Jumpy, Patient Frog Named Sally And A Math Problem About Her Life-II

A frog, namely, Sally, is jumping about the vertices of a hexagon, A B C D E F ABCDEF , each time jumping to an adjacent vertex. In how many ways can she get from A A to C C in 212 212 moves, assuming that there is some plutonium on D D and Sally cannot jump there?

As a bonus, can you generalise for n n jumps?

This problem is a part of my froggy, soggy set .


The answer is 1.252E+50.

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1 solution

Let c k c_k be the number of ways leading from A A to C C , of length 2 k 2k . Let b k b_k be the number of ways of length 2 k 1 2k-1 leading from A A to B B . Since, in two jumps, there are two ways of returning from B B to B B and one way of getting to F F , b k + 1 = 3 b k b_{k+1}=3b_k .

However, c k = b k c_k=b_k , meaning that

c k + 1 = 3 c k c_{k+1}=3c_k when k > 0 k>0 . c 1 c_1 is still 1 1 , and by induction, the generalized formula is c k = 3 k 1 c_k=3^{k-1} .

We would like to express this in terms of n n . As k k is equal to half of n n (the paths being of length n = 2 k n=2k ), we have that there are 3 n 2 1 3^{\frac{n}{2}-1} ways for an even n n . It is impossible to reach C C from A A in an odd number of jumps.

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