Chapter One: Recurrent Problems
Warmups
3. Show that, in the process of transferring a tower under the restrictions of the preceding exercise, we will actually encounter every properly stacked arrangement of n disks on three pegs.
Here, the preceding exercise is referred to the exercise in Chapter One warmups exercise 2. In my opinion, every properly stacked arrangement of n disks is referred to as the following kinds of arrangements: (we mark n disks as d1, d2, ..., dn)
1) arrangements compose of 1 disk, such as: (d1), (d2), ..., (dn). There are n arrangements of such kind.
2) arrangements compose of 2 disks, such as: (d1, d2), (d1, d3), ..., (d1, dn), (d2, d3), (d2, d3), ..., (dn-1, dn). There are C(n,2) arrangements of such kind, where C(n,2) is the number of compositions of 2 elements selected from n elements.
3) arrangements compose of 3 disks, such as: (d1, d2, d3), (d1, d2, d4), ..., (dn-2, dn-1, dn). There are C(n, 3) arrangements of such kind.
4) as of 3), but instead arrangements compose of 4 disks.
......
n) arrangements compose of n disks, only (d1, d2, ..., dn).
The total number of arrangements is T(n) = C(n, 1) + C(n, 2) + C(n, 3) + ...... + C(n, n). And the question is all these T(n) arrangements would occur at least once on three pegs during the process of transferring disks in the preceding exercise.
At the first glance, it seems hard to prove because there are T(n) arrangements!! After thinking of several hours, inspired by the previous horse problem, why not use induction? Yes, induction is a suitable tool here.
a) First, we prove the basis. Let n = 1, this should be the trivial case, for d1 would appear on three pegs.
b) Now, we assume that for n >= 1, the proposition in the question holds, and we proceed to prove the proposition holds for n+1 disks. As the steps described in the previous exercise 2 can be divided into three phrase:
ba) When the dn+1 disk is in A, we would need to move n disk above dn+1 from A to B. During this process, according to the assumption, the T(n) arrangements would appear at least once in A, and since dn+1 is at the bottom in A, the T(n+1)-T(n) arrangements would appear at least once in A, and T(n) arrangements would appear at least once in C, and T(n) arrangements would appear at least once in B.
bb) When the dn+1 disk is in C, we would need to move n disk from B to A. During this process, according to the assumption, the T(n) arrangements would appear at least once in C, and since dn+1 is at the bottom in C, the T(n+1)-T(n) arrangements would appear at least once in C, and T(n) arrangements would appear at least once in A and B.
bc) Similarly, when the dn+1 disk is in B, we would need to move n disk from A to B. During this process, according to the assumption, the T(n) arrangements would appear at least once in B, and since dn+1 is at the bottom in B, the T(n+1)-T(n) arrangements would appear at least once in B, and T(n) arrangements would appear at least once in A and C.
bd) Take all analysis above, during the process of moving n+1 disks from A to B, we have T(n+1) arrangements appear at least once in A and B and C, so the proposition holds for n+1.
c) QED