 = p1
            if x1 > -box and x1 < box and y1 > -box and y1 < box:
                if plot_lattice_points:
                    G += point(p1, color = 'green', size = 70)
                if plot_lattice_lines:
                    G += line([p1, p1 + v2], linestyle = '--', alpha = .20, color = 'green')
                    G += line([p1, p1 + v1], linestyle = '--', alpha = .20, color = 'green')
        if plot_lattice_lines:
            G+= arrow((0, 0), v1, color = 'green', arrowsize = 2)
            G+= arrow((0, 0), v2, color = 'green', arrowsize = 2)
            G+= text('v1', v1 + .2 * v1, color = 'green')
            G+= text('v2', v2 + .2 * v2, color = 'green')
        
        # qary
        for p1 in vecs_:
            p1 = p1
            x1, y1 = p1
            if x1 > -box and x1 < box and y1 > -box and y1 < box:
                if plot_qary_points:
                    G += point(p1, color = 'red', size = 25)
                if plot_qary_lines:
                    G += line([p1, p1 + v2_], linestyle = '--', alpha = .20, color = 'red')
                    G += line([p1, p1 + v1_], linestyle = '--', alpha = .20, color = 'red')
        if plot_qary_lines:
            G+= arrow((0, 0), v1_, color = 'purple', arrowsize = 2)
            G+= arrow((0, 0), v2_, color = 'purple', arrowsize = 2)
            G+= text('v1\'', v1_.change_ring(ZZ) + .2 * v1, color = 'purple')
            G+= text('v2\'', v2_.change_ring(ZZ) + .2 * v2, color = 'purple')
        
        G.show(axes = False, figsize = (10, 10))
    from Crypto.Util.number import long_to_bytes
    
    def wiener(e, n):
        # Convert e/n into a continued fraction
        cf = continued_fraction(e/n)
        convergents = cf.convergents()
        for kd in convergents:
            k = kd.numerator()
            d = kd.denominator()
            # Check if k and d meet the requirements
            if k == 0 or d%2 == 0 or e*d % k != 1:
                continue
            phi = (e*d - 1)/k
            # Create the polynomial
            x = PolynomialRing(RationalField(), 'x').gen()
            f = x^2 - (n-phi+1)*x + n
            roots = f.roots()
            # Check if polynomial as two roots
            if len(roots) != 2:
                continue
            # Check if roots of the polynomial are p and q
            p,q = int(roots[0][0]), int(roots[1][0])
            if p*q == n:
                return d
        return None
    # Test to see if our attack works
    if __name__ == '__main__':
        n = 6727075990400738687345725133831068548505159909089226909308151105405617384093373931141833301653602476784414065504536979164089581789354173719785815972324079
        e = 4805054278857670490961232238450763248932257077920876363791536503861155274352289134505009741863918247921515546177391127175463544741368225721957798416107743
        c = 5928120944877154092488159606792758283490469364444892167942345801713373962617628757053412232636219967675256510422984948872954949616521392542703915478027634
        d = wiener(e,n)
        assert not d is None, "Wiener's attack failed :("
        print(long_to_bytes(int(pow(c,d,n))).decode())
    #!/usr/bin/env python3
    import owiener
    from Crypto.Util.number import long_to_bytes
    
    #--------Data--------#
    
    N = <N>
    e = <e>
    c = <c>
    
    #--------Wiener's attack--------#
    
    d = owiener.attack(e, N)
    
    if d:
        m = pow(c, d, N)
        flag = long_to_bytes(m).decode()
        print(flag)
    else:
        print("Wiener's Attack failed.")
    # We can find the shortest vector using the LLL algorithm
    M = matrix([[-1, 2], [-2, 3]])
    B = M.LLL()
    print(B[0])
    # (0, -1)
    
    # Or we can use the Integer Lattice class
    L = IntegerLattice(M)
    L.shortest_vector()
    # (-1, 0)
    M = matrix([[-1, 2], [-2, 3]])
    L = IntegerLattice(M)
    
    w = vector([1.8, 1.5])
    L.closest_vector(w)
    # (2.00000000000000, 2.00000000000000)
    Humans are fascinated with symmetries. The guiding star of theoretical physics is the study of dualities. How much one thing can be said to be another leads to strange and beautiful links between areas of mathematics that appear to be totally distinct.

    A cruical piece of building this understanding is how one can map between objects which share structure. When we learn about topology, we are given the fun: "A doughnut is the same as a mug"; a statement which says within topology, we identify objects related by continuous functions.

    Elliptic curves are beautiful mathematical objects. The fact that a geometric comes hand-in-hand with a algebraic group law is, to me, incredible. The study of isogenies is the study of maps (morphisms) between elliptic curves which preserves the origin. This condition is enough to preserve the group scheme of the elliptic curve.

    In short, isogenies allow us to map between curves preserving their geometric properties (as projective varieties) and algebraic properties (the group associated with point addition).

    hashtag
    Isogenies of Elliptic Curves

    Definition: We say an isogeny ϕ:E1→E2\phi : E_1 \to E_2ϕ:E1​→E2​ between elliptic curves defined over a field kkkis a surjective morphism of curves which includes a group homomorphism E1(kˉ)→E1(kˉ)E_1(\bar{k}) \to E_1(\bar{k})E1​(kˉ)→E1​(kˉ)

    hashtag
    References

    • https://arxiv.org/pdf/1711.04062.pdfarrow-up-right

    • https://math.mit.edu/classes/18.783/2019/LectureNotes5.pdfarrow-up-right

    • https://doc.sagemath.org/html/en/reference/arithmetic_curves/sage/schemes/elliptic_curves/ell_curve_isogeny.htmlarrow-up-right

    bib_ibi​
    LLL
    B\mathcal BB
    iii
    bib_ibi​
    bi∗b_i^*bi∗​
    bib_ibi​
    B∗\mathcal B^*B∗
    iii
    bi∗b_i^*bi∗​
    μi,j=⟨bi,bj∗⟩⟨bj∗,bj∗⟩\mu_{i,j}=\frac{\langle b_i,b_j^*\rangle}{\langle b_j^*,b_j^*\rangle}μi,j​=⟨bj∗​,bj∗​⟩⟨bi​,bj∗​⟩​
    λi(L)\lambda_i(L)λi​(L)
    iii
    LLL
    ed≡1mod  ϕ(n),    ⇒    ed=1+kϕ(n)ed \equiv 1 \mod \phi(n), \;\; \Rightarrow \;\; ed = 1 + k\phi(n)ed≡1modϕ(n),⇒ed=1+kϕ(n)
    k∈Zk \in \mathbb{Z}k∈Z
    mmm
    cd≡(me)dmod  n≡medmod  n≡mkϕ+1mod  n≡mkϕmmod  n≡(mϕ)kmmod  n≡1kmmod  n≡mmod  n\begin{align} c^d &\equiv (m^e)^d &&\mod n \\ &\equiv m^{ed} &&\mod n \\ &\equiv m^{k\phi + 1} &&\mod n \\ &\equiv m^{k\phi} m &&\mod n \\ &\equiv (m^\phi)^km &&\mod n \\ &\equiv 1^km &&\mod n \\ &\equiv m &&\mod n \\ \end{align}cd​≡(me)d≡med≡mkϕ+1≡mkϕm≡(mϕ)km≡1km≡m​​modnmodnmodnmodnmodnmodnmodn​​
    gcd⁡(m,n)=1\gcd(m, n) = 1gcd(m,n)=1
    ppp
    qqq
    gcd⁡(m,n)=n,\gcd(m, n) = n,gcd(m,n)=n,
    c≡m≡0mod  nc \equiv m \equiv 0 \mod nc≡m≡0modn
    m=k⋅p,m = k\cdot p,m=k⋅p,
    c≡m≡0mod  pc \equiv m \equiv 0 \mod pc≡m≡0modp
    cd≡(me)d(modq).c^d \equiv (m^e)^d \pmod q.cd≡(me)d(modq).
    gcd⁡(m,q)=q,\gcd(m, q) = q,gcd(m,q)=q,
    gcd⁡(m,q)=1,\gcd(m, q) = 1,gcd(m,q)=1,
    qqq
    mℓϕ(q)≡1mod  qm^{\ell\phi(q)} \equiv 1 \mod qmℓϕ(q)≡1modq
    ϕ\phiϕ
    mℓnϕ(n)=mℓϕ(q)≡1mod  q.m^{\ell_n\phi(n)} = m^{\ell\phi(q)} \equiv 1 \mod q.mℓn​ϕ(n)=mℓϕ(q)≡1modq.
    m≡0mod  pm≡1ℓmmod  q\begin{align*} m &\equiv 0 &&\mod p \\ m &\equiv 1^\ell m &&\mod q\\ \end{align*}mm​≡0≡1ℓm​​modpmodq​
    me⋅d≡mmod  n.m^{e\cdot d} \equiv m \mod n.me⋅d≡mmodn.
    m=k⋅qm = k\cdot qm=k⋅q

    Probability Theory

    such as
  • By using the Extended Euclidean algorithm, we compute the invert d\mathbb{d}d of emod  n\mathbb{e \mod n}emodn :d≡e−1mod  ϕ(n)d \equiv e^{-1} \mod \phi(n)d≡e−1modϕ(n) which is our private exponent.

  • Public key: n,en, en,e

  • Private key: n,dn, dn,d

  • Now, chose a message m\mathbb{m}mthat you convert into integers

  • We can encrypt this plaintext mmm and receive a ciphertext c≡memod  nc \equiv m^e \mod nc≡memodn

  • We can decrypt a ciphertext ccc with m≡cdmod  nm \equiv c^d \mod nm≡cdmodn

  • NN N
    NN N
    ppp
    qqq
    ppp
    qqq
    n=p×qn = p\times qn=p×q
    ϕ(n)=(p−1)×(q−1)\phi(n) = (p-1) \times (q-1)ϕ(n)=(p−1)×(q−1)
    e\mathbb{e}e
    eee
    ddd
    ed≡1mod  ϕ(n)ed \equiv 1\mod \phi(n)ed≡1modϕ(n)
    mmm
    s≡mdmod  ns \equiv m^d \mod ns≡mdmodn
    sss
    mmm
    sss
    m≡semod  nm \equiv s^e \mod nm≡semodn
    mmm
    sss
    public-key cryptosystemarrow-up-right
    acronymarrow-up-right
    Ron Rivestarrow-up-right
    Adi Shamirarrow-up-right
    Leonard Adlemanarrow-up-right
    GCHQarrow-up-right
    signals intelligencearrow-up-right
    Clifford Cocksarrow-up-right
    declassifiedarrow-up-right
    trapdoor functionsarrow-up-right
    hardness of factoring integersarrow-up-right
    Shor's algorithmarrow-up-right
    gcd(e,ϕ(n))=1\mathbb{gcd(e, \phi(n)) = 1}gcd(e,ϕ(n))=1
    These two statements are mathematically equivalent, but one is easier to implement in code:
    (Mi,Ci) for i∈[1,∞](M_i,C_i) \text{ for } i \in [1,\infty] (Mi​,Ci​) for i∈[1,∞]
    C=Me (mod N)C = M^{e} \text{ } (mod\text{ }N)C=Me (mod N)
    a≡b (mod N)a=b+kN for some k∈Za \equiv b\text{ }(mod\text{ } N)\\ a = b + kN \text{ for some } k \in \mathbb{Z} a≡b (mod N)a=b+kN for some k∈Z
    a−b≡0 (mod N)a−b=kNa - b \equiv 0\text{ } (mod \text{ } N)\\ a - b = kNa−b≡0 (mod N)a−b=kN
    m,cm, cm,c
    eee
    Ci−Mie≡0 (mod N)Ci−Mie=kiNC_i - M_i^e \equiv 0\text{ } (mod \text{ } N)\\ C_i - M_i^e = k_iNCi​−Mie​≡0 (mod N)Ci​−Mie​=ki​N
    kNkNkN
    NNN
    NNN
    NNN
    N=gcd(C1−M1e,C2−M2e)N = gcd(C_1 - M_1^e, C_2 - M_2^e)N=gcd(C1​−M1e​,C2​−M2e​)
    gcd(k1,k2)=1gcd(k_1, k_2) = 1gcd(k1​,k2​)=1
    k1k_1k1​
    k2k_2k2​
    gcd(k1,k2)≠1gcd(k_1, k_2) \ne 1gcd(k1​,k2​)=1
    aN=gcd(C1−M1e,C2−M2e) s.t. 1≠a∈ZaN = gcd(C_1 - M_1^e, C_2 - M_2^e) \text{ s.t. } 1 \ne a \in \mathbb{Z}aN=gcd(C1​−M1e​,C2​−M2e​) s.t. 1=a∈Z
    NNN
    Pr(a≠1)→0 as k→∞Pr(a \ne1) \rightarrow 0 \text{ as } k\rightarrow \inftyPr(a=1)→0 as k→∞
    N=lim⁡k→∞gcd(C1−M1e,C2−M2e,...,Ck−Mke)N = \lim_{k \rightarrow \infty} gcd(C_1 - M_1^e, C_2 - M_2^e, ..., C_k - M_k^e)N=k→∞lim​gcd(C1​−M1e​,C2​−M2e​,...,Ck​−Mke​)
    gcd(a,b,c,d,...)=gcd(a,gcd(b,gcd(c,gcd(d,...))))gcd(a, b, c, d, ...) = gcd(a, gcd(b, gcd(c, gcd(d, ...))))gcd(a,b,c,d,...)=gcd(a,gcd(b,gcd(c,gcd(d,...))))

    Alice's public key is gamod  pg^a \mod pgamodp

  • Bob's public key is gbmod  pg^b \mod pgbmodp

  • Their shared key is gab≡(ga)b≡(gb)a(modp)g^{ab} \equiv (g^a)^b \equiv (g^b)^a \pmod pgab≡(ga)b≡(gb)a(modp)

  • ppp
    g∈Fpg \in \mathbb{F}_pg∈Fp​
    a∈Zp−1a \in \mathbb{Z}_{p-1}a∈Zp−1​
    b∈Zp−1b \in \mathbb{Z}_{p-1}b∈Zp−1​
    p,g,ga,gbp, g, g^a, g^bp,g,ga,gb
    gabg^{ab}gab
    ggg
    gag^aga
    aaa
    Discrete Log Problem
    herearrow-up-right
    fff
    f(x)=x8+x4+x3+x+1.f(x) = x^8 + x^4 + x^3 + x + 1.f(x)=x8+x4+x3+x+1.
    bbb
    (b7,…,b0)2(b_7, \ldots, b_0)_2(b7​,…,b0​)2​
    b7x7+b7x6+⋯+b1x+b0.b_7x^7 + b_7 x^6 + \cdots + b_1 x + b_0.b7​x7+b7​x6+⋯+b1​x+b0​.
    (0,0,1,1,1,0,1,0)2(0, 0, 1, 1, 1, 0, 1, 0)_2(0,0,1,1,1,0,1,0)2​
    0⋅x7+0⋅x6+1⋅x5+1⋅x4+1⋅x3+0⋅x2+1⋅x+0=x5+x4+x3+x.0\cdot x^7 + 0\cdot x^6 + 1\cdot x^5 + 1\cdot x^4 + 1\cdot x^3 + 0\cdot x^2 + 1\cdot x + 0 = x^5 + x^4 + x^3 + x.0⋅x7+0⋅x6+1⋅x5+1⋅x4+1⋅x3+0⋅x2+1⋅x+0=x5+x4+x3+x.
    f(x)=0f(x) = 0f(x)=0
    x8=x4+x3+x+1x^8 = x^4 + x^3 + x + 1x8=x4+x3+x+1
    x8=−x4−x3−x−1x^8 = - x^4 - x^3 - x - 1x8=−x4−x3−x−1
    F2\mathbf F_2F2​
    −1-1−1
    111
    fff
    F2[x]/f(x)\mathbf{F}_2[x]/f(x)F2​[x]/f(x)
    ∑i=07aixi+∑i=07bixi=∑i=07(ai+bi)xi.\sum_{i=0}^7 a_ix^i + \sum_{i=0}^7 b_ix^i = \sum_{i=0}^7(a_i + b_i)x^i.i=0∑7​ai​xi+i=0∑7​bi​xi=i=0∑7​(ai​+bi​)xi.
    F2\mathbf{F}_2F2​
    xxx
    b7x7+⋯+b1x+b0b_7x^7 + \cdots + b_1x + b_0b7​x7+⋯+b1​x+b0​
    xxx
    x×(b7x7+⋯+b1x+b0)=b7x8+b6x7+⋯b1x2+b0x.x\times (b_7x^7 + \cdots + b_1x + b_0) = b_7x^8 + b_6x^7 + \cdots b_1 x^2 + b_0x.x×(b7​x7+⋯+b1​x+b0​)=b7​x8+b6​x7+⋯b1​x2+b0​x.
    b7b_7b7​
    000
    b7b_7b7​
    111
    x8x^8x8
    x4+x3+x+1x^4 + x^3 + x + 1x4+x3+x+1
    x×(x7+b6x6+⋯+b1x+b0)=x8+b6x7+⋯b1x2+b0x=(x4+x3+x+1)+b6x7+⋯+b1x2+b0x\begin{align*} x\times (x^7 + b_6x^6 + \cdots + b_1x + b_0) & = x^8 + b_6x^7 + \cdots b_1 x^2 + b_0x \\ & = (x^4 + x^3 + x + 1) + b_6x^7 + \cdots + b_1x^2 + b_0x \end{align*}x×(x7+b6​x6+⋯+b1​x+b0​)​=x8+b6​x7+⋯b1​x2+b0​x=(x4+x3+x+1)+b6​x7+⋯+b1​x2+b0​x​
    xxx
    bbb
    111
    Karatsuba algorithmarrow-up-right
    into the message
    or returns an error (
    ).
    . Sometimes written as
    M\mathcal MM
    K\mathcal KK
    C\mathcal CC
    E\mathcal EE
    (KGen, Enc, Dec)(\text{KGen, Enc, Dec})(KGen, Enc, Dec)
    \text{KGen}: \diamond \xrightarrow $ \mathcal K
    Enc:M×K↦C\text{Enc}: \mathcal M \times \mathcal K \mapsto \mathcal CEnc:M×K↦C
    mmm
    kkk
    c=Enc(m,k)c = \text{Enc}(m, k)c=Enc(m,k)
    c=Enck(m)c = \text{Enc}_k(m)c=Enck​(m)
    Dec:C×K↦M×{⊥}\text{Dec}: \mathcal C \times \mathcal K \mapsto \mathcal M \times \{ \bot\}Dec:C×K↦M×{⊥}
    ccc
    E\mathcal EE
    Dec(Enc(m,k),k)=m;∀m,k\text{Dec}(\text{Enc}(m,k), k) = m; \forall m,kDec(Enc(m,k),k)=m;∀m,k
    kkk
    mmm
    ⊥\bot⊥
    m=Dec(c,k)m = \text{Dec}(c, k)m=Dec(c,k)
    m=Deck(c)m = \text{Dec}_k(c)m=Deck​(c)
    hashtag
    XOR as a One-Time Pad

    XOR(addition modulo 2) can be used as an encryption scheme as follows: The message space is M⊆{0,1}n\mathcal M \subseteq \{0, 1\}^nM⊆{0,1}n(i.e.: length n bit strings), the key space is K={0,1}\mathcal K = \{0, 1\}K={0,1} and the ciphertext space is also {0,1}\{0,1\}{0,1}

    • Encryption: Enc(m,k)=m⊕k\text{Enc}(m,k) = m \oplus kEnc(m,k)=m⊕k

    • Decryption:Dec(c,k)=c⊕k\text{Dec}(c,k) = c \oplus kDec(c,k)=c⊕k

    circle-check

    The correctness of the schemes is easily verifiable. If the encryption produces c=m⊕kc = m \oplus kc=m⊕k, then the decryption produces m′=c⊕k=m⊕k⊕k=mm' = c \oplus k = m \oplus k \oplus k = mm′=c⊕k=m⊕k⊕k=m.

    circle-info

    In the Python snippet below with use to os module to generate random bits.

    As seen above the receiver with access to the same key can recover the message.

    hashtag
    Generalized One-Time Pad

    Although XOR is commonly used for the OTP, OTP can be made out of more general objects. If fact We can define an OTP if the message and the keys are objects from a set with a group like structure. (see GroupTheorySection #TODO)

    import gmpy2
    
    """
    @param pairings
        list: [(pt1, ct1), (pt2, ct2), ..., (ptk, ctk)]
    @param e
        int : encryption exponent
    @return
        int : recovered N
    """
    def recover_n(pairings, e):
        pt1, ct1 = pairings[0]
        N = ct1 - pow(pt1, e)
        
        # loop through and find common divisors
        for pt,ct in pairings:
            val = gmpy2.mpz(ct - pow(pt, e))
            N = gmpy2.gcd(val, N)
        
        return N
        
    import Crypto.Util.number as cun
    import Crypto.Random.random as crr
    
    
    class DiffieHellman:
        def __init__(self, p: int):
            self.p = p
            self.g = 5
            self.private_key = crr.randrange(2, p-1)
    
        def public_key(self) -> int:
            return pow(self.g, self.private_key, self.p)
    
        def shared_key(self, other_public_key: int) -> int:
            return pow(other_public_key, self.private_key, self.p)
    
    
    p = cun.getPrime(512)
    alice = DiffieHellman(p)
    bob = DiffieHellman(p)
    
    shared_key = bob.shared_key(alice.public_key())
    assert shared_key == alice.shared_key(bob.public_key())
    F2 = GF(2)
    K.<x> = F2[]
    f = x^8 + x^4 + x^3 + x + 1
    f.is_irreducible()
    # True
    (x^8 + x^6 + x^4 + 1) % f
    # x^6 + x^3 + x
    R = K.quotient(f)
    R(x^8 + x^6 + x^4 + 1)
    # xbar^6 + xbar^3 + xbar
    F.<x> = GF(2^8, modulus=x^8 + x^4 + x^3 + x + 1)
    x^8 + x^6 + x^4 + 1
    # x^6 + x^3 + x
    def F_to_hex(a):
        return ZZ(a.integer_representation()).hex()
    
    b = x^4 + x^3 + x + 1
    F_to_hex(b)
    # '1b'
    b = x^7 + x^5 + x^4 + x^2 + 1
    F_to_hex(b)
    # 'b5'
    (2*0xb5 & 0xff) ^^ 0x1b).hex() == F_to_hex(x*b)    # the xor in Sage is "^^"
    # True
    import os
    
    def xor(a,b):
        res = bytes([x^y for (x,y) in zip(a,b)])
        return res
        
    message = b"YELLOW SUBMARINE"
    key = os.urandom(len(message))
    ciphertext = xor(message, key)
    recovered = xor(ciphertext, key)
    print(f"Message: {message}\nKey: {key}\nCiphertext: {ciphertext}\nrecovered: {recovered}")
    # A possible ouput might be as below
    # Message: b'YELLOW SUBMARINE'
    # Key: b'\x8e<3\xc9\x8d\xbaD\x16Zb\x1b\xbb\xb3\x0c@<'
    # Ciphertext: b'\xd7y\x7f\x85\xc2\xeddE\x0f V\xfa\xe1E\x0ey'
    # recovered: b'YELLOW SUBMARINE'

    0f

    00

    63

    7c

    77

    7b

    f2

    6b

    6f

    c5

    30

    01

    67

    2b

    fe

    d7

    ab

    10

    ca

    82

    c9

    7d

    fa

    59

    47

    f0

    ad

    d4

    a2

    af

    9c

    a4

    72

    20

    b7

    fd

    93

    26

    36

    3f

    f7

    cc

    34

    a5

    e5

    f1

    71

    d8

    31

    30

    04

    c7

    23

    c3

    18

    96

    05

    9a

    07

    12

    80

    e2

    eb

    27

    b2

    40

    09

    83

    2c

    1a

    1b

    6e

    5a

    a0

    52

    3b

    d6

    b3

    29

    e3

    2f

    50

    53

    d1

    00

    ed

    20

    fc

    b1

    5b

    6a

    cb

    be

    39

    4a

    4c

    58

    60

    d0

    ef

    aa

    fb

    43

    4d

    33

    85

    45

    f9

    02

    7f

    50

    3c

    9f

    70

    51

    a3

    40

    8f

    92

    9d

    38

    f5

    bc

    b6

    da

    21

    10

    ff

    f3

    80

    cd

    0c

    13

    ec

    5f

    97

    44

    17

    c4

    a7

    7e

    3d

    64

    5d

    19

    90

    60

    81

    4f

    dc

    22

    2a

    90

    88

    46

    ee

    b8

    14

    de

    5e

    0b

    a0

    e0

    32

    3a

    0a

    49

    06

    24

    5c

    c2

    d3

    ac

    62

    91

    95

    e4

    b0

    e7

    c8

    37

    6d

    8d

    d5

    4e

    a9

    6c

    56

    f4

    ea

    65

    7a

    ae

    c0

    ba

    78

    25

    2e

    1c

    a6

    b4

    c6

    e8

    dd

    74

    1f

    4b

    bd

    8b

    d0

    70

    3e

    b5

    66

    48

    03

    f6

    0e

    61

    35

    57

    b9

    86

    c1

    1d

    e0

    e1

    f8

    98

    11

    69

    d9

    8e

    94

    9b

    1e

    87

    e9

    ce

    55

    28

    f0

    8c

    a1

    89

    0d

    bf

    e6

    42

    68

    41

    99

    2d

    0f

    b0

    54

    bb

    (a0,a1,a2,a3)(a_0, a_1, a_2, a_3)(a0​,a1​,a2​,a3​)
    MMM
    xxx
    x+1x+1x+1
    NNN
    M×N=N×M=IdM\times N = N \times M = \textrm{Id}M×N=N×M=Id

    00

    01

    02

    03

    04

    05

    06

    07

    08

    09

    0a

    0b

    0c

    0d

    sbox(b1⊕b2)≠sbox(b1)⊕sbox(b2)\texttt{sbox}(\texttt{b1} \oplus \texttt{b2}) \neq \texttt{sbox}(\texttt{b1}) \oplus \texttt{sbox}(\texttt{b2})sbox(b1⊕b2)=sbox(b1)⊕sbox(b2)
    aaa
    b=a−1b = a^{-1}b=a−1
    000
    000
    aaa
    [1000111111000111111000111111000111111000011111000011111000011111]⋅[b0b1b2b3b4b5b6b7]+[11000110]\begin{bmatrix} 1 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 0 & 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 0 & 0 & 0 & 1 & 1 \\ 1 & 1 & 1 & 1 & 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} b_0 \\ b_1 \\ b_2 \\ b_3 \\ b_4 \\ b_5 \\ b_6 \\ b_7 \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 1 \\ 0 \end{bmatrix}​11111000​01111100​00111110​00011111​10001111​11000111​11100011​11110001​​⋅​b0​b1​b2​b3​b4​b5​b6​b7​​​+​11000110​​
    Avalanche effectarrow-up-right
    AK: each byte is xored with a byte of the round key in the same position.
    MC: each column is multiplied by a matrix.
    SR: each is shifted by a different offset to the left.

    0e

    MC: MixColumn.

    (p0,…,p15)(p_0,\ldots,p_{15})(p0​,…,p15​)
    4×44 \times 44×4
    (c0,…,c15)(c_0,\ldots,c_{15})(c0​,…,c15​)

    Key length

    Number of rounds

    128

    10

    192

    12

    256

    14

    AES encryption.
    Rounds of AES.

    The incredible fact that paths within isogeny graphs commute (with the help of torsion points)

  • Supersingular ℓ\ellℓ isogeny graphs are (ℓ+1)(\ell + 1)(ℓ+1)regular Ramanujan graphs

  • Using paths through these graphs has spawned a whole bunch of protocols

    • SIKE / BIKE / ...

    • Hashes

    • ...

  • First we will look at the fundementals that allow these protocols to be expected as good candidates for post-quantum

  • Then we look at these protocols in more detail. Hopefully with SageMath implementations for each

  • hashtag
    References I plan to use

    • Introduction by Craig Costello

      • https://eprint.iacr.org/2019/1321.pdfarrow-up-right

    • Mathematics of Isogeny Based Cryptography by Luca De Feo

      • -

    • The Arithmetic of Elliptic Curves by Joseph H. Silverman

    Motivation
    • Breaking a hash function (insert story)

    hashtag
    The birthday paradox

    circle-exclamation

    *insert story / introduction about why it's called a paradox + use*

    hashtag
    Birthday version

    Question 1

    What is the probability that 1 person has the same birthday as you?

    Solution

    • Let AAA be the event that someone has the same birthday as you and Aˉ\bar{A}Aˉ be the complementary event

      • The events are mutually exclusive => Pr(A)=1−Pr(Aˉ)Pr(A) = 1 - Pr(\bar{A})Pr(A)=1−Pr(Aˉ)

    • Let EiE_iEi​ be the events that person iii does not have your birthday

    Then

    • Pr(A)=1−Pr(Aˉ)=1−∏i=1nPr(Ei)=1−(364365)nPr(A) = 1 - Pr(\bar{A}) = 1 - \prod_{i=1}^n Pr(E_i) = 1 - \left( \dfrac {364} {365}\right)^nPr(A)=1−Pr(Aˉ)=1−∏i=1n​Pr(Ei​)=1−(365364​)n

    circle-check

    Reminder: Pr(A,B)=Pr(A)⋅Pr(B)Pr(A, B) = Pr(A) \cdot Pr(B)Pr(A,B)=Pr(A)⋅Pr(B)if A,BA, BA,Bare independent events

    Question 2

    What is the probability that 2 out of nnn people in a room share the same birthday?

    • Suppose the birthdays are distributed independently and uniformly

    Solution

    • Let AAA be the event that 2 people have the same birthday, let Aˉ\bar{A}Aˉ be the complementary event (no 2 people have the same birthday)

    • Event 1 = Person 1 is born => Pr(E1)=365365Pr(E_1) = \dfrac {365} {365}Pr(E1​)=365365​

    • Event 2 = Person 2 is born on a different day than Person 1 => Pr(E2)=364365Pr(E_2) = \dfrac {364} {365}Pr(E2​)=365364​

    • Event n = Person n is born on a different day than Person

    Pr(Aˉ)=Pr(E1)⋅Pr(E2)⋅⋯⋅Pr(En)=365365⋅364365⋅⋯⋅365−(n−1)365=(1365)n⋅365!(365−n)!=∏i=1n−1(1−i365)Pr(\bar{A}) = Pr(E1) \cdot Pr(E_2) \cdot \dots \cdot Pr(E_n) = \dfrac {365} {365} \cdot \dfrac {364} {365} \cdot \dots \cdot \dfrac {365-(n-1)} {365} = \left( \dfrac {1} {365} \right) ^{n} \cdot \dfrac {365!} {(365-n)!} = \prod_{i=1}^{n-1} \left(1 - \dfrac i {365}\right)Pr(Aˉ)=Pr(E1)⋅Pr(E2​)⋅⋯⋅Pr(En​)=365365​⋅365364​⋅⋯⋅365365−(n−1)​=(3651​)n⋅(365−n)!365!​=∏i=1n−1​(1−365i​)

    hashtag
    General case

    Question 1

    • Instead of 365365365 days we have d⇒1−(d−1d)nd \Rightarrow \boxed{1 - \left( \dfrac {d-1} {d}\right)^n}d⇒1−(dd−1​)n​

    Question 2

    • Instead of 365365365 days we have d⇒1−∏i=1n−1(1−id)d \Rightarrow \boxed{1 - \prod_{i=1}^{n-1} \left(1 - \dfrac i {d}\right)}d⇒1−i=1∏n−1​(1−di​)​

    Code examples

    hashtag
    An useful approximation

    From the Taylor approximation we know ex=1+x+x22!+⋯=>ex≈1+xe^x = 1 + x + \dfrac {x^2} {2!} + \dots => e_x\approx 1 + xex=1+x+2!x2​+⋯=>ex​≈1+x for x≪1x \ll 1x≪1 Apply for each event: ⇒x=−ad⇒e−a/d≈1−ad⇒Pr(A)=1−∏i=1n−1e−i/d=1−e−n(n−1)/2d≈1−e−n2/2d\Rightarrow x = -\dfrac a d \Rightarrow e^{ -a /d} \approx 1- \dfrac a d \Rightarrow Pr(A) = 1 - \prod_{i=1}^{n-1}e^{-i/d} = 1-e^{-n(n-1) /{2d}} \approx 1-\boxed{e^{-{n^2} / {2d}}}⇒x=−da​⇒e−a/d≈1−da​⇒Pr(A)=1−∏i=1n−1​e−i/d=1−e−n(n−1)/2d≈1−e−n2/2d​

    If we want to solve for nnn knowing Pr(A)Pr(A)Pr(A) we take the ln⁡\lnln => n≈2dln⁡(11−Pr(A))\boxed{n \approx \sqrt{2d \ln \left(\dfrac 1 {1-Pr(A)}\right)}}n≈2dln(1−Pr(A)1​)​​

    hashtag
    Finding a collision

    • Let H:M⟶TH:\mathcal{M} \longrightarrow \mathcal{T}H:M⟶T be a hash function with ∣M∣>>∣T∣|\mathcal{M}| >> |T|∣M∣>>∣T∣

    • Let's denote N=∣T∣N = |\mathcal{T}|N=∣T∣

    Algorithm

    1. Choose s≈Ns \approx \sqrt{N}s≈N​ random distinct messages in M\mathcal{M}M

    2. Compute ti=H(mi)t_i = H(m_i)ti​=H(mi​) for 1≤i≤N1\leq i \leq \sqrt{N}1≤i≤N​

    3. Look for (ti=tj)→(t_i = t_j) \to(ti​=tj​)→ If not found go to step 1

    Example:

    Consider the following hash function:

    We make the following function to find the hashes:

    We use it as follows:

    hashtag
    Eliminating Storage Requirements with Pollard's Rho

    While the above algorithm works to find a hash collision in time O(N)O(\sqrt N )O(N​), it also requires storing O(N)O(\sqrt N )O(N​)hash values. As such, it represents a classic time-space tradeoff over the naive approach, which involves randomly selecting pairs of inputs until they hash to the same value. While the naive approach does not require any additional storage, it does have runtime O(N)O(N )O(N).

    However, there is a better approach combining the best of both worlds: constant storage requirements and O(N)O(\sqrt N)O(N​)runtime. This approach is based on Pollard's Rho algorithm, which is better-known for its application to solving discrete logarithms. The core insight behind the algorithm is that by the Birthday Paradox, we expect to encounter a hash collision after trying O(N)O(\sqrt N)O(N​)random inputs. However, it is possible to detect whether a collision has occurred without needing to store all of the inputs and hashes if the inputs are chosen in a clever way.

    This is done by choosing the next input based on the hash of the previous input, according to the following sequence:

    Where H:M⟶TH:\mathcal{M} \longrightarrow \mathcal{T}H:M⟶T is our hash function and g:T⟶Mg: \mathcal{T} \longrightarrow \mathcal{M} g:T⟶M is a "sufficiently random" function which takes a hash value and produces a new. We define the composition of the functions to bef=H∘g:T⟶Tf = H \circ g : \mathcal{T} \longrightarrow \mathcal{T} f=H∘g:T⟶T.

    By the Birthday Paradox, we expect that the sequence will have a collision (where xi=xjx_i = x_jxi​=xj​ for two distinct values i,ji,ji,j) after O(N)O(\sqrt N)O(N​)values. But once this occurs, then the sequence will begin to cycle, because xi+1=g(H(xi))=g(H(xj))=xj+1x_{i+1} = g(H(x_i)) = g(H(x_j)) = x_{j+1}xi+1​=g(H(xi​))=g(H(xj​))=xj+1​.

    Therefore, we can detect that a collision has occurred by using standard cycle-detection algorithms, such as Floyd's tortoise and harearrow-up-right!

    And finally, we can locate the first place in the sequence where the collision occurred, which will let us determine what the colliding inputs to the hash function are. This is done by determining how many iterations apart the colliding inputs are, and then stepping together one iteration at a time until the collision occurs.

    For Floyd's tortoise and hare, this is done by noting that when we found our collision after nnn iterations, we were comparing H(xn)=H(x2n)H(x_{n}) = H(x_{2 n})H(xn​)=H(x2n​). And because the sequence is cyclic, if the first colliding input is xix_ixi​, then it collides withH(xi)=H(xi+n)H(x_{i}) = H(x_{i+n})H(xi​)=H(xi+n​). So we define the new sequence yi=xn+iy_i = x_{n+i}yi​=xn+i​, and step through the xix_ixi​ and yiy_iyi​ sequences together until we find our collision!

    This is implemented in the following code snippet.

    Finally, there is a third algorithm for finding hash collisions in O(N)O(\sqrt N)O(N​)time, namely the van Oorschot-Wiener algorithmarrow-up-right based on Distinguished Points. While it does have additional storage requirements over Pollard's Rho, the main advantage of this algorithm is that it parallelizes extremely well, achieving O(Nm)O(\frac{\sqrt N}{m})O(mN​​)runtime when run on mmm processors.

    hashtag
    Resources

    • https://en.wikipedia.org/wiki/Birthday_problemarrow-up-right - wiki entry

    • https://en.wikipedia.org/wiki/Birthday_attackarrow-up-right - wiki entry

    • https://www.youtube.com/watch?v=ofTb57aZHZsarrow-up-right - vsauce2 video

    x0=g(seed)x1=g(H(x0))x2=g(H(x1))…xn+1=g(H(xn))x_0 = g(seed) \\ x_1 = g(H(x_0)) \\ x_2 = g(H(x_1)) \\ \dots \\ x_{n+1} = g(H(x_n)) \\x0​=g(seed)x1​=g(H(x0​))x2​=g(H(x1​))…xn+1​=g(H(xn​))

    Hashes are efficient algorithms to check if two files are the same based on the data they contain. The slightest change (a single bit) would change the hash completely.

    circle-check

    On the internet, when you download, files you often see a number near the download button called the hash of that file. If you download that file, recalculate the hash locally and obtain the same hash you can be sure that the data you downloaded is the intended one.

    circle-check

    Another use for hashes is storing passwords. We don't want to store plaintext passwords because in case of a breach the attacker will know our password. If we hash them he will have to reverse the hash (or find a collision) to use our password. Luckily the hashes are very hard to reverse and collision resistant by definition and construction.

    triangle-exclamation

    Note that hashes need a secure channel for communication. Alice must have a secure way to send her hash to Bob. If Eve intercepts Alice's message and hash she can impersonate Alice by changing the file, computing the hash and sending them to Bob. Hashes do not provide authenticity.

    hashtag
    Definitions and Formalism

    Definition - Hash

    A hash is an efficient deterministic function that takes an arbitrary length input and produces a fixed length output (digest, hash). Let H:M⟶TH:\mathcal{M} \longrightarrow \mathcal{T}H:M⟶T be a function where

    • M\mathcal{M}M = message space

    • T\mathcal{T}T = digest space

    Desired proprieties

    • Deterministic

    • Fast to compute

    • Small changes change the hash completely

    • Preimage, second preimage and collision resistance (Explained below)

    How to use a hash:

    • Suppose you want to check if Alice and Bob have the same version of some file (File integrity)

      • They compute H(a),H(b)H(a), H(b)H(a),H(b)

      • They check if H(a)=H(b)H(a) = H(b)H(a)=H(b)

    Figure 1.

    hashtag
    Proprieties

    • Preimage Image Resistance

    • Second Preimage resistance

    • Resistant to collisions

    hashtag
    1. Preimage Resistance

    The hash function must be a one way function. Given t∈Tt \in \mathcal{T}t∈T find m∈Mm \in \mathcal{M}m∈M s.t H(m)=tH(m) = tH(m)=t

    circle-info

    Intuition

    It should be unfeasible to reverse a hash function (O(2l)\mathcal{O}(2^l)O(2l) time where lll is the number of output bits)

    This propriety prevents an attacker to find the original message from a hash

    hashtag
    2. Second Preimage Resistance

    Given mmm it should be hard to find m′≠mm' \neq mm′=m with H(m′)=H(m)H(m') = H(m)H(m′)=H(m)

    Attack game

    An adversary A\mathcal{A}A is given a message mmm and outputs a message m′≠mm' \neq mm′=m.

    A\mathcal{A}A wins the game if he finds H(m)=H(m′)H(m) = H(m')H(m)=H(m′)

    His advantage is Pr[A finds a second preimage]Pr[\mathcal{A} \text{ finds a second preimage}]Pr[A finds a second preimage] where Pr(⋅)Pr(\cdot)Pr(⋅)is a probability

    Figure 2. Security game - second preimage resistance
    • In practice a hash function with lll bits output should need 2l2^l2l queries before one can find a second preimage

    • This propriety prevents an attacker to substitute a message with another and get the same hash

    hashtag
    3. Hash Collisions

    circle-info

    Intuition

    A hash collision happens when we have two different messages that have the same hash

    Why do we care about hash collisions?

    • Since hashes are used to fastly verify a message integrity if two messages have the same hash then we can replace one with another => We can play with data

    • Now, we want to hash big files and big messages so ∣M∣>>∣T∣|\mathcal{M}| >> |\mathcal{T}|∣M∣>>∣T∣ => It would appear that hash collisions are possible

    • Natural collisions are normal to happen and we consider them improbable if T\mathcal{T}T is big enough (SHA256⇒T=\text{SHA256} \Rightarrow T =SHA256⇒T= {0,1}256\{0,1\}^{256}{0,1}256)

    • Yet, we don't want hash collisions to be computable

      • We don't want an attacker to be able to craft collisions or find collisions given a message

    hashtag
    Let's throw some definitions

    Attack game

    An adversary A\mathcal{A}A outputs two messages m0≠m1m_0 \neq m_1m0​=m1​

    A\mathcal{A}A wins the game if he finds H(m0)=H(m1)H(m_0) = H(m_1)H(m0​)=H(m1​)

    His advantage is Pr[Adversary finds a collision]Pr[\text{Adversary finds a collision}]Pr[Adversary finds a collision]

    Figure 3. Security game - Collision resistance

    Security

    A hash function HHH is collision resistant if for all efficient and explicit adversaries the advantage is negligible

    circle-info

    Intuition

    We know hash collisions exist (therefore an efficient adversary must exist) and that is easy to prove therefore we request an explicit algorithm that finds these collisions

    circle-info

    This propriety makes it difficult for an attacker to find 2 input values with the same hash

    hashtag
    Difference from 2nd preimage

    • There is a fundamental difference in how hard it is to break collision resistance and second-preimage resistance.

      • Breaking collision-resistance is like inviting more people into the room until the room contains 2 people with the same birthday.

      • Breaking second-preimage resistance is like inviting more people into the room until the room contains another person with your birthday.

    • One of these fundamentally takes longer than the other

    Implications

    Lemma 1

    Assuming a function HHH is preimage resistant for every element of the range of HHH is a weaker assumption than assuming it is either collision resistant or second preimage resistant.

    Note

    • Provisional implication

    • https://crypto.stackexchange.com/questions/10602/why-does-second-pre-image-resistance-imply-pre-image-resistance?rq=1arrow-up-right

    • https://crypto.stackexchange.com/questions/9684/pre-image-resistant-but-not-2nd-pre-image-resistantarrow-up-right

    Lemma 2

    Assuming a function is second preimage resistant is a weaker assumption than assuming it is collision resistant.

    hashtag
    Resources

    • https://en.wikipedia.org/wiki/Cryptographic_hash_functionarrow-up-right - Wikipedia entry

    • https://www.youtube.com/watch?v=b4b8ktEV4Bgarrow-up-right - Computerphile

    • https://www.tutorialspoint.com/cryptography/cryptography_hash_functions.htmarrow-up-right

    • - Good read for more details

    hashtag
    Bibliography

    • Figure 1 - Wikipedia

    ϕ(n)\phi(n)ϕ(n)
    x2−(n−ϕ(n)+1)x+nx^2 - (n - \phi(n) + 1)x + nx2−(n−ϕ(n)+1)x+n
    ppp
    qqq
    pq=npq = npq=n
    kkk
    ddd
    ddd
    ed≡1mod  ked \equiv 1 \mod ked≡1modk
    ϕ(n)=ed−1k\phi(n) = \frac{ed-1}{k}ϕ(n)=ked−1​
    x2−(n−ϕ(n)+1)x+nx^2 - (n - \phi(n) + 1)x + nx2−(n−ϕ(n)+1)x+n
    ddd
    GitHub - orisano/owiener: A Python3 implementation of the Wiener attack on RSAGitHubchevron-right
    owiener
    d<Nid < N^{i}d<Ni
    1=ed+kϕ(N)21=ed + \frac{k \phi(N)}{2}\\ 1=ed+2kϕ(N)​
    ϕ(N)=(p−1)(q−1)=N−q−p+1 \phi(N) = (p-1)(q-1) = N-q-p+1ϕ(N)=(p−1)(q−1)=N−q−p+1
    1=ed+k(N+12−p+q2)1 = ed + k(\frac{N+1}{2}-\frac{p+q}{2} )1=ed+k(2N+1​−2p+q​)
    x=N+12x = \frac{N+1}{2} x=2N+1​
    y=p+q2y = \frac{p+q}{2} y=2p+q​
    1=ed+k(x+y)1 = ed + k(x+y)1=ed+k(x+y)
    ddd
    xxx
    yyy
    f(x,y)≡k(x+y)≡1 (mod e)k(x+y)≡1( mod e)f(x,y) \equiv k(x+y) \equiv 1 \space (mod \space e) \\ k(x + y) \equiv 1 (\space mod \space e)f(x,y)≡k(x+y)≡1 (mod e)k(x+y)≡1( mod e)
    s=−p+q2s = -\frac{p+q}{2}s=−2p+q​
    A=N+12A = \frac{N+1}{2}A=2N+1​
    ϕ(N)\phi(N) ϕ(N)
    k(A+s)≡1( mod e)k(A+s) \equiv 1 (\space mod \space e)k(A+s)≡1( mod e)
    e=Nαe=N^{\alpha}e=Nα
    α\alphaα
    α\alphaα
    111
    eee
    NNN
    e=N∼1−e=N^{\sim 1^{-}}e=N∼1−
    ∣s∣<2N=p+q2<2N=2e12α∼e∣k∣<2edϕ(N)≤3edN<3e1+α+1α<ei| s | < 2\sqrt{N} = \\ \frac{p+q}{2} < 2\sqrt{N} =2e^{\frac{1}{2\alpha}} \sim \sqrt{e} \\ |k| < \frac{2ed}{\phi{(N)}} \leq\frac{3ed}{N}< 3e^{1+\frac{\alpha+1}{\alpha}} < e^{i}∣s∣<2N​=2p+q​<2N​=2e2α1​∼e​∣k∣<ϕ(N)2ed​≤N3ed​<3e1+αα+1​<ei

    Isogeny and Ramanujan Graphs

    triangle-exclamation

    I (Jack) know nothing about this. At all. But it will need to be talked about.

    • Isogeny graph (general definition, degree, duals...).

    • Starting vertex (Bröker's algorithm).

    • Isogeny Volcanos: Sutherland might be a good source .

    • Supersingular isogeny graphs

      • Size, everything is defined over GF(p^2). (*as long as the degree divides (p+1)^2 or (p-1)^2).

      • Random walks are probably the best motivation to define Ramanujanness, and are directly applicable to cryptography. A (perhaps too large) source is Hoory-Linial-Wigderson.

    # we construct the matrix
    # (we use the finite field F constructed in the previous page)
    M = Matrix.circulant([x, x + 1, 1, 1])
    M
    # [    x x + 1     1     1]
    # [    1     x x + 1     1]
    # [    1     1     x x + 1]
    # [x + 1     1     1     x]
    
    # We multiply a column
    col1 = vector([F.random_element() for i in range(4)])
    col1
    # (x^6 + x + 1, x^7 + x^4 + x^2, x^6 + x^4 + x^3 + x^2, x^7 + x^6 + x^4 + x^3 + x^2)
    col2 = M*col1
    col2
    # (x^7 + x^5 + 1, x^6 + x^3, x^4, x^7 + x^5 + x^3 + x^2 + x)
    N = M^-1
    N
    # [x^3 + x^2 + x   x^3 + x + 1 x^3 + x^2 + 1       x^3 + 1]
    # [      x^3 + 1 x^3 + x^2 + x   x^3 + x + 1 x^3 + x^2 + 1]
    # [x^3 + x^2 + 1       x^3 + 1 x^3 + x^2 + x   x^3 + x + 1]
    # [  x^3 + x + 1 x^3 + x^2 + 1       x^3 + 1 x^3 + x^2 + x]
    
    # hexadecimal representation
    for row in N:
        print([F_to_hex(a) for a in row])
    # ['e', 'b', 'd', '9']
    # ['9', 'e', 'b', 'd']
    # ['d', '9', 'e', 'b']
    # ['b', 'd', '9', 'e']
    
    col3 = N*col2
    col3 == col1
    # True
    def construct_sbox():
        mat = Matrix.circulant(vector(GF(2), [1,0,0,0,1,1,1,1]))
        sbox = []
        for a in range(256):
            # convert a byte value to field element
            b = sum(((a >> i) & 1)*x^i for i in range(8))
            # first step: map to inverse
            if b != 0:
            b = b^-1
            # second step:
            # affine transformation using a vector space over GF(2)
            b = mat*vector(b) + vector(GF(2), [1, 1, 0, 0, 0, 1, 1, 0])
            # convert the vector as an integer in [0, 255]
            sbox.append(sum(ZZ(b[i]) << i for i in range(8)))
        return sbox
    
    sbox = construct_sbox()
    for i in range(16):
        print([f'{a:02x}' for a in sbox[16*i : 16*(i + 1)]])
    # ['63', '7c', '77', '7b', 'f2', '6b', '6f', 'c5', '30', '01', '67', '2b', 'fe', 'd7', 'ab', '76']
    # ['ca', '82', 'c9', '7d', 'fa', '59', '47', 'f0', 'ad', 'd4', 'a2', 'af', '9c', 'a4', '72', 'c0']
    # ['b7', 'fd', '93', '26', '36', '3f', 'f7', 'cc', '34', 'a5', 'e5', 'f1', '71', 'd8', '31', '15']
    # ['04', 'c7', '23', 'c3', '18', '96', '05', '9a', '07', '12', '80', 'e2', 'eb', '27', 'b2', '75']
    # ['09', '83', '2c', '1a', '1b', '6e', '5a', 'a0', '52', '3b', 'd6', 'b3', '29', 'e3', '2f', '84']
    # ['53', 'd1', '00', 'ed', '20', 'fc', 'b1', '5b', '6a', 'cb', 'be', '39', '4a', '4c', '58', 'cf']
    # ['d0', 'ef', 'aa', 'fb', '43', '4d', '33', '85', '45', 'f9', '02', '7f', '50', '3c', '9f', 'a8']
    # ['51', 'a3', '40', '8f', '92', '9d', '38', 'f5', 'bc', 'b6', 'da', '21', '10', 'ff', 'f3', 'd2']
    # ['cd', '0c', '13', 'ec', '5f', '97', '44', '17', 'c4', 'a7', '7e', '3d', '64', '5d', '19', '73']
    # ['60', '81', '4f', 'dc', '22', '2a', '90', '88', '46', 'ee', 'b8', '14', 'de', '5e', '0b', 'db']
    # ['e0', '32', '3a', '0a', '49', '06', '24', '5c', 'c2', 'd3', 'ac', '62', '91', '95', 'e4', '79']
    # ['e7', 'c8', '37', '6d', '8d', 'd5', '4e', 'a9', '6c', '56', 'f4', 'ea', '65', '7a', 'ae', '08']
    # ['ba', '78', '25', '2e', '1c', 'a6', 'b4', 'c6', 'e8', 'dd', '74', '1f', '4b', 'bd', '8b', '8a']
    # ['70', '3e', 'b5', '66', '48', '03', 'f6', '0e', '61', '35', '57', 'b9', '86', 'c1', '1d', '9e']
    # ['e1', 'f8', '98', '11', '69', 'd9', '8e', '94', '9b', '1e', '87', 'e9', 'ce', '55', '28', 'df']
    # ['8c', 'a1', '89', '0d', 'bf', 'e6', '42', '68', '41', '99', '2d', '0f', 'b0', '54', 'bb', '16']
    def my_birthday(n, d):
        return 1 - pow((d-1)/d , n)
    
    def same_birthday(n, d):
        p = 1
        for i in range(1, n): #1 -> n-1
            p*=(1-i/d)
        return 1 - p
    
    print(same_birthday(23, 365), same_birthday(32, 365), same_birthday(100, 365))
    # (0.5072972343239854, 0.7533475278503207, 0.9999996927510721)
    def approx_same_birthday(n, d):
        return 1 - pow(e, -pow(n, 2) / (2*d)).n()
        
    print(approx_same_birthday(23, 365)) 
    print(approx_same_birthday(32, 365))
    print(approx_same_birthday(100, 365))
    
    # 0.515509538061517
    # 0.754077719532824
    # 0.999998876014983
    import hashlib
    import random
    from Crypto.Util.number import long_to_bytes, bytes_to_long
    
    def small_hash(m, hash_bits):
        '''
        Arguments
            m: bytes -- input
            hash_bits: int -- number of bits of the hash
        Returns:
            {bytes} - hash of the message of dimension `hash_bits`
            '''
        assert hash_bits > 0, "no negative number of bits"
        
        mask = (1 << hash_bits) - 1 # Mask of bits
        t = hashlib.sha256(m).digest() # the hash in bytes
        t = bytes_to_long(t)
        t = t & mask # get the last 12 bits
        return long_to_bytes(t)
    
    def small_hash_colision(M_bits, T_bits):
        '''
        Arguments
            M_bits: int -- dimension of the message space
            T_bits: int -- dimension of the hash space
        Returns:
            {(bytes, bytes, bytes)} -- message1, message2, hash
            or
            {(b'', b'', b'')} -- if no collision found
        '''
        N = 1<<T_bits 
        print('Hash size: ', N)
        # This is `s`
        num_samples = 1 * isqrt(N)
        num_samples += num_samples//5 + 1 # num_samples = 1.2 * isqrt(N) + 1
        print(f'Making a list of {num_samples} hashes')
        print(f'Probability of finding a collision is {same_birthday(num_samples, N)}')
        
        m_list = []
        t_list = []
        
        # Make a list of hashes
        for i in range(num_samples):
            m = random.getrandbits(M_bits) # Get random message
            #m = random.randint(0, M_bits-1) 
            m = long_to_bytes(m)
            
            t = small_hash(m, T_bits) # Hash it
            if m not in m_list:
                m_list.append(m)
                t_list.append(t)
                
        # Check for collisionn
        for i in range(len(t_list)):
            for j in range(i+1, len(t_list)):
                if t_list[i] == t_list[j]:
                    print('Collision found!')
                    return m_list[i], m_list[j], t_list[i]
        else:
            print('Collision not found :(')
            return b"", b"", b""
    M_bits = 30
    T_bits = 20
    
    m1, m2, t = small_hash_colision(M_bits, T_bits)
    print(m1, m2, t)
    print(small_hash(m1, T_bits) == small_hash(m2, T_bits))
    
    # Hash size:  1048576
    # Making a list of 1229 hashes
    # Probability of finding a collision is 0.513213460854798
    # Collision found!
    # b'!\xafB\xc5' b'4F\xb6w' b'\x01y\xf7'
    # True
    
    # TODO:
    # have the two hash functions used in this chapter be the same
    
    from Crypto.Hash import SHA256
    from Crypto.Util.number import bytes_to_long, long_to_bytes
    
    from tqdm.autonotebook import tqdm
    
    # bits in hash output
    n = 30
    
    # H
    def my_hash(x):
        x = bytes(x, 'ascii')
        h = SHA256.new()
        h.update(x)
        y = h.digest()
        y = bytes_to_long(y)
        y = y % (1<<n)
        y = int(y)
        return y
    
    # g
    def get_message(r):
        x = "Crypto{" + str(r) + "}"
        return x
    
    # f
    def f(r):
        return my_hash(get_message(r))
    
    """
    initialization
    
    This routine will find a hash collision in sqrt time with constant space.
    """
    
    seed = 0
    
    x0 = get_message(seed)
    x = x0
    y = f(x)
    
    """
    detect collision
    using Floyd / tortoise and hare cycle finding algorithm
    
    expected number of iterations is ~ sqrt(pi/2) * 2^(n/2),
    we run for up to 4 * 2^(n/2) iterations
    """
    for i in tqdm(range(4 << (n//2))):
            
        if f(x) == f(y):
            break
            
        x = f(x)
        
        y = f(y)
        y = f(y)
        
    
    """
    locate collision
    """
    x = x0
    y = f(y)
    
    
    for j in tqdm(range(i)):
        if f(x) == f(y):
            break
        
        x = f(x)
        
        y = f(y)
        
    
    m1 = get_message(x)
    m2 = get_message(y)
    
    assert my_hash(m1) == f(x)
    assert my_hash(m2) == f(y)
    
    print("[+] seeds for messages: {}, {}".format(x, y))
    print("[+] messages: {}, {}".format(m1, m2))
    print("[+] collided hashes of messages: {}, {}".format(my_hash(m1), my_hash(m2)))
    
    
    # 31%    40391/131072 [00:03<00:08, 10666.20it/s]
    # 30%    12032/40391 [00:00<00:02, 13112.15it/s]
    # 
    # [+] seeds for messages: 404842900, 254017312
    # [+] messages: Crypto{404842900}, Crypto{254017312}
    # [+] collided hashes of messages: 1022927209, 1022927209
    from hashlib import sha256
    m1 = b"Some file"
    m2 = b"Some file"
    sha256(m1).digest() == sha256(m2).digest() # -> True
    https://arxiv.org/pdf/1711.04062.pdfarrow-up-right
    https://www.springer.com/gp/book/9780387094939arrow-up-right
    ⋮\vdots⋮
    1,...,n−1⇒1,...,n-1 \Rightarrow1,...,n−1⇒
    ⇒Pr(En)=365−(n−1)365\Rightarrow Pr(E_n) = \dfrac {365-(n-1)} {365}⇒Pr(En​)=365365−(n−1)​
    van Oorschot-Wiener Parallel Collision Search with Cryptanalytic Applicationsarrow-up-right
    http://www.cs.umd.edu/~jkatz/imc/hash-erratum.pdfarrow-up-right
    https://crypto.stackexchange.com/questions/3295/how-does-a-birthday-attack-on-a-hashing-algorithm-work?rq=1arrow-up-right

    Consequence from Ramanujan + random walk convergence: O(log p) diameter.

    https://arxiv.org/abs/1208.5370arrow-up-right

    76

    c0

    15

    75

    84

    cf

    a8

    d2

    73

    db

    79

    08

    8a

    9e

    df

    16

    https://www.cs.ucdavis.edu/~rogaway/papers/relates.pdfarrow-up-right
    RSA-and-LLL-attacks/boneh_durfee.sage at master · mimoo/RSA-and-LLL-attacksGitHubchevron-right
    Boneh-Durfee attack Sage implementation
    Logo
    Logo