Lattices of interest

Needs review.

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.

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.

Dual lattice

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

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\}

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

Examples:

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

  2. Scaling: (k⋅L)∨=1k⋅L(k \cdot L)^\vee = \dfrac 1 k \cdot 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^\vee 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)^\vee

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

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

Basis of the dual lattice

We will now focus on the problem of finding the basis B∨B^\vee of the dual lattice L∨L^\veegiven the lattice LL and its basis BB.

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

We have the following equivalences:

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*}

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

B∨=(B−1)T∈GLn(R)B^\vee = (B^{-1})^T \in GL_n(\mathbb{R})
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

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

Properties

  1. L1⊆L2  ⟺  L2∨⊆L1∨{L}_1 \subseteq {L}_2 \iff {L}^\vee_2 \subseteq {L}^\vee_1

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

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

  4. For x∈L,y∈L∨x \in {L}, y \in {L}^\veeconsider the vector dot product and addition - x⋅y∈Zx \cdot y \in \mathbb{Z} - x+yx + y has no geometric meaning, they are in different spaces

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 LL and its dual L∨L^\vee. Let's see what can we uncover about them.

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

Reminder: We defined the successive minima of a lattice LLas 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)

Claim 1:

λ1(L)⋅λ1(L∨)≤n\lambda_1(L) \cdot \lambda_1(L^\vee) \leq n

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} 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}}. By multiplying them we get the desired result.

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

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

Claim 2:

λ1(L)⋅λn(L∨)≥1\lambda_1(L) \cdot \lambda_n(L^\vee) \geq 1

Proof:

Let x∈Lx∈L be such that ∥x∥=λ1(L)\|x\|=λ_1(L). Then take any set (y1,...,yn)(y_1, . . . , y_n) of nn linearly independent vectors in L∨L^\vee. Not all of them are orthogonal to xx. Hence, there exists an ii such that yi⋅x≠0y_i \cdot x \neq 0 . By the definition of the dual lattice, we have yi⋅x∈Zy_i \cdot x \in \mathbb 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^\vee

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

Geometry + Partitioning

// TODO

Q-ary lattices

We've seen that in cryptography we don't like to work with infinite sets (like Z\mathbb Z) and we limit them to some finite set using the  mod \bmod operation (Z→Z/qZ\mathbb Z \to \mathbb Z/ q\mathbb{Z}). 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 3we call a lattice q-ary if

qZn⊆L⊆Znq\mathbb{Z}^n \subseteq {L} \subseteq \mathbb{Z}^n

Intuition:

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

  • We use arithmetic  mod  q\bmod \ 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} be a matrix with m>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:Ay=0 mod q}⊂ZmL^\perp_q(A) = \{y \in \mathbb Z^m : Ay = 0 \bmod q \} \subset \mathbb{Z^m}

Intuition:

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

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

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

Claim:

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

Proof:

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

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

  • 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' for some x∈Zn, z′∈Zmx \in \mathbb Z^n, \ z' \in \mathbb Z^m

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')

⇒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

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))^\vee

Resources

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