We often see polished proofs and solutions. They have no mistakes, no signs of exploration… I will not do that here, we are going to explore. Apologies for not having images and videos, they would’ve been of much help.

Exercise 1.11

A Double Tower of Hanoi contains \(2n\) disks of \(n\) different sizes, two of each size. As usual, we’re required to move only one disk at a time, without putting a larger one over a smaller one.

  1. How many moves does it take to transfer a double tower from one peg to another, if disks of equal size are indistinguishable from each other?
  2. What if we are required to reproduce the original top-to-bottom order of all the equal-size disks in the final arrangement? [Hint: This is difficult—it’s really a “bonus problem.”]


We will always have even number of disks. Let’s run the double tower manually. Pick up some objects and transfer the tower, write the results down:

\(\textbf{n}\) \(\textbf{D(n)}\)
2 2
4 6
6 22
8 30

That is odd, why is it jumping at 6? If we disregard \(D(6)\) then there seems to be a pattern. We need to play more with this. Let’s note the results again (I was indeed making mistake with \(D(6)\) for some reason!):

\(\textbf{n}\) \(\textbf{D(n)}\)
2 2
4 6
6 14
8 30

Let’s remind ourselves of the original Tower of Hanoi problem. The results were:

\(\textbf{n}\) \(\textbf{T(n)}\)
1 1
2 3
3 7
4 15
5 31
6 63

The recurrence was:

\[\begin{equation} T_n = 2T_{n-1} + 1 \end{equation}\]

There does not seem to be any correlation between \(T(n)\)’s results and \(D(n)\). You can try various things, such as checking if adding or subtracting between \(T(n)\) and \(D(n)\) results into someting interesting. Or perhaps some offset of \(n\) might work. I couldn’t find anything.

Going forward, should we think in terms of \(n\) disks or \(2n\) disks? Not sure… Anyway, let’s still move on.

To move a tower of \(4\) disks, first \(2\) disks must be transferred to the middle peg. Then \(2\) more moves to transfer bottom two disks. Then \(2\) more moves to transfer \(2\) upper disks.

To transfer \(n\) disks, \(n-2\) disks must first be transferred to the middle peg. Requiring \(D_{n-2}\) moves. Next, \(2\) moves for the bottom \(2\) disks. Finally we need \(D_{n-2}\) more moves.

It seems that

\[\begin{equation} D_n = 2D_{n-2} + 2 \quad \text{for $n>1$.} \end{equation}\]

Does it fit our data?

\[\begin{align*} D_2 & = 2D_0 + 2 \\ & = 2 \\ D_4 & = 2D_2 + 2 \\ & = 6 \\ D_6 & = 2D_4 + 2 \\ & = 14 \\ D_8 & = 2D_6 + 2 \\ & = 30 \end{align*}\]

Yes! It fits, and is already shown to be minimum (perhaps it could be a bit more rigorous).

Now we have to solve the recurrence. Instead of trying induction, let’s try “non-clairvoyant” ways similar to “adding \(1\)” that was tried in the chapter. In the graffito, we find an interesting remark: “Interesing: We get rid of the \(+1\) in \((1.1)\) by adding, not subtracting.” Hmm, we would like to get rid of the \(+2\) and so let’s add \(2\) and see where it goes.

\[\begin{align*} D_n & = 2D_{n-2} + 2 \\ \therefore D_n + 2 & = 2D_{n-2} + 4 \end{align*}\]

Let \(U_n = D_n + 2\).

\[\begin{align*} \therefore U_n & = 2(D_{n-2}+2) \\ \therefore U_n & = 2U_{n-2} \end{align*}\]

Does it take a genius to discover that this is \(2^n - 2\)? Probably not.