Projects. Probabilistically Checkable Proofs... Probabilistically - PowerPoint PPT Presentation

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. Probabilistically Checkable Proof System:( r ( n ) , q ( n ) ) Rest will be done this afternoon. 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 1 , G 2 ) . n is size of φ . There is no way to permute the vertices of G 1 to get G 2 . n is size of φ . Proof: π . Proof: π . Give a PCP(poly ( n ) , 1). (Weaker than PCP( O ( log n ) , 3). Polytime V( π ) Polytime V( π ) Exponential sized proof is allowed. Using r ( n ) random bits, look at q ( n ) places in π Using r ( n ) random bits, look at q ( n ) places in π Can only look at 1 place, though. If φ is satisfiable, there is a π , If φ is satisfiable, there is a π , where Verifier says yes. Any thoughts? where Verifier says yes. If φ is not satisfiable, for all π , If φ is not satisfiable, for all π , Design a proof format: π Pr[V( π ) says yes] ≤ 1 / 2. Pr[V( π ) says yes] ≤ 1 / 2. A language is in NP if it there is a polynomial time proof system. For every labelled graph, H , What is the maximum size of a proof? π [ H ] = 1 if permutation of G 1 A language is in PCP ( r ( n ) , q ( n )) if it has a probabilistically checkable (A) O ( q ( n )) . π [ H ] = 2 if permutation of G 1 proof system: r ( n ) , q ( n ) . 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 x to get H (C) O ( r ( n ) 2 q ( n )) . Only checks 3 places !!! Is π ( H ) equal to x ? (D) O ( q ( n ) 2 r ( n )) . How is this possible? For not equal, π exists. If G 1 = G 2 , any prover has 1 / 2 probability of mistake! (D) Well..its not easy. Only 2 r ( 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 ( log n ) bits and 3 query bits.

Another view. Gap CSP Two views of PCP . q -Constraint satisfaction problem: Formula. Theorem A: There is a constant where 1 Clauses of length q . 2 -Gap q -CSP is NP-complete q -ary truth table for each constraint. Theorem B: PCP ( O ( log n ) , O ( 1 )) ⊆ NP Theorem: There are constants q and ρ < 1 where ρ -Gap q -CSP is Is the formula satisfiable. NP-complete Note: A = ⇒ B . Variable types can be.. Binary Variables are [ 0 , 1 ] Just another NP-complete problem... There is 1 2 -Gap q -CSP is in PCP ( O ( log n ) , O ( 1 )) . Large alphabet variables are [ 0 ,..., k ] Well, a bit more. Given φ . Not only is CSP NP-complete, Proof format: assignment to variables. val ( φ ) is fraction of satisfiable clauses. it is NP-hard to approximate within factor of 1? 1 / ρ ? ρ ? if satisfiable π ( φ ) = is satisfying assignment. val ( φ ) = 1 ≡ φ satisfiable Within a factor of 1 / ρ Verifier checks random clause and looks up variables. Example: if there is π where Verifier accepts with probability ≥ 1 / 2. ...and Theorem 1 is same statement that NP ⊆ PCP ( O ( log n ) , O ( 1 )) . 3SAT. q = 3, and truth table for “or” of 3 variables. val ( φ ) ≥ 1 2 The ρ -Gap q -CSP Problem: φ Plus Theorem A any problem in NP in PCP ( O ( log n ) , 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. Theorem A: There are constants q and ρ where 1 2 -Gap q -CSP is NP-complete Theorem B’: PCP ( O ( log n ) , O ( 1 )) ⊆ NP CSP is NP-complete. ⇒ A ′ B = Recall 3SAT is a CSP . Is x ∈ L ? Construct formula φ x . Cook/Levin Theorem: SAT is NP-complete. GAP-CSP? PCP proof system: Proof Idea: π = π 0 ... π s , s = O ( poly ( n )) . Given a CSP instance, φ , produce an instance of GAP CSP , φ G . Any problem A has polytime verifier. Verifier: yes/no based on constant number bits. Write out circuit for program for input x ∈ A . φ satisfiable → φ G 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 i , π j , π k ,... φ is not satifiable → val ( φ G ) < 1 − ∆ . Gap: ∆ . for each random string. 2 r ( n ) different random strings Grow ∆ >> 1 / µ . poly ( n ) clauses. For any NO answer, a bad run of the verifier. → Assignment with val ( φ x ) gives proof π Pr [ V ( x ) = yes ] = val ( φ x ) for every π → x ∈ L , φ x is satisfiable. x �∈ L , val ( φ x ) < 1 / 2.

Detour: Expanders. Expanders and gap amplification. Dinur’s amplification: breakthrough alert. Given a coloring instance, G , produce an instance of coloring, G . G has a 3-coloring, or at least one edge is not correctly colored in any Consider “GAP” k independent set. coloring Either there is an independent set of size 2 k . For a graph G . → val ( φ ) ≤ 1 − 1 / m . Gap: 1 / m . or every set is of size at most k . This is NP-complete. Pr ( u , v ) ∈ E [ u ∈ S , v ∈ S ] ≤ | S | Step 1: Produce G ′ | V | ( 1 2 + λ 2 ( G )) 3-colorable G → 3 t -colorable G ′ Lemma: Given F , with indendepent set size α ( F ) , there is a polytime G ℓ has an edge for every ℓ -step path in G . G has gap of ∆ → G ′ has a gap Ω( t ∆) for any 3 t -coloring. algorithm to find G where since adjacency matrix of G ℓ is A ℓ ( α ( F ) − 2 λ ) log n ≤ α ( G ) ≤ ( α ( F )+ 2 λ ) log n . Construction: given graph G , add expander H on same node set. Make graph G ′ connecting t -length paths in graph. where A is adjacency matrix of G . Construction: Add graph on F with eigenvalue λ . → λ 2 ( G ℓ ) = λ ℓ Construct log n length paths in F , 2 ( G ) Idea: bad edges spread out and make many bad edges. Connect paths x 1 ,..., x log n with y 1 ,..., y log n if any ( x i , y j ) in F . � 1 Pr ( u , v ) ∈ E [ u ∈ S , v ∈ S ] ≤ | S | 2 + λ 2 ( G ℓ ) � infection of uncolorability. | V | Argument idea: √ Use expansion. | S | Ramanujan Graphs: there exists degree d graphs with λ < 2 / d . Leave ind. set. | S | w/prob 2 | V | ( 1 + λ ) in each step. No bad edges, no infection. → max size of independent set is a power of log n . Gap amplification. However.... Changed problem! 3-coloring to 3 t -coloring. Some details to work through... Alphabet size from 3 to 3 t . 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. PCP - proof, check a few places to catch errors with some probability. → Mind the gap! preserve the gap. How? CSP - formula, which is satisfiable or far from satisfiable. Remember: Equivalent! Proof contains a 3 t 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. Blow up the gap. Idea: Verifier is a polynomial sized circuit. (blows up the alphabet.) Are the two colors the same? Use PCP view to reduce the alphabet. (Exponential blowup for alphabet size.) Recursively construct PCP for verifier with smaller alphabet! (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.