COL106: Data Structures and Algorithms (IIT Delhi, Semester-II )

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1 1 Solve the following recurrence relations giving a Θ bound for each of the cases 1 : (a) T (n) = 2T (n/3) + 1; T (1) = 1 (Assume n is a power of 3) (b) T (n) = 5T (n/4) + n; T (1) = 1 (Assume n is a power of 4) (c) T (n) = 7T (n/7) + n; T (1) = 1 (Assume n is a power of 7) (d) T (n) = 9T (n/3) + n 2 ; T (1) = 1 (Assume n is a power of 3) (e) T (n) = 8T (n/2) + n 3 ; T (1) = 1 (Assume n is a power of 2) (f) T (n) = 49T (n/25) + n 3/2 log n; T (1) = 1 (Assume n is a power of 25) (g) T (n) = T (n 1) + 2; T (1) = 1 (h) T (n) = T (n 1) + n c ; T (1) = 1, where c 1 is a constant (i) T (n) = T (n 1) + c n ; T (1) = 1, where c 1 is a constant (j) T (n) = 2T (n 1) + 1; T (1) = 1 Solution: We give the solution for one of the recurrence relations above The unrolling procedure for the others will be similar T (n) = 2T (n/3) + 1 = 2 (2 T (n/3 2 ) + 1) + 1 = 2 2 T (n/3 2 ) + (2 + 1) = 2 2 (2 T (n/3 3 ) + 1) + (2 + 1) = 2 3 T (n/3 3 ) + ( ) = 2 i T (n/3 i ) + (2 i i ) = 2 log 3 n T (n/3 log 3 n ) + (2 log 3 n i ) = 2 log 3 n T (1) + (2 log 3 n 1) = 2 n log Som T (n) = Θ(n log 3 2 ) 1 This is a question from the Papadimitiou, Dasgupta, and Vazirani book 1 of 5

2 2 Show that the following recursive algorithm for computing the n th Fibonacci number has running time Ω(2 n/2 ) RFib(n) - if (n = 0 or n = 1)return(n) - return(rfib(n 1) + RFib(n 2)) Solution: Let us focus on getting a bound on the number of addition operations within the return statement Let A(n) denote this number Note that the running time T (n) A(n) We can write the following recurrence relation for A(n): A(n) = A(n 1) + A(n 2) + 1; A(0) = 0; A(1) = 0 To get a lower bound on A(n), we use A(n) 2 A(n 2) + 1 Using unrolling, we get the following: This gives T (n) = Ω(2 n/2 ) A(n) 2 A(n 2) A(n 4) + (2 + 1) 2 i A(n 2i) + (2 i ) 2 n/2 A(n 2 n/2 ) + (2 n/2 1) = 2 n/2 1 2 of 5

3 3 Suppose you are given an array A that contains n integers each of which are either 0 or 1 Furthermore, it is known that there is some index i such that A[1] = A[2] = = A[i] = 0 and A[i + 1] = A[i + 1] = = A[n] = 1 Design an algorithm that takes as input such an array and output this index at which the transition from 0 to 1 happens For example, if the array is A = [0, 0, 0, 0, 1, 1], then your algorithm should output 5 Give pseudocode and analyse the running time of your algorithm Solution: Here is the pseudocode for the algorithm This code should be called with FindTransition(A, 1, n) FindTransition(A, i, j) - mid (i + j)/2 - if (A[mid] = 0) - if (A[mid + 1] = 1) return(mid + 1) - else return(findtransition(a, mid + 1, j)) - return(findtransition(a, i, mid)) We can write the following recurrence relation for the running time T (n) of the above recursive algorithm: T (n) T ( n/2 ) + c; T (2) = d, for some constants c and d Since this is the same recurrence relation as for binary search which we have already solved The running time T (n) = O(log n) 3 of 5

4 4 Analyse the running time for the recursive algorithm below LIS(A, n) - If (n = 1) return(1) - max 1 - For j = (n 1) to 1 - if (A[j] A[n]) - s LIS(A, j) - if (max < s + 1) max s return(max) Solution: The recurrence relation for the running time T (n) of the above program is as follows: T (n) = T (n 1) + T (n 2) + + T (1) + cn; T (1) = d Try unrolling the recursion: T (n) = T (n 1) + T (n 2) + + T (1) + cn = 2 (T (n 2) + + T (1)) + cn + c(n 1) = 2 2 (T (n 3) + T (1)) + cn + c(n 1) + 2c(n 2) = 2 3 (T (n 4) + T (1)) + cn + c(n 1) + 2c(n 2) c(n 3) = 2 n 2 T (1) + cn + c(n 1) + 2c(n 2) n 3 c 2 So, some simple statements that we can make are T (n) = Ω(2 n ) and T (n) = O(n 2 n ) 4 of 5

5 5 You are given a sorted array A with n integers and an integer w and you want to determine whether there exists two distinct indices i, j in the array such that A[i] + A[j] = w Design a recursive algorithm for this problem Write pseudocode and discuss running time You should also argue the correctness of your recursive algorithm Solution: Here is the pseudocode for the algorithm This is called with inputs (A, w, 1, n) CheckSum(A, w, i, j) - if (j i)return( No ) - if (A[i] + A[j] = w)return( Yes ) - if (A[i] + A[j] > w)return(checksum(a, w, i, j 1)) - else return(checksum(a, w, i + 1, j)) Running time: The recurrence relation for the running time T (n) is given by: T (n) = T (n 1) + Θ(1); T (1) = Θ(1) Use unrolling of recursion (skipped in this solution but you should do this) to get that T (n) = Θ(n) We can prove correctness of the algorithm using induction Let P (k) denote the proposition that for all i < j such that j i = k the algorithm correctly determines if there are two indices in the subarray A[i]A[j] which sum to w Base case: k = 0 is trivial since the algorithm will correctly return No Inductive step: Assume that P (0),, P (k) are true Let us try proving P (k +1) Consider arbitrary indices i < j such that j i = k + 1 If A[i] + A[j] = w, then the algorithm correctly outputs yes in line 2 If A[i] + A[j] > w, then there cannot exist index i > i such that A[i ] + A[j] = w, so a suitable index pair in subarray A[i]A[j] exists if and only if a suitable index pair in subarray A[i]A[j 1] exists So, again the algorithm in this case returns the correct answer because of our induction hypothesis The final case A[i] + A[j] < w is symmetric to the previous case 5 of 5