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6 Pages

### ps1sol

Course: COMS 12100, Fall 2009
School: East Los Angeles College
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Word Count: 1773

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to Introduction Algorithms Massachusetts Institute of Technology Professors Erik D. Demaine and Charles E. Leiserson September 30, 2005 6.046J/18.410J Handout 8 Problem Set 1 Solutions Problem 1-1. Asymptotic Notation For each of the following statements, decide whether it is always true, never true, or sometimes true for asymptotically nonnegative functions f and g. If it is always true or never true, explain...

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to Introduction Algorithms Massachusetts Institute of Technology Professors Erik D. Demaine and Charles E. Leiserson September 30, 2005 6.046J/18.410J Handout 8 Problem Set 1 Solutions Problem 1-1. Asymptotic Notation For each of the following statements, decide whether it is always true, never true, or sometimes true for asymptotically nonnegative functions f and g. If it is always true or never true, explain why. If it is sometimes true, give one example for which it is true, and one for which it is false. (a) f (n) = O(f (n)2 ) Solution: Sometimes true: For f (n) = n it is true, while for f (n) = 1/n it is not true. (The statement is always true for f (n) = (1), and hence for most functions with which we will be working in this course, and in particular all time and space complexity functions). (b) f (n) + g(n) = (max (f (n), g(n))) Solution: Always true: max(f (n), g(n)) f (n) + g(n) 2 max(f (n), g(n)). (c) f (n) + O(f (n)) = (f (n)) Solution: Always true: Consider f (n) + g(n) where g(n) = O(f (n)) and let c be a constant such that 0 g(n) < cf (n) for large enough n. Then f (n) f (n) + g(n) (1 + c)f (n) for large enough n. (d) f (n) = (g(n)) and f (n) = o(g(n)) (note the little-o) Solution: Never true: If f (n) = (g(n)) then there exists positive constant c and n such that for all n > n , cg(n) f (n). But if f (n) = o(g(n)), then for any positive constant c, there exists no (c) such that for all n > no (c), f (n) < cg(n). If f (n) = (g(n)) and f (n) = o(g(n)), we would have that for n > max(n , no (c )) it should be that f (n) < c g(n) f (n) which cannot be. (e) f (n) = O(g(n)) and g(n) = O(f (n)) Solution: Sometimes true: For f (n) = 1 and g(n) = n sin(n) it is true, while for any f (n) = O(g(n)), e.g. f (n) = g(n) = 1, it is not true. 2 Problem 1-2. Recurrences Handout 8: Problem Set 1 Solutions Give asymptotic upper and lower bounds for T (n) in each of the following recurrences. Assume that T (n) is constant for n 3. Make your bounds as tight as possible, and justify your answers. (a) T (n) = 2T (n/3) + n lg n Solution: By Case 3 of the Master Method, we have T (n) = (n lg n). (b) T (n) = 3T (n/5) + lg2 n Solution: By Case 1 of the Master Method, we have T (n) = (nlog5 (3) ). (c) T (n) = T (n/2) + 2n Solution: Case 3 of masters theorem, (check that the regularity condition holds), (2n ). (d) T (n) = T ( n) + (lg lg n) Solution: Change of variables: let m = lg n. Recurrence becomes S(m) = S(m/2) + (lg m). Case 2 of masters theorem applies, so T (n) = ((lg lg n)2 ). (e) T (n) = 10T (n/3) + 17n1.2 Solution: Since log3 9 = 2, so log3 10 > 2 > 1.2. Case 1 of masters theorem applies, (nlog3 10 ). (f) T (n) = 7T (n/2) + n3 Solution: By Case 3 of the Master Method, we have T (n) = (n3 ). (g) T (n) = T (n/2 + n) + 6046 Solution: By induction, T (n) a monotonically increasing function. Thus, for large is enough n, T (n/2) T (n/2 + n) T (3n/4). At each stage, we incur constant cost 6046, but we decrease the problem size to atleast one half and at most three-quarters. Therefore T (n) = (lg n). Handout 8: Problem Set 1 Solutions (h) T (n) = T (n 2) + lg n Solution: T (n) = (n log n). This is T (n) = i=1 lg 2i i=1 lg i (n/4)(lg n/4) = (n lg n). For the upper bound, note that T (n) S(n), where S(n) = S(n1)+lg n, which is clearly O(n lg n). (i) T (n) = T (n/5) + T (4n/5) + (n) Solution: Masters theorem doesnt apply here. Draw recursion tree. At each level, do (n) work. Number of levels is log5/4 n = (lg n), so guess T (n) = (n lg n) and use the substitution method to verify guess. In the f (n) = (n) term, let the constants for (n) and O(n) be n0 , c0 and c1 , respectively. In other words, let for all n n0 , we have c0 n f (n) c1 n. First, we show T (n) = O(n). For the base case, we can choose a sufciently large constant d1 such that T (n) < d1 n lg n. For the inductive step, assume for all k < n, that T (k) < d1 n lg n. Then for k = n, we have T (n) T 4n n +T + c1 n 5 5 n 4n n 4n d1 lg + d1 lg + c1 n 5 5 5 5 4d1n 5 d1 n + c1 n lg 5 lg = d1 n lg n 5 5 4 lg 5 + 4 lg(5/4) = d1 n lg n n d 1 c1 . 5 n/2 n/2 3 The residual is negative as long as we pick d1 > 5c1 /(lg 5+4 lg(5/4)). Therefore, by induction, T (n) = O(n lg n). To show that T (n) = (n), we can use almost the exact same math. For the base case, we choose a sufciently small constant d0 such that T (n) > d0 n lg n. For the inductive step, assume for all k < n, that T (k) > d0 n lg n. Then, for k = n, we have T (n) T n 4n +T + c0 n 5 5 4n n 4n n + d0 lg + c0 n d0 lg 5 5 5 5 lg 5 + 4 lg(5/4) d0 . d0 = n lg n + n c0 5 4 Handout 8: Problem Set 1 Solutions The residual is positive as long as d0 < 5c0 /(lg 5 + 4 lg(5/4)). Thus, T (n) = (n lg n). (j) T (n) = n T ( n) + 100n Solution: Masters theorem doesnt apply here directly. Pick S(n) = T (n)/n. The recurrence becomes S(n) = S( n) + 100. The solution of this recurrece is S(n) = (lg lg n). (You can do this by a recursion tree, or by substituting m = lg n again.) Therefore, T (n) = (n lg lg n). Problem 1-3. Unimodal Search An array A[1 . . n] is unimodal if it consists of an increasing sequence followed by a decreasing sequence, or more precisely, if there is an index m {1, 2, . . . , n} such that A[i] < A[i + 1] for all 1 i < m, and A[i] > A[i + 1] for all m i < n. In particular, A[m] is the maximum element, and it is the unique locally maximum element surrounded by smaller elements (A[m 1] and A[m + 1]). (a) Give an algorithm to compute the maximum element of a unimodal input array A[1 . . n] in O(lg n) time. Prove the correctness of your algorithm, and prove the bound on its running time. Solution: Notice that by the denition of unimodal arrays, for each 1 i < n either A[i] < A[i + 1] or A[i] > A[i + 1]. The main idea is to distinguish these two cases: 1. By the denition of unimodal arrays, if A[i] < A[i + 1], then the maximum element of A[1..n] occurs in A[i + 1..n]. 2. In a similar way, if A[i] > A[i + 1], then the maximum element of A[1..n] occurs in A[1..i]. This leads to the following divide and conquer solution (note its resemblance to binary search): 1 a, b 1, n 2 while a < b 3 do mid (a + b)/2 4 if A[mid] < A[mid + 1] 5 then a mid + 1 6 if A[mid] > A[mid + 1] 7 then b mid 8 return A[a] Handout 8: Problem Set 1 Solutions The precondition is that we are given a unimodal array A[1..n]. The postcondition is that A[a] is the maximum element of A[1..n]. For the loop we propose the invariant The maximum element of A[1..n] is in A[a..b] and a b. When the loop completes, a b (since the loop condition failed) and a b (by the loop invariant). Therefore a = b, and by the rst part of the loop invariant the maximum element of A[1..n] is equal to A[a]. We use induction to prove the correctness of the invariant. Initially, a = 1 and b = n, so, the invariant trivially holds. Suppose that the invariant holds at the start of the loop. Then, we know that the maximum element of A[1..n] is in A[a..b]. Notice that A[a..b] is unimodal as well. If A[mid] < A[mid + 1], then the maximum element of A[a..b] occurs in A[mid + 1..b] by case 1. Hence, after a mid + 1 and b remains unchanged in line 4, the maximum element is again in A[a..b]. The other case is symmetric. To complete the proof, we need to show that the secon...

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0: ; This tests out every instruction 0: ; 0: LOD 1000 1: ADD 1001 2: ADD 1002 3: STD 1003 4: A2S 5: LDI 1001 6: S2A 7: ; 7: ; Now test out loading indirectly thro
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LOD XSTD Y
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LDI 17 SHL STD X NOP HLTX: NUM 44
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0: LOD X 1: A2S 2: LDS 3: HLT 4: 4: X: NUM 6000
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