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Complexity Theory Lecture 4 Lecturer: Moni Naor

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Recap Last week: Space Complexity Savitch’s Theorem: NSPACE(f(n)) µ SPACE(f 2 (n)) –Collapse of NPSPACE to PSPACE PSPACE Completeness –TQBF, Games Logarithmic space – deterministic and non Sublogarithmic space Non-deterministic space is closed under complementation This week: Probabilistic Space Complexity

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The Central Questions of Complexity Theory LogSpace µ NL µ P µ NP µ PSPACE Are any of the containments proper? All we can say: NL ( PSPACE Since NL µ Space(log 2 n) ( PSPACE from diagonalization (space hierarchy) Is NP = Co-NP? Is P = NP Å Co-NP ? Have not seen yet power of: Oracles Interaction Randomization If P = NP Å Co-NP then factoring is easy

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2-SAT 2-SAT : given a formula = C 1 Æ C 2 … Æ C m where C i =( Ç ) where 2 {x j, : x j } and 2 {x k, : x k } for some j and k. Is there a satisfying assumption for Let G( ) be a directed graph with –Nodes: literals –Edge ( , iff clause ( : Ç ) exists Claim : is satisfiable iff for no x i are there both a path from x i to : x i and a path from : x i to x i in G( ) Corollary : 2-SAT is in NL easy to see 2-unsat in NL and apply Immerman- - Szelepcsenyi Represent the implication the constraint puts

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2-SAT is Complete for NL Reduce from (un)reachability of DAGs –This is NL-Complete as well Given a DAG G, source s and target t construct (G): –For each node x create variable x –For edge (x,y) 2 G: add clause ( : x Ç y) –Add clauses (s) and ( : t) Claim : (G) is satisfiable iff there is no path from s to t in G Log space construction

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Algorithm for 2-SAT Start with arbitrary assignment While the assignment is not satisfying –Choose an arbitrary unsatisfied clause C i =( Ç ) –Choose at random from { , and flip the assignment to that variable –If unsuccessful for many step stop and declare ` unsat ’ Analysis: let A 2 {0,1} n be any satisfying assignment –With probability at least ½ distance to A is reduced –With probability at most ½ distance to A is increased 0n

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Analysis of Algorithm Distance can never be larger than n Want to compute how long it takes a pebble to get with high probability to 0 if it starts at some 0 · i · n Dominated by a walk where –With probability exactly ½ distance to A is reduced –With probability exactly ½ distance to A is increased –When pebble hits 0 it is absorbed –When pebble hits n it is reflected 0n Dominated : for all t probability exact process takes more than t steps is at least as the original process

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Analysis of Random Walks Would like to be able to say: Expected time to visit 0 For what time period can we say that there high probability?

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Probabilistic Turing Machines Probabilistic TM: transition function X Q X Q X {left,right} may be Probabilistic. A probability distribution on the set of moves – Accuracy Alternative view: in addition to input tape, random coins tape – How do we count it in space bounded computation head movement is one-way When is the PTM M considered to recognize a language L: Two sided error : For all x 2 L we have Pr[M stops with `yes ’]>2/3 For all x 2 L we have Pr[M stops with `no ’]>2/3 One-sided error For all x 2 L we have Pr[M stops with `yes ’]>1/2 For all x 2 L we have Pr[M stops with `no ’]=1 zero error: M never stops with the wrong answer but might never stop Want to consider expected consumption of resources Always stops Monte Carlo Vs. Las Vegas

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Random LogSpace (RL) A probabilistic Turing Machine with worktape size O(log |X|) –Long enough to count and assure stopping in time RL : one-sided Error For all x 2 L we have Pr[M stops with `yes ’]>1/2 For all x 2 L we have Pr[M stops with `no ’]=1 BPL : two-sided Error For all x 2 L we have Pr[M stops with `yes ’]>2/3 For all x 2 L we have Pr[M stops with `no ’]>2/3 ZPL : No Error but stopping time is only expected polynomial time Show: ZPL= RL Å Co-RL

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Undirected Connectivity and RL Given an undirected graph G=(V,E) –for nodes s and t, is there a path from s to t. – is the graph connected Homework : the two versions are log space equivalent via oracle reductions Idea of algorithm: Perform random walk for certain amount of time If node t is visited declare connected Ow unconnected Random walk: If at node u with degree d u Choose uniformly at random a neighbor and move to it

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Markov Chains A discrete time stochastic process X 1, X 2, … X t … –defined over a set of states S Finite or countably infinite –in terms of transition matrix P Entry P ij is probability next state is j given current state is i: Pr[X t+1 =j|X t =i] –For all i,j 2 S we have 0 · P ij · 1 and j 2 S P ij = 1 Memoryless Property : Pr[X t+1 =j|X t =i]= Pr[X t+1 =j|X t =i, X t-1 =i t-1, … X 0 =i 0 ] –what matter is where I am, not how I got here Example: pebble walk on the line S ={0,1, …,n} For all 1<i<n we have P ii+1 = ½ and P ii-1 = ½ P 00 = 1 and P nn-1 = 1 all other P ij = 0 0n

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Markov Chains Initial state distribution q (0) =(q 1 (0),q 2 (0), … q n (0) ) State probability vector q (t) =(q 1 (t),q 2 (t), … q n (t) ) distribution of the chain at time t q (t+1) =q (t) P and q (t) = q (0) P t Stationary distribution : for Markov Chain with transition matrix P is distribution such that = P If a Markov Chain is in a/the stationary distribution at step t it remains so forever Steady state behavior There can be unique, several or no stationary distribution Depending on the initial distribution

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Markov Chains For states i,j 2 S first transition probability r ij (t) is probability that given X 0 =i first time state j occurs is at step t r ij (t) =Pr[X t =j Æ X s ≠j, 1 · s < t| X 0 =i] Probability that there is such a visit f ij = t >0 r ij (t) Hitting Time : Let h ij be the expected number of steps to reach state j starting from state i. If f ij <1 then h ij = 1 h ij = t >0 t r ij (t) converse not necessarily true in infinite chains If f ij =1 then state i is called persistent

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Markov Chains Underlying directed graph of a Markov Chain: Edge (i,j) iff P ij > 0 Markov Chain is irreducible if its underlying graph consists of a single strong connected component Periodicity of a state i : GCD of the lengths of the path from i back to itself –Bipartite graphs: periodicity 2 –State is aperiodic if it has period 1

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Fundamental Theorem of Markov chains For any irreducible, finite and aperiodic Markov Chain: All states are ergodic ( h ii < 1 ) There is a unique Stationary distribution For all i 2 S we have – i > 0 –f ij = 1 –h ii = 1/ i

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Famous Markov Chain: PageRank algorithm [Brin and Page 98] Good authorities should be pointed by good authorities Random walk on the web graph –pick a page at random –with probability α follow a random outgoing link –with probability 1- α jump to a random page Rank according to the stationary distribution

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Random Walks on Graphs Undirected graph G=(V,E), |E|=m, |V|=n The transition matrix P uv = 1/d u if (u,v) 2 E and 0 ow. Claim: is G=(V,E) is connected and non-bipartite then for all v 2 V v =d u /2m Conclusion: h uu = 2m/d u How to make the graph non-bipartite? Add self-loops

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Cover and Commute time Commute time C uv : expected time for a random walk starting at u to reach v and come back C uv = h uv +h vu Cover time C u (G): expected length of a walk starting at u, visiting every node at least once. C(G) =max u C u (G). Lemma: for any edge (u,v): C uv = h uv +h vu · 2m Consider Markov chain of (directed) edges. Stationary distribution: uniform (each edge+direction) 1/2m Time to get from u to v and back on edge (v,u): 2m Not true for non-edges

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Upper bound on Cover Time Theorem: C u (G) · 2m (n-1) Proof: consider any spanning tree T of G. There is a traversal of T v 0, v 1, … v 2n-2 where each edge of T is traversed once in each direction. Consider random walk starting and ending at v 0, that traverses each of the edge of T once in each direction Upper bound on walk is upper bound on cover time C u (G) · j=0 2n-3 h v j v j+1 = (u,w) 2 T C uw · (n-1) 2m Conclusion: taking 4n 3 as the bound in the random walk suffices for constant probability of success At most m, from Lemma

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Universal Traversal Sequences G is a regular graph of degree d Labeling: the edges neighboring a node are labeled with {1,2 …d} in an arbitrary manner –Port number Can we find a polynomial sized sequence of labels s 1, s 2, … s k, s i 2 {1,2 …d} such that following it assures visiting all nodes of the graph Yes, it exists. Use the probabilistic method. Fix a graph an labeling Compute probability that sequence is not good for graph Sum over all graphs ( n O(dn) ) A log space construction puts connectivity in log space 7 4 7

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Hot off the press news Omer Reingold, Undirected ST-Connectivity in Log-Space, Available: Electronic Colloquium on Computational Complexity, Report TR04-094 Important Web Resources on Complexity: ECCC: http://www.eccc.uni-trier.de http://www.eccc.uni-trier.de Lance Fortnow’s Computational Complexity Web Log: http://fortnow.com/lance/complog/

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What happens in directed graphs There are strongly connected directed graphs with an exponential covering time Can count to doubly exponential space probabilistically So if we do not insist on always stopping may do directed connectivity in random log space (but exponential time) –uninteresting

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Probabilistic Variants of P RP : one-sided Error For all x 2 L we have Pr[M stops with `yes ’]>1/2 For all x 2 L we have Pr[M stops with `no ’]=1 BPP : two-sided Error For all x 2 L we have Pr[M stops with `yes ’]>2/3 For all x 2 L we have Pr[M stops with `no ’]>2/3 ZPP : No Error but stopping time is only expected polynomial time

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The Schwartz-Zippel Algorithm/Theorem Theorem : Let Q(x 1, x 2, …, x n ) 2 F(x 1, x 2, …, x n ) be a non-zero muitivariate polynomial of total degree d. Fix any finite set S µ F and choose r 1, r 2, …, r n 2 R S. Then Pr[Q(r 1, r 2, …, r n )=0] · d/|S| Proof : by induction on n Useful when Q is not given explicitly and want to test equality to 0 Determinant of a matrix

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Matching in Bipartite Graphs Let G=(V,U,E) b e a bipartite graph, |V|=|U|=n Define A to be the n £ n matrix with |E| variables A ij = x ij if (i,j) 2 E and 0 otherwise Theorem (Edmonds): G has a perfect matching iff det(A)≠0 Proof : det(A)= 2 S n sgn( ) A 1, (1) A 1, (2) … A 1, (n)

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Algorithm for deciding whether a matching exists in bipartite graphs Fix a prime P larger than 2n Choose {r ij } (i,j) 2 E 2 R GF[P] Compute det(A({r ij } (i,j) )) If non-zero declare matching. Ow, no matching In general computing det(A) more expensive than running combinatorial algorithm for matching Not true in parallel computation

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