# Send out b a has another birthday but you cannot say

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send out B, A has another birthday, but you cannot say anything about the relationship between A’s birthday and B’s birthday. 4 First let’s prove that we can indeed move n disks in 2 n - 1 moves. We shall do this by induction. We have: P ( n ) 2 n - 1 Base case: If we have only one disk, our conjecture claims that we can move it in 2 1 - 1 = 1 moves. And indeed, we can move it in one move. Hence, the base case is true. Induction step: Now, let’s assume P ( k ) i.e. if we have k disks, we can move them in 2 k - 1 moves. Then, consider the case when we have k + 1 disks. In other words, we have added another disk to our pile. Then let’s move them to tower B in the following order. 1. Move the top k disks on tower A to tower C. This will take 2 k - 1 moves as per P ( k ). 2. Move the remaining disk on tower A to tower B. This will take 1 move. 3. Move the k disks on tower C to tower B. This will again take 2 k - 1 moves as per P ( k ). 1 or more 3

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Hence we had moves totalling to: 2 k - 1 + 1 + 2 k - 1 = 2 . 2 k - 1 = 2 k +1 - 1 Therefore, we have shown that if we can move k disks in 2 k - 1 moves, then we can move k + 1 disks in 2 k +1 - 1 moves, and one disk can be moved in 2 1 - 1 moves. Hence, by mathematical induction we claim that n disks can be moved in 2 n - 1 moves. So now we want to prove that we cannot do it in better than 2 n - 1 steps. The argument is again inductive, and a minor variation on the first argument. The base case is obvious. Assume true for
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