What is the upper bound in the context of this question? (Tower of Hanoi)

Question

Consider the Double Tower of Hanoi game. It contains $2n$ disks of $n$ different sizes, arranged in decreasing size from bottom to top on the left most peg. The rules and solving criteria from the original game apply. Let $T$ be the number of moves needed to solve the Tower of Hanoi problem and $D$ the number of moves needed to solve the Double Tower game. Explain why $2T$ is an upper bound for $D$

I get stuck at what they mean by upper bound, as the original game takes a minimum of 7 moves. Which is computed with the algorithm $2^{n} - 1~where~n=3$. So $T = 7$ but if I do $2 * 7 = 14$, it is far less than the number the algorithm would get you which is 63... am I missing the point of what the question is asking? If so what is the point?

Answer 1

You are thinking about making each pair of identical discs into one that is slightly larger and one that is slightly smaller, thus reducing the double tower problem into the single tower problem for twice as many disks. (Is "reducing" the right word in this context?)

If you instead pretend that each pair of same-sized disks is a single disk that costs 2 to move, then a solution to the single Hanoi tower problem with each move substituted for two moves will solve the double problem.

Answer 2

As given I don't think this is true. If we have $n=2$ and start a double tower game with discs of size $(2,1,1,1)$, it will take $7$ moves which is more than twice the $3$ moves for the simple $n=2$ game.

I think it is worst-case to concentrate the extra discs in the smallest size, since those are the ones moved most often.

There may of course be an assumption that there are $2$ discs of each size, but that is not stated if so.