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6. Now let D, represent the number of domino tilings of a 3 x n checkerboard. It is considerably trickier to come up with a recurrence relation in this case, though the ideas involved are similar. (a) List (by drawing them) three possibilities for how the rightmost column of a 3 x n board can be covered.

6. Now let D, represent the number of domino tilings of a 3 x n checkerboard. It is considerably trickier to come up with a recurrence relation in this case, though the ideas involved are similar. (a) List (by drawing them) three possibilities for how the rightmost column of a 3 x n board can be covered.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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6. Now let Dn represent the number of domino tilings of a 3 x n checkerboard. It is considerably trickier to come up with a recurrence relation in this case, though the ideas involved are similar. (a) List (by drawing them) three possibilities for how the rightmost column of a 3 x n board can be covered. (b) Let C, be the number of a domino tilings of a 3 x n board with one missing corner. By considering the options you gave in part (a), explain why D, = Dn-2+ 2C-1. (c) Now start with a board missing a corner (say the top right corner). Again consider the possibilities for how to cover the rightmost column, and use your analysis to show that Cn = Dr-1+ Cn-2. (d) Combine the results of (b) and (c) to show that D, = 4D-2 - Dn-4. (e) Find initial conditions Do and D2. What is D, for any odd value of n?
Transcribed Image Text:6. Now let Dn represent the number of domino tilings of a 3 x n checkerboard. It is considerably trickier to come up with a recurrence relation in this case, though the ideas involved are similar. (a) List (by drawing them) three possibilities for how the rightmost column of a 3 x n board can be covered. (b) Let C, be the number of a domino tilings of a 3 x n board with one missing corner. By considering the options you gave in part (a), explain why D, = Dn-2+ 2C-1. (c) Now start with a board missing a corner (say the top right corner). Again consider the possibilities for how to cover the rightmost column, and use your analysis to show that Cn = Dr-1+ Cn-2. (d) Combine the results of (b) and (c) to show that D, = 4D-2 - Dn-4. (e) Find initial conditions Do and D2. What is D, for any odd value of n?
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