Introduction / overview

PreviousRound Transformations NextThe Birthday paradox / attack

Last updated 4 years ago

Was this helpful?

Introduction / overview

Authors: Zademn Reviewed by:

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?

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.

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.

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.

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.

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 $H:\mathcal{M} \longrightarrow \mathcal{T}$ be a function where
$\mathcal{M}$ = message space
$\mathcal{T}$ = 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)

from hashlib import sha256
m1 = b"Some file"
m2 = b"Some file"
sha256(m1).digest() == sha256(m2).digest() # -> True

Proprieties

Preimage Image Resistance
Second Preimage resistance
Resistant to collisions

1. Preimage Resistance

Intuition

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

2. Second Preimage Resistance

Attack game

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

3. Hash Collisions

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
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

Let's throw some definitions

Attack game

Security

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

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

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

Note

Provisional implication

Lemma 2

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

Resources

Bibliography

Figure 1 - Wikipedia

PreviousRound Transformations NextThe Birthday paradox / attack

Last updated 4 years ago

Was this helpful?

Authors: Zademn Reviewed by:

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.

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.

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 $H:\mathcal{M} \longrightarrow \mathcal{T}$ be a function where
$\mathcal{M}$ = message space
$\mathcal{T}$ = 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 $H(a), H(b)$
- They check if $H(a) = H(b)$

from hashlib import sha256
m1 = b"Some file"
m2 = b"Some file"
sha256(m1).digest() == sha256(m2).digest() # -> True

Proprieties

Preimage Image Resistance
Second Preimage resistance
Resistant to collisions

1. Preimage Resistance

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

Intuition

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

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

2. Second Preimage Resistance

Given $m$ it should be hard to find $m' \neq m$ with $H(m') = H(m)$

Attack game

An adversary $\mathcal{A}$ is given a message $m$ and outputs a message $m' \neq m$ .
$\mathcal{A}$ wins the game if he finds $H(m) = H(m')$
His advantage is $Pr[\mathcal{A} \text{ finds a second preimage}]$ where $Pr(\cdot)$ is a probability

In practice a hash function with $l$ bits output should need $2^l$ queries before one can find a second preimage
This propriety prevents an attacker to substitute a message with another and get the same hash

3. Hash Collisions

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 $|\mathcal{M}| >> |\mathcal{T}|$ => It would appear that hash collisions are possible
Natural collisions are normal to happen and we consider them improbable if $\mathcal{T}$ is big enough ( $\text{SHA256} \Rightarrow T =$ $\{0,1\}^{256}$ )
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

Let's throw some definitions

Attack game

An adversary $\mathcal{A}$ outputs two messages $m_0 \neq m_1$
$\mathcal{A}$ wins the game if he finds $H(m_0) = H(m_1)$
His advantage is $Pr[\text{Adversary finds a collision}]$

Security

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

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

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

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

Note

Provisional implication

Lemma 2

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

Resources

- Wikipedia entry
- Computerphile
- Good read for more details

Bibliography

Figure 1 - Wikipedia

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

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

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

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

Given $m$ it should be hard to find $m' \neq m$ with $H(m') = H(m)$

An adversary $\mathcal{A}$ is given a message $m$ and outputs a message $m' \neq m$ .

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

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

In practice a hash function with $l$ bits output should need $2^l$ queries before one can find a second preimage

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

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

An adversary $\mathcal{A}$ outputs two messages $m_0 \neq m_1$

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

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

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

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