6.842 Randomness and Computation March 3, Lecture 8

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1 6.84 Randomness and Computation March 3, 04 Lecture 8 Lecturer: Ronitt Rubinfeld Scribe: Daniel Grier Useful Linear Algebra Let v = (v, v,..., v n ) be a non-zero n-dimensional row vector and P an n n matrix. We say v is an eigenvector of P with corresponding eigenvalue λ iff vp = λv. The L -norm of v (denoted v ) is n v i. The L -norm of v (denoted v ) is n v i. The inner product of two vectors v and w (denoted v w) is n v iw i. { We say vectors v (), v (),... v (m) are orthonormal iff v (j) if i = j = 0 if i j Suppose P is an n n matrix with positive entries, eigenvectors v (), v (),..., v (n), and eigenvalues λ, λ,..., λ n. Let α R. Using the above definitions we derive the following facts: Fact αp has eigenvectors v (), v (),..., v (n) and eigenvalues αλ, αλ,..., αλ n Proof (αp ) = α( P ) = αλ i. Fact P + I has eigenvectors v (), v (),..., v (n) and eigenvalues λ +, λ +,..., λ n + Proof (P + I) = P + I = λ i + = (λ i + ). Fact 3 P k has eigenvectors v (), v (),..., v (n) and eigenvalues λ k, λ k,..., λ k n Proof P k = ( P )P k = λ i P k = λ i v(i) P k =... = λ k i v(i). Fact 4 If P is stochastic, then λ i for all i. Proof For all i, let I = {j j > 0}. Notice that we can force I to be non-empty. If had all nonpositive entries, we could let. Instead of trying to find a bound directly on λ i, we attempt to find a bound on λ i j I v(i) j. λ i j = n k P kj (select only the columns that produce positive value) j I j I k= k P kj (since P has only positive entries) j,k I = k I k I k k j I P kj (since P is stochastic) This implies that λ i. Notice, however, that in forcing I to be non-empty we could have negated the value of the corresponding eigenvalue. Thus, what we should really conclude is that λ i.

2 Theorem 5 Suppose P is a symmetric n n transition matrix. P has eigenvectors v (), v (),..., v (n) and corresponding eigenvalues λ, λ,..., λ n such that the eigenvectors are an orthonormal basis of R n, = λ λ λ 3... λ n, and v () = n (,,..., ). The power of this theorem will be evident later when we use λ eigenvalues (besides λ ). to bound the size of all other Mixing Times of Markov Chains For ɛ > 0, the mixing time T (ɛ) of a Markov chain with transition matrix P and stationary distribution Π is the minimum t such that Π Π 0 P t < ɛ for all initial distributions Π 0. We say that a Markov chain is rapidly mixing if T (ɛ) = poly(log n, log ɛ ) where n is the number of states. Theorem 6 Suppose P is the transition matrix of an undirected, nonbipartite, d-regular, connected Markov chain with starting distribution Π 0. The stationary distribution of the Markov chain is unique and equal to n (,,..., ). Furthermore, Π0 P t Π λ t where λ is the eigenvalue corresponding to the eigenvectors obtained from Theorem 5. Before we prove this theorem, it might help to take a moment to decipher what it tells us. First, we know that any ergodic Markov chain has a unique stationary distribution. However, the above Markov chain does not necessarily need to be ergodic, but it still has a unique (known) stationary distribution. For instance, the cycle of length k for any k falls into this category. As we will see later, this theorem provides an important method of determining how quickly a Markov chain converges to its stationary distribution. For example, when λ is a constant less than, we have that the Markov chain is rapidly mixing (actually, t only depends on ɛ). Proof Since P is undirected and d-regular, P is symmetric. Thus, P is real and symmetric, justifying our use of Theorem 5 to produce eigenvectors v (), v (),..., v (n) with corresponding eigenvectors = λ > λ λ 3... λ n. Since these eigenvector form an orthonormal basis of R n, we can express Π 0 as a linear combination of the s. So, Π 0 = = Π 0 P t = = n α i n α i P t n α i λ t i (using Fact 3) = α λ t v () + = α v () + n α i λ t i n α i λ t i

3 Using the orthonormality of the basis, we can find the value of α. Recall from Theorem 5 that v () = n (,,..., ). Π 0 v () = α v () v () + n Π 0 (,,..., ) = α n α i λ t i v () (since the v (), v (),..., v (n) are orthonormal) n = α (since Π 0 is a probability distribution) So, α v () = n (,,..., ). We claim now that this is fact the stationary distribution of the Markov chain. That is, Π 0 P t n n (,,..., ) = α i λ t i = n α i λ t i v(i) = n α i λt i (using above calculations) n α i λ t i v(i) (by orthonormality of basis vectors) λ t n αi (since λ > λ i ) λ t Π 0 since n αi = Π0 λ t Π 0 (since L -norm is at least L -norm when entries at most ) = λ t We now state (without proof) that the nonbipartite property of P ensures that λ <. Thus, λ t goes to 0 as t goes to infinity. Thus, n (,,..., ) must be the stationary distribution for Π0! Since there is no dependence on Π 0, we conclude that this is the unique stationary distribution for any starting distribution. 3 Using Markov Chains to Reduce Randomness Recall our previous methods for reducing error for problems in RP. By repeating the algorithm k times, we used O(k r) bits of randomness. Using ideas from pairwise independence, we were able to reduce to the randomness further to O(k + r). We now give an approach using random walks on Markov chains that uses r + O(k) bits of randomness. We concern ourselves with problems that have one-sided error. That is, for algorithm A deciding language L we have. x L, Pr[A(x) = ]

4 . x L, Pr[A(x) = 0] = The idea is to associate all (random) strings in {0, } n with nodes of a graph G. If x L, we do not care which path we take on the graph because A will never accept. However, if x L, we wish to design G in such a way that a random walk starting from a random node is likely to arrive at any random string for which the algorithm accepts. Lemma 7 There exists a graph G on r nodes with the following properties constant degree d-regular, connected, nonbipartite transition matrix for random walk on G has λ 0. uniform stationary distribution (since d-regular) Algorithm RP error reduction algorithm w {0, } r repeat w neighbor of w in G Run A(x) using randomness w if A(x) outputs x L then return x L and halt end if until A does not accept k times return x L Use Algorithm to reduce error for RP problems, and note that all assignments are done uniformly at random. Examining the algorithm, r bits of randomness are used to choose the initial w and log d bits of randomness are used to choose a random neighbor on each iteration. Thus, the total amount of randomness used in the algorithm is r + k log d = r + O(k) since d is constant. Claim 8 Probability of error of Algorithm is at most 5 k for x L. If x L, probability of error is 0. Proof If x L, then A never accepts, so probability of error is 0. If x L, then at least r choices of random bits have accepting paths in A. Let B be the set capturing those random strings which are bad. That is, B = {w A(x) with randomness w rejects}. By the above observation, B r 00. To use the linear algebraic properties of G, we need a linear algebraic way to describe the random walks that stay within B. We define N as a r r diagonal matrix (i.e. the only non-zero elements are on the diagonal). The ith diagonal of N is if i B and is 0 otherwise. Let Π be any probability distribution. We arrive at the following ideas. ΠN = Pr [A(x) rejects] w Π ΠP N = Pr [A(x) rejects after taking a random step] w Π. Π(P N) i = Pr [A(x) rejects on each of i random steps] w Π Notice that the expression Π(P N) i ignores the possibility that the initial w drawn from Π is a bad random string. However, since this only hurts our estimates, we are okay to ignore it. Lemma 9 For all Π (not necessarily probability distributions), ΠP N 5 Π. 4

5 Before we prove the lemma, let us see how it implies the theorem. Let Π be the uniform distribution on {0, } r r. The L -norm of the uniform distribution is ( ) = r. r Pr[Algorithm incorrect] Π(P N) k r Π(P K) k (by Cauchy-Schwarz) r Π 5 k (applying lemma k times) = 5 k (using above calculation of norm of uniform distribution) So Algorithm only uses r + O(k) bits of randomness and still guarantees that the error probability decreases exponentially in k. Proof (of Lemma 9) This is where we use the nice linear algebraic properties of G. Since G is real and symmetric we can apply Theorem 5 to the transition matrix P of G. Let v, v,..., v r be the eigenvectors of P. Since v = (,,..., ), v r =. Use the fact that the eigenvectors form a basis to write Π = r α iv i. So, ΠP N = = r r α i v i P N α i λ i v i N α λ v N + r α i λ i v i N (by triangle inequality) We will proceed by bounding each term separately. Intuitively, the first term should be small because we are unlikely to draw a bad string drawing uniformly from {0, } r. The second term should be small because the eigenvalues are small. α λ v N = α v N (since λ i = ) = α ( ) (since v = (,,..., )) i B r r = α α 0 Π 0 B r ( since B r ) 00 since Π = r α i 5

6 r α i λ i v i N r α i λ i v i = r (α i λ i ) r Π 0 α i since vn = vi r vi = v ( ) (since λ i 0 0 ) i B So, ΠP N Π 5. 6