Stat 155 Fall 2009: Solutions to Homework 3 (was due September 24, 2009) 1. This is a sum of two subtraction games. Using the notation from class, we can call the subtraction sets S 4 and S 5 . Then we know that if the two piles have n and m chips, then g 4 ( n ) = n mod 5 and g 5 ( m ) = m mod 6. For ( n,m ) to be a P-position, we must have g (( n,m )) = g 4 ( n ) ⊕ g 5 ( m ) = 0. This is true if and only if g 4 ( n ) = g 5 ( m ), that is: n mod 5 = m mod 6. (Also see Example 2.13.) 2. • The loops in the ﬂower become stalks, and since 1 ⊕ 1 ⊕ 1 = 1, the ﬂower has SG value 2. • In the case of the little girl, fuse the two vertices in her head, since there are two edges, this circuit reduces to a single vertex, as does the circuit that forms her skirt (since it has 4 edges). Proceeding in this way, and using the Colon principle, we see that the SG value of the girl is 3. • Keep in mind that the ground is considered a single vertex, so the legs actually form a circuit with two vertices and two edges. Using
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This note was uploaded on 10/02/2009 for the course STAT 87531 taught by Professor Muralistoyanov during the Spring '09 term at Berkeley.