IITM-CS6845: Theory Toolkit February 3, 2012

Transcription

1 IITM-CS6845: Theory Toolkit February 3, 2012 Lecture 4 : Derandomizing the logspace algorithm for s-t connectivity Lecturer: N S Narayanaswamy Scribe: Mrinal Kumar Lecture Plan:In this lecture, we will see a derandomization of the logspace algorithm for s-t connectivity based upon random walks that we saw in the last lecture. The derandomization comes at the cost of more space. Our new algorithm uses log 2 (n) space and polynomial time. 1 Recall In the last lecture, we looked at an randomized log space algorithm for the following problem. Our input is an undirected graph G and a pair of vertices s, t V (G), and the question is to find out whether s is connected to t or not. Our algorithm was based on random walks and therefore had access to random bits at each step. Moreover, the algorithm ran in polynomial time and displayed one sided error. The aim of this lecture is to try and obtain a deterministic algorithm for this problem which uses the same amount of resources, space and time. The algorithm described here runs in polynomial time but uses poly logarithmic space. In the last lecture we also proved, ( using the Chebychev s inequality) the following lemma. Lemma 1. Let A {0, 1} m and B {0, 1} m and let H be the family of linear functions from {0, 1} m {0, 1} m defined by H = {ax + b a, b {0, 1} m } for some positive integer m. Let ɛ > 0. Then, P r h H [ P r x {0,1} m[x A h(x) B] ρ(a)ρ(b) ɛ] 1 2 m ɛ 2, where ρ(a) = A /2 m and ρ(b) = B /2 m. It should be observed that unlike in the case of deterministic log-space algorithms, which always run in polynomial time, a randomized logspace algorithm has considerably more computation power if it is allowed more than polynomial time. 4-1

2 2 Strategy Consider any bounded space randomized algorithm A which uses space s and runs in time t. We will use the word algorithm and machine alternatively for A. Let the algorithm use a maximum of r random bits in any run. Now, let us fix an input x and try to observe the run of the algorithm on this input as a function of random bits it uses. Since the machine uses a maximum of s space, the number of configurations that the machine can get into is bounded by 2 s. It is also worth noting that t 2 s as the algorithm always terminates. Again without loss of generality we can assume that the algorithm runs in exactly 2 s steps. Since our algorithm right now uses r random bits, so an obvious approach to derandomization would be to iterate over all bit strings of length r and keep a count of the number of time the algorithm accepts. Unfortunately, this blows up the running time exponentially. So, we try to construct a psedurandom source, which itself takes O(log(r)) random bits as input and in turn generates r 1 r bits, which our construction uses. Finally, we iterate over the choice of this log(r) bits to totally derandomize the procedure. The running time blow up is therefore controlled. To ensure the correctness of the construction, we need to prove that both the algorithms, namely A and our construction accept and reject the same set of strings. For this purpose, it is convenient to visualize the working of a bounded space algorithm in the following way. Consider the set of all possible configurations of this algorithm on the fixed input x as C. Now, construct the following 2 s 2 s square matrix M whose rows and columns are indexed by the elements in C. The element M i,j represents the probability of the algorithm going from configuration i to configuration j on when it is given access to a random bit x chosen uniformly at random. Without loss of generality, it can be assumed that the starting, accepting and rejecting configurations of A on input x are unique and let them be labelled as in, ac and rej respectively. The probability that the A accepts x in one step is given by M in,ac. Similarly, it can also be observed that the probability of A acepting x is two steps will be given by Min,ac 2. Inductively proceeding like this, for any t, the probability that M accepts x in t steps is given by Min,ac t. The above matrix multiplication model gives us an alternative way of looking at the run of any bounded space randomized algorithm and we will use precisely this model for our derandomization. Whether or not x is accepted by the algorithm is given by the entries of the matrix Min,ac t. If we provide some alternative distribution of random bits to A, the entries of the matrix described above will change. Let us say that A instead of being provided a string chosen uniformly at random from {0, 1} r is provided a string from some other distribution such that we obtain a matrix D at step t such that D and M t are not too different from each other in terms of their entries. Hence, we will use D to solve the problem deterministically. We should recall that since A is a randomized algorithm making one sided error, it accepts x with a probability bounded away from half when x is in the language and rejects with probability 1 otherwise. So, in trying to come up with D, if we manage to make sure that the individual elements of M t and D differ only in a way that the entries which retain their location on the real line with respect 4-2

3 to 1/2, the output of D and the output of A remain the same. 3 Construction 3.1 Notation Let us fix the following notation for this section. 1. The norm of a vector x R n is defined as x = i x i and for a real square matrix of dimension n, the norm of matrix M denoted by M is defined as M = max Mx x R n x where x is a non null vector. We will be using the following properties of the matrix norm at various places during the course of analysis. For two vectors x 1 and x 2 in R n, x 1 + x 2 x 1 + x 2. The proof follows from definition and an easy application of the properties of the real numbers under modulas. M + N M + N Proof. (M + N)x M + N = max x R n x max x R n( Mx x (1) + Nx ) (2) x M + N (3) Here (1) and (3) follow from the definiton of the norm as defined above while (2) follows due to property 1 proved above. MN M N Proof. Let n be the norm of N and m be the norm of M. Now, let x be an arbitrary vector in R n such that Nx = y. Now, MNx = My m y, by definition of norm. Similarly, y = N x n x. coupling the two together, we have MNx mn x. Therefore, MN mn. For a stochastic matrix, where the rows sum up to 1, M =

4 Proof. To show the above statement, we show the following. For any vector x = (x 1, x 2... x n ) and a stochastic matrix M, we will show that the norm is upper bounded by 1. Then, if we can come up with a vector x such that Mx / x = 1, we are done. It is easy to verify that the all ones vetctor (1, 1,... 1) is such a vector. For the first part, consider the i th element of the vector Mx. It is equal to the sum j M ijx j j M ij x j j M ij x x. Hence, Mx / x 1. For a matrix with each entry bounded by ɛ, M nɛ where M is a n n matrix. Proof. For any vector x and any matrix M whose each entry is bounded above by ɛ, Mx i = j M ijx j j ɛx j j ɛ x j j ɛ x nɛ x. Hence, the norm of M is bounded above by nɛ. 2. For a transition probability matrix M for any randomized algorithm and a given distribution D, M D [i, j] is the probability that the machine goes from the configuration i to configuration j when given access to a string randomly chosen acording to D. 3. U r represents the uniform distribution on {0, 1} r. 4. M represents the transition probability matrix over the configurations of the randomized algorithm that we are trying to derandomize. We will ideally try to estimate the entries of the matrix M t where the machine runs in time t and the random bits are chosen uniformly at random. 3.2 Pseudorandom generator We use the following generator to obtain the pseudorandom string used in our simulation. Definition 2. For positive integers m, k, we define the function G k : {0, 1} m H k {0, 1} m2k where H is the family of all linear functions {0, 1} m {0, 1} m defined by H = {ax+b a, b {0, 1} m }. The function is recursively defined as follows. Base case: G 0 (x) = x Recursion: G k (x, h 1, h 2... h k ) = G k 1 (x, h 1... h k 1 )og k 1 (h k (x), h 1... h k 1 ), where o represents concatenation of two strings. The first few strings computed by the function above will be as follows. 1. G 0 (x) = x 4-4

5 2. G 1 (x, h 1 (x)) = xh 1 (x) 3. G 2 (x, h 1 (x), h 2 (x)) = xh 1 (x)h 2 (x)h 1 (h 2 (x)) Intuitively, G k will provide us the pseudorandom string to be used in the k th step of our recursive construction. The idea is to carefully chose the hash functions h 1, h 2... h k such that our randomized algorithm can not distinguish between a random string from U m2 k and the one provided by the generator G k. We need two more definitions before we can proceed to the algorithm. Definition 3. For h 1, h 2... h k H, D h1,h 2...h k denotes the probability distribution of G k (x, h 1, h 2... h + k), when x is chosen uniformly at random from {0, 1} m. We denote M Dh1,h 2...h k by M h1,h 2...h k. Definition 4. Given a transition probability matrix M h1,h 2...h k 1 and a hash function h H, h is said to be ɛ good for h 1, h 2... h k, if Mh 2 1,h 2...h k 1 M h1,h 2...h k ɛ. The above definition essentially denotes the error in the transition probabilities whih will occur if the machine is run for two steps by providing it with half the number of random bits than required. 3.3 Algorithms The main loop 1. Input: The transition probabiltiy matrix of a log space randomised algorithm denoted by M (M is an n n square matrix) which uses r random bits where r n. The algorithm runs in time t n for any input. 2. Output: To obtain a matrix N, such that N h1,h 2...h k M t ɛ for some given constant ɛ. Let K = logt, δ = ɛ 2 K, m = O(log(n)), t 1 = m2 K For k {1, 2... K} : (a) Call the find hash function routine to compute a hash function h k which is δ good for the h 1, h 2... h k 1. For all pairs of configurations i, j, compute N[i, j] (a) Call the compute value routine to compute the value of N[i, j] = M h1,h 2...h K [i, j] 4-5

6 Find Hash Function 1. Input: Hash functions h 1, h 2... h k 1 and a rational number δ > Output: A hash function h k which is δ good for h 1, h 2... h k 1 For every h H do the following: (a) For each pair of configurations i, j, compute λ = M h1,h 2,h 3...h k [i, j] Mh 2 1,h 2...h k 1 [i, j] using calls to the routine Compute Values described below. (b) If λ > δ n, then this choice of h is a bad choice. Go to the next h H. Else return h. Compute Values 1. Input: hash functions h 1, h 2... h k, a pair of configurations i, j. 2. Output: M h1,h 2...h k [i, j] For every string x {0, 1} m, simulate the algorithm starting with configuration i and random bits being read from the string G k (x, h 1, h 2... h k ). Keep a count of strings x such that the final state is j (say λ). Return M h1,h 2...h k[i, j] = λ 2 m. It is worth noting that the length of the string G k (x, h 1, h 2... h k ) is polynomial in n and hence can notbe stored on the tape as it would violate our desired space bound of log 2 (n). Therefore, we do not compute the whole string and write it down. We compute a bit only when it is needed during the run of the algorithm. It is easy to observe that this computation can be done m bit at a time using the recursive definition of G k as defined in the previous section. 4 Resources used by the algorithm The space utilized by the algorithm canbe bounded by O(log(t)m) as that is the amount of space required to store the hash functions h 1, h 2... h K. All other space used in the algorithm can be bounded by O(m). Hence the total space used by the algorithm is at most O(log 2 (n)). For the ananlysis of running time, we observe the following. 1. The main loop runs K = O(n) times. 4-6

7 2. Each call to the routine Find hash function takes time O(2 2m n 2 n (time taken to call the function Compute values)). 3. each call to the routine Compute values takes time which is O(2 m n m 2 2 k ). 4. Therefore, the total time taken by the algorithm takes time which can be bounded by some sufficiently large polynomial in n. Thus our algorithm runs in polynomial time and takes log 2 (n) space. 5 The proof of correctness We begin with proving the following claim. Claim 5. For any sequence of functions h 1, h 2... h k 2, there always exists a function h H such that for every i and j: M 2 h 1,h 2...h k 1 [i, j] M h1,h 2...h k 1,h[i, j] δ n Proof. Let R denote the stochastic matrix M h1,h 2...h k 1. Let the machine use an m bit random string in one step. An entry of R say R[l, f] is esentially the fraction of random strings which take the probabilistic machine from state l to state f. Similarly an entry, R 2 [l, f] essentially denotes the fraction of strings which takes from some state a to some state b in two steps. This, by definition is equal to the product of fraction of string which takes to the machine to some intermediate state a from l in one step and then takes it from l to f for any intermediate state l. Let us call the set of strings as good strings. Now, the probabiltiy of the machine going from state l to state f in two steps is essentially equal to the probabiltiy that the machine gets a good string in the first step and it gets a good string in the second step. If the set of good strings for the first step are denoted y A 1 and the set of good strings for the second step are denoted by A 2, then the entry R 2 [l, f] = A 1 A 2 = ρ(a)ρ(b). We formalize this argument now. Let, for a pair 2 2m of configurations (or states) i and j, A i,j = {x {0, 1} m G k (x, h 1, h 2... h k 1 ) takes the machine from state i to state j }. Therefore, M h1,h 2...h k 1 [i, j] = ρ(a ij ). Similarly, from the definition of G k M h1,h 2...h k 1,h[i, j] = l P r x {0,1} m [x A il and h(x) A l j]. For any fixed triplet of states i, j, l, using the lemma 1 we know that for a h which is chosen uniformly at random from the set H and for any epsilon, the following holds 4-7

8 P r h H [ P r x {0,1} m[x A h(x) B] ρ(a)ρ(b) ɛ] 1 2 m ɛ 2 Therefore, by substituting ɛ = δ, for a random h, with a probabiltiy at least 1 n4, the n 2 2 m δ 2 following is true. P r x {0,1} m[x A h(x) B] ρ(a)ρ(b) δ n 2 ] By using the union bound over all triplets i, j, l, we can conclude that a random h satisfies the above expression with probabiltiy at least 1 2. Summing over the intermediate states l, which are at most n in number and using the probabilisti method, the claim follows. Using the claim proved above and the properties of the matrix norm earlier defined, we prove the following cruicial lemma. Lemma 6. The routine Find Hash Function always returns a function h which is δ good for h 1, h 2... h k 1. Proof. From the definition of a function being δ good, we need to bound the norm of the matrix M 2 h 1,h 2...h k 1 M h1,h 2...h k by δ. Now, if we see the claim above, every entry of this matrix is bounded by δ n. Therefore, from the property of the norm defined, the norm of the matrix is at most n δ n which is equal to δ. We now prove the final lemma which ensures that the matrix M h1,h 2...h k closely approximates the transition probability matrix MU t, which exxentially completes our proof. m2 k Lemma 7. If for every 1 z k, h z is a δ good for h 1, h 2... h r 1, then M h1,h 2...h k M t U m2 k (2k 1)δ. Proof. The proof follows by induction on k. The base case is trivial as we iterate over all x {0, 1} m. For the induction step, we assume that the statement holds for k 1. Using the triangle inequality which our matrix norm satisfies, M h1,h 2...h k M M Um2 k h 1,h 2...h k Mh 2 1,h 2...h k 1 + Mh 2 1,h 2...h k 1 MU t. Since h m2 k k is δ good for h 1, h 2... h k 1, the first term in the sum above is bounded by δ. Now we bound the second term of the sum. We can observe t/2 = (MU )2. So, M 2 m2 k 1 h 1,h 2...h k 1 MU t M m2 k h 1,h 2...h k 1 M h1,h 2...h k 1 that MU t m2 k M t/2 U + M h m2 k 1 1,h 2...h k 1 M t/2 U m2 k 1 t/2 MU. Since each of these matrices are stochastic m2 k 1 in nature, their individual norms are bounded by 1 and the difference term occuring in the expansion above is bounded by (2 k 1 1)δ from the induction hypothesis. Substituting these bounds back, the lemma follows. 4-8

9 ɛ Since the value of δ was chosen to be, substituting back into the above lemma with 2 K k = K, we have the following theorem. Theorem 8. The algorithm computes a matrix N such that N M t U m2 k ɛ. 6 Tools used We used the following major tools that we have seen in previous lectures of this course, at different places in this construction and the proof of correctness. 1. Chebychev s Inequality. 2. Linearity of Expectations. 3. The Union Bound 7 References Noam Nisan: RL SC STOC 1992 : Noam Nisan: Psuedorandom Generators for Space-Bounded Computation STOC 1990 :