stat155hw3soln

stat155hw3soln - Stat 155 Fall 2009 Solutions to Homework...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/02/2009 for the course STAT 87531 taught by Professor Muralistoyanov during the Spring '09 term at Berkeley.

Ask a homework question - tutors are online