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6 Pages

### hw5-sol

Course: RXC 064000, Fall 2009
School: Dallas
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Word Count: 1200

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5 Homework Solutions 1. 15-1 (p. 364) Bitonic Euclidean Traveling Salesman Problem 1. Sort the cities using an O(nlogn) sort from the smallest to largest xcoordinate. 2. Suppose city vn is adjacent to city vi. Then we need the shortest path from vi to v1 and then to vn, and then back to vi that includes every other city as well. 3. We define t(i ,j), i &lt; j, to be the shortest path starting from vi, reaching...

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5 Homework Solutions 1. 15-1 (p. 364) Bitonic Euclidean Traveling Salesman Problem 1. Sort the cities using an O(nlogn) sort from the smallest to largest xcoordinate. 2. Suppose city vn is adjacent to city vi. Then we need the shortest path from vi to v1 and then to vn, and then back to vi that includes every other city as well. 3. We define t(i ,j), i < j, to be the shortest path starting from vi, reaching v1, and going back to vj. t(i, j) is computed as follows: t(i, j) = t(i, j-1) + d(vj-1, vj) if j > i + 2 t(i, j) = min{ t(k, i) + d(vk, vj) | if j = i+1 for 1 < k < i 1 d(a, b) is the distance between cities a and b. Note: Because the Traveling Salesman Problem is NP-complete, this yields only an approximate solution. This algorithm is not guaranteed to find the best solution. 2. 9-2 (p. 194) Weighted Median a.) Argue that the median of x1, ... , xn is the weighted median of the xi with weights wi = 1/n. Since all xi have the same weight, the median of x1, ... , xn, the median will be xceiling(n/2). If n is even, then n = 2k. Summing the first k xi-values followed by the second k values yields the same weighted sum for each set: k/n = k/2k = 1/2 The weighted lower median is thus xk = xn/2. If n is odd, then n = 2k + 1. Summing the first k xi-values followed by the second set of k + 1 xi-values yields the following two equations: First k xi: Second k+1 xi: k/n = k/(2k+1) < 1/2 (k+1)/n = (k+1)/(2k+1) > (k+1)/(2k+2) = 1/2 The weighted lower median is thus thus xk = xn/2 = xceiling(n/2) since n is odd. b.) Show how to compute the weighted median of n elements in O(nlogn) worst-case time using sorting. 1. Sort the xi such that w1< ... < wn. 2. sum 0 3. for j 1 to n a. sum sum + wj. i. if sum > 1. return xj-1 Complexity: Sorting can be accomplish in O(nlogn) time using your favorite O(nlogn) sorting algorithm. Initializing the sum variable can be done in O(1) time. In the worst case, the for loop iterates n times. All of the operations inside of the for loop are O(1). Therefore, the total time complexity is: T(n) = O(nlogn) + O(1) + O(n)O(1) = O(nlogn) d.) Argue that the weighted median is a best solution for the one-dimensional post office location problem, in which points are simply real numbers and the distance between points a and b is d(a, b) = |a b|. Note: In the original weighted mean problem, element xi had one attribute: weight wi. In the one-dimensional weighted median problem, post office (element) pi has two attributes: weight wi, and one-dimensional position coordinate pi. Weighted median is a property that depends only on weights. Our goal is to find a value for p that minimizes: wi d ( p, pi ) = wi | pi - p | i =1 i =1 n n This value indicates the optimal coordinate at which to place the new post office p. We are asked to show that the p that minimizes the objective function is the weighted median. The weighted median one of the pi, which we can see from the definition of weighted median. Suppose the weighted median of the post offices is pk. Assume to the contrary that the value for p that minimizes the weighted distance is not pk. We must consider two cases: (1) (2) p = pk + p = pk Let D be the weighted distance when p = pk, D1 be the weighted distance when p = pk + , and D2 be the weighted distance when p = pk . D = wi pi - p k = w1 ( p1 - p k ) + w2 ( p 2 - p k ) + ... wn + ( p n - p k ) D1 = wi pi - ( p k + ) = w1 ( p1 - p k - ) + w2 ( p 2 - p k - ) + ... + wn ( p n - p k - ) D2 = wi pi - ( p k - ) = w1 ( p1 - p k + ) + w2 ( p 2 - p k + ) + ... + wn ( p n - p k + ) i =1 i =1 n i =1 n n k n D1 - D = wi - wi = (w1 + ... + wi ) - (wi +1 + ... + wn ) i =1 i = k +1 n k D2 - D = wi - wi = (wi +1 + ... + wn ) - (w1 + ... + wi ) i = k +1 i =1 Since pk is weighted median, 1 k wi = (w1 + ... + wi ) 2 i =1 n 1 wi = (wi +1 ... + ...wn ) 2 i = k +1 Therefore: D1 D > 0 D2 D < 0 D is a better solution than either D1 or D2. Therefore = 0 is the best solution. In both cases, p = pk. 3. 16.1-3 (pg. 379) Suppose that we have a set of activities to schedule among a large number of lecture halls. We wish to schedule all activities using a few lecture halls as possible. Give an efficient greedy algorithm to determine which activity should use which hall. Note: Although the problem didn't explain this, both the scheduling problem and the graph coloring problem are NP-Complete. Therefore, no perfect solution to either problem can be guaranteed in polynomial time. Our algorithm works as follows: 1. Sort the activities in terms of length, such that the longest activity is at the front of the list. 2. Select the longest activity from the front of the list. Place it in a lecture hall. 3. Process each activity in the list, stating with the second-longest activity. 4. If it is possible to schedule this activity in one of the already-allocated lecture halls, schedule it there. Otherwise, open a new lecture hall and schedule the activity in the new hall. In the worst case, each activity must be placed in its own hall. Therefore, the kth largest activity will have to determine that it cannot fit into any of the k-1 lecture halls that have already been appropriated. Therefore, the worst-case complexity is O(n2) if a total of n activities must be scheduled. 16.2-5 (pg. 384) Describe an efficient algorithm that, given a s...

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Dallas - RXC - 064000
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Penn State - JTK - 412
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