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1. Messages from the organisers Thursday afternoon: workworkwork Confirm dinner tonight Wear your name tags!

2. Improving Exhaustive Search Implies Superpolynomial Lower Bounds Ryan Williams IBM Almaden TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AAAAAAAAAA

3. Two Difficult Areas of Research Faster Algorithms for NP Given:Verifier V(x, y) that reads w(|x|) bits of witness y, and runs in t(|x|) time. Find: Deterministic algorithm which: Runs in less than2w(|x|) t(|x|) time Given any input x, finds a witness y s.t. V(x,y) accepts (or conclude there is none) • Circuit Lower Bounds • Given: Any NP problem • (or EXPNP problem!) • Find: Sequence of algorithms {An} such that: • |An| ·nk +k • On all inputs x of length n, An(x) correctly solves the problem in O(nk) time. • (Alternatively, prove that no such algorithms exist!)

4. One Seems Easier Than The Other... Faster Algorithms for NP Given:Verifier V(x, y) that reads w(|x|) bits of witness y, and runs in t(|x|) time. Find: Deterministic algorithm which: Runs in less than2w(|x|) t(|x|) time Given any input x, finds a witness y s.t. V(x,y) accepts (or conclude there is none) • Circuit Lower Bounds • Given: Any NP problem • (or EXPNP problem!) • Find: Sequence of algorithms {An} such that: • |An| ·nk +k • On all inputs x of length n, An(x) correctly solves the problem in O(nk) time. • (Alternatively, prove that no such algorithms exist!)

5. One Seems Easier Than The Other... • Faster Algorithms for NP • 3SAT: O(1.321n) time • kSAT: O(2n - n/k) • Hamiltonian Path: O(1.66n) • Max Indep. Set: O(1.21n) • on degree-3 graphs: O(1.16n) • 3-Coloring: O(1.33n) • k-Coloring: O(2n) • Non-Uniform Circuits for NP • We don’t know how to get non-uniform algs that are better than the uniformalgs • - Best lower bound known: • There is a function in NP that requires circuits of size 5n + o(n) • Could be that even EXPNP has poly(n) size circuits… • Circuit Lower Bounds • Given: Any NP problem • (or EXPNP problem!) • Find: Sequence of algorithms {An} such that: • |An| ·nk +k • On all inputs x of length n, An(x) correctly solves the problem in O(nk) time. • (Alternatively, prove that no such algorithms exist!)

6. One Seems Easier Than The Other... • Faster Algorithms for NP • 3SAT: O(1.321n) time • kSAT: O(2n - n/k) • Hamiltonian Path: O(1.66n) • Max Indep. Set: O(1.21n) • on degree-3 graphs: O(1.16n) • 3-Coloring: O(1.33n) • k-Coloring: O(2n) • Circuit Lower Bounds • We don’t know how to get non-uniform algorithms that are better than these uniform ones • - Best lower bound known: • There is a function in NP that requires circuits of size 5n + o(n) • Could be that even EXPNP has poly(n) size circuits…

7. Faster Algorithms ⟹ Lower Bounds! • Faster Algorithms • Deterministic algorithm for: • Circuit SAT in O(2n/a(n)) (n inputs and nk gates) • Formula SAT in O(2n/a(n)) • ACC SAT in O(2n/3/a(n)) • Given a circuit that’s either unsatisfiable, or has at least2n-1 satisfying assignments, determine which is the case inO(2n/a(n)) time • (This problem is in BPP!) • Circuit Lower Bounds • Would imply : • NEXP * P/poly • EXPNP*non-uniformNC1 • EXPNP*non-uniformACC • NEXP * P/poly

8. Further Improvements?... Faster Algorithms An algorithm for: Circuit SAT with n inputs and nk gates that runs in O(2n/a(n)) time and O(2n/a(n)) space Non-Uniform Circuits Would imply : NEXP * P/poly But the trivial algorithm uses O(2n) time and O(nk) space…What happens if we can always improve on exhaustive search in O(2.99n) time and O(poly(nk)) space?

9. Major Consequences… Theorem Let Π be a problem with log n length witnesses that can be verified in O(na) time and poly(log n) space. A trivial deterministic algorithm for Π takesO(na+1) time, poly(log n) space to find a witness Suppose every such problem Π can be solved in O(na+.999) time and poly(log n) space. Then every such Π can be solved in Nondeterministic O(n2) time. Corollary Either LOGSPACE  NP, or the trivial exhaustive search algorithm is optimal for some search problem Π.

10. Fast Circuit SAT ) NEXP Lower Bounds Theorem 1 Let a(n) ¸ n!(1) Circuit SAT in O(2n/a(n)) ) NEXP * P/poly. Proof Idea: Show that if we had both Circuit SAT in O(2n/a(n)) and NEXP µ P/poly then NTIME[2n] µNTIME[o(2n)], contradicting the nondeterministic time hierarchy. For every L 2NTIME[2n], use small circuits for L to design a quick nondeterministic simulation of L that makes one call to Circuit SAT on n + O(log n) variables.

11. Theorem 1 Let a(n) ¸ n!(1). Circuit SAT in O(2n/a(n)) ) NEXP * P/poly. For every L 2NTIME[2n], use small circuits for L to design a quick nondeterministic simulation of L that makes one call to Circuit SAT on n + O(log n) variables. Succinct 3SAT:Given a circuit C, let F be the string obtained by evaluating C on all possible assignments in lexicographical order. Does F encode a satisfiable 3CNF formula? Theorem [PY]Succinct 3SAT is NEXP-complete. Def: Succinct 3SAT has succinct satisfying assignmentsif for every C that encodes a satisfiable 3CNF F, there is another circuit D of poly(|C|) size such that D encodes an assignment A which satisfies F. Evaluating D on all of its possible assignments generates A.

12. Theorem 1 Let a(n) ¸ n!(1). Circuit SAT in O(2n/a(n)) ) NEXP * P/poly. For every L 2NTIME[2n], use small circuits for L to design a quick nondeterministic simulation of L that makes one call to Circuit SAT on n + O(log n) variables. Two Lemmas: Lemma 1 [Follows from IKW’02] If NEXP µ P/poly then Succinct 3SAT has succinct satisfying assignments. The proof applies results on “hardness versus randomness” 1. If NEXP µ P/poly then NEXP = MA. 2. If Succinct 3SAT does not have succinct witnesses then can derandomize MA infinitely often with n bits of advice: NEXP = MA µi.o.-NTIME[2n]/n µi.o.-SIZE(nk)(a contradiction)

13. Theorem 1 Let a(n) ¸ n!(1). Circuit SAT in O(2n/a(n)) ) NEXP * P/poly. For every L 2NTIME[2n], use small circuits for L to design a quick nondeterministic simulation of L that makes one call to Circuit SAT on n + O(log n) variables. Two Lemmas: Lemma 1 [Follows from IKW’02] If NEXP µ P/poly then Succinct 3SAT has succinct satisfying assignments. RemarkThe lemma is easier if we replace “NEXP µ P/poly” with “EXPNP µ P/poly”. The following can be done in EXPNP: On input (C, i), find the lexicographically first satisfying assignment to F encoded by C, then output its i-th bit. Hence there are poly(|C|) circuits that take (C, i) and output the ith bit of a satisfying assignment to F encoded by C.

14. Theorem 1 Let a(n) ¸ n!(1). Circuit SAT in O(2n/a(n)) ) NEXP * P/poly. For every L 2NTIME[2n], use small circuits for L to design a quick nondeterministic simulation of L that makes one call to Circuit SAT on n + O(log n) variables.Two Lemmas: Lemma 1 [Follows from IKW’02] If NEXP µ P/poly then Succinct 3SAT has succinct satisfying assignments. Lemma 2 [FLvMV ’05] For all L 2NTIME(2n), there is a polytime reduction RL from L to Succinct 3SAT such that: - x 2 L ⇔ RL(x) = Cx encodes a satisfiable 3CNF formula - Cx has size at most poly(n) - Cx has at most n + O(log n) inputs

15. Theorem 1 Let a(n) ¸ n!(1). Circuit SAT in O(2n/a(n)) ) NEXP * P/poly. Proof: Let L 2NTIME[2n] be arbitrary. We want to produce a faster NTIME algorithm for L. - By Lemma 2, there is a reduction RL from L to Succinct 3SAT. Given x, let Cx be the circuit generated by RL(x), and let φx be the exponentially long formula encoded by Cx - By Lemma 1, Cx has succinct satisfying assignments: there is a poly(|x|) size circuit Y such that Y(i) outputs the i-th bit of a satisfying assignment for φx. Now we will give an NTIME[o(2n)] algorithm for L

16. A o(2n) algorithm for L 2NTIME[2n] Given input x of length n, Guess a circuit Y of poly(n) size (Y(j) is supposed to output the j-th bit of a satisfying assignment for φx ) Construct the following circuit D which has polynomial size: D outputs 1 iff the assignment encoded by Y does not satisfy the i-th clause of φx Constant size circuit Outputs assignments to the variables x, y, z Y Y Y s x t y u z Cx Outputs the ith clause of φx in + O(log n) input bits Using CircuitSAT algorithm: determine satisfiability of D in o(2n) time

17. Weak Derandomization Suffices Theorem 2 Let a(n) ¸ n!(1). Suppose we are given a circuit C and are promised that it is either unsatisfiable, or at least ½ of its assignments are satisfying. Determine which.If this problem is in O(2n/a(n)) time then NEXP * P/poly. Proof Idea: Same as Theorem 1, but replace the succinct reduction RL from L to 3SAT with a succinct PCP reduction Lemma 3[BGHSV’05]For all L 2NTIME(2n),there is a reduction SL from L to MAX CSP such that: x 2 L ) All constraints of SL(x) are satisfiable x  L ) At most ½ of the constraints are satisfiable 1. |SL(x)| = 2n poly(n) 2. The i-th constraint of SL(x) is computable in poly(n) time.

18. A Major Consequence of Improving Exhaustive Search Theorem 3 Let Π be a problem with log n length solutions that can be verified in O(na) time and poly(log n) space. Suppose for all a, every such problem Π can be solved in O(na+.999) time and poly(log n) space. Then every such Π can be solved in Nondeterministic O(n3) time. Corollary Either LOGSPACE  NP, or the trivial algorithm is optimal for some Π.

19. Why the Corollary is True Corollary Either LOGSPACE  NP, or the trivial algorithm is optimal for some Π. (Recall: Π is a problem with log n length solutions that can be verified in O(na) time and poly(log n) space.) Proof:Suppose LOGSPACE = NP, and that every such Π can be solved in O(na+.999) time and poly(log n) space. By Theorem 3, every Π2NTIME[n3]. But this would imply NP = LOGSPACE µ NTIME[n2], Which contradicts the nondeterministic time hierarchy.

20. Theorem 3 Let Π be a problem with log n length solutions that can be verified in O(na) time and poly(log n) space Suppose every such Π is O(na+.999) time, poly(log n) space Then every such Π is in nondeterministic O(n2) time Proof Idea:Use tools from SAT time-space lower bounds. Every such Π can be simulated by a §2 machine which existentially guesses n bits, then universally guesses log n bits, then runs in O(na-1) time and poly(log n) space. By the assumption, this machine can be simulated by a nondeterministic algorithm which guesses n bits then runs in O(na-1+.999) = O(na-.001) time and poly(log n) space Recurse!Apply the same observation to this slightly faster computation, reducing its runtime by a tiny amount each time. After finitely many times, get down to O(n2).

21. Open Problems • Could new lower bounds follow from algorithms for other problems, e.g., Subgraph Isomorphism? • Is it possible that circuit lower bounds could help provide faster algorithms for NP? • Can we prove one of: - LOGSPACE  NP - Exhaustive search is optimal for some problems

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