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Tiling with Triominoes (Posted on 2023-09-22) Difficulty: 4 of 5
There are two triominoes. One is simple a 1x3 rectangle and the other is a small L-shaped piece.

Using any combination of these two triominoes and in any orientation, how many ways can a 3 by N rectangle be tiled?
Rotations and reflections of an asymmetrical solution are to be considered distinct.

No Solution Yet Submitted by Brian Smith    
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re: soln | Comment 3 of 6 |
(In reply to soln by Steven Lord)

Very nice job, Steve.  I can help you over the last hurdle going from your summation of f(n) and s(n) to the finite recursion f(n).


s(n) eventually reduces to a cyclic sequence of period 3.  So that motivates me to compare f(n) and f(n-3) when expanded:

f(n) = f(n-1) + 2*f(n-2) + 5*f(n-3) + 2*f(n-4) + 2*f(n-5) + 4*f(n-6) + 2*f(n-7) + 2*f(n-8) + 4*f(n-9) + ....
f(n-3) = f(n-4) + 2*f(n-5) + 5*f(n-6) + 2*f(n-7) + 2*f(n-8) + 4*f(n-9) + ....

Now it is easy to see that the expansions for f(n) and f(n-3) are identical for terms f(n-7) and later.  Then subtracting the two series will have the long tails cancel each other.

f(n)-f(n-3) = [f(n-1) + 2*f(n-2) + 5*f(n-3) + 2*f(n-4) + 2*f(n-5) + 4*f(n-6)] - [f(n-4) + 2*f(n-5) + 5*f(n-6)]

Now just some basic simplification and we get f(n) = f(n-1) + 2*f(n-2) + 6*f(n-3) + f(n-4) - f(n-6)

  Posted by Brian Smith on 2023-11-08 10:58:57
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