the unique games conjecture with entangled provers is false julia kempe tel aviv university oded...
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The Unique Games The Unique Games Conjecture Conjecture
with Entangled Provers is with Entangled Provers is FalseFalseJulia KempeJulia Kempe
Tel Aviv UniversityTel Aviv University
Oded RegevOded RegevTel Aviv UniversityTel Aviv University
Ben TonerBen TonerCWI, AmsterdamCWI, Amsterdam
• The verifier sends one question to each prover• Each prover responds with an answer from {1,…,k}
(no communication allowed)• The verifier decides whether to accept or reject• The value of a game is the maximum success
probability the provers can achieve, and is denoted by (G)– The provers can have shared randomness but it can’t help
them
Two-Prover One-round Games
VerifierVerifier
AliceAlice BobBob
acceptacceptrejectreject
ss ttaa bb
• We say that a game is unique if for each answer of the first prover there is exactly one good answer of the second prover and vice versa [CaiCondonLipton90, FeigeLovász92]
– In other words, the verifier accepts answers a,b iff b=(a) where is some permutation on k elements
Unique Games
1 2 3 4 51 2 3 4 5 …… kk
1 2 3 4 51 2 3 4 5 …… kk
1 2 3 4 51 2 3 4 5 …… kk
1 2 3 4 51 2 3 4 5 …… kk
11
33
44
99
44
33
11
kk
• The CHSH game: [ClauserHorneShimonyHolt69]
– The verifier sends a random bit to each prover– Each prover responds with a bit (so k=2 here)– The verifier accepts iff the XOR of the answers
is equal to the AND of the questions– Unique game
• The value of this game is 3/4
Example: the CHSH game
====
==
≠≠
11
11
11
11
• In 2002, Khot conjectured that estimating the value of unique games is also hard:
• Conjecture: [Khot02] >0 k such that it is NP-hard to determine whether a
given unique game G with answer size k has (G)1- or (G)<
Remarks:– It is crucial that >0 since otherwise easy– The PCP theorem + parallel repetition shows
this without the “unique” requirement
Unique Games Conjecture (UGC)
Implications of the UGCImplications of the UGC• Based on the UGC, several tight hardness Based on the UGC, several tight hardness
results were shown:results were shown:• VertexCover to within 1.9999 VertexCover to within 1.9999 [[KhotRegev03KhotRegev03]]• MaxCut to within 0.8789 MaxCut to within 0.8789
[[KhotKindlerMosselO’Donnell04KhotKindlerMosselO’Donnell04]]• SparsestCut to within any constant (and SparsestCut to within any constant (and
beyond) beyond) [[ChawlaKrauthgamerKumarRabaniSivakumar05, ChawlaKrauthgamerKumarRabaniSivakumar05, KhotVishnoi05KhotVishnoi05]]
• Coloring 3-colorable graphs with any Coloring 3-colorable graphs with any constant number of colors constant number of colors [[DinurMosselRegev05DinurMosselRegev05]]
• And more…And more…
• The UGC is The UGC is currently one of most important currently one of most important open questions in theoretical computer scienceopen questions in theoretical computer science
Algorithmic Results on Unique Algorithmic Results on Unique GamesGames
• Lots of algorithmic work on Lots of algorithmic work on approximating approximating for unique games for unique games [[Trevisan05, CharikarMakarychevM06, Trevisan05, CharikarMakarychevM06, GuptaTalwar06, ChlamtacMakarychevM06GuptaTalwar06, ChlamtacMakarychevM06]]– Can be seen as attempts to disprove the UGCCan be seen as attempts to disprove the UGC
• One of the best known results is One of the best known results is [CharikarMakarychevM06][CharikarMakarychevM06]. Given any unique game . Given any unique game G, their algorithm outputs a value G, their algorithm outputs a value s.t.: s.t.:
1-O((1-O((logk)logk)½½) ) (G) (G) 1-1-– This does not contradict the UGC, but instead This does not contradict the UGC, but instead
tells us how big k=k(tells us how big k=k(,,) needs to be in order ) needs to be in order for the conjecture to make sensefor the conjecture to make sense
• These games are as before, except the provers are allowed to share an arbitrary entangled quantum state
• Originate in the works of [EinsteinPodolskyRosen35, Bell64,…]
Games with Entangled Provers
VerifierVerifier
AliceAlice BobBob
acceptacceptrejectreject
ss ttaa bb
• The entangled value of a game is the maximum success probability that entangled provers can achieve, and is denoted by (G)– For instance, (CHSH)=0.8536…, which is
strictly greater than (CHSH)=0.75.
• This remarkable ability of entanglement to create correlations that are impossible to obtain classically (something Einstein referred to as “spooky”) is one of the most peculiar aspects of quantum mechanics
Games with Entangled Provers
• Why study this model?– It was here first – If we ever want to ‘really’ use proof
systems, then there is no physical way to guarantee that the provers don’t share entanglement
– It might give us new insight on (non-entangled) games
• Despite considerable work, our understanding of this model is still quite limited
Games with Entangled Provers
• One of most important results is that of [Tsirelson80] who showed that for the special case of unique games with k=2, the entangled value is given exactly by an SDP and can therefore be computed efficiently (see also [CleveHøyerTonerWatrous04])– This is in contrast to the (non-entangled)
value of unique games with k=2 which is NP-hard to approximate (by Håstad’s hardness result for MaxCut)
– This SDP is used to determine that (CHSH)=0.8536…
Games with Entangled Provers
• The only other known result is by [Masanes05] who shows how to compute for games with two possible questions to each prover and k=2
• In all other cases, no method is known to compute or even approximate
Games with Entangled Provers
• Theorem: There exists an efficient algorithm that, given any unique game G outputs a value s.t.
1-6 (G) 1-
• This gives for the first time a way to approximate for games with k>2
• It shows that the analogue of the UGC for entangled provers is false
• Notice that our lower bound is independent of k, whereas in the non-entangled case, the lower bound is 1-f(,k)
Our Results
• We prove our main theorem in two steps:1.We formulate an SDP relaxation of *
•Surprisingly, this is essentially identical to the Feige-Lovász SDP, often used as a relaxation of
2.We then show how to take a solution to the SDP and transform it into a strategy for entangled provers•We call this ‘quantum rounding’ in analogy
with the rounding technique used in SDPs
Techniques
The ProofThe Proof
• If Alice and Bob share the n-dimensional maximally entangled state then they can perform a measurement as follows:– Each party chooses an orthonormal basis of Rn
– Each party obtains an outcome in {1,...,n}
– If Alice uses the orthonormal basis (x1,…,xn) and Bob uses the orthonormal basis (y1,…,yn), the output has the joint distribution given by
• Notice that each party’s marginal is uniform
Quantum Correlations
• For example, here’s how to get (cos/8)2=0.8536… success probability in the CHSH game:– The provers share a 2-dimensional
maximally entangled state– They perform a measurement in a basis
depending on their input:
Quantum Correlations
xx11
xx00
xx00
xx11
Input = 0Input = 0Input = 1Input = 1
yy00
yy11
yy00
yy11
AliceAlice BobBob
• For simplicity, let’s consider unique games for which there exists an optimal strategy in which each prover’s answer distribution is uniform over {1,…,k}
• Theorem: There exists an efficient algorithm that, given any ‘uniform’ unique game G, outputs a value s.t.
1-4 (G) 1-• Proof: Start by writing an SDP relaxation:
• I will not show why this is a relaxation of the entangled value (the proof is not difficult)
• Let the value of this SDP be 1-and output • Our goal is to show that there exists a strategy that
achieves success probability 1-4
The Proof
• A solution consists of k orthonormal vectors in Rn for each question (for some large n)
• In a good solution, the vectors should be ‘aligned’ according to the permutation
The SDP Relaxation
1 2 3 4 51 2 3 4 5 …… kk
1 2 3 4 51 2 3 4 5 …… kk
1 2 3 4 51 2 3 4 5 …… kk
1 2 3 4 51 2 3 4 5 …… kk
• Main Idea: On input s, Alice performs the measurement given by . Similarly for Bob using the v vectors.
• However, is not a basis of Rn !• Instead, the provers complete their k
vectors to an orthonormal basis of Rn in an arbitrary way
Quantum Rounding – The Idea
• New problem: the probability that Alice obtains one of the first k outcomes is only k/n; otherwise, with probability 1-k/n, her outcome is meaningless
• Luckily, if Alice gets a meaningless answer, then with high probability Bob also gets a meaningless answer
• This allows us to solve the problem by having both parties repeat the measurement process until they get a meaningful answer
Quantum Rounding – The Idea
Quantum Rounding
• Alice’s strategy: On input s, performs the measurement given by the completion of to an orthonormal basis. – If obtains an outcome in {1,…,k}, returns it. – Otherwise, repeats the measurement again
(with a fresh maximally entangled state)
• Bob’s strategy is similar.
• Fix some question pair (s,t)• Assume for simplicity that the
permutation st is the identity permutation.
• The contribution to the SDP is
• We will show that Alice and Bob have probability at least 1-4’ to output the same value i
Quantum Rounding – Analysis
• Alice and Bob have k+1 possible outcomes in one measurement, with k+1 signifying ‘try again’.
• The joint probability distribution is given by
Quantum Rounding – Analysis
1 2 … j … k k+11 2 … j … k k+1
11ii
kk
k+1k+1
k/nk/n
k/nk/n* * * * * * * *
**
**
• The probability that in one measurement, both Alice and Bob output the same value is
• The probability that both of them ‘try again’ is therefore at least
• The probability of success is therefore at least
Quantum Rounding – Analysis
• We showed that the entangled value of unique games can be well approximated (up to factor 6)
• We extend our results to d-to-d games
• Our result also implies a parallel repetition theorem for the entangled value of unique games
Conclusions
• Unique games:– Improve our factor 6, or even compute *
exactly– So far we only know how to do it for k=2 by
[Tsirelson80]
• General games:– Can one compute * exactly?
• Probably not. [KempeKobayashiMatsumotoTonerVidick07]
show this for games with 3 provers, and for games with quantum communication.
– But what about approximating * ?• Mostly open (even with many provers, quantum
communication, etc.)• The most important open question in this area
Open Questions