R2 = Iso Lef T R1

ALGORITHMS -CSE202-SPRING 2000 Divide and Conquer and Sorting, Some Solutions Name Tree Isomorphism 0 T Problem Give an efficient algorithm to decide , given two rooted binary trees T and , 0 T whether T and are isomorphic as rooted trees. Solution 0 0 0 0 T r oot T L; R ; L ;R Let r oot be the root of and be the root of , and let be the left and 0 0 T T T f right subtrees of T and respectively. If embeds in , then the isomorphism must 0 0 r oot L L map r oot to , and must map every member of either entirely to or entirely to 0 Iso R and vice versa. Thus, we can define the isomorphism relationship on sub-trees as NIL;NIL=True Isox; N I L=IsoN I L; x=False x 6= NIL follows: Iso , if and otherwise Isor ;r = 1 2 IsoLef tr ; Lef tr AN D I soRig htr ; R ig htr 1 2 1 2 OR IsoLef tr ; R ig htr AN D I soRig htr ; R ig htr : 1 2 1 2 2 n The above immediately translates into a program, and the time, I claim, is O . 0 2 T; y 2 T x; y The reason is that for any nodes x , the above calls itself on at most y P once (in fact, precisely when x and are at the same depths of their trees.) Let be 0 0 x P R oot y the path from R oot to and be that from to ; in a tree there is exactly one’ such path. Then, of the four recursive calls at any level, at most one involves nodes from 0 P both P and . Inductively, at any recursion depth, at most one call compares a sub-tree y d containing x to one containing . Since all sub-trees called at depth of the recursion are depth d in their respective trees, the claim follows. Therefore the total number of 2 n calls is at most O . 1 PROBLEM 1(8-3IN CLR): Stooge sort Solution: [1 :::n] a) We will prove by induction that Stooge sort correctly sorts the input array A . = 1 n = 2 n = 1 i = j i +1 > j The base cases are for n and .If then , and thus ; [i] n =2 then, line 4 returns A (which is correct). If then lines 1 and 2 guarantee that a bigger element will be returned in a position following that of a smaller element. Thus, n =1 n =2 Stooge sort correctly sorts A when and when . 3 The inductive case happens when n . The inductive hypothesis is that Stooge sort correctly sorts arrays of length <n. We must show that Stooge sort correctly sorts arrays of length n. The input array A is essentially split into three pieces: the first third and the last n 2k k = bn=3c third are both of length k , and the second third is of length ,where . i +1 = n Note 1: j in the first or “top” call to Stooge sort. Note 2: Observe that the = bn=3c n=3 middle third is at least as long as each of the other thirds (since k ,but 2k = n 2bn=3c n2n=3 = n=3 n ; this is actually necessary for the algorithm to work). In line 6, Stooge sort is applied to the first two-thirds of A (i.e., the subarray of length k A n 3 n consisting of the first and middle thirds of ). We are assuming that ,so = bn=3c 1 n k n k and is strictly less than . This means that we can invoke the inductive hypothesis on this subarray (i.e., we assume by inductive hypothesis that the 0 A first two-thirds of A will be sorted in the first two-thirds of the new array ). 0 A n k A After line 6, the largest k elements of will be in the last positions of (i.e., 0 x they will be in the middle third and/or the last third of A ). Here’s why: Let be any one A x A of the k largest elements in .Case1:If is in the last third of , then it will be in the 0 x last third of A , and our claim is rather trivially true. Case 2: Assume is in the first two- x k A n k thirds of A.Since is one of the largest elements in , at least other elements x k n k in A are “smaller” (actually, ) than .Atmost, of these elements can be in the n 2k last third of A. Therefore, at least of these elements are in the first two-thirds of n 2k A A. We can conclude that there are at least elements in the first two-thirds of n 2k = n 2bn=3cn2n=3 = n=3 bn=3c = k that are smaller than x.But . 0 n 2k A Because k , the first k positions of will consist of elements that are all 0 x A x smaller than x. That means that will be in the middle third of . In either case, will 0 be in the last two-thirds of A . 0 In line 7, Stooge sort is applied to the last two-thirds of A (i.e., the subarray of length 0 k A n 3 n consisting of the middle and last thirds of ). We are assuming that ,so = bn=3c 1 n k n k and is strictly less than . This means that we can invoke the inductive hypothesis on this subarray (i.e., we assume by inductive hypothesis that the 0 00 A last two-thirds of A will be sorted in the last two-thirds of the new array ). A Recall that after line 6, the largest k elements of will be in the last two-thirds (i.e., 0 0 k A k A A the last n positions) of . After line 7, the largest elements of (or ) will make 00 up the last third of A , and they will be in sorted order. 00 In line 8, Stooge sort is applied to the first two-thirds of A (i.e., the subarray of 00 k A length n consisting of the first and middle thirds of ). We are assuming that 3 k = bn=3c 1 n k n n ,so and is strictly less than . This means that we can invoke the inductive hypothesis on this subarray (i.e., we assume by inductive hypothesis 00 B that the first two-thirds of A will be sorted in the first two-thirds of the new array ). 2 00 A A Recall that after line 7, the largest k elements of will be in the last third of k in sorted order. After line 8, the other elements (i.e., the n smallest elements) will make up the first two-thirds of B , and they will be in sorted order; also, the last third of 00 A B B will be the same as the last third of . Thus, , which is the final array returned by the algorithm, will consist of all the elements of A in sorted order. n = 3T 2n=3 + O 1 b,c) The recurrence is T , since we split arrays into three smaller arrays, each of size 2/3 the size of the big array (don’t worry about floor and 1 ceiling operators here); the nonrecursive portion of the code is O (i.e., the complexity is bounded by a constant). Its solution is, by Case 1 of the Master Theorem (with log 3 3=2 = 3 b = 3=2 = log 3 T n n log 3 a , ,and ,say), . Now, to compute , 3=2 3=2 log 3 e 2:71 :71 > 2:7 observe that it is equal to . Thus, since 2 , we can say that log 3=2 e 2:7 n n T . The worst-case running time of Stooge sort is worse than all worst-case 2 2 n n n n running times of insertion sort ,mergesort lg , quicksort and n n heapsort lg . Begone, Stooges! Note: You could also look at the corresponding recursion tree, which will have depth log n n = 1 n = 2 (remember that and are both base cases). For simplicity, let’s 3=2 n say that the depth of all leaves in the recursion tree is log . Then, when we add up 3=2 the number of subproblems at each recursive level, we get: log n+1 log n+1 3=2 3=2 log n 13 3 1 0 1 3=2 3 +3 +::: +3 = = 13 2 (Think: “first in” - “first out” .) 1 common base log n 3=2 For simplicity, let’s consider 3 . By Logarithm Property 6 on p.443 in Appendix log n log 3 3=2 3=2 = n O 1 A.5, 3 (look familiar?). Remember that work is being done at each node in the recursion tree. 3 PROBLEM N.1:Stooge sort (problem 8-3, pag. 169 of CLR, questions a,b,c). Solution: [1 :::n] a) We will prove by induction that Stooge sort correctly sorts the input array A . = 2 The base case is for n . In this case the first two lines of Stooge sort guarantee that a bigger element will be returned in a position following that of a smaller element. n = 2 Thus, Stooge sort correctly sorts A when . The inductive case happens when 3 n .

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