Question

Problem 5 (Counting triominos) [20 marks] We saw in class that every 2 n × 2 n board, with one square removed, could be covered with triominos. Determine a formula counting the number of triominos covering such a truncated 2 n × 2 n board. Prove this formula by induction.

I have seen solutions for this question posted but they seem to use iteration rather than induction and use notation that I don't understand.

Answer 1

Solution: Let P(n)be the proposition that every 2n×2n checkerboard with one square removed

can be tiled using right triominoes. We can use

mathematical induction to prove that P(n) is true for all positive integers n.

BASIS STEP:

P(1) is

true.because each of the four 2×2checkerboardswith one square removed can be tiled using one right triomino.

INDUCTIVE STEP: The inductive hypothesis is the assumption that P(k)is true for the positive

integer kk; that is, it is the assumption that every 2k×2kcheckerboard with one square removed

can be tiled using right triominoes.

It must be shown that under the assumption of the inductive hypothesis, P(k+1) must also be true; that is, any 2k+1×2(k+1) checkerboard with

one square removed can be tiled using right

triominoes. To see this, consider a 2k+1×2k+1checkerboardd with one square removed. Split this checkerboard into four checkerboards of size 2k×2k by dividing it in half

in both directions. This is illustrated in Figure )No square has been removed from three of these four checkerboards. The fourth 2k×2kcheckerboard has one square removed, so we now use the inductive hypothesis to conclude that it can be covered by right triominoes. Now temporarily remove the square from each of the other three 2k×2kheckerboards that has the center of the original, larger checkerboard as one of its corners. By the inductive hypothesis, each of these three 2k×2kheckerboards with a square removed can be tiled by right triominoes. Furthermore, the three squares that were temporarily removed can be covered by one right triomino. Hence, the entire 2k+1×2k+1 checkerboard can be tiled with right triominoes. We have completed the basis step and the inductive step.

Therefore, by mathematical induction P(n) is true for all positive integers nn. This shows that we can tile every 2n×2ncheckerboard, where n is a positive integer, with one square removed, using right triominoes.