Algorithms and Complexity by Herbert S. Wilf

By Herbert S. Wilf

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In an easy case like this, we can write out the first few xs and then guess the answer. We find, successively, that x1 = b1 x0 , then x2 = b2 x1 = b2 b1 x0 and x3 = b3 x2 = b3 b2 b1 x0 etc. At this point, we can guess that the solution is: xn = x0 n Y bi (n = 0, 1, 2, . ). 27) i=1 Since that wasn’t hard enough, we’ll raise the ante a step further. Suppose we want to solve the first-order inhomogeneous (because xn = 0 for all n is not a solution) recurrence relation: xn+1 = bn+1 xn + cn+1 (n ≥ 0; x0 given).

That is indeed a recursive call, because we can just change the ‘n’ to ‘i − 1’ and call Quicksort. The second recursive call is the problem. It wants to sort a portion of the array that doesn’t begin at the beginning of the array. The routine Quicksort as written so far doesn’t have enough flexibility to do that. So we will have to give it some more parameters. Instead of trying to sort all of the given array, we will write a routine that sorts only the portion of the given array x that extends from x[left] to x[right], inclusive, where left and right are input parameters.

We’re back where we started. Indeed, if not, then we arrived at v 0 one more time than we departed from it, each time using a new edge, and finding no edges remaining at the end. Thus, there was an odd number of edges of G incident with v 0 , a contradiction. Hence, we are indeed back at our starting point when the walk terminates. Let W denote the sequence of edges along which we have so far walked. If W includes all edges of G, then we have found an Euler tour and we are finished. Else there are edges of G that are not in W .

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