Write a divide-and-conquer algorithm for the Towers of Hanoi problem. The Towers of Hanoi problem consists of three pegs and \(n\) disks of different sizes. The object is to move the disks that are stacked, in decreasing order of their size, on one of the three pegs to a new peg using the third one as a temporary peg. The problem should be solved according to the following rules: (1) when a disk is moved, it must be placed on one of the three pegs; (2) only one disk may be moved at a time, and it must be the top disk on one of the pegs; and (3) a larger disk may never be placed on top of a smaller disk. a. Show for your algorithm that \(S(n)=2^{n}-1\). (Here \(S(n)\) denotes the number of steps (moves), given an input of \(n\) disks.) b. Prove that any other algorithm takes at least as many moves as given in part (a).

Step by step solution

01

Understanding the problem

The Towers of Hanoi is a classic problem of recursion. The objective is to move all \(n\) disks from one peg to another, obeying the rules that only one disk can be moved at a time and a larger disk can never be put on top of a smaller disk. An additional peg is provided to facilitate the moves.

02

Formulate the algorithm recursively

The task can be solved by recursively moving \(n-1\) disks to the auxiliary peg, moving the \(n\)th disk to the target peg, and finally moving the \(n-1\) disks from the auxiliary peg to the target peg.\n\nAlgorithm Towers_of_Hanoi(n, source, auxiliary, target)\n\n IF n > 0 THEN\n Towers_of_Hanoi(n-1, source, target, auxiliary)\n move disk from source to target\n Towers_of_Hanoi(n-1, auxiliary, source, target)

03

Derive the number of moves for your algorithm

The recursive nature of the problem leads to the recursion relation \(S(n) = 2S(n - 1) + 1\). By solving this recurrence relation, we can find that \(S(n) = 2^{n} - 1\) steps.

04

Prove that any other algorithm takes at least as many moves

To prove this, let's suppose there exists an algorithm \(A'\) that could solve the problem in less than \(2^{n} - 1\) steps for an \(n\) disk problem. It means there would exist a situation where we can remove the largest disk and still manage to transfer all the remaining \(n-1\) disks from source to destination peg in less than \(2^{n-1} - 1\) steps, contradicting our earlier claim. Therefore, no such algorithm \(A'\) exists, and the minimum number of moves required to solve the Tower of Hanoi problem with n disks is \(2^{n} - 1\).

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