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AlgorithmsBy Mohamed Gamal© Mohamed Gamal 2024

a)Let 𝑀(𝑛) be the number of multiplications made by the algorithm.Recurrence relation equation: 𝑀(𝑛) = 𝑀(𝑛 − 1) + 1 and 𝑀 0 = 0• Using forward substitution:𝑀 0 = 0𝑀 1 = 𝑀 0 + 1 = 1𝑀 2 = 𝑀 1 + 1 = 2𝑀 3 = 𝑀 2 + 1 = 3𝑀 𝑛 = 𝑛Basic Operation

a) Another approachLet 𝑀(𝑛) be the number of multiplications made by the algorithm.Recurrence relation equation: 𝑀(𝑛) = 𝑀(𝑛 − 1) + 1 and 𝑀 0 = 0• Using backward substitution:𝑀 𝑛 = 𝑀 𝑛 − 1 + 1= 𝑀 𝑛 − 2 + 1 + 1 = 𝑀 𝑛 − 2 + 2= 𝑀 𝑛 − 3 + 1 + 2 = 𝑀 𝑛 − 3 + 3= 𝑀 𝑛 − 4 + 1 + 3 = 𝑀 𝑛 − 4 + 4𝑀 𝑛 = 𝑀 𝑛 − 𝑖 + 𝑖= 𝑀 0 + 𝑛= 𝑛𝑛 − 𝑖 = 0𝑖 = 𝑛

Example: The Tower of Hanoi PuzzleProblem: move disks from source peg A to destination peg C using an auxiliary (temp) peg B.Constraints:– Only one disk can be moved at a time.– A larger disk cannot be placed on top of a smaller disk.Source DestinationAuxiliary

ALGORITHM: Hanoi(n, src, dest, temp)// Input: no. disks, source, destination, temp// Output: Moves to solve the Tower of Hanoi problemif n = 1 thenMOVE disk from src to dest directlyelseHanoi(n-1, src, temp, dest) // Step 1: from src to tempMOVE disk n from src to dest // Step 2: nth disk from src to destHanoi(n-1, temp, dest, src) // Step 3: from temp to dest

Let 𝑀(𝑛) be the number of moves made by the algorithm.Recurrence relation equation: 𝑀 𝑛 = 𝑀 𝑛 − 1 + 1 + 𝑀(𝑛 − 1) and 𝑀 1 = 1= 2𝑀 𝑛 − 1 + 1• Using forward substitution:𝑀(1) = 1𝑀(2) = 2𝑀(1) + 1 = 3𝑀(3) = 2𝑀(2) + 1 = 7𝑀(𝑛) = 2𝑛 − 1Basic Operation

Let 𝑀(𝑛) be the number of moves made by the algorithm.Recurrence relation equation: 𝑀 𝑛 = 2𝑀 𝑛 − 1 + 1 and 𝑀 1 = 1• Using backward substitution:𝑀(𝑛) = 2𝑀 𝑛 − 1 + 1= 2 2𝑀 𝑛 − 2 + 1 + 1 = 4𝑀(𝑛 − 2) + 3= 4 2𝑀 𝑛 − 3 + 1 + 3 = 8𝑀(𝑛 − 3) + 7𝑀(𝑛) = 2𝑖 𝑀(𝑛 − 𝑖) + (2𝑖 − 1)= 2𝑛−1𝑀(1) + (2𝑛−1− 1)= 2𝑛− 1𝑛 − 𝑖 = 1𝑖 = 𝑛 − 1Basic Operation

Determine what this algorithm computes?𝑄(1) = 1𝑄(2) = 𝑄(1) + 2 ∗ 2 − 1 = 1 + 3 = 4𝑄(3) = 𝑄(2) + 2 ∗ 3 − 1 = 4 + 5 = 9𝑄(4) = 𝑄(3) + 2 ∗ 4 − 1 = 9 + 7 = 16𝑄(𝑛) = 𝑛2Then this algorithm computes 𝑛2for each positive 𝑛.Recurrence relation equation: 𝑄(𝑛 − 1) + 2𝑛 − 1 for 𝑛 > 1a)

b)Let 𝑀(𝑛) be the number of multiplications made by the algorithm.Recurrence relation equation: 𝑀(𝑛) = 𝑀(𝑛 − 1) + 1 and 𝑀 1 = 0• Using backward substitution:𝑀 𝑛 = 𝑀 𝑛 − 1 + 1= 𝑀 𝑛 − 2 + 1 + 1 = 𝑀 𝑛 − 2 + 2= 𝑀 𝑛 − 3 + 1 + 2 = 𝑀 𝑛 − 3 + 3= 𝑀 𝑛 − 4 + 1 + 3 = 𝑀 𝑛 − 4 + 4𝑀 𝑛 = 𝑀 𝑛 − 𝑖 + 𝑖= 𝑀 1 + (𝑛 − 1)= (𝑛 − 1)𝑛 − 𝑖 = 1𝑖 = 𝑛 − 1

c)Let 𝐶(𝑛) be the number of additions and subtractions made by the algorithm.Recurrence relation equation: 𝐶(𝑛) = 𝐶(𝑛 − 1) + 2 and 𝐶 1 = 0• Using backward substitution:𝐶 𝑛 = 𝐶 𝑛 − 1 + 2= 𝐶 𝑛 − 2 + 2 + 2 = 𝐶 𝑛 − 2 + 4= 𝐶 𝑛 − 3 + 2 + 4 = 𝐶 𝑛 − 3 + 6= 𝐶 𝑛 − 4 + 2 + 6 = 𝐶 𝑛 − 4 + 8𝐶(𝑛) = 𝐶(𝑛 − 𝑖) + 2 ∗ 𝑖= 𝐶 1 + 2 ∗ 𝑛 − 1= 2(𝑛 − 1)𝑛 − 𝑖 = 1𝑖 = 𝑛 − 1

a)Let 𝑀(𝑛) be the number of multiplications made by the algorithm.Recurrence relation equation: 𝑀(𝑛) = 𝑀(𝑛 − 1) + 2 and 𝑀 1 = 0• Using forward substitution:𝑀(1) = 0𝑀(2) = 𝑀(1) + 2 = 2𝑀(3) = 𝑀(2) + 2 = 4𝑀(4) = 𝑀(3) + 2 = 6𝑀(5) = 𝑀(4) + 2 = 8𝑀 𝑛 = 2 𝑛 − 1Basic Operation

a) Another approachLet 𝑀(𝑛) be the number of multiplications made by the algorithm.Recurrence relation equation: 𝑀(𝑛) = 𝑀(𝑛 − 1) + 2 and 𝑀 1 = 0• Using backward substitution:𝑀(𝑛) = 𝑀(𝑛 − 1) + 2= [ 𝑀(𝑛 − 2) + 2 ] + 2 = 𝑀(𝑛 − 2) + 4= [ 𝑀(𝑛 − 3) + 2 ] + 4 = 𝑀(𝑛 − 3) + 6= [ 𝑀(𝑛 − 4) + 2 ] + 6 = 𝑀(𝑛 − 4) + 8𝑀(𝑛) = 𝑀 𝑛 − 𝑖 + 2𝑖= 𝑀 1 + 2(𝑛 − 1)= 2 𝑛 − 1Basic Operation𝑛 − 𝑖 = 1𝑖 = 𝑛 − 1

ALGORITHM: S(n)// Input: A positive integer n// Output: The sum of the first n cubesS ← 1for i ← 2 to n doS ← S + i * i * ireturn SThe number of multiplications made by this algorithm will be෍𝑖=2𝑛2 =𝑛 − 2 + 12× (2 + 2) = 2(𝑛 − 1)This is exactly the same number as in the recursive version, but the no recursive versiondoesn't carry the time and space overhead associated with the recursion's stack.b) Comparing with the non-recursive algorithm

b)Let 𝐴(𝑛) be the number of additions made by the algorithm.Recurrence relation equation: 𝐴 𝑛 = 𝐴 𝑛 − 1 + 𝐴(𝑛 − 1) + 1 and 𝐴 0 = 0= 2𝐴 𝑛 − 1 + 1• Using forward substitution:𝐴(0) = 0𝐴(1) = 2𝐴(0) + 1 = 1𝐴(2) = 2𝐴(1) + 1 = 3𝐴(3) = 2𝐴(2) + 1 = 7𝐴(4) = 2𝐴(3) + 1 = 15𝐴 𝑛 = 2𝑛 − 1Basic Operation

b) Another approachLet 𝐴(𝑛) be the number of additions made by the algorithm.Recurrence relation equation: 𝐴(𝑛) = 2𝐴(𝑛 − 1) + 1 and 𝐴 0 = 0• Using backward substitution:𝐴(𝑛) = 2𝐴(𝑛 − 1) + 1= 2[ 2𝐴(𝑛 − 2) + 1 ] + 1 = 4𝐴(𝑛 − 2) + 3= 4[ 2𝐴(𝑛 − 3) + 1 ] + 3 = 8𝐴(𝑛 − 3) + 7= 8[ 2𝐴(𝑛 − 4) + 1 ] + 7 = 16𝐴(𝑛 − 4) + 15𝐴 𝑛 = 2𝑖𝐴 𝑛 − 𝑖 + (2𝑖 − 1)= 2𝑛 𝑀 0 + (2𝑛 − 1)= 2𝑛 − 1Basic Operation𝑛 − 𝑖 = 0𝑖 = 𝑛

d. It’s a very bad algorithm because it is vastly inferior to the algorithm that simplymultiplies an accumulator by 2 𝑛 times. O(2𝑛)

a) The algorithm computes the value of the smallest element in a given array.Let A = [9, 5, 8, 2]▪ n = 4 ≠ 1▪ Call Riddle([9, 5, 8]) (subarray A[0..2]):▪ For Riddle([9, 5, 8]):▪ n = 3 ≠ 1▪ Call Riddle([9, 5]) (subarray A[0..1]):▪ For Riddle([9, 5]):▪ n = 2 ≠ 1▪ Call Riddle([9]) (subarray A[0..0]):▪ n = 1, so it returns 9.▪ Compare temp = 9 with A[1] = 5:▪ 9 > 5, so return 5.▪ Riddle([9, 5]) returns 5.▪ Compare temp = 5 with A[2] = 8:▪ 5 <= 8, so return 5.▪ Riddle([9, 5, 8]) returns 5.▪ Compare temp = 5 with A[3] = 2:▪ 5 > 2, so return 2.

b)Let 𝐶(𝑛) be the number of the number of key comparisons made by the algorithm.Recurrence relation equation: 𝐶(𝑛) = 𝐶(𝑛 − 1) + 1 and 𝐶 1 = 0• Using backward substitution:𝐶 𝑛 = 𝐶(𝑛 − 1) + 1= [ 𝐶(𝑛 − 2) + 1 ] + 1 = 𝐶(𝑛 − 2) + 2= [ 𝐶(𝑛 − 3) + 1 ] + 2 = 𝐶(𝑛 − 3) + 3= [ 𝐶(𝑛 − 4) + 1 ] + 3 = 𝐶(𝑛 − 4) + 4𝐶 𝑛 = 𝐶 𝑛 − 𝑖 + 𝑖= 𝐶 1 + (𝑛 − 1)= 𝑛 − 1Basic Operation𝑛 − 𝑖 = 1𝑖 = 𝑛 − 1

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