Question: In the euclidean traveling-salesman problem, we are given a set of n points in the plane, and we wish to find the shortest closed tour

In the euclidean traveling-salesman problem, we are given a set of n points in the plane, and we wish to find the shortest closed tour that connects all n points. Figure 15.11(a) shows the solution to a 7-point problem. The general problem is NP-hard, and its solution is therefore believed to require more than polynomial time.

J. L. Bentley has suggested that we simplify the problem by restricting our attention to bitonic tours, that is, tours that start at the leftmost point, go strictly rightward to the rightmost point, and then go strictly leftward back to the starting point. Figure 15.11(b) shows the shortest bitonic tour of the same 7 points. In this case, a polynomial-time algorithm is possible.

Describe an O(n2)-time algorithm for determining an optimal bitonic tour. You may assume that no two points have the same x-coordinate and that all operations on real numbers take unit time. Scan left to right, maintaining optimal possibilities for the two parts of the tour.

Figure 15.11

> (depth, (k;) + 1) · pi +>(depth7 (d;) + 1) · qi E [search cost in T] i=0 i=1 п п = 1+> depth7 (k¡) · Pi + > dept

> (depth, (k;) + 1) pi +>(depth7 (d;) + 1) qi E [search cost in T] i=0 i=1 = 1+> depth7 (k) Pi + > depth7 (d;) qi , i=0 i=1

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Taking the books hint we sort the points by xcoordinate left to right in On lg n time Let the sorted points be left to right p 1 p 2 p 3 p n Thereforep 1 is the leftmost point and p n is the rightmost ... View full answer

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