1. CS151 Complexity Theory Lecture 13 May 11, 2004

2. CS151 Lecture 132 Outline Natural complete problems for PH and PSPACE proof systems interactive proofs and their power

3. May 11, 2004CS151 Lecture 133 Simpler version of MIN DNF Theorem (U): MIN DNF is Σ 2 -complete. we’ll consider a simpler variant: –IRREDUNDANT: given DNF φ, integer k; is there a DNF φ’ consisting of at most k terms of φ computing same function φ does?

4. May 11, 2004CS151 Lecture 134 Simpler version of MIN DNF analogy with an NP-complete problem: –SET COVER: given subsets S 1,S 2,...,S m  U, integer k, is there a collection of at most k sets that cover U. U sets S i {0,1} n φ -1 (0) φ -1 (1) terms of φ

5. May 11, 2004CS151 Lecture 135 SSC is Σ 2 -complete “sets” in IRREDUNDANT lie in an exponentially larger universe; they are represented succinctly by terms of φ helpful intermediate problem: –SUCCINCT SET COVER (SSC): given 3- DNFs S = {φ 1, φ 2, φ 3,..., φ m } on n variables, integer k; is there a collection S’  S of size at most k for which  φ  S’  1 (S’ is a cover)?

6. May 11, 2004CS151 Lecture 136 SSC is Σ 2 -complete Theorem: SSC is Σ 2 -complete. Proof: –in Σ 2 (why?) “  S’  S  x [  φ  S’ (x) = 1]” –reduce from QSAT 2 –instance:  A  B φ(A, B) = 1 –assume |A| = |B| = n

7. May 11, 2004CS151 Lecture 137 SSC is Σ 2 -complete  A  B φ(A, B) = 1 Proof (continued): –2 new sets of variables S, T –|S| = |T| = n –Define: wt(S, T) = # of 1s in S and T together –“(S,T) encodes A” means  i (s i =a i )  (t i =  a i )

8. May 11, 2004CS151 Lecture 138 SSC is Σ 2 -complete –From φ(A, B) we define function f(S, T, B): 0 if wt(S, T) < n 0 if wt(S, T) = n and (S,T) does not encode any A 0 if wt(S, T) = n and (S,T) encodes A for which φ(A, B) = 0 1 if wt(S, T) = n and (S,T) encodes A for which φ(A, B) = 1 1 if wt(S, T) > n –verify: poly(n) size circuit C computes f –verify: f is monotone in S and T

9. May 11, 2004CS151 Lecture 139 SSC is Σ 2 -complete Proof (continued): –produce an instance of SSC: # of sets m = 2n + 1 φ i = (  s i ) φ i+n = (  t i ) from C get a 3-DNF φ m (S,T,B,W): f(S,T,B) = 1   W φ m (S,T,B,W) = 1 0 if wt(S, T) < n 0 if wt(S, T) = n and (S,T) does not encode any A 0 if wt(S, T) = n and (S,T) encodes A : φ(A,B) = 0 1 if wt(S, T) = n and (S,T) encodes A : φ(A,B) = 1 1 if wt(S, T) > n

10. May 11, 2004CS151 Lecture 1310 SSC is Σ 2 -complete Claim:  A  B φ(A, B) = 1 implies S’ = {φ i | a i = 1}  {φ i+n | a i = 0}  {φ m } is a cover of size n+1. Proof: consider each fixed point (S,T,B,W) if there exists (S’,T’) that encodes A and we have (S’,T’)  (S,T) then φ m (S,T,B,W) = 1 else,  i (a i = 1 and s i = 0) or (a i = 0 and t i = 0) 0 if wt(S, T) < n 0 if wt(S, T) = n and (S,T) does not encode any A 0 if wt(S, T) = n and (S,T) encodes A : φ(A,B) = 0 1 if wt(S, T) = n and (S,T) encodes A : φ(A,B) = 1 1 if wt(S, T) > n =(  s i ) =(  t i )

11. May 11, 2004CS151 Lecture 1311 SSC is Σ 2 -complete Claim: cover S’ of size n+1 implies  A  B φ(A, B) = 1 Proof: –S’ must contain φ m ; otherwise it fails to cover the all-ones point –consider the pair (S*, T*) for which: s i = 1 if φ i  S’ and 0 otherwise t i = 1 if φ i+n  S’ and 0 otherwise –must have:  B  W φ m (S*, T*, B, W) = 1 –implies:  B f(S*, T*, B) = 1

12. May 11, 2004CS151 Lecture 1312 SSC is Σ 2 -complete –defined the pair (S*, T*) as follows: s i = 1 if φ i  S’ and 0 otherwise t i = 1 if φ i+n  S’ and 0 otherwise –concluded:  B f(S*, T*, B) = 1 –Note: wt(S*, T*) = n –(S*,T*) must encode A s.t.  B φ(A, B) = 1 –Conclude:  A  B φ(A, B) = 1 0 if wt(S, T) = n and (S,T) does not encode any A 0 if wt(S, T) = n and (S,T) encodes A : φ(A,B) = 0 1 if wt(S, T) = n and (S,T) encodes A : φ(A,B) = 1

13. May 11, 2004CS151 Lecture 1313 IRR is Σ 2 -complete Recall: IRREDUNDANT: given DNF φ, integer k; is there a DNF φ’ consisting of at most k terms of φ computing same function φ does? Theorem: IRR is Σ 2 -complete. Proof: –in Σ 2 : “  φ’  x [φ’(x) = φ(x)]”

14. May 11, 2004CS151 Lecture 1314 IRR is Σ 2 -complete –reduce from SSC –instance: S = {φ 1, φ 2, φ 3,..., φ m } –may assume φ 1, φ 2,..., φ m-1 single literals φ m necessary in any cover S is a cover –write out terms: φ m = t 1  t 2  t 3  …  t n –produce an instance of IRR: φ =  i=1…n (z 1 …z i-1 z i+1 …z n t i )   j=1…m-1 (z 1 …z n φ j )

15. May 11, 2004CS151 Lecture 1315 IRR is Σ 2 -complete S = {φ 1, φ 2, φ 3,..., φ m = t 1  …  t n } φ =  i=1…n (z 1 …z i-1 z i+1 …z n t i )   j=1…m-1 (z 1 …z n φ j ) Proof (continued): Claim: if S’  S is a cover of size k then φ’=  i=1…n (z 1 …z i-1 z i+1 …z n t i )   j<m, φ j  S’ (z 1 z 2 …z n φ j ) is equivalent to φ and has k+n-1 terms. Proof: by cases

16. May 11, 2004CS151 Lecture 1316 IRR is Σ 2 -complete S = {φ 1, φ 2, φ 3,..., φ m = t 1  …  t n }; S’  S φ =  i=1…n (z 1 …z i-1 z i+1 …z n t i )   j=1…m-1 (z 1 …z n φ j ) φ’=  i=1…n (z 1 …z i-1 z i+1 …z n t i )   j<m, φ j  S’ (z 1 z 2 …z n φ j ) –more than one z variable 0: both φ’, φ are 0 –z i 0, other z’s 1: φ’, φ equivalent to t i –all z’s 1: φ’ equivalent to  φ j  S’ (z 1 z 2 …z n φ j ) φ’ equivalent to  φ j  S (z 1 …z n φ j ) S’ is a cover implies both equivalent to 1

17. May 11, 2004CS151 Lecture 1317 IRR is Σ 2 -complete S = {φ 1, φ 2, φ 3,..., φ m = t 1  …  t n } φ =  i=1…n (z 1 …z i-1 z i+1 …z n t i )   j=1…m-1 (z 1 …z n φ j ) Proof (continued): Claim: if φ’  φ uses k+n-1 terms of φ, then there exists a cover S’ of size k Proof: –each “t i term” of φ must be present

18. May 11, 2004CS151 Lecture 1318 IRR is Σ 2 -complete S = {φ 1, φ 2, φ 3,..., φ m = t 1  …  t n } φ =  i=1…n (z 1 …z i-1 z i+1 …z n t i )   j=1…m-1 (z 1 …z n φ j ) φ’=  i=1…n (z 1 …z i-1 z i+1 …z n t i )  ??? (k+n-1 terms total) –other k-1 terms all involve some φ j –let S’ be these φ j together with φ m (  φ j  S’ φ j )  φ’ z  11…1  φ z  11…1  (  φ j  S φ j )  1 –conclude S’ is a cover of size k

19. May 11, 2004CS151 Lecture 1319 PSPACE General phenomenon: many 2-player games are PSPACE-complete. –2 players I, II –alternate pick- ing edges –lose when no unvisited choice pasadena athens auckland san francisco oakland davis GEOGRAPHY = {(G, s) : G is a directed graph and player I can win from node s}

20. May 11, 2004CS151 Lecture 1320 PSPACE Theorem: GEOGRAPHY is PSPACE- complete. Proof: –in PSPACE easily expressed with alternating quantifiers –PSPACE-hard reduction from QSAT

21. May 11, 2004CS151 Lecture 1321 PSPACE  x 1  x 2  x 3 …  x n φ(x 1, x 2, …, x n )? truefalse variable gadget for x i C 1 C 2 C m clause choice gadget …

22. May 11, 2004CS151 Lecture 1322 PSPACE  x 1  x 2  x 3 …(  x 1  x 2  x 3 )  (  x 3  x 1 )  …  (x 1  x 2 ) true false truefalse truefalse … x1x1 x2x2 x3x3 I II I I I I alternately pick truth assignment pick a clause I II I pick a true literal?

23. May 11, 2004CS151 Lecture 1323 Proof systems L = { (A, 1 k ) : A is a true mathematical assertion with a proof of length k} New topic: What is a “proof”? complexity insight: meaningless unless can be efficiently verified

24. May 11, 2004CS151 Lecture 1324 Proof systems given language L, goal is to prove x  L proof system for L is a verification algorithm V –completeness: x  L   proof, V accepts (x, proof) “true assertions have proofs” –soundness: x  L   proof*, V rejects (x, proof*) “false assertions have no proofs” –efficiency:  x, proof, V(x, proof) runs in polynomial time in |x|

25. May 11, 2004CS151 Lecture 1325 Classical Proofs previous definition: “classical” proof system recall: L  NP iff expressible as L = { x |  y, |y| < |x| k, (x, y)  R } and R  P. NP is the set of languages with classical proof systems (R is the verifier)

26. May 11, 2004CS151 Lecture 1326 Interactive Proofs Two new ingredients: –randomness: verifier tosses coins, errs with some small probability –interaction: rather than “reading” proof, verifier interacts with computationally unbounded prover NP proof systems lie in this framework: prover sends proof, verifier does not use randomness

27. May 11, 2004CS151 Lecture 1327 Interactive Proofs interactive proof system for L is an interactive protocol (P, V) ProverVerifier...... common input: x accept/ reject # rounds = poly(|x|)

28. May 11, 2004CS151 Lecture 1328 Interactive Proofs interactive proof system for L is an interactive protocol (P, V) –completeness: x  L  Pr[V accepts in (P, V)(x)]  2/3 –soundness: x  L   P* Pr[V accepts in (P*, V)(x)]  1/3 –efficiency: V is p.p.t. machine repetition: can reduce error to any ε

29. May 11, 2004CS151 Lecture 1329 Interactive Proofs IP = {L : L has an interactive proof system} Observations/questions: –philosophically interesting: captures more broadly what it means to be convinced a statement is true –clearly NP  IP. Potentially larger. How much larger? –if larger, randomness is essential (why?)

30. May 11, 2004CS151 Lecture 1330 Graph Isomorphism graphs G 0 = (V, E 0 ) and G 1 = (V, E 1 ) are isomorphic (G 0  G 1 ) if exists a permutation π:V  V for which (x, y)  E 0  (π(x), π(y))  E 1 1 2 3 4 1 2 4 3 1 4 3 2

31. May 11, 2004CS151 Lecture 1331 Graph Isomorphism GI = {(G 0, G 1 ) : G 0  G 1 } –in NP –not known to be in P, or NP-complete GNI = complement of GI –not known to be in NP Theorem (GMW): GNI  IP –indication IP may be more powerful than NP

32. May 11, 2004CS151 Lecture 1332 GNI in IP interactive proof system for GNI: ProverVerifier input: (G 0, G 1 ) flip coin c  {0,1}; pick random π H = π(G c ) if H  G 0 r = 0, else r = 1 r accept iff r = c

33. May 11, 2004CS151 Lecture 1333 GNI in IP completeness: –if G 0 not isomorphic to G 1 then H is isomorphic to exactly one of (G 0, G 1 ) –prover will choose correct r soundness: –if G 0  G 1 then prover sees same distribution on H for c = 0, c = 1 –no information on c  any prover P* can succeed with probability at most 1/2