homework-01 - Student Yu Cheng(Jade Math 475 Homework#1...

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Student: Yu Cheng (Jade) Math 475 Homework #1 January 20, 2010 Section 1.8 Exercise 1: Show that an g1865 -by- g1866 chessboard has a perfect cover by dominoes if and only if at least one of g1865 and g1866 is even. Answer: Step 1 . Suppose one of g1865 and g1866 is even and the other one is odd. Since we can cover rows with even number of blocks, we can cover the entire chessboard. For example, a 4-by-3 chessboard: Step 2 . Suppose both g1865 and g1866 are even. Again since we can cover rows with even number of blocks, we can cover the entire chessboard. For example, a 4-by-4 chessboard: Step 3 . Assume both g1865 and g1866 are odd, and the chessboard has a perfect cover. Since the numbers of two color blocks are different, and one domino covers exactly one of the each, the chessboard can’t have a perfect cover. This is a conflict. Therefore, the assumption is wrong. When both g1865 and g1866 are odd, the chess board doesn’t have a perfect cover. In other words, if the chessboard has a perfect cover, at least one of g1865 and g1866 is even. For example, a 3-by-3 chessboard:
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Combining Step 1 , Step 2 , and Step 3 , we’ve shown both side of “if and only if”. Therefore an g1865 - by- g1866 chessboard has a perfect cover by dominoes is the necessary and sufficient condition for at least one of g1865 and g1866 is even, and vice versa. Exercise 2: Consider an g1865 -by- g1866 chessboard with g1865 and g1866 both odd. To fix the notation, suppose that the square in the upper left-hand corner is colored white. Show that if a white square is cut out anywhere on the board, the resulting pruned board has a perfect cover by dominoes. Answer: If we index the blocks as the two dimensional array, the sum of indexes for white blocks is an even number, giving both g1865 and g1866 are odd. For example, a 5-by3 chessboard: Suppose we have white block g1849 g3036,g3037 cut out. We can picture the chessboard as below. It consists four rectangular boards and a missing white block. As we discussed in the first paragraph, the indexes g1861,g1862 are either both even or both odd. If they are both even, then in rectangular g1827 , we have one edge ( g1861 ) being an even number; in rectangular g1828 , we have one edge ( g1865g3398g1862g33971 ) being an even number; in rectangular g1829 , we have one edge ( g1866g3398g1861g33971 ) being an even number; in rectangular g1830 , we have one edge ( g1862 ) being an even number. If the indexes g1861,g1862 are both odd, then the other set of edges in each rectangular are even. They are shown as below. They orange edges are even. Both g1861,g1862 are even Btoh g1861,g1862 are odd
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We’ve proved in Exercise 1 that an g1865 -by- g1866 chessboard has a perfect cover by dominoes if and only if at least one of g1865 and g1866 is even. Therefore all generated rectangular ( g1827,g1828,g1829,g1830 ) have prefect covers. In other words, an g1865 -by- g1866 chessboard with g1865 and g1866 both odd and with one extra white block anywhere on the board cut off would end with a board that can has perfect cover by dominoes.
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