Approximating E3-LIN is NP-Hard

This lecture, which I gave for my 18.434 seminar, focuses on the MAX-E3LIN problem. We prove that approximating it is NP-hard by a reduction from LABEL-COVER.

1. Introducing MAX-E3LIN

In the MAX-E3LIN problem, our input is a series of linear equations {\pmod 2} in {n} binary variables, each with three terms. Equivalently, one can think of this as {\pm 1} variables and ternary products. The objective is to maximize the fraction of satisfied equations.

Example 1 (Example of MAX-E3LIN instance)

\displaystyle \begin{aligned} x_1 + x_3 + x_4 &\equiv 1 \pmod 2 \\ x_1 + x_2 + x_4 &\equiv 0 \pmod 2 \\ x_1 + x_2 + x_5 &\equiv 1 \pmod 2 \\ x_1 + x_3 + x_5 &\equiv 1 \pmod 2 \end{aligned}

\displaystyle \begin{aligned} x_1 x_3 x_4 &= -1 \\ x_1 x_2 x_4 &= +1 \\ x_1 x_2 x_5 &= -1 \\ x_1 x_3 x_5 &= -1 \end{aligned}

A diligent reader can check that we may obtain {\frac34} but not {1}.

Remark 2

We immediately notice that

  • If there’s a solution with value {1}, we can find it easily with {\mathbb F_2} linear algebra.
  • It is always possible to get at least {\frac{1}{2}} by selecting all-zero or all-one.

The theorem we will prove today is that these “obvious” observations are essentially the best ones possible! Our main result is that improving the above constants to 51% and 99%, say, is NP-hard.

Theorem 3 (Hardness of MAX-E3LIN)

The {\frac{1}{2}+\varepsilon} vs. {1-\delta} decision problem for MAX-E3LIN is NP-hard.

This means it is NP-hard to decide whether an MAX-E3LIN instance has value {\le \frac{1}{2}+\varepsilon} or {\ge 1-\delta} (given it is one or the other). A direct corollary of this is approximating MAX-SAT is also NP-hard.

Corollary 4

The {\frac78+\varepsilon} vs. {1-\delta} decision problem for MAX-SAT is NP-hard.

Remark 5

The constant {\frac78} is optimal in light of a random assignment. In fact, one can replace {1-\delta} with {\delta}, but we don’t do so here.

Proof: Given an equation {a+b+c=1} in MAX-E3LIN, we consider four formulas {a \lor \neg b \lor \neg c}, {\neg a \lor b \lor \neg c}, {a \lor \neg b \lor \neg c}, {a \lor b \lor c}. Either three or four of them are satisfied, with four occurring exactly when {a+b+c=0}. One does a similar construction for {a+b+c=1}. \Box

The hardness of MAX-E3LIN is relevant to the PCP theorem: using MAX-E3LIN gadgets, Ha}stad was able to prove a very strong version of the PCP theorem, in which the verifier merely reads just three bits of a proof!

Theorem 6 (Hastad PCP)

Let {\varepsilon, \delta > 0}. We have

\displaystyle \mathbf{NP} \subseteq \mathbf{PCP}_{\frac{1}{2}+\varepsilon, 1-\delta}(3, O(\log n)).

In other words, any {L \in \mathbf{NP}} has a (non-adaptive) verifier with the following properties.

  • The verifier uses {O(\log n)} random bits, and queries just three (!) bits.
  • The acceptance condition is either {a+b+c=1} or {a+b+c=0}.
  • If {x \in L}, then there is a proof {\Pi} which is accepted with probability at least {1-\delta}.
  • If {x \notin L}, then every proof is accepted with probability at most {\frac{1}{2} + \varepsilon}.

2. Label Cover

We will prove our main result by reducing from the LABEL-COVER. Recall LABEL-COVER is played as follows: we have a bipartite graph {G = U \cup V}, a set of keys {K} for vertices of {U} and a set of labels {L} for {V}. For every edge {e = \{u,v\}} there is a function {\pi_e : L \rightarrow K} specifying a key {k = \pi_e(\ell) \in K} for every label {\ell \in L}. The goal is to label the graph {G} while maximizing the number of edges {e} with compatible key-label pairs.

Approximating LABEL-COVER is NP-hard:

Theorem 7 (Hardness of LABEL-COVER)

The {\eta} vs. {1} decision problem for LABEL-COVER is NP-hard for every {\eta > 0}, given {|K|} and {|L|} are sufficiently large in {\eta}.

So for any {\eta > 0}, it is NP-hard to decide whether one can satisfy all edges or fewer than {\eta} of them.

3. Setup

We are going to make a reduction of the following shape:

reduction

In words this means that

  • “Completeness”: If the LABEL-COVER instance is completely satisfiable, then we get a solution of value {\ge 1 - \delta} in the resulting MAX-E3LIN.
  • “Soundness”: If the LABEL-COVER instance has value {\le \eta}, then we get a solution of value {\le \frac{1}{2} + \varepsilon} in the resulting MAX-E3LIN.

Thus given an oracle for MAX-E3LIN decision, we can obtain {\eta} vs. {1} decision for LABEL-COVER, which we know is hard.

The setup for this is quite involved, using a huge number of variables. Just to agree on some conventions:

Definition 8 (“Long Code”)

A {K}-indexed binary string {x = (x_k)_k} is a {\pm 1} sequence indexed by {K}. We can think of it as an element of {\{\pm 1\}^K}. An {L}-binary string {y = (y_\ell)_\ell} is defined similarly.

Now we initialize {|U| \cdot 2^{|K|} + |V| \cdot 2^{|L|}} variables:

  • At every vertex {u \in U}, we will create {2^{|K|}} binary variables, one for every {K}-indexed binary string. It is better to collect these variables into a function

    \displaystyle f_u : \{\pm1\}^K \rightarrow \{\pm1\}.

  • Similarly, at every vertex {v \in V}, we will create {2^{|L|}} binary variables, one for every {L}-indexed binary string, and collect these into a function

    \displaystyle g_v : \{\pm1\}^L \rightarrow \{\pm1\}.

Picture:

edge

Next we generate the equations. Here’s the motivation: we want to do this in such a way that given a satisfying labelling for LABEL-COVER, nearly all the MAX-E3LIN equations can be satisfied. One idea is as follows: for every edge {e}, letting {\pi = \pi_e},

  • Take a {K}-indexed binary string {x = (x_k)_k} at random. Take an {L}-indexed binary string {y = (y_\ell)_\ell} at random.
  • Define the {L}-indexed binary {z = (z_\ell)_\ell} string by {z = \left( x_{\pi(\ell)} y_\ell \right)}.
  • Write down the equation {f_u(x) g_v(y) g_v(z) = +1} for the MAX-E3LIN instance.

Thus, assuming we had a valid coloring of the graph, we could let {f_u} and {g_v} be the dictator functions for the colorings. In that case, {f_u(x) = x_{\pi(\ell)}}, {g_v(y) = y_\ell}, and {g_v(z) = x_{\pi(\ell)} y_\ell}, so the product is always {+1}.

Unfortunately, this has two fatal flaws:

  1. This means a {1} instance of LABEL-COVER gives a {1} instance of MAX-E3LIN, but we need {1-\delta} to have a hope of working.
  2. Right now we could also just set all variables to be {+1}.

We fix this as follows, by using the following equations.

Definition 8 (Equations of reduction)

For every edge {e}, with {\pi = \pi_e}, we alter the construction and say

  • Let {x = (x_k)_k} be and {y = (y_\ell)_\ell} be random as before.
  • Let {n = (n_\ell)_\ell} be a random {L}-indexed binary string, drawn from a {\delta}-biased distribution ({-1} with probability {\delta}). And now define {z = (z_\ell)_\ell} by

    \displaystyle z_\ell = x_{\pi(\ell)} y_\ell n_\ell .

    The {n_\ell} represents “noise” bits, which resolve the first problem by corrupting a bit of {z} with probability {\delta}.

  • Write down one of the following two equations with {\frac{1}{2}} probability each:

    \displaystyle \begin{aligned} f_u(x) g_v(y) g_v(z) &= +1 \\ f_u(x) g_v(y) g_v(-z) &= -1. \end{aligned}

    This resolves the second issue.

This gives a set of {O(|E|)} equations.

I claim this reduction works. So we need to prove the “completeness” and “soundness” claims above.

4. Proof of Completeness

Given a labeling of {G} with value {1}, as described we simply let {f_u} and {g_v} be dictator functions corresponding to this valid labelling. Then as we’ve seen, we will pass {1 - \delta} of the equations.

5. A Fourier Computation

Before proving soundness, we will first need to explicitly compute the probability an equation above is satisfied. Remember we generated an equation for {e} based on random strings {x}, {y}, {\lambda}.

For {T \subseteq L}, we define

\displaystyle \pi^{\text{odd}}_e(T) = \left\{ k \in K \mid \left\lvert \pi_e^{-1}(k) \cap T \right\rvert \text{ is odd} \right\}.

Thus {T} maps subsets of {L} to subsets of {K}.

Remark 9

Note that {|\pi^{\text{odd}}(T)| \le |T|} and that {\pi^{\text{odd}}(T) \neq \varnothing} if {|T|} is odd.

Lemma 10 (Edge Probability)

The probability that an equation generated for {e = \{u,v\}} is true is

\displaystyle \frac{1}{2} + \frac{1}{2} \sum_{\substack{T \subseteq L \\ |T| \text{ odd}}} (1-2\delta)^{|T|} \widehat g_v(T)^2 \widehat f_u(\pi^{\text{odd}}_e(T)).

Proof: Omitted for now\dots \Box

6. Proof of Soundness

We will go in the reverse direction and show (constructively) that if there is MAX-E3LIN instance has a solution with value {\ge\frac{1}{2}+2\varepsilon}, then we can reconstruct a solution to LABEL-COVER with value {\ge \eta}. (The use of {2\varepsilon} here will be clear in a moment). This process is called “decoding”.

The idea is as follows: if {S} is a small set such that {\widehat f_u(S)} is large, then we can pick a key from {S} at random for {f_u}; compare this with the dictator functions where {\widehat f_u(S) = 1} and {|S| = 1}. We want to do something similar with {T}.

Here are the concrete details. Let {\Lambda = \frac{\log(1/\varepsilon)}{2\delta}} and {\eta = \frac{\varepsilon^3}{\Lambda^2}} be constants (the actual values arise later).

Definition 11

We say that a nonempty set {S \subseteq K} of keys is heavy for {u} if

\displaystyle \left\lvert S \right\rvert \le \Lambda \qquad\text{and}\qquad \widehat{f_u}(S) \ge \varepsilon^2.

Note that there are at most {\varepsilon^{-2}} heavy sets by Parseval.

Definition 12

We say that a nonempty set {T \subseteq L} of labels is {e}-excellent for {v} if

\displaystyle \left\lvert T \right\rvert \le \Lambda \qquad\text{and}\qquad S = \pi^{\text{odd}}_e(T) \text{ is heavy.}

In particular {S \neq \varnothing} so at least one compatible key-label pair is in {S \times T}.

Notice that, unlike the case with {S}, the criteria for “good” in {T} actually depends on the edge {e} in question! This makes it easier to keys than to select labels. In order to pick labels, we will have to choose from a {\widehat g_v^2} distribution.

Lemma 13 (At least {\varepsilon} of {T} are excellent)

For any edge {e = \{u,v\}}, at least {\varepsilon} of the possible {T} according to the distribution {\widehat g_v^2} are {e}-excellent.

Proof: Applying an averaging argument to the inequality

\displaystyle \sum_{\substack{T \subseteq L \\ |T| \text{ odd}}} (1-2\delta)^{|T|} \widehat g_v(T)^2 \left\lvert \widehat f_u(\pi^{\text{odd}}(T)) \right\rvert \ge 2\varepsilon

shows there is at least {\varepsilon} chance that {|T|} is odd and satisfies

\displaystyle (1-2\delta)^{|T|} \left\lvert \widehat f_u(S) \right\rvert \ge \varepsilon

where {S = \pi^{\text{odd}}_e(T)}. In particular, {(1-2\delta)^{|T|} \ge \varepsilon \iff |T| \le \Lambda}. Finally by \Cref{rem:po}, we see {S} is heavy. \Box

Now, use the following algorithm.

  • For every vertex {u \in U}, take the union of all heavy sets, say

    \displaystyle \mathcal H = \bigcup_{S \text{ heavy}} S.

    Pick a random key from {\mathcal H}. Note that {|\mathcal H| \le \Lambda\varepsilon^{-2}}, since there are at most {\varepsilon^{-2}} heavy sets (by Parseval) and each has at most {\Lambda} elements.

  • For every vertex {v \in V}, select a random set {T} according to the distribution {\widehat g_v(T)^2}, and select a random element from {T}.

I claim that this works.

Fix an edge {e}. There is at least an {\varepsilon} chance that {T} is {e}-excellent. If it is, then there is at least one compatible pair in {\mathcal H \times T}. Hence we conclude probability of success is at least

\displaystyle \varepsilon \cdot \frac{1}{\Lambda \varepsilon^{-2}} \cdot \frac{1}{\Lambda} = \frac{\varepsilon^3}{\Lambda^2} = \eta.

(Addendum: it’s pointed out to me this isn’t quite right; the overall probability of the equation given by an edge {e} is {\ge \frac{1}{2}+\varepsilon}, but this doesn’t imply it for every edge. Thus one likely needs to do another averaging argument.)

 

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s