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★ See also the **PDF version of this chapter** (better formatting/references) ★

# Lattice based cryptography

Lattice based public key encryption (and its cousins known as knapsack and coding based encryption) have almost as long a history as discrete logarithm and factoring based schemes. Already in 1976, right after the Diffie-Hellman key exchange was discovered (and before RSA), Ralph Merkle was working on building public key encryption from the NP hard *knapsack* problem (see Diffie’s recollection). This can be thought of as the task of solving a linear equation of the form \(Ax = y\) (where \(A\) is a given matrix, \(y\) is a given vector, and the unknown are \(x\)) over the real numbers but with the additional constraint that \(x\) must be either \(0\) or \(1\). His proposal evolved into the Merkle-Hellman system proposed in 1978 (which was broken in 1984).

McEliece proposed in 1978 a system based on the difficulty of the decoding problem for general linear codes. This is the task of solving *noisy linear equations* where one is given \(A\) and \(y\) such that \(y=Ax+e\) for a “small” error vector \(e\), and needs to recover \(x\). Crucially, here we work in a finite field, such as working modulo \(q\) for some prime \(q\) (that can even be \(2\)) rather than over the reals or rationals. There are special matrices \(A^*\) for which we know how to solve this problem efficiently: these are known as efficiently decodable error correcting codes. McEliece suggested a scheme where the key generator lets \(A\) be a “scrambled” version of a special \(A^*\) (based on the Goppa algebraic geometric code). So, someone that knows the scrambling could solve the problem, but (hopefully) someone that doesn’t know it wouldn’t. McEliece’s system has so far not been broken.

In a 1996 breakthrough, Ajtai showed a *private key* scheme based on integer lattices that had a very curious property- its security could be based on the assumption that certain problems were only hard in the *worst case*, and moreover variants of these problems were known to be NP hard. This re-ignited the hope that we could perhaps realize the old dream of basing crypto on the mere assumption that \(P\neq \ensuremath{\mathit{NP}}\). Alas, we now understand that there are fundamental barriers to this approach.

Nevertheless, Ajtai’s work attracted significant interest, and within a year both Ajtai and Dwork, as well as Goldreich, Goldwasser and Halevi came up with lattice based constructions for *public key* encryption (the former based also on *worst case* assumptions). At about the same time, Hoffstein, Pipher, and Silverman came up with their NTRU public key system which is based on stronger assumptions but offers better performance, and they started a company around it together with Daniel Lieman.

You may note that I haven’t yet said what *lattices* are; we will do so later, but for now if you simply think of questions involving linear equations modulo some prime \(q\), you will get enough of the intuition that you need. (The lattice viewpoint is more geometric, and we’ll discuss it more below; it was first used to *attack* cryptosystems and in particular break the Merkle-Hellman knapsack scheme and many of its variants.)

Lattice based cryptography has captured a lot of attention recently from both theory and practice. In the theory side, many cool new constructions are now based on lattice based cryptography, and chief among them fully homomorphic encryption, as well as indistinguishability obfuscation (though the latter’s security’s foundations are still far less solid). On the applied side, the steady advances in the technology of quantum computers have finally gotten practitioners worried about RSA, Diffie Hellman and Elliptic Curves. While current constructions for quantum computers are nowhere near being able to, say, factor larger numbers that can be done classically (or even than can be done by hand), given that it takes many years to develop new standards and get them deployed, many believe the effort to transition away from these factoring/dlog based schemes should start today (or perhaps should have started several years ago). The NSA has suggested that it plans to initiate the process to “transition to quantum resistant algorithms in the not too distant future”; see also this very interesting FAQ on this topic.

Cryptography has the peculiar/unfortunate feature that if a machine is built that can factor large integers in 20 years, it can still be used to break the communication we transmit *today*, provided this communication was recorded. So, if you have some data that you expect you’d want still kept secret in 20 years (as many government and commercial entities do), you might have reasons to worry. Currently lattice based cryptography is the only real “game in town” for potentially quantum-resistant public key encryption schemes.

Lattice based cryptography is a huge area, and in this lecture and this course we only touch on few aspects of it. I highly recommend Chris Peikert’s Survey for a much more in depth treatment of this area.

## A world without Gaussian elimination

The general approach people used to get a public key encryption is to obtain a hard computational problem with some mathematical *structure*. We’ve seen this in the *discrete logarithm* problem, where the task is to invert the map \(a \mapsto g^a \pmod{p}\), and the integer factoring problem, where the task is to invert the map \(a,b \mapsto a\cdot b\). Perhaps the simplest structure to consider is the task of solving linear equations.

Pretend that we didn’t know of Gaussian elimination,^{1} and that if we picked a “generic” matrix \(A\) then the map \(x \mapsto Ax\) would be hard to invert. (Here and elsewhere, our default interpretation of a vector \(x\) is as a *column* vector, and hence if \(x\) is \(n\) dimensional and \(A\) is \(m\times n\) then \(Ax\) is \(m\) dimensional. We use \(x^\top\) to denote the row vector obtained by *transposing* \(x\).) Could we use that to get a public key encryption scheme?

Here is a concrete approach. Let us fix some prime \(q\) (think of it as polynomial size, e.g., \(q\) is smaller than \(1024\) or so, though people can and sometimes do consider \(q\) of exponential size), and all computation below will be done modulo \(q\). The secret key is a vector \(x\in\Z_q^n\), and the public key is \((A,y)\) where \(A\) is a random \(m\times n\) matrix with entries in \(\Z_q\) and \(y=Ax\). Under our assumption, it is hard to recover the secret key from the public key, but how do we use the public key to encrypt?

The crucial observation is that even if we don’t know how to solve linear equations, we can still combine several equations to get new ones. To keep things simple, let’s consider the case of encrypting a single bit.

If you have a CPA secure public key encryption scheme for single bit messages then you can extend it to a CPA secure encryption scheme for messages of any length. Can you see why?

We think of the public key as the set of equations \(\langle a_1,x \rangle=y_1,\ldots, \langle a_m,x \rangle=y_m\) in the unknown variables \(x\). The idea is that to encrypt the value \(0\) we will generate a new *correct* equation on \(x\), while to encrypt the value \(1\) we will generate an *incorrect* equation. To decrypt a ciphertext \((a,\sigma)\in \Z_q^{n+1}\), we think of it as an equation of the form \(\langle a,x \rangle=\sigma\) and output \(1\) if and only if the equation is correct.

How does the encrypting algorithm, that does not know \(x\), get a correct or incorrect equation on demand? One way would be to simply take two equations \(\langle a_i,x \rangle=y_i\) and \(\langle a_j,x \rangle=y_j\) and add them together to get the equation \(\langle a_i+a_j,x \rangle=y_i+y_j\). This equation is correct and so one can use it to encrypt \(0\), while to encrypt \(1\) we simply add some fixed nonzero number \(\alpha\in\Z_q\) to the right hand side to get the incorrect equation \(\langle a_i+a_j,x \rangle= y_i+y_j + \alpha\). However, even if it’s hard to solve for \(x\) given the equations, an attacker (who also knows the public key \((A,y)\)) can try itself all pairs of equations and do the same thing.

Our solution for this is simple- just add more equations! If the encryptor adds a random subset of equations then there are \(2^m\) possibilities for that, and an attacker can’t guess them all. Thus, at least intuitively, the following encryption scheme would be “secure” in the Gaussian-elimination free world of attackers that haven’t taken freshman linear algebra:

Scheme LwoE-ENC:Public key encryption under the hardness of “learning linear equations without errors”.

Key generation: Pick random \(m\times n\) matrix \(A\) over \(\Z_q\), and \(x\leftarrow_R\Z_q^n\), the secret key is \(x\) and the public key is \((A,y)\) where \(y=Ax\).

Encryption: To encrypt a message \(b\in\{0,1\}\), pick \(w\in\{0,1\}^m\) and output \(w^\top A,\langle w,y \rangle+\alpha b\) for some fixed nonzero \(\alpha\in\Z_q\).

Decryption:To decrypt a ciphertext \((a,\sigma)\), output \(0\) iff \(\langle a,x \rangle=\sigma\).

Please stop here and make sure that you see why this is a valid encryption, and this description corresponds to the previous one; as usual all calculations are done modulo \(q\).

## Security in the real world.

Like it or not (and cryptographers typically don’t) Gaussian elimination *is* possible in the real world and the scheme above is completely insecure. However, the Gaussian elimination algorithm is extremely *brittle*.

Errors tend to be amplified when you combine equations. This is usually thought of as a bad thing, and numerical analysis is much about dealing with issue. However, from the cryptographic point of view, these errors can be our saving grace and enable us to salvage the security of the ridiculous scheme above.

To see why Gaussian elimination is brittle, let us recall how it works. Think of \(m=n\) for simplicity. Given equations \(Ax=y\) in the unknown variables \(x\), the goal of Gaussian elimination is to transform them into the equations \(Ix = y'\) where \(I\) is the identity matrix (and hence the solution is simply \(x=y'\)). Recall how we do it: by rearranging and scaling, we can assume that the top left corner of \(A\) is equal to \(1\), and then we add the first equation to the other equations (scaled appropriately) to zero out the first entry in all the other rows of \(A\) (i.e., make the first column of \(A\) equal to \((1,0,\ldots,0)\)) and continue onwards to the second column and so on and so forth.

Now, suppose that the equations were *noisy*, in the sense that we added to \(y\) a vector \(e\in\Z_q^m\) such that \(|e_i|<\delta q\) for every \(i\).^{2} Even ignoring the effect of the scaling step, simply adding the first equation to the rest of the equations would typically tend to increase the relative error of equations \(2,\ldots,m\) from \(\approx \delta\) to \(\approx 2\delta\). Now, when we repeat the process, we increase the error of equations \(3,\ldots,m\) from \(\approx 2\delta\) to \(\approx 4\delta\), and we see that by the time we’re done dealing with about \(n/2\) variables, the remaining equations have error level roughly \(2^{n/2}\delta\). So, unless \(\delta\) was truly tiny (and \(q\) truly big, in which case the difference between working in \(\Z_q\) and simply working with integers or rationals disappears), the resulting equations have the form \(Ix = y' + e'\) where \(e'\) is so big that we get no information on \(x\).

The *Learning With Errors (LWE)* conjecture is that this is *inherent*:

Conjecture (Learning with Errors, Regev 2005):Let \(q=q(n)\) and \(\delta=\delta(n)\) be some functions. TheLearning with Error (LWE) conjecture with respect to \(q,\delta\), is that for every polynomial-time adversary \(E\) and \(m=poly(n)\), the probability that \(E(A,Ax+e)=x\) is negligible, where \(A\) is a random \(m\times n\) matrix in \(\Z_q\), \(x\) is random in \(\Z_q^n,\) and \(e \in \Z_q^m\) is a random noise vector with magnitude \(\delta q\).^{3}The

LWE conjectureis that for every polynomial \(p(n)\) there is some polynomial \(q(n)\) such that LWE holds with respect to \(q(n)\) and \(\delta(n)=1/p(n)\).^{4}

## Search to decision

It turns out that if the LWE is hard, then it is even hard to distinguish between random equations and nearly correct ones:

If the LWE conjecture is true then for every \(q=poly(n)\) and \(\delta=1/poly(n)\) and \(m=poly(n)\), the following two distributions are computationally indistinguishable:

\(\{ (A,Ax+e) \}\) where \(A\) is random \(m\times n\) matrix in \(\Z_q\), \(x\) is random in \(\Z_q^n\) and \(e\in \Z_q^m\) is random noise vector of magnitude \(\delta\).

\(\{ (A,y) \}\) where \(A\) is random \(m\times n\) matrix in \(\Z_q\) and \(y\) is random in \(\Z_q^m\).

Suppose that we had a decisional adversary \(D\) that succeeds in distinguishing the two distributions above with bias \(\epsilon\). For example, suppose that \(D\) outputs \(1\) with probability \(p+\epsilon\) on inputs from the first distribution, and outputs \(1\) with probability \(p\) on inputs from the second distribution.

We will show how we can use this to obtain a polynomial-time algorithm \(S\) that on input \(m\) noisy equations on \(x\) and a value \(a\in\ Z_q\), will learn with high probability whether or not the first coordinate of \(x\) equals \(a\). Clearly, we can repeat this for all the possible \(q\) values of \(a\) to learn the first coordinate exactly, and then continue in this way to learn all coordinates.

Our algorithm \(S\) gets as input the pair \((A,y)\) where \(y=Ax+e\) and we need to decide whether \(x_1 = a\). Now consider the instance \((A+(r\|0^m\|\cdots \|0^m),y+ar)\), where \(r\) is a random vector in \(\Z_q^m\) and the matrix \((r\|0^m\|\cdots \|0^m)\) is simply the matrix with first column equal to \(r\) and all other columns equal to \(0\). If \(A\) is random then \(A+(r\|0^m\|\cdots \|0^m)\) is random as well. Now note that \(Ax + (r|0^m\cdots \|0^m)x = Ax + x_1 r\) and hence if \(x_1 = a\) then we still have an input of the same form \((A',A'x+e)\).

In contrast, we claim that if if \(x_1 \neq a\) then the distribution \((A',y')\) where \(A'=A+(r\|0^m\|\cdots \|0^m)\) and \(y'= Ax + e + ar\) is identical to the uniform distribution over a random uniformly chosen matrix \(A'\) and a random and independent uniformly chosen vector \(y'\). Indeed, we can write this distribution as \((A',y')\) where \(A'\) is chosen uniformly at random, and \(y'= A'x + e + (a-x_1)r\) where \(r\) is a random and independent vector. (Can you see why?) Since \(a-x_1 \neq 0\), this amounts to adding a random and independent vector \(r'\) to \(y'\), which means that the distribution \((A',y')\) is uniform and independent.

Hence if we send the input \((A',y')\) to our the decision algorithm \(D\), then we would get \(1\) with probability \(p+\epsilon\) if \(x_1=a\) and an output of \(1\) with probability \(p\) otherwise.

Now the crucial observation is that if our decision algorithm \(D\) requires \(m\) equations to succeed with bias \(\epsilon\), we can use \(100mn/\epsilon^2\) equations (which is still polynomial) to invoke it \(100n/\epsilon^2\) times. This allows us to distinguish with probability \(1-2^{-n}\) between the case that \(D\) outputs \(1\) with probability \(p+\epsilon\) and the case that it outputs \(1\) with probability \(p\) (this follows from the Chernoff bound; can you see why?). Hence by using polynomially more samples than the decision algorithm \(D\), we get a search algorithm \(S\) that can actually recover \(x\).

## An LWE based encryption scheme

We can now show the secure variant of our original encryption scheme:

LWE-based encryption LWEENC:

Parameters:Let \(\delta(n)=1/n^4\) and let \(q=poly(n)\) be a prime such that LWE holds w.r.t. \(q,\delta\). We let \(m = n^2\log q\).

Key generation:Pick \(x\in\Z_q^n\). The private key is \(x\) and the public key is \((A,y)\) with \(y=Ax+e\) with \(e\) a \(\delta\)-noise vector and \(A\) a random \(m\times n\) matrix.

Encrypt:To encrypt \(b\in\{0,1\}\) given the key \((A,y)\), pick \(w\in\{0,1\}^m\) and output \(w^\top A, \langle w,y \rangle+b\floor{q/2}\) (all computations are done in \(\Z_q\)).

Decrypt:To decrypt \((a,\sigma)\), output \(0\) iff \(|\langle a,x \rangle-\sigma|<q/10\).

The scheme LWEENC is also described in Figure 13.2 with slightly different notation. I highly recommend you stop and verify you understand why the two descriptions are equivalent.

Unlike our typical schemes, here it is not immediately clear that this encryption is valid, in the sense that the decrypting an encryption of \(b\) returns the value \(b\). But this is the case:

With high probability, the decryption of the encryption of \(b\) equals \(b\).

\(\langle w^\top A,x \rangle = \langle w,Ax \rangle\). Hence, if \(y=Ax+e\) then \(\langle w,y \rangle = \langle w^\top A,x \rangle + \langle w,e \rangle\). But since every coordinate of \(w\) is either \(0\) or \(1\), \(|\langle w,e \rangle|<\delta m q < q/10\) for our choice of parameters.^{5} So, we get that if \(a= w^\top A\) and \(\sigma = \langle w,y \rangle+b\floor{q/2}\) then \(\sigma - \langle a,x \rangle = \langle w,e \rangle + b\floor{q/2}\) which will be smaller than \(q/10\) iff \(b=0\).

We now prove security of the LWE based encryption:

If the LWE conjecture is true then LWEENC is CPA secure.

For a public key encryption scheme with messages that are just bits, CPA security means that an encryption of \(0\) is indistinguishable from an encryption of \(1\), even given the public key. Thus Theorem 13.3 will follow from the following lemma:

Let \(q,m,\delta\) be set as in LWEENC. Then, assuming the LWE conjecture, the following distributions are computationally indistinguishable:

\(D\): The distribution over four-tuples of the form \((A,y,w^\top A,\langle w,y \rangle)\) where \(A\) is uniform in \(\Z_q^{m\times n}\), \(x\) is uniform in \(\Z_q^n\), \(e \in Z_q\) is chosen with \(e_i \in \{-\delta q,\ldots,+\delta q\}\), \(y=Ax+e\), and \(w\) is uniform in \(\{0,1\}^m\).

\(\overline{D}\): The distribution over four-tuples \((A,y',a,\sigma)\) where all entries are uniform: \(A\) is uniform in \(\Z_q^{m\times n}\), \(y'\) is uniform in \(\Z_q^m\), \(a\) is uniform in \(\Z_q^n\) and \(\sigma\) is uniform in \(\Z_q\).

You should stop here and verify that **(i)** You understand the statement of Lemma 13.4 and **(ii)** you understand why this lemma implies Theorem 13.3. The idea is that Lemma 13.4 shows that the concatenation of the public key and encryption of \(0\) is indistinguishable from something that is completely random. You can then use it to show that the concatenation of the public key and encryption of \(1\) is indistinguishable from the same thing, and then finish using the hybrid argument.

We now prove Lemma 13.4, which will complete the proof of Theorem 13.3.

Define \(D\) to be the distribution \((A,y,w^\top A,\langle w,y \rangle)\) as in the lemma’s statement (i.e., \(y=Ax+e\) for some \(x\), \(e\) chosen as above). Define \(D'\) to be the distribution \((A,y',w^\top A, \langle w,y' \rangle)\) where \(y'\) is chosen uniformly in \(\Z_q^m\). We claim that \(D'\) is computationally indistinguishable from \(D\) under the LWE conjecture. Indeed by Theorem 13.1 (search to decision reduction) this conjecture implies that the distribution \(X\) over pairs \((A,y)\) with \(y=Ax+e\) is indistinguishable from the distribution \(X'\) over pairs \((A,y')\) where \(y'\) is uniform. But if there was some polynomial-time algorithm \(T\) distinguishing \(D\) from \(D'\) then we can design a randomized polynomial-time algorithm \(T'\) distinguishing \(X\) from \(X'\) with the same advantage by setting \(T'(A,y)=T(A,y,w^\top A,\langle w,y \rangle)\) for random \(w \leftarrow_R \{0,1\}^m\).

We will finish the proof by showing that the distribution \(D'\) is *statistically indistinguishable* (i.e., has negligible total variation distance) from \(\overline{D}\). This follows from the following claim:

**CLAIM:** Suppose that \(m > 100 n \log q\). If \(A'\) is a random \(m\times n+1\) matrix in \(\Z_q^m\), then with probability at least \(1-2^{-n}\) over the choice of \(A'\), the distribution \(Z_{A'}\) over \(\Z_q^n\) which is obtained by choosing \(w\) at random in \(\{0,1\}^m\) and outputting \(w^\top A'\) has at most \(2^{-n}\) statistical distance from the uniform distribution over \(\Z_q^{n+1}\).

Note that the randomness used for the distribution \(Z_{A'}\) is only obtained by the choice of \(w\), and *not* by the choice of \(A'\) that is fixed. (This passes a basic “sanity check” since \(w\) has \(m\) random bits, while the uniform distribution over \(\Z_q^n\) requires \(n \log q \ll m\) random bits, and hence \(Z_{A'}\) at least has a “fighting chance” in being statistically close to it.) Another way to state the same claim is that the pair \((A',w^\top A')\) is statistically indistinguishable from the uniform distribution \((A',z)\) where \(z\) is a vector chosen independently at random from \(\Z_q^n\).

The claim completes the proof of the theorem, since letting \(A'\) be the matrix \((A|y)\) and \(z=(a,\sigma)\), we see that the distribution \(D'\), as the form \((A',z)\) where \(A'\) is a uniformly random \(m\times (n+1)\) matrix and \(z\) is sampled from \(Z_{A'}\) (i.e., \(z=w^\top A'\) where \(w\) is uniformly chosen in \(\{0,1\}^m\)). Hence this means that the statistical distance of \(D'\) from \(\overline{D}\) (where all elements are uniform) is \(O(2^{-n})\). (Please make sure you understand this reasoning!)

We will not do the whole proof of the claim (which uses the mod \(q\) version of the leftover hash lemma which we mentioned before and is also “Wikipedia-able”) but the idea is simple. For every \(m\times (n+1)\) matrix \(A'\) over \(\Z_q\), define \(h_{A'}:\Z_q^m \rightarrow \Z_q^n\) to be the map \(h_{A'}(w)=w^\top A'\). This collection can be shown to be a “good” hash function collection in some specific technical sense, which in particular implies that for every distribution \(D\) with much more than \(n\log q\) bits of min-entropy, with all but negligible probability over the choice of \(A'\), \(h_{A'}(D)\) is statistically indistinguishable from the uniform distribution. Now when we choose \(w\) at random in \(\{0,1\}^m\), it is coming from a distribution with \(m\) bits of entropy. If \(m \gg (n+1)\log q\), then because the output of this function is so much smaller than \(m\), we expect it to be completely uniform, and this is what’s shown by the leftover hash lemma.

The proof of Theorem 13.3 is quite subtle and requires some re-reading and thought. To read more about this, you can look at the survey of Oded Regev, “On the Learning with Error Problem” Sections 3 and 4.

## But what are lattices?

You can think of a lattice as a discrete version of a subspace. A lattice \(L\) is simply a discrete subset of \(\mathbb{R}^n\) such that if \(u,v\in L\) and \(a,b\) are integers then \(au+bv\in L\).^{6} A lattice is given by a basis which simply a matrix \(B\) such that every vector \(u\in L\) is obtained as \(u=Bx\) for some vector of integers \(x\). It can be shown that we can assume without loss of generality that \(B\) is full dimensional and hence it’s an \(n\) by \(n\) invertible matrix. Note that given a basis \(B\) we can generate vectors in \(L\), as well as test whether a vector \(v\) is in \(L\) by testing if \(B^{-1}v\) is an integer vector. There can be many different bases for the same lattice, and some of them are easier to work with than others (see Figure 13.3).

Some classical computational questions on lattices are:

*Shortest vector problem:*Given a basis \(B\) for \(L\), find the nonzero vector \(v\) with smallest norm in \(L\).*Closest vector problem:*Given a basis \(B\) for \(L\) and a vector \(u\) that is*not*in \(L\), find the closest vector to \(u\) in \(L\).*Bounded distance decoding:*Given a basis \(B\) for \(L\) and a vector \(u\) of the form \(u=v+e\) where \(v\) is in \(L\), and \(e\) is a particularly short “error” vector (so in particular no other vector in the lattice is within distance \(\|e\|\) to \(u\)), recover \(v\). Note that this is a special case of the closest vector problem.

In particular, if \(V\) is a linear subspace of \(\Z_q^n\), we can think of it also as a lattice \(\hat{V}\) of \(\mathbb{R}^n\) where we simply say that that a vector \(\hat{u}\) is in \(\hat{V}\) if all of \(\hat{u}\)’s coordinates are integers and if we let \(u_i = \hat{u}_i \pmod{q}\) then \(u\in V\). The learning with error task of recovering \(x\) from \(Ax+e\) can then be thought of as an instance of the bounded distance decoding problem for \(\hat{V}\).

A natural algorithm to try to solve the *closest vector* and *bounded distance decoding* problems is that to take the vector \(u\), express it in the basis \(B\) by computing \(w = B^{-1}u\), and then round all the coordinates of \(w\) to obtain an integer vector \(\tilde{w}\) and let \(v=B\tilde{w}\) be a vector in the lattice. If we have an extremely good basis \(L\) for the lattice then \(v\) may indeed be the closest vector in the lattice, but in other more “skewed” bases it can be extremely far from it.

## Ring based lattices

One of the biggest issues with lattice based cryptosystem is the key size. In particular, the scheme above uses an \(m\times n\) matrix where each entry takes \(\log q\) bits to describe. (It also encrypts a single bit using a whole vector, but more efficient “multi-bit” variants are known.) Schemes using *ideal lattices* are an attempt to get more practical variants. These have very similar structure except that the matrix \(A\) chosen is not completely random but rather can be described by a single vector. One common variant is the following: we fix some polynomial \(p\) over \(\Z_q\) with degree \(n\) and then treat vectors in \(\Z_q^n\) as the coefficients of \(n-1\) degree polynomials and always work modulo this polynomial \(p()\). (By this I mean that for every polynomial \(t\) of degree at least \(n\) we write \(t\) as \(ps+r\) where \(p\) is the polynomial above, \(s\) is some polynomial and \(r\) is the “remainder” polynomial of degree \(<n\); then \(t \pmod{p} = r\).) Now for every fixed polynomial \(t\), the operation \(A_t\) which is defined as \(s \mapsto ts \pmod{p}\) is a linear operation mapping polynomials of degree at most \(n-1\) to polynomials of degree at most \(n-1\), or put another way, it is a linear map over \(\Z_q^n\). However, the map \(A_d\) can be described using the \(n\) coefficients of \(t\) as opposed to the \(n^2\) description of a matrix. It also turns out that by using the Fast Fourier Transform we can evaluate this operation in roughly \(n\) steps as opposed to \(n^2\). The ideal lattice based cryptosystem use matrices of this form to save on key size and computation time. It is still unclear if this structure can be used for attacks; recent papers attacking principal ideal lattices have shown that one needs to be careful about this.

One ideal-lattice based system is the “New Hope” cryptosystem (see also paper) that has been experimented with by Google. People have also made highly optimized general (non ideal) lattice based constructions, see in particular the “Frodo” system (paper here, can you guess what’s behind the name?). Both New Hope and Frodo have been submitted to the NIST competition to select a “post quantum” public key encryption standard.

Despite the name, Gaussian elimination has been known to Chinese mathematicians since 150BC or so, and was popularized in the west through the 1670 notes of Isaac Newton.

Over \(\Z_q\), we can think of \(q-1\) also as the number \(-1\), and so on. Thus if \(a\in\Z_q\), we define \(|a|\) to be the minimum of \(a\) and \(q-a\). This ensures the absolute value satisfies the natural property of \(|a|=|-a|\).

One can think of \(e\) as chosen by simply letting every coordinate be chosen at random in \(\{ -\delta q, -\delta q + 1 , \ldots, +\delta q \}\). For technical reasons, we sometimes consider other distributions and in particular the

*discrete Gaussian*distribution which is obtained by letting every coordinate of \(e\) be an independent Gaussian random variable with standard deviation \(\delta q\), conditioned on it being an integer. (A closely related distribution is obtained by picking such a Gaussian random variable and then rounding it to the nearest integer.)People sometimes also consider variants where both \(p(n)\) and \(q(n)\) can be as large as exponential.

In fact, due to the fact that the

*signs*of the error vector’s entries are different, we expect the errors to have significant cancellations and hence we would expect \(|\langle w,e \rangle|\) to only be roughly of magnitude \(\sqrt{m}\delta q\), but this is not crucial for our discussions.By discrete we mean that points in \(L\) are isolated. One formal way to define it is that there is some \(\epsilon>0\) such that every distinct \(u,v \in L\) are of distance at least \(\epsilon\) from one another.

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