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| + | 42. a) To solve this problem, divide how you can place the upper right dominoe into two cases | ||
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| + | Case 1: Vertical-if you place a vertical tile in the upper right of the board, you can only place another vertical tile on the left to fill up the open space. After doing this, you have a 2 x (n-2) space to fill. (a_n-2) | ||
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| + | Case 2: Horizontal-if you place a horizontal tile at the top of the board, you are left with a 2 x (n-1) space to fill. (a_n-1) | ||
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| + | Adding these two cases, we find: a_n = a_n-1 + a_n-2 | ||
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| + | b) There is one way to fill a 2 x 1 board with a 2 square tile, so a1 = 1. A 2 x 2 board can be filled with two horizontal tiles or two vertical tiles, so a2 = 2. | ||
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| + | c) To find a17 just repeat the recursion, starting with the initial conditions, until a17 = a16 + a15 is reached. | ||
Revision as of 20:48, 4 March 2009
42. a) To solve this problem, divide how you can place the upper right dominoe into two cases
Case 1: Vertical-if you place a vertical tile in the upper right of the board, you can only place another vertical tile on the left to fill up the open space. After doing this, you have a 2 x (n-2) space to fill. (a_n-2)
Case 2: Horizontal-if you place a horizontal tile at the top of the board, you are left with a 2 x (n-1) space to fill. (a_n-1)
Adding these two cases, we find: a_n = a_n-1 + a_n-2
b) There is one way to fill a 2 x 1 board with a 2 square tile, so a1 = 1. A 2 x 2 board can be filled with two horizontal tiles or two vertical tiles, so a2 = 2.
c) To find a17 just repeat the recursion, starting with the initial conditions, until a17 = a16 + a15 is reached.
