What is the least number of moves needed to solve the 4-peg Towers of Hanoi problem?

We prove that the solutions to the k-peg Tower of Hanoi problem given by Frame and Stewart are minimal. This paper solves the problem of finding the least number of moves needed to transfer a Tower of Hanoi of n disks, from an initial peg to another one of the k− 1 other pegs. This problem generalizes a well known puzzle proposed and solved in [5] for the case of three pegs. The generalization to the case of k pegs was proposed in [6], and can be phrased as follows: “Given k pegs and n disks initially stacked on one peg in decreasing order of size (i.e. no larger disk can be on top of a smaller one), how many moves are needed to transfer the stack of disks from the initial peg to another peg, assuming that you can move only one disk at a time and it is not allowed to place a larger disk on top of a smaller disk?” Two solutions to this problem were proposed in [2] and [7] using methods that have been shown to be equivalent in [4]. However, as already observed in [1], both presumed solutions make use of a special assumption, which up to date is still unproven, and restricts their proofs of optimality only to algorithms of a certain scheme. These two models of solution can be however regarded as empirically optimal, as verified for up to 20 disks in [3]. In this paper we prove Theorem 1 The solutions to the k-peg Tower of Hanoi problem given in [2] and [7] are minimal. We introduce now some preliminary terminology and notation. We label n disks with the number 1 . . . n, with the convention that the disk j is larger than the disk i if and only if j > i. We use the symbol ∞ to indicate a free peg. The triple (j,i,t), with 1 ≤ j < i ≤ ∞ and j < t ≤ ∞, denotes that the disk j moves from being on the disk i to be placed on the disk t. We say that j ≤ n is freed on a peg, when it moves for the first time (i.e. when we find for the first time the move (j, j+1,∞) or, in the case j = n the move (j,∞,∞)). ∗[email protected]