ma017-2010-sol3

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Unformatted text preview: MA017 2010 HOMEWORK 3 (1) Start with two piles of chips of sizes x , y . Two players alternately take any number of chips from one pile or the same number of chips from each pile, and the winner is the one who takes the last chip. Assuming both players play perfectly, who wins? We identify the losing positions ( x,y ) using the method discussed in class. Note first that ( x,y ) and ( y,x ) are indistinguishable positions, so without loss of generality let us consider positions ( x,y ) for which x ≤ y . Let us find the first few losing positions by hand. In what follows, n ranges over the positive integers n = 1 , 2 , 3 ,... . By assumption, (0 , 0) is losing; thus ( n,n ) and (0 ,n ) are winning. Then (1 , 2) is losing, since all the positions it can reach are winning; thus (1 + n, 2 + n ), (1 , 2 + n ) and (2 , 1 + n ) are winning. Then (3 , 5) is losing, so (3 + n, 5 + n ), (3 , 5 + n ) and (5 , 3 + n ) are winning. Then (4 , 7) is losing, so (4 + n, 7 + n ), (4 , 7 + n ) and (7 , 4 + n ) are winning. Continuing, we see that the losing positions are (0 , 0), (1 , 2), (3 , 5), (4 , 7), (6 , 10), (8 , 13), (9 , 15), (11 , 18), (12 , 20), (14 , 23), (16 , 26), (17 , 28), (19 , 31), (21 , 34), (22 , 36), (24 , 39), (25 , 41), (27 , 44), (29 , 47), (30 , 49), (32 , 53), (33 , 55), and so on. From this we reach the following conjecture: the losing positions ( x,y ) with ( x ≤ y ) are of the form ( a n ,b n ) for n = 0 , 1 , 2 , 3 ,... , where • a n is the smallest positive integer not of the form a k or b k for some k = 0 , 1 ,...,n- 1 , and • b n = a n + n . This is probably obvious to anyone who has worked out extensive numerical examples as above, but let us prove it by induction. First, order the piles lexicographically: if x ≤ y and r ≤ s , let us say that ( x,y ) < ( r,s ) iff either x < r , or x = r and y < s . Then the allowed moves are all monotonic for this ordering: if we perform the move (...
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