This the first lecture for CSE 599S: Analytical and geometric methods in the theory of computation. Today we’ll consider the gap amplification problem for 2-prover games, and see how it’s intimately related to some high-dimensional isoperimetric problems about foams. In the next lecture, we’ll use spectral techniques to find approximately optimal foams (which will then let us cheat at repeated games).
The PCP Theorem, 2-prover games, and parallel repetition
For a 3-CNF formula , let denote the maximum fraction of clauses in which are simultaneously satisfiable. For instance, is satisfiable if and only if . One equivalent formulation of the PCP Theorem is that the following problem is NP-complete:
Given a 3-CNF formula , answer YES if and NO if (any answer is acceptable if neither condition holds).
We can restate this result in the language of 2-prover games. A 2-prover game consists of four finite sets , where and are sets of questions, while and are sets of answers to the questions in and , respectively. There is also a verifier which checks the validity of answers. For a pair of questions and answers , the verifier is satisfied if and only if . The final component of is a probability distribution on .
Now a strategy for the game consists of two provers and who map questions to answers. The score of the two provers is precisely
where is drawn from . This is just the probability that the verifier is happy with the answers provided by the two provers. The value of the game is now defined as
i.e. the best-possible score achievable by any two provers.
Now we can again restate the PCP Theorem as saying that the following problem is NP-complete:
Given a 2-prover game with with , answer YES if and answer NO if .
To see that Formulation 1 implies Formulation 2, consider, for any 3-CNF formula , the game defined as follows. is the set of clauses in , is the set of variables in , while and . Here represents the set of eight possible truth assignments to a three-variable clause, and represents the set of possible truth assignments to a variable.
The distribution is defined as follows: Choose first a uniformly random clause , and then uniformly at random one of the three variables which appears in . An answer is valid if the assignment makes true, and if and are consistent in the sense that they give the same truth value to the variable . The following statement is an easy exercise:
For every 3-CNF formula , we have .
The best strategy is to choose an assignment to the variables in . plays according to , while plays according to unless he is about to answer with an assignment that doesn’t satisfy the clause . At that point, he flips one of the literals to make his assignment satisfying (in this case, the chance of catching cheating is only , the probability that is sent the variable that flipped).
This completes our argument that Formulation 1 implies Formulation 2.
A very natural question is whether the constant in Formulation 2 can be replaced by (or an arbitrarily small constant). A natural way of gap amplification is by “parallel repetition” of a given game. Starting with a game , we can consider the game , where is just the product distribution on . Here, we choose pairs of questions i.i.d. from and the two provers then respond with answers and . The verifier is satisfied if and only if for every .
Clearly because given a strategy for , we can play the same strategy in every coordinate, and then our probability to win is just the probability that we simultaneously win independent games. But is there a more clever strategy that can do better? Famously, early papers in this arena assumed it was obvious that .
In fact, there are easy examples of games where . (Exercise: Show that this is true for the following game devised by Uri Feige. The verifier chooses two independent random bits , and sends to and to . The answers of the two provers are from the set . The verifier accepts if both provers answer or both provers answer .)
Nevertheless, in a seminal work, Ran Raz proved the Parallel Repetition Theorem, which states that the value of the repeated game does, in fact, drop exponentially.
Theorem 1.1: For every 2-prover game , there exists a constant such that if , then for every ,
The exponent 3 above is actually due to an improvement of Holenstein (Raz’s original paper can be mined for an exponent of 32).
Special games and the unique games conjecture
There are two special times of games which should be emphasized, depending on the structure of the mapping . A projection game is one in which, for every and , there is at most one value for which . In other words, after fixing an answer to the first question, there is at most a single answer to the second question which the verifier accepts. The 3-CNF game above is an example (given an assignment to the variables in the clause , the second prover has to give the consistent assignment to the variable ). Recently, Anup Rao gave an improved parallel repetition theorem for this special case.
Theorem 1.2: For every 2-prover projection game , with , and for every ,
Notice that in addition to the improved exponent of , there is no dependence on . (This dependence is known to be necessary for general games.)
Why do we care about the exponent?
Our main focus in this lecture will actually be on the exponent of in the preceding theorems, so it seems like a good time to discuss its importance. For that, we need to introduce unique games. This is a game where , , and the verifier behaves as follows. Given a pair of questions , there exist a bijection such that if and only if . In other words, after fixing a pair of questions, for every answer of one prover there exists a unique answer of the other prover that satisfies the verifier. (This is almost the same as satisfying the projection property from both sides, except that here we have enforced that after fixing and , there exists exactly one satisfying instead
of at most one.)
Now we state Khot’s unique games conjecture (UGC), which has far-reaching consequences in hardness of approximation (and is the central open question in the field).
Conjecture (UGC): For every , there exists a such that the following problem is NP-complete: Given a unique game with , answer YES if and answer NO if .
At present, there are very few good ideas on how to attack this conjecture, and there are differing beliefs about its truth. Observe that, unlike Formulations 1 and 2 above, in the YES instance, we now only require . Of course, this problem is harder than distinguishing from . In fact, this is fundamental: It is easy to design an efficient algorithm which checks whether for a unique game. (Exercise: Verify this.)
We now outline one possible approach to proving the UGC, in which the exponent of becomes crucial. For an undirected graph , define
where denotes the set of edges from to its complement. Consider the following conjecture.
Conjecture (MAX-CUT conjecture): There exists a constant such that for every , the following problem is NP-complete: Given a graph , distinguish between the two cases
Note that this conjectured hardness ( vs. ) is tight, as shown by the Goemans-Williamson algorithm. See also a recent spectral algorithm of Luca Trevisan that achieves the same bound. From work of Khot, Kindler, Mossel, and O’Donnell (and the subsequent Majority is Stablest theorem, which we’ll get to later in the course), we know that the UGC implies the MAX-CUT conjecture. Might we prove that the two conjectures are actually equivalent?
To address this, let’s first observe that we can turn the MAX-CUT problem into a 2-prover unique game in a straightforward way. Given a graph , consider the game where and .
The distribution on questions is as follows. With probability each: (1) Choose a uniformly random vertex and ask the questions , or (2) choose a uniformly random edge and ask the questions . The verifier is satisfied in case (1) only if the two provers give the same answer for . The verifier is satisfied in case (2) only if the two provers give different answers. It only takes a moment’s though to see that the two provers should fix a maximum cut in the graph, and play according to it, yielding
Now we are ready to see that the two conjectures are equivalent if we can obtain a good-enough exponent in parallel repetition of unique games. To this end, let be the infimal value of such that there exists a constant so that for every unique game , we have
for every . From Theorem 1.2, we know that . A little better would show the equivalence of these two conjectures.
Claim 1.3: If , then the the MAX-CUT conjecture implies the UGC.
Proof: The proof is straightforward: If , then , which implies that . On the other hand, if , then , and for some ,
for . Taking , we get in the first case, and in the other. Since is a unique game whenever is unique, this completes the proof.
MAX-CUT on a cycle
To this end, a number of authors have asked whether a strong parallel repetition theorem holds, i.e. whether (even for general games).
The rest of the lecture follows Feige, Kindler, and O’Donnell. We’ll consider a very simple unique game: MAX-CUT on an odd cycle, though we’ll slightly change the probabilities of our earlier verifier for the sake of convenience. We define the game as follows: , , and is specified by first choosing uniformly at random, and then choosing where is chosen uniformly at random. As before, when , accepts only if the answers satisfy , and otherwise accepts only if the answers satisfy .
It is easy to check that for odd, we have
since any cut in an odd cycle must have some some edge which doesn’t cross it (e.g. the edge with two green endpoints above). The point now is to understand the behavior of .
Toward this end, consider the graph whose vertex set is and which has an edge whenever for (this is also known as the -fold AND-product of an -cycle with itself). For instance, here is , where the dashed edges are meant to wrap around.
There is a natural embedding of into the -dimensional torus , which endows oriented cycles in with a homotopy class from .
Given such a cycle , we use to denote the corresponding element of the fundamental group. if wraps around the torus times in the th direction.
A cycle is said to be topologically non-trivial if . This is the same as saying that cannot be continuously contracted to a point in . The cycle is topologically odd if is odd for some . Here are two topologically odd cycles on the 2-torus, corresponding to the classes and .
In the following pictures, the red cycle is topologically trivial, the blue cycle is simply a shift of the red cycle, and the purple cycle is topologically non-trivial (indeed, it is topologically odd).
Finally, consider the double cover , which is a bipartite graph with vertex set , and with an edge whenever is an edge in or . Oriented cycles in inherit the notions of topological non-triviality and oddness from the projection . A curve in is said to be non-trivial or odd if its projection is. Now consider the following problem.
Odd-cycle elimination problem in :
Remove the minimum-possible fraction of edges from so as to delete all topologically odd cycles.
For example, here is a natural set of blocking edges (projected from onto ).
It turns out that this is just in disguise. For example, the above set of blocked edges corresponds to a strategy where and play each coordinate independently. More generally, we have the following:
Claim 1.4: .
Consider two provers for the repeated game . Edges in correspond to pairs of questions . Let be the set of edges corresponding to questions on which answer incorrectly, i.e. edges for which . Clearly
where we recall that is the number of edges in .
Say that an arbitrary set of edges is blocking if removing leaves no topologically odd cycles. We need to show that every set of the form is blocking, and that from every blocking set , we can recover two provers with .
First, assume that is not blocking. Then there exists a cycle in and a coordinate such that is odd. But if we consider the projection of onto coordinate , then we get an odd cycle on which play perfectly, which is clearly impossible.
On the other hand, consider an arbitrary blocking solution . From any connected component of , choose an arbitrary vertex, say , and put . Once this is set, since we cannot violate any edges in , there is a unique greedy extension of to the rest of . For instance, if , then we must put as well. If and and differ by 1 in the th coordinate, then , where the 1 occurs in the th coordinate. It is easy to see that this greedy assignment cannot fail precisely because there are no topologically odd cycles remaining in (every bipartite graph has a cut which contains all edges). Note that topologically trivial cycles are inconsequential since the coordinate projection of such a cycle results in a path. This completes the proof.
Now define to be the minimum be number of edges which need to be removed from so as to eliminate all topologically non-trivial cycles. It is easy to see that because we can remove all topologically non-trivial cycles in (and, in particular, all topologically odd cycles) by taking a set of edges which do this for and using the edges in .
Thus in order to “cheat” at the odd-cycle game, we only need to find small sets of edges whose removal blocks all non-trivial cycles. In this next lecture, we will see how this is intimately related to foams which tile , and how Dirichlet eigenfunctions help us find good foams.