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1 The Monte Carlo method

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2 (0,0) (1,1) (-1,-1) (-1,1) (1,-1) 1 Z= 1 If X 2 +Y 2 1 0 o/w (X,Y) is a point chosen uniformly at random in a 2 2 square centered at the origin (0,0). P(Z=1)= /4.

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3 Assume we run this experiment m times, with Z i being the value of Z at the ith run. If W= i Z i, then W ’ =4W/m is an estimate of .

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4 By Chernoff bound, Def: A randomized algorithm gives an ( , )- approximation for the value V if the output X of the algorithm satisfies Pr[|X-V| V] 1- .

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5 The above method for estimating gives an ( , )-approximation, as long as <1 and m large enough.

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6 Thm1: Let X 1, …,X m be independent and identically distributed indicator random variables, with =E[X i ]. If m (3 ln (2/ ))/ 2 , then I.e. m samples provide an ( , )-approximation for . Pf: Exercise!

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7 Def: FPRAS: fully polynomial randomized approximation scheme. A FPRAS for a problem is a randomized algorithm for which, given an input x and any parameters and with 0< , <1, the algorithm outputs an ( , )-approximation to V(x) in time poly(1/ , ln -1,|x|).

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8 Def: DNF counting problem: counting the number of satisfying assignments of a Boolean formula in disjunctive normal form (DNF). Def: a DNF formula is a disjunction of clauses C 1 C 2 … C t, where each clause is a conjunction of literals. Eg. (x 1 x 2 x 3 ) (x 2 x 4 ) (x 1 x 3 x 4 )

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9 Counting the number of satisfying assignments of a DNF formula is actually #P- complete. Counting the number of Hamiltonian cycles in a graph and counting the number of perfect matching in a bipartite graph are examples of #P-complete problems.

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10 A na ï ve algorithm for DNF counting problem: Input: A DNF formula F with n variables. Output: Y = an approximation of c(F) The number of satisfying Assigments of F. 1. X 0. 2. For k=1 to m, do: (a) Generate a random assignment for the n variables, chosen uniformly at random. (b) If the random assignment satisfies F, then X X+1. 3. Return Y (X/m)2 n.

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11 Analysis Pr[X k =1]=c(F)/2 n. Let X= and then E[X]=mc(F)/2 n. Xk=Xk= 1 If the k-th iteration in the algorithm generated a satisfying assignment; 0 o/w.

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12 Analysis By Theorem 1, X/m gives an ( , )- approximation of c(F)/2 n, and hence Y gives an ( , )-approximation of c(F), when m 3 2 n ln (2/ ))/ 2 c(F). If c(F) 2 n /poly(n), then this is not too bad, m is a poly. But, if c(F)=poly(n), then m=O(2 n /c(F))!

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13 Analysis Note that if C i has l i literals then there are exactly 2 n- l i satisfying assignments for C i. Let SC i denote the set of assignments that satisfy clause i. U={(i,a): 1 i t and a SC i }. |U|= Want to estimate Define S={(i,a): 1 i t and a SC i, a SC j for j<i}.

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14 DNF counting algorithm II: Input: A DNF formula F with n variables. Output: Y: an approximation of c(F). 1. X 0. 2. For k=1 to m do: (a) With probability choose, uniformly at random an assignment a SC i. (b) If a is not in any SC j, j<i, then X X+1. 3. Return Y

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15 DNF counting algorithm II: Note that |U| t|S|. Why? Let Pr[i is chosen]= Then Pr[(i,a) is chosen] =Pr[i is chosen] Pr[a is chosen|i is chosen] =

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16 DNF counting algorithm II: Thm: DNF counting algorithm II is an FPRAS for DNF counting problem when m= (3t/ 2 ) ln (2/ ) . Pf: Step 2(a) chooses an element of U uniformly at random. The probability that this element belongs to S is at least 1/t. Fix any , >0, and let m= (3t/ 2 ) ln (2/ ) . poly(t,1/ , ln (1/ ))

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17 DNF counting algorithm II: The processing time of each sample is poly(t). By Thm1, with m samples, X/m gives an ( , )- approximation of c(F)/|U| and hence Y gives an ( , )-approximation of c(F).

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18 Counting with Approximate Sampling Def: w: the output of a sampling algorithm for a finite sample space . The sampling algorithm generates an -uniform sample of if, for any subset S of , |Pr[w S]-|S|/| || .

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19 Def: A sampling algorithm is a fully polynomial almost uniform sampler (FPAUS) for a problem if, given an input x and >0, it generates an - uniform sample of (x) in time poly(|x|, ln (1/ )). Consider an FPAUS for independent sets would take as input a graph G=(V,E) and a parameter . The sample space: the set of all independent sets in G.

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20 Goal: Given an FPAUS for independent sets, we construct an FPRAS for counting the number of independent sets. Assume G has m edges, and let e 1, …,e m be an arbitrary ordering of the edges. E i : the set of the first i edges in E and let G i =(V,E i ). (G i ): denote the set of independent sets in G i.

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21 | (G 0 )|=2 n. Why? To estimate | (G)|, we need good estimates for

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22 Let be estimate for r i, then To evaluate the error, we need to bound the ratio To have an ( , )-approximation, we want Pr[|R-1| ] 1- .

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23 Lemma: Suppose that for all i, 1 i m, is an ( /2m, /m)-approximation for r i. Then Pr[|R-1| ] 1- . Pf: For each 1 i m, we have

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24 Equivalently,

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25 Estimating r i : Input: Graph G i-1 =(V,E i-1 ) and G i =(V,E i ) Output: = an approximation of r i. 1. X 0 2. Repeat for M= (1296m 2 / 2 ) ln (2m/ ) independent trials: (a) Generate an ( /6m)-uniform sample from (G i-1 ). (b) If the sample is an independent set in G i, then X X+1. 3. Return X/M.

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26 Lemma: When m 1 and 0< 1, the procedure for estimating r i yields an ( /2m, /m)- approximation for r i. Pf: Suppose G i-1 and G i differ in that edge {u,v} is in G i but not in G i-1. (G i ) (G i-1 ). An independent set in (G i-1 )\ (G i ) contains both u and v.

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27 Associate each I (G i-1 )\ (G i ) with an independent set I\{v} (G i ). Note that I ’ (G i ) is associated with no more than one independent set I ’ {v} (G i-1 )\ (G i ), thus | (G i-1 )\ (G i )| | (G i )|. It follows that

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28 Let X k =1 if the k-th sample is in (G i ) and 0 o/w. Because our samples are generated by an ( /6m)-uniform sampler, by definition,

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29 By linearity of expectations, Since r i 1/2, we have

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30 If M 3 ln (2m/ )/( /12m) 2 (1/3)=1296m 2 -2 ln (2m/ ), then Equivalently, with probability 1- /m, -----(1)

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31 As we have Using, r i 1/2, then -----(2)

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32 Combining (1) and (2), with probability 1- /m, This gives the desired ( /2m, /m)-approximation. Thm: Given FPAUS for independent sets in any graph, we can construct an FPRAS for the number of independent sets in a graph G.

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33 The Markov Chain Monte Carlo Method The Markov Chain Monte Carlo Method provides a very general approach to sampling from a desired probability distribution. Basic idea: Define an ergodic Markov chain whose set of states is the sample space and whose stationary distribution is the required sampling distribution.

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34 Lemma: For a finite state space and neighborhood structure {N(x)|x }, let N=max x |N(x)|. Let M be any number such that M N. Consider a Markov chain where If this chain is irreducible and aperiodic, then the stationary distribution is the uniform distribution. P x,y = 1/M if x y and y N(x) 0 if x y and y N(x) 1-N(x)/M if x=y.

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35 Pf: For x y, if x = y, then x P x,y = y P y,x, since P x,y =P y,x =1/M. It follows that the uniform distribution x =1/| | is the stationary distribution by the following theorem.

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36 Thm: P: transition matrix of a finite irreducible and ergodic Markov chain. If there are nonnegative numbers =( 0,.., n ) such that and if, for any pair of states i,j, i P i,j = j P j,i, then is the stationary distribution corresponding to P. Pf: Since, it follows that is the unique stationary distribution of the Markov Chain.

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37 Eg. Markov chain with states from independent sets in G=(V,E). 1. X 0 is an arbitrary independent set in G. 2. To compute X i+1 : (a) choose a vertex v uniformly at random from V; (b) if v X i then X i+1 =X i \{v}; (c) if v X i and if adding v to X i still gives an independent set, then X i+1 =X i {v}; (d) o/w, X i+1 =X i.

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38 The neighbors of a state X i are independent sets that differ from X i in just one vertex. Since every state is reachable from the empty set, the chain is irreducible. Assume G has at least one edge (u,v), then the state {v} has a self-loop (P v,v >0), thus aperiodic. When x y, it follows that P x,y =1/|V| or 0, by the previous lemma, the stationary distribution is the uniform distribution.

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39 The Metropolis Algorithm (When stationary distribution is nonuniform) Lemma: For a finite state space and neighborhood structure {N(x)|x }, let N=max x |N(x)|. Let M be and number such that M N. For all x , let x >0 be the desired probability of state x in the stationary distribution. Consider a Markov chain Then, if this chain is irreducible and aperiodic, then the stationary distribution is given by x. P x,y = (1/M)min(1, y / x ) if x y and y N(x) 0 if x y and y N(x) 1- y x P x,y if x=y.

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40 Pf: For any x y, if x y, then P x,y =1/M and P y,x =(1/M)( x / y ). It follows that P x,y =1/M=( y / x )P y,x. x P x,y = y P y,x. Similarly, for x > y. Again, by the previous theorem, x ’ s form the stationary distribution.

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41 Eg. Create a Markov chain, in the stationary distribution, each independent set I has probability proportional to |I|, for some >0. I.e. x = |I x | /B, where I x is the independent set corresponding to state x and B= x |I x |. Note that, when =1, this is the uniform distribution.

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42 1. X 0 is an arbitrary independent set in G. 2. To compute X i+1 : (a) choose v uniformly at random from V; (b) if v X i then X i+1 =X i \{v} with probability min(1,1/ ) ; (c) if v X i and X i {v} still gives an independent set, then put X i+1 =X i {v} with probability min(1, ) ; (d) o/w, set X i+1 =X i.

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