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Randomized Advance Algorithms Randomized Advance Algorithms
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A randomized algorithm for game tree evaluation can in fact be viewed as a probability distribution over deterministic algorithms, because the length of the computation as well as the number of choices at each step are both finite. We may imagine that all of these coins are tossed before the beginning of the execution. The tree T 2 ,k is equivalent to a balanced binary tree all of whose leaves are at distance 2k from the root, and all of whose internal nodes compute the NOR function.
(x 1 ∨ y 1 ) ∧ (x 2 ∨ y 2 ) = (x 1 ∨ y 1 ) ∨ (x 2 ∨ y 2 ) Consider a tree of NOR’s of depth 2k. Let p = (3 −
5)/2 which is a solution of (1 − p)^2 = p. Each leaf of the tree is independently set to 1 with probability p. If each input to a NOR node is independently set to 1 with probability p, then the probability that its output is 1 is the probability that
both its inputs are 0, which is (1 − p)^2 = p. The value of every node of the NOR tree is 1 with probability p. And the value of a node is independent of the values of all the other nodes at same level. Let T be a NOR tree each of whose leaves is independently set to 1 with probability q for a fixed value of q ∈ [0, 1]. Let W (T ) denote the minimum over all deterministic algorithms of the expected number of steps to evaluate T. Then, there is a depth-first pruning algorithm (inspects leaves visited in a depth-first-search tree; sub-trees of a node v are pruned after the value of v is determined), whose expected number of steps to evaluate T is W (T ). For a depth-first pruning algorithm evaluating this tree, let W (h) be the expected number of leaves it inspects in determining the value of a node at distance h from the leaves: W (h) = p × W (h − 1) + (1 − p) × (2 × W (h − 1)) = (2 − p) × W (h − 1) Where the first term represents the work done in evaluating only one of the sub-trees of the node, and the second term represents the work done in evaluating both the sub-trees (which will be necessary if the first sub-tree returns the value 0, an event occurring with probability 1 − p). Letting h be log 2 n and solving we get W (h) ≥ n^0.^694 : W (h) = (2 − p)h^ = nlog^2 n×logn(2−p)^ = nlog^2 n×(log^2 (2−p)/^ log^2 n)^ = nlog^2 (2−p)^ = n^0.^694