SLIDE 1
Projects.
5 minute presentations. Class times over RRR week. Will post schedule. Tests submitted by Monday (mostly) graded. Rest will be done this afternoon.
Probabilistically Checkable Proofs...
A real theory. A tiny introduction... ...and one idea ...using random walks.
Probabilistically Checkable Proofs
What’s a proof? Statement: A formula φ is satisfiable. Proof: assignment, a. Property: Can check proof in polynomial time. Polytime Verifier: V (π) There is π for satisfiable φ where V(π) says yes. For unsatisfiable φ, V(π) always says no. Can prove any statement in NP . 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
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. What is the maximum size of a proof? (A) O(q(n)). (B) O(q(n)r(n)). (C) O(r(n)2q(n)). (D) O(q(n)2r(n)). (D) Only 2r(n) different runs. Look at q(n) bits in each possible run. 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. A language is in NP if it there is a polynomial time proof system. A language is in PCP(r(n),q(n)) if it has a probabilistically checkable proof system: r(n),q(n). Theorem: NP ⊂ PCP(O(log(n)),3). Only checks 3 places !!! How is this possible? Well..its not easy.
Example.
Graph Not Isomorphism: (G1,G2). There is no way to permute the vertices of G1 to get G2. Give a PCP(poly(n), 1). (Weaker than PCP(O(logn), 3). Exponential sized proof is allowed. Can only look at 1 place, though. Any thoughts? Design a proof format: π For every labelled graph, H, π[H] = 1 if permutation of G1 π[H] = 2 if permutation of G1 Don’t care if either. Verifier: randomly choose x ∈ {1,2}, permute Gx to get H Is π(H) equal to x? For not equal, π exists. If G1 = G2, any prover has 1/2 probability of mistake! Cool.......but exponential ..not so interesting of a proof. Lots of ideas to get to O(logn) bits and 3 query bits.
SLIDE 2 Another view.
q-Constraint satisfaction problem: Formula. Clauses of length q. q-ary truth table for each constraint. Is the formula satisfiable. Variable types can be.. Binary Variables are [0,1] Large alphabet variables are [0,...,k] val(φ) is fraction of satisfiable clauses. val(φ) = 1 ≡ φ satisfiable Example:
- 3SAT. q = 3, and truth table for “or” of 3 variables.
The ρ-Gap q-CSP Problem: φ 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
Gap CSP
Theorem: There are constants q and ρ < 1 where ρ-Gap q-CSP is NP-complete Just another NP-complete problem... Well, a bit more. Not only is CSP NP-complete, it is NP-hard to approximate within factor of 1? 1/ρ? ρ? Within a factor of 1/ρ ...and Theorem 1 is same statement that NP ⊆ PCP(O(logn),O(1)).
Two views of PCP .
Theorem A: There is a constant where 1
2-Gap q-CSP is NP-complete
Theorem B: PCP(O(logn),O(1)) ⊆ NP Note: A = ⇒ B. There is 1
2-Gap q-CSP is in PCP(O(logn),O(1)).
Given φ. Proof format: assignment to variables. if satisfiable π(φ) = is satisfying assignment. Verifier checks random clause and looks up variables. if there is π where Verifier accepts with probability ≥ 1/2. val(φ) ≥ 1
2
Plus Theorem A any problem in NP in PCP(O(logn),O(1)).
Reminder: Cook/Levin theorem.
Cook/Levin Theorem: SAT is NP-complete. Proof Idea: Any problem A has polytime verifier. Write out circuit for program for input x ∈ A. Convert it to a formula. ish
The other way: PCP → CSP
Theorem A: There are constants q and ρ where 1
2-Gap q-CSP is
NP-complete Theorem B’: PCP(O(logn),O(1)) ⊆ NP B = ⇒ A′ Is x ∈ L? Construct formula φx. PCP proof system: π = π0 ...πs, s = O(poly(n)). Verifier: yes/no based on constant number bits. Verifer: constant sized clause
- n constant bits, πi,πj,πk,...
for each random string. 2r(n) different random strings poly(n) clauses. 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, φx is satisfiable. x ∈ L, val(φx) < 1/2.
Gap amplification.
CSP is NP-complete. Recall 3SAT is a CSP . GAP-CSP? Given a CSP instance, φ, produce an instance of GAP CSP , φG. φ satisfiable → φG satisfiable. Note: φ is not satifiable, val(φ) < 1 → val(φ) ≤ 1−1/m. Gap: 1/m. φ is not satifiable → val(φG) < 1−∆. Gap: ∆. Grow ∆ >> 1/µ.
SLIDE 3 Detour: Expanders.
For a graph G. Pr(u,v)∈E[u ∈ S,v ∈ S] ≤ |S|
|V|( 1 2 +λ2(G))
Gℓ has an edge for every ℓ-step path in G. since adjacency matrix of Gℓ is Aℓ where A is adjacency matrix of G. → λ2(Gℓ) = λ ℓ
2(G)
Pr(u,v)∈E[u ∈ S,v ∈ S] ≤ |S|
|V|
1
2 +λ2(Gℓ)
- Ramanujan Graphs: there exists degree d graphs with λ < 2/
√ d.
Expanders and gap amplification.
Consider “GAP” k independent set. Either there is an independent set of size 2k.
- r every set is of size at most k.
This is NP-complete. Lemma: Given F, with indendepent set size α(F), there is a polytime algorithm to find G where (α(F)−2λ)logn ≤ α(G) ≤ (α(F)+2λ)logn. Construction: Add graph on F with eigenvalue λ. Construct logn length paths in F, Connect paths x1,...,xlogn with y1,...,ylogn if any (xi,yj) in F. Argument idea: Leave ind. set. |S| w/prob
|S| 2|V| (1+λ) in each step.
→ max size of independent set is a power of logn. Some details to work through...
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 coloring → val(φ) ≤ 1−1/m. Gap: 1/m. Step 1: Produce G′ 3-colorable G → 3t-colorable G′ G has gap of ∆ → G′ has a gap Ω(t∆) for any 3t-coloring. Construction: given graph G, add expander H on same node set. Make graph G′ connecting t-length paths in graph. Idea: bad edges spread out and make many bad edges. infection of uncolorability. Use expansion. No bad edges, no infection. Gap amplification. However.... Changed problem! 3-coloring to 3t-coloring. Alphabet size from 3 to 3t. Need to reduce alphabet size. A, B, C! 1, 2, 3!
Alphabet reduction.
Clauses (colorability) on variables with lots of possible values. → formula on variables with few possible values. → Mind the gap! preserve the gap. How? Remember: Proof contains a 3t coloring. Verifier chooses random edge and checks colors. Cheating proof can only get away ∆ times. (Dude, the gap!) Idea: Verifier is a polynomial sized circuit. Are the two colors the same? Recursively construct PCP for verifier with smaller alphabet! Lose some of the gap. This PCP? Prover: Dude, there is a circuit that checks! Here is a proof that the circuit works. 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.
Summary
PCP - proof, check a few places to catch errors with some probability. CSP - formula, which is satisfiable or far from satisfiable. Equivalent! Proving both are NP-hard Start with CSP Blow up the gap. (blows up the alphabet.) Use PCP view to reduce the alphabet. (Exponential blowup for alphabet size.) (Ok, since number of colors are constant.) Lots of cool ideas here. 276: Next spring. Prasad Raghavendra.
See you ...
Next week. ...for projects.
