Mid Term 1 Solution

COT5405 Analysis of Algorithms Midterm 1 Solution Summer 2012 June 11 In all cases explain clearly and as succinctly as possible. Problem 1 10 Pts Answer n T (n) = 2T ( n ) + pound n 2 2 n = 4T ( n ) + logn n + log n 4 2 2 2 n = 4T ( n ) + log nn? 1 + log n 4 2 2 = ?log2 n 1 = nT (1) + n i=1 i ? n Since i=1 1 ln n, T (n) ? ?(n log log n) i Problem 2 20 Pts Answer The general liking is to use the technique similar to quick sort, by doing naval division on some(prenominal) lids and cups.First we pick a cup randomly, and use it to partition the lids into twain sub impersonates those lids smaller than the size of that cup, and those larger than the size of the cup. We can also ? nd the correspondent lid for that chosen cup. Second we use that lid to partition the cups and divide them into two sets. We keep on repeating this procedure on each subset of cups/lids until all the cups/lids are paired. The overall time complexity is O(n log n) (Worst case O(n2 )). Problem 3 20 Pts Ans wer In this line of work we are more interested in ? ding the median quite of the momentuteimum/oozeimum section. The ? n ? th element in a min/ pocket gage is not the median. 2 In this case, we should erupt a new type of heap to adapt this problem. Problem 3 2 The solution is to use two heaps a min heap and a max heap. Suppose the total number of elements is n, we set the restriction that the max heap should contain ? n ? 2 elements. Correspondingly, the min heap contains n ? ? n ? elements. 2 When we hive away an element, we always stash away it into the max heap.If the number of elements in the max heap exceeds ? n ? , we lease the maximum element in the 2 max heap (the expelle), and insert it into the minimum heap. During this procedure, we need to do heapify to maintain the heap structure for both heaps. Under this setting, it is easy to see that all the elements in the max heap are less than those in the min heap, and the two elements at the root of both heaps repr esent the ? n ? th 2 and (? n ? + 1)th element. 2 Suppose the median is de? ned to be the ? n ? th element over all n elements.When 2 we redact the median, we just delete the root of the max heap, and the following two cases might occur (1) If the max heap contains ? n? 1 ? elements, then we do delete-max to the max 2 heap. (2) If the max heap contains ? n? 1 ? ? 1 elements, we take step up the root of the min 2 heap and set it to be the root in the max heap (because it is larger than all the elements in the max heap), then we do delete-min to the min heap. It is straightforward to see that the time complexity for both insert and delete-median is O(log n). COT5405 Analysis of Algorithms HW 2