by Victor Bryant
Published Friday June 14 2024
The famous “Tower of Hanoi” puzzle consists of a number of different-sized rings in a stack on a pole with no ring above a smaller one. There are two other poles on which they can be stacked and the puzzle is to move one ring at a time to another pole (with no ring ever above a smaller one) and to end up with the entire stack moved to another pole.
Unfortunately I have a faulty version of this puzzle in which two of the rings are of equal size. The minimum number of moves required to complete my puzzle (the two equal rings may end up in either order) consists of four different digits in increasing order.
What is the minimum number of moves required?
Sunday Times Teasers Discussion Group 
If there are (n+1) rings with the duplicate rings at the m’th level from the bottom, then one first needs to find the minimum number of moves for the basic problem with n different-sized rings. Then consider the number of moves which involve moving the ring at level m – these particular moves have to be replicated, as the two equal sized rings at this level simply need to be moved as a pair one after another.
Long ago I worked out for myself the maths behind the Tower of Hanoi puzzle. This made this teaser, which has a very small solution space, straightforward
I played with a 7-ring Hanoi Tower.
The minimum number of moves is obtained with a pair of the largest possible rings satisfying the condition that the number has 4 digits in increasing order.
I found that, In a standard tower with n different rings, the contributions of largest to smallest ring in are 2^0, 2^1,…2^(n-1).
Choosing n-1 different sizes and doubling a contribution led me to a valid number of moves.
I did it by subtraction.
It is clear what the range of n must be to produce 4-digit answers.
Work out the additional moves needed for the two disks below each n, and subtract these.
There are three 4-digit values possible, only one of which has four unique digits.
I considered n rings in stack positions 1 (smallest) to n (largest). The upper ring in the identical sized pair can be in positions 1 to (n-1). The number of moves required to move the stack from one pole to another is M(n,m). By observing trends in the solution of stacks with small n (eg n = 2, 3, 4 …) it is possible to fairly quickly derive an analytic expression for M(n,m). This shows that only 3 values of n lead to 4-digit moves where the first digit must be in the range 1 to 6. The unique answer is then easily found.
I enjoyed this as I remember having one of these as a kid with coloured wooden squares. Can’t remember how many layers it had or the colours, maybe about 7 and possibly the rainbow.
Reading others comments now it looks like I did it the same way, working out the simple formula for n disks all different then examining how it changes for two the same in different possible positions. Yes there are 3 potential n to calculate for so not much effort to find the answer with 4 digits in increasing order. Fun teaser.
If the posts are at the corners of a triangle, or on a circle, then the odd-numbered pieces always travel in one direction, and even-numbered in the other. An easy way to remember what to do is that the smallest ring moves on every other move, and always in the same direction.
For the standard version, if 1is smallest, 2 next smallest etc the recipe for 1,2,3….rings is
1
1,2,1
1,2.1,3.1,2,1,
1,2,1.3.1,2,1,4,1,2,1,3,1,2,1 etc
Just for completeness, the analytic expression (as mentioned above) for the minimum number of moves needed to transpose a stack of n rings where the upper ring of an identical pair is in a position between 1 and (n-1) is given by: M(n,m) = 2^(n-1) – 2^(n-m-1) – 1