Solutions_HW4_547 (1)

# C note given an element y y the algorithm 37 nds k y

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Unformatted text preview: T . If G is the union of disjoint directed cycles of length at most T . Then Z = ∅. On the other hand, the case where Z will contain the highest number of elements is when G is the union of disjoint directed cycles of length T + 1. In this case two elements of each cycle will have to be in Z in order for the remaining elements of the cycle to be within distance T from an element in Z . Since the number of cycles of length T + 1 is less or equal to TN , we get |Z | = 2 · TN . Thus, for a general G, +1 +1 |Z | ≤ 2 · N 2N ≤ , T +1 T so |Z | is O(N/T ). (c) Note: Given an element y ∈ Y , the algorithm 3.7 ﬁnds K ∈ Y (recall that P = C = K = Y ) such that g (K ) = eK (x) = y , that is, considering the fact that G is the union of disjoint directed cycles, K is such that (K, y ) is an edge in G. Thus, if y is a member of a cycle of length 1 (that is (y, y ) is an edge), then the algorithm won’t enter the while loop and K = y . If g (y ) ￿= y ; We know that, by construction of Z , either y is a member of a loop with length at most T , or there exists zj ￿= y such that the distance between y and zj is at most T . • Suppose y is a member of a cycle with length at mos...
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## This document was uploaded on 02/24/2014.

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