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Computability & Complexity II Chris Umans Caltech
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June 25, 2004CBSSS2 Complexity Theory Classify problems according to the computational resources required –running time –storage space –parallelism –randomness –rounds of interaction, communication, others… Attempt to answer: what is computationally feasible with limited resources?
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June 25, 2004CBSSS3 The central questions Is finding a solution as easy as recognizing one? P = NP? Is every sequential algorithm parallelizable? P = NC? Can every efficient algorithm be converted into one that uses a tiny amount of memory? P = L? Are there small Boolean circuits for all problems that require exponential running time? EXP P/poly? Can every randomized algorithm be converted into a deterministic algorithm one? P = BPP?
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June 25, 2004CBSSS4 Central Questions We think we know the answers to all of these questions … … but no one has been able to prove that even a small part of this “world-view” is correct. If we’re wrong on any one of these then computer science will change dramatically
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June 25, 2004CBSSS5 Outline We’ll look at three of these problems: –P vs. NP “the power of nondeterminism” –L vs. P “the power of sequential computation” –P vs. BPP “the power of randomness”
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June 25, 2004CBSSS6 P vs. NP
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June 25, 2004CBSSS7 Completeness In an ideal world, given language L 1. state an algorithm deciding 2. prove that no algorithm does better we are pretty good at part 1 we are currently completely helpless when it comes to part 2, for most problems that we care about
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June 25, 2004CBSSS8 Completeness in place of part 2 we can –relate the difficulty of problems to each other via reductions –prove that a problem is a “hardest” problem in a complexity class via completeness powerful, successful surrogate for lower bounds
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June 25, 2004CBSSS9 Recall: reductions “many-one” reduction: –now, require f computable in polynomial time yes no yes no AB reduction from language A to language B f f
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June 25, 2004CBSSS10 Completeness complexity class C (set of languages) language L is C-complete if –L is in C –every language in C reduces to L formalizes “L is hardest problem in complexity class C”: –L in P implies every language in C is in P related concept: language L is C-hard if –every language in C reduces to L
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June 25, 2004CBSSS11 Some problems we care about 3SAT = { : 3-CNF is satisfiable} TSP = { : there is a tour of all nodes of G with weight ≤ k} SUBSET-SUM = { : some subset of {x 1, x 2, … x n } sums to B} IS= { : there is a subset of at least k vertices of G with no edges between them} Only know exponential time algorithms for these problems
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June 25, 2004CBSSS12 From last week… We have defined the complexity classes P (polynomial time), EXP (exponential time) all languages decidable RE HALT EXP P some language
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June 25, 2004CBSSS13 Why not EXP 3-SAT, TSP, SUBSET-SUM, IS, … Why not show these are EXP-complete? all possess positive feature that some problems in EXP probably don’t have: a solution can be recognized in polynomial time
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June 25, 2004CBSSS14 NP Definition: NP = all languages decidable by a Nondeterministic TM in Polynomial time Theorem: language L is in NP if and only if it is expressible as: L = { x | y, |y| ≤ |x| k, (x, y) R } where R is a language in P. poly-time TM M R deciding R is a “verifier” “witness” or “certificate” efficiently verifiable
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June 25, 2004CBSSS15 Poly-time verifiers Example: 3SAT expressible as 3SAT = {φ : φ is a 3-CNF formula for which assignment A for which (φ, A) R} R = {(φ, A) : A is a sat. assign. for φ} –satisfying assignment A is a “witness” of the satisfiability of φ (it “certifies” satisfiability of φ) –R is decidable in poly-time
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June 25, 2004CBSSS16 all languages EXP decidable NP P P NP EXP believe all containments proper assuming P ≠ NP, can prove a problem is hard by showing it is NP-complete
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June 25, 2004CBSSS17 NP-complete problems L is NP-complete if –L is in NP –every problem in NP reduces to L Theorem (Cook): 3SAT is NP-complete. –opened door to showing thousands of problems we care about are NP-complete
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June 25, 2004CBSSS18 NP-complete problems Example reduction: –3SAT = { φ : φ is a 3-CNF Boolean formula that has a satisfying assignment } (3-CNF = AND of OR of ≤ 3 literals) –IS = { | G is a graph with an independent set V’ V of size ≥ k } (ind. set = set of vertices no 2 of which are connected by an edge)
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June 25, 2004CBSSS19 Ind. Set is NP-complete The reduction f: given φ = (x y z) ( x w z) … (…) we produce graph G φ : x y zz x x wz one triangle for each of m clauses edge between every pair of contradictory literals set k = m …
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June 25, 2004CBSSS20 Ind. Set is NP-complete φ = (x y z) ( x w z) … (…) Claim: φ has a satisfying assignment if and only if G has an independent set of size at least k –Proof? IS NP. Reduction shows NP-complete. x y zz x x wz …
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June 25, 2004CBSSS21 Interlude: Transforming Turing Machine computations into Boolean circuits
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June 25, 2004CBSSS22 Configurations Useful convention: Turing Machine configurations. Any point in computation represented by string: C = x 1 x 2 … x i q x i+1 x i+2 … x m start configuration for single-tape TM on input x: q start x 1 x 2 …x n x1x1... state = q x2x2 …xixi x i+1 …xmxm
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June 25, 2004CBSSS23 Turing Machine Tableau Tableau (configurations written in an array) for machine M on input x=x 1 …x n : x1/qsx1/qs x2x2 …xnxn _ … x1x1 x2/q1x2/q1 …xnxn _ … x1/q1x1/q1 a…xnxn _ … _/q a _…__ …............ height = time taken width = space used
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June 25, 2004CBSSS24 Turing Machine Tableau Important observation: contents of cell in tableau determined by 3 others above it: a/q 1 ba b/q 1 a a a aba b
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June 25, 2004CBSSS25 Turing Machine Tableau Can build Boolean circuit STEP –input (binary encoding of) 3 cells –output (binary encoding of) 1 cell ab/q 1 a a STEP each output bit is some function of inputs can build circuit for each size is independent of size of tableau
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June 25, 2004CBSSS26 Turing Machine Tableau w copies of STEP compute row i from i-1 x1/qsx1/qs x2x2 …xnxn _ … x1x1 x2/q1x2/q1 …xnxn _ …............ width w tableau for M on input x … … STEP
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June 25, 2004CBSSS27 x 1/ q s x2x2 …xnxn _ … STEP............ 1 iff cell contains q accept ignore these Circuit C built from TM M C(x) = 1 iff M accepts input x x1x1 x2x2 xnxn Size = O(width x height)
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June 25, 2004CBSSS28 Back to NP-completeness
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June 25, 2004CBSSS29 Circuit-SAT is NP-complete Circuit SAT: given a Boolean circuit (gates , , ), with variables y 1, y 2, …, y m is there some assignment that makes it output 1? Theorem: Circuit SAT is NP-complete. Proof: –clearly in NP
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June 25, 2004CBSSS30 NP-completeness –Given L NP of form L = { x | y such that (x, y) R } x1x1 x2x2 …xnxn y1y1 y2y2 …ymym circuit produced from TM deciding R 1 iff (x,y) R –hardwire input x; leave y as variables
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June 25, 2004CBSSS31 3SAT is NP-complete Idea: auxiliary variable for each gate in circuit x 1 x 2 x 3 x 4 …x n gigi (g i z) ( z g i ) (z g i ) z gigi ( z 1 g i ) ( z 2 g i ) ( g i z 1 z 2 ) (z 1 z 2 g i ) z 1 z 2 gigi ( g i z 1 ) ( g i z 2 ) ( z 1 z 2 g i ) (z 1 z 2 g i ) z 1 z 2
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June 25, 2004CBSSS32 P vs. NP Most “hard” problems we care about have turned out to be NP-complete –L is in NP –every problem in NP reduces to L Not known that P ≠ NP arguably most significant open problem in Computer Science and Math.
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June 25, 2004CBSSS33 L vs. P
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June 25, 2004CBSSS34 Multitape TMs A useful variant: k-tape TM 0 11001 1 10 1 00 q0q0 read-only input tape finite control … k read/write heads 0 11001 k-1 “work tapes” … 0 11001 1 10 1 00 … 0 … …
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June 25, 2004CBSSS35 Space complexity SPACE(f(n)) = languages decidable by a multi-tape TM that touches at most f(n) squares of its work tapes, where n is the input length, and f :N N Important space class: L = SPACE (O(log n))
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June 25, 2004CBSSS36 L and P configuration graph: nodes are configurations, edge (C, C’) iff C yields C’ in one step # configurations for a TM that runs in space O(log n) is n k for some constant k can determine if reach q accept or q reject from start configuration by exploring configuration graph (e.g. use DFS) Conclude: L P
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June 25, 2004CBSSS37 all languages EXP decidable NP P L P NP EXP believe all containments proper L
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June 25, 2004CBSSS38 Logspace A question: –Boolean formula with n nodes –evaluate using O(log n) space? 101 depth-first traversal requires storing intermediate values idea: short-circuit ANDs and ORs when possible
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June 25, 2004CBSSS39 Logspace 10 101 Can we evaluate an n node Boolean circuit using O(log n) space?
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June 25, 2004CBSSS40 A P-complete problem We don’t know how to prove L ≠ P But, can identify problems in P least likely to be in L using P-completeness. need stronger reduction (why?) yes no yes no L1L1 L2L2 f f
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June 25, 2004CBSSS41 A P-complete problem logspace reduction: f computable by TM that uses O(log n) space Theorem: If L 2 is P-complete, then L 2 in L implies L = P
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June 25, 2004CBSSS42 A P-complete problem Circuit Value (CVAL): given a variable-free Boolean circuit (gates , , , 0, 1), does it output 1? Theorem: CVAL is P-complete. Proof: –already argued in P
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June 25, 2004CBSSS43 A P-complete problem –L arbitrary language in P –TM M decides L in n k steps x1x1 x2x2 x3x3 …xnxn circuit produced from TM deciding L 1 iff x L –hardwire input x –poly-size circuit; logspace reduction
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June 25, 2004CBSSS44 L vs. P Not known that L ≠ P Interesting connection to parallelizability: –small-space algorithm automatically gives efficient parallel algorithm –efficient parallel algorithm automatically gives small-space algorithm P-complete problems least likely to be parallelizable
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June 25, 2004CBSSS45 P vs. BPP
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June 25, 2004CBSSS46 Communication complexity Goal: compute f(x, y) while communicating as few bits as possible between Alice and Bob count number of bits exchanged (computation free) at each step: one party sends bits that are a function of held input and received bits so far two parties: Alice and Bob function f:{0,1} n x {0,1} n {0,1} Alice holds x {0,1} n ; Bob holds y {0,1} n
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June 25, 2004CBSSS47 Communication complexity simple function (equality): EQ(x, y) = 1 iff x = y simple protocol: –Alice sends x to Bob (n bits) –Bob sends EQ(x, y) to Alice (1 bit) –total: n + 1 bits
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June 25, 2004CBSSS48 Communication complexity Can we do better? –deterministic protocol? –probabilistic protocol? at each step: one party sends bits that are a function of held input and received bits so far and the result of some coin tosses required to output f(x, y) with high probability over all coin tosses
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June 25, 2004CBSSS49 Communication complexity Theorem: no deterministic protocol can compute EQ(x, y) while exchanging fewer than n+1 bits. Proof: –“input matrix”: X = {0,1} n Y = {0,1} n f(x,y)
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June 25, 2004CBSSS50 Communication complexity –assume without loss of generality 1 bit sent at a time –A sends 1 bit depending only on x: X = {0,1} n Y = {0,1} n inputs x causing A to send 1 inputs x causing A to send 0
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June 25, 2004CBSSS51 Communication complexity –B sends 1 bit depending only on y and received bit: X = {0,1} n Y = {0,1} n inputs y causing B to send 1 inputs y causing B to send 0
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June 25, 2004CBSSS52 Communication complexity –at end of protocol involving k bits of communication, matrix is partitioned into at most 2 k combinatorial rectangles –bits sent in protocol are the same for every input (x, y) in given rectangle –conclude: f(x,y) must be constant on each rectangle
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June 25, 2004CBSSS53 Communication complexity –any partition into combinatorial rectangles with constant f(x,y) must have 2 n + 1 rectangles –protocol that exchanges ≤ n bits can only create 2 n rectangles, so must exchange at least n+1 bits. X = {0,1} n Y = {0,1} n 1 1 1 1 0 0 Matrix for EQ:
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June 25, 2004CBSSS54 Communication complexity protocol for EQ employing randomness? –Alice picks random prime p in {1...4n 2 }, sends: p (x mod p) –Bob sends: (y mod p) –players output 1 if and only if: (x mod p) = (y mod p)
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June 25, 2004CBSSS55 Communication complexity –O(log n) bits exchanged –if x = y, always correct –if x ≠ y, incorrect if and only if: p divides |x – y| –# primes in range is ≥ 2n –# primes dividing |x – y| is ≤ n –probability incorrect ≤ 1/2 Randomness gives an exponential advantage!!
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June 25, 2004CBSSS56 BPP model: probabilistic Turing Machine –deterministic TM with additional read-only tape containing “coin flips” BPP (Bounded-error Probabilistic Poly-time) –L BPP if there is a p.p.t. TM M: x L Pr y [M(x,y) accepts] ≥ 2/3 x L Pr y [M(x,y) rejects] ≥ 2/3 –“p.p.t” = probabilistic polynomial time
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June 25, 2004CBSSS57 all languages EXP decidable P L P BPP EXP L BPP
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June 25, 2004CBSSS58 P vs. BPP Theorem: If E ( = TIME(2 O(n) )) contains a hard problem (L that requires exponential-size Boolean circuits) then P = BPP. “for decision problems, randomness probably does not help”
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