Mid Term 1 Solution

COT5405 Analysis of Algorithms Midterm 1 root word Summer 2012 June 11 In wholly cases explain clearly and as succinctly as possible. Problem 1 10 Pts get along n T (n) = 2T ( n ) + log 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 dissolvent The general idea is to engage the technique similar to quick sort, by doing partition on both lids and cups.First we pick a cup randomly, and mapping it to partition the lids into both sub embeds those lids smaller than the size of that cup, and those larger than the size of the cup. We screwing 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 reiterate this procedure on each subset of cups/lids until all the cups/lids are paired. The overall cartridge clip complexity is O(n log n) (Worst case O(n2 )). Problem 3 20 Pts Answer In this problem we are more interested in ? dingdong the median instead of the minimum/ pocketimum atom. The ? n ? th element in a min/max jackpot is not the median. 2 In this case, we should develop a new type of nap to set this problem. Problem 3 2 The solution is to use two loads a min heap and a max heap. Suppose the heart and soul 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 enfold an element, we always insert it into the max heap.If the number of elements in the max heap exceeds ? n ? , we remove the maximum element in the 2 max heap (the root), 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 verify 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 rep resent 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 delete the median, we just delete the root of the max heap, and the following two cases qualification 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 out 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