|
|
Line 16: |
Line 16: |
| | | |
| b) Prove that such a tasteful tiling is unique. | | b) Prove that such a tasteful tiling is unique. |
− |
| |
− | == Solution by DGUPTA714 ==
| |
− | Prove (a): To prove that any chessboard polygon that can be tiled by dominoes can be done so tastefully, we will use mathematical induction. We will show that any rectangular polygon can be tiled tastefully, and then show that any polygon that can be divided into smaller rectangular polygons can also be tiled tastefully.
| |
− |
| |
− | Base case: A 1 x 1 rectangular polygon can be tiled tastefully with a single domino.
| |
− | Induction hypothesis: Let us assume that any rectangular polygon of area n can be tiled tastefully with dominoes.
| |
− | Induction step: Let us consider a rectangular polygon P of area n + 1. If P is already tastefully tiled, then we are done. Otherwise, there must be a vertical or horizontal domino that violates the tasteful tiling condition. Without loss of generality, let us assume that there is a vertical domino dividing P into two subpolygons P1 and P2.
| |
− |
| |
− | Since P1 and P2 have smaller areas than P, by the induction hypothesis they can be tiled tastefully. We can then combine these tasteful tilings to obtain a tasteful tiling of P by removing the vertical domino and replacing it with two horizontal dominoes.
| |
− | By a similar argument, if P can be divided into smaller rectangular polygons, each of which can be tiled tastefully, then P itself can be tiled tastefully.
| |
− |
| |
− |
| |
− |
| |
− | Prove (b): To prove that a tasteful tiling is unique, we will assume the opposite: that there are two different tasteful tilings of the same polygon.
| |
− | Let us consider the first location where the two tilings differ. WLOG, let us assume that in one tiling there is a horizontal domino and in the other tiling there is a vertical domino at this location.
| |
− | Since the two tilings are tasteful, neither of them can have the two configurations of dominoes shown on the left in the problem statement. Therefore, the horizontal domino must be part of a larger horizontal segment of dominos, and the vertical domino must be part of a larger vertical segment of dominos.
| |
− | Now let us consider the four sub polygons formed by the intersection of these two segments. Each of these subpolygons is a rectangular chessboard polygon, and each of them can be tiled tastefully because they are smaller than the original polygon. Therefore, we can combine these tasteful tilings to obtain a tasteful tiling of the original polygon that differs from both of the original tilings. This contradicts our assumption that the two original tilings were different.
| |
− | Therefore, there can be no two different tasteful tilings of the same polygon, and any tasteful tiling is unique.
| |
− |
| |
− | Not sure if my proofs make any sense, but don't attack me if it's wrong.
| |
| | | |
| == See also == | | == See also == |