The Tower of Hanoi is a classical problem in computer science and artificial intelligence that demonstrates recursion, state-space representation, and systematic problem solving. The problem consists of three pegs, usually named A, B, and C, and a number of disks of different sizes placed on peg A. The disks are arranged in decreasing order of size, with the largest disk at the bottom and the smallest disk at the top. The objective is to move all the disks from the source peg A to the destination peg C using peg B as an auxiliary peg. The Tower of Hanoi problem follows three important rules. First, only one disk can be moved at a time. Second, only the top disk of any peg can be moved. Third, a larger disk can never be placed on top of a smaller disk. These rules define the legal states and transitions of the problem. In artificial intelligence, the problem can therefore be represented as a state-space tree, where every node represents a particular arrangement of disks on the three pegs and every edge represents one legal disk movement. For a simple state-space representation, consider three disks. Let the initial state be represented as: Initial State: (ABC, –, –) Here, ABC indicates that all three disks are on peg A, with disk 3 being the largest and disk 1 being the smallest. The state can also be written as: A = {3, 2, 1}, B = { }, C = { } The goal state is: Goal State: (–, –, ABC) That is: A = { }, B = { }, C = {3, 2, 1} The state-space tree can be developed by generating only legal moves from every state. The complete solution for three disks requires seven moves. A simplified state-space tree showing the successful path is given below. ```text (ABC, –, –) | A → C | (BC, –, A) | A → B | (C, B, A) | C → B | (C, AB, –) | A → C | (–, AB, C) | B → A | (B, A, C) | B → C | (B, –, AC) | A → C | (–, –, ABC) GOAL ``` The above representation shows one valid sequence of movements. A more detailed state-space tree can be represented by considering all possible legal moves at each stage. For three disks, the first state is: ```text (ABC, –, –) / \ A→B A→C / \ (BC, A, –) (BC, –, A) / \ A→C A→B / \