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Projects. Probabilistically Checkable Proofs... Probabilistically Checkable Proofs What’s a proof? Statement: A formula φ is satisfiable. Proof: assignment, a. Property: 5 minute presentations. Can check proof in polynomial time. A real theory. Polytime Verifier: V (π) Class times over RRR week. A tiny introduction... There is π for satisfiable φ where V(π) says yes. For unsatisfiable φ, V (π) always says no. Will post schedule. ...and one idea Can prove any statement in NP. Tests submitted by Monday (mostly) graded. ...using random walks. Rest will be done this afternoon. Probabilistically Checkable Proof System:(r(n),q(n)) n is size of φ. Proof: π. Polytime V(π) Using r(n) random bits, look at q(n) places in π If φ is satisfiable, there is a π, where Verifier says yes. If φ is not satisfiable, for all π, Pr[V(π) says yes] 1/2. ≤ PCP Example. Probabilistically Checkable Proof System:(r(n),q(n)) Probabilistically Checkable Proof System:(r(n),q(n)) Graph Not Isomorphism: (G ,G ). n is size of φ. 1 2 n is size of φ. There is no way to permute the vertices of G1 to get G2. Proof: π. Proof: π. Polytime V(π) Give a PCP(poly(n), 1). (Weaker than PCP(O(logn), 3). Polytime V(π) Using r(n) random bits, look at q(n) places in π Exponential sized proof is allowed. Using r(n) random bits, look at q(n) places in π If φ is satisfiable, there is a π, Can only look at 1 place, though. If φ is satisfiable, there is a π, where Verifier says yes. where Verifier says yes. Any thoughts? If φ is not satisfiable, for all π, If φ is not satisfiable, for all π, Pr[V(π) says yes] 1/2. Design a proof format: π Pr[V(π) says yes] 1/2. ≤ ≤ A language is in NP if it there is a polynomial time proof system. What is the maximum size of a proof? For every labelled graph, H, A language is in PCP(r(n),q(n)) if it has a probabilistically checkable π[H] = 1 if permutation of G1 (A) O(q(n)). proof system: r(n),q(n). π[H] = 2 if permutation of G1 Don’t care if either. (B) O(q(n)r(n)). Theorem: NP PCP(O(log(n)),3). ⊂ Verifier: randomly choose x 1,2 , permute G to get H (C) O(r(n)2q(n)). ∈ { } x Only checks 3 places !!! Is π(H) equal to x? (D) O(q(n)2r (n)). How is this possible? For not equal, π exists. (D) Well..its not easy. If G1 = G2, any prover has 1/2 probability of mistake! Only 2r (n) different runs. Cool.......but exponential ..not so interesting of a proof. Look at q(n) bits in each possible run. Lots of ideas to get to O(logn) bits and 3 query bits. Another view. Gap CSP Two views of PCP. q-Constraint satisfaction problem: Formula. 1 Clauses of length q. Theorem A: There is a constant where 2 -Gap q-CSP is NP-complete q-ary truth table for each constraint. Theorem: There are constants q and ρ < 1 where ρ-Gap q-CSP is Theorem B: PCP(O(logn),O(1)) NP Is the formula satisfiable. ⊆ NP-complete Variable types can be.. Note: A = B. ⇒ Binary Variables are [0,1] Just another NP-complete problem... 1 There is 2 -Gap q-CSP is in PCP(O(logn),O(1)). Large alphabet variables are [0,...,k] Well, a bit more. Given φ. Not only is CSP NP-complete, val(φ) is fraction of satisfiable clauses. Proof format: assignment to variables. it is NP-hard to approximate within factor of 1? 1/ρ? ρ? val(φ) = 1 φ satisfiable if satisfiable π(φ) = is satisfying assignment. ≡ Within a factor of 1/ρ Verifier checks random clause and looks up variables. Example: ...and Theorem 1 is same statement that NP PCP(O(logn),O(1)). if there is π where Verifier accepts with probability 1/2. 3SAT. q = 3, and truth table for “or” of 3 variables. ⊆ 1 ≥ val(φ) 2 The ρ-Gap q-CSP Problem: φ ≥ Plus Theorem A any problem in NP in PCP(O(logn),O(1)). Is val(φ) = 1 or is val(φ) < ρ? Theorem: There are constants q and ρ < 1 where ρ-Gap q-CSP is NP-complete Reminder: Cook/Levin theorem. The other way: PCP CSP Gap amplification. → 1 Theorem A: There are constants q and ρ where 2 -Gap q-CSP is NP-complete Theorem B’: PCP(O(logn),O(1)) NP ⊆ B = A CSP is NP-complete. ⇒ 0 Is x L? Construct formula φx . Recall 3SAT is a CSP. Cook/Levin Theorem: SAT is NP-complete. ∈ GAP-CSP? Proof Idea: PCP proof system: = ... = ( ( )) Any problem A has polytime verifier. π π0 πs, s O poly n . Given a CSP instance, φ, produce an instance of GAP CSP, φG. Write out circuit for program for input x A. Verifier: yes/no based on constant number bits. φ satisfiable φG satisfiable. ∈ → Convert it to a formula. ish Verifer: constant sized clause Note: φ is not satifiable, val(φ) < 1 val(φ) 1 1/m. Gap: 1/m. on constant bits, π ,π ,π ,... → ≤ − i j k φ is not satifiable val(φ ) < 1 ∆. Gap: ∆. for each random string. → G − 2r(n) different random strings poly(n) clauses. Grow ∆ >> 1/µ. For any NO answer, a bad run of the verifier. → Assignment with val(φx ) gives proof π Pr[V (x) = yes] = val(φx ) for every π x L, φ is satisfiable. → ∈ x x L, val(φ ) < 1/2. 6∈ x Detour: Expanders. Expanders and gap amplification. Dinur’s amplification: breakthrough alert. Given a coloring instance, G, produce an instance of coloring, G. Consider “GAP” k independent set. G has a 3-coloring, or at least one edge is not correctly colored in any Either there is an independent set of size 2k. coloring For a graph G. val(φ) 1 1/m. Gap: 1/m. or every set is of size at most k. → ≤ − S 1 This is NP-complete. Pr(u,v) E [u S,v S] | | ( + λ2(G)) Step 1: Produce G0 ∈ ∈ ∈ ≤ V 2 | | Lemma: Given F, with indendepent set size α(F), there is a polytime 3-colorable G 3t-colorable G0 ` → G has an edge for every `-step path in G. G has gap of ∆ G has a gap Ω(t∆) for any 3t-coloring. algorithm to find G where → 0 since adjacency matrix of G` is A` (α(F) 2λ)logn α(G) (α(F) + 2λ)logn. − ≤ ≤ Construction: given graph G, add expander H on same node set. where A is adjacency matrix of G. Construction: Add graph on F with eigenvalue λ. Make graph G0 connecting t-length paths in graph. ` ` λ2(G ) = λ2(G) Construct logn length paths in F, → Connect paths x ,...,x with y ,...,y if any (x ,y ) in F. Idea: bad edges spread out and make many bad edges. S 1 ` 1 logn 1 logn i j Pr(u,v) E [u S,v S] |V| 2 + λ2(G ) infection of uncolorability. ∈ ∈ ∈ ≤ | | Argument idea: Use expansion. S Ramanujan Graphs: there exists degree d graphs with λ < 2/√d. Leave ind. set. S w/prob | | (1 + λ) in each step. No bad edges, no infection. | | 2 V max size of independent| | set is a power of logn. → Gap amplification. However.... Changed problem! Some details to work through... 3-coloring to 3t-coloring. Alphabet size from 3 to 3t. Need to reduce alphabet size. A, B, C! 1, 2, 3! Alphabet reduction. Summary See you ... Clauses (colorability) on variables with lots of possible values. formula on variables with few possible values. → Mind the gap! preserve the gap. PCP - proof, check a few places to catch errors with some probability. → How? CSP - formula, which is satisfiable or far from satisfiable. Remember: Equivalent! Proof contains a 3t coloring. Proving both are NP-hard Verifier chooses random edge and checks colors. Next week. Cheating proof can only get away ∆ times. (Dude, the gap!) Start with CSP ...for projects. Idea: Verifier is a polynomial sized circuit. Blow up the gap. (blows up the alphabet.) Are the two colors the same? Use PCP view to reduce the alphabet. Recursively construct PCP for verifier with smaller alphabet! (Exponential blowup for alphabet size.) (Ok, since number of colors are constant.) Lose some of the gap. Lots of cool ideas here. This PCP? Prover: Dude, there is a circuit that checks! Here is a proof that the circuit works. 276: Next spring. Prasad Raghavendra. Verifier: I am lazy, will only check a few of the gates. Goal: Format requires errors to be everywhere! Ideas from error correcting codes: Hadamard codes..
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