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Unformatted text preview: Solution to HW 2 Ning SUN July 28, 2009 2.2#4 (f) One Hamilton path is: d e c b a g f h i n k j l m . No Hamilton circuit. Method 1: The graph is symmetric. If c d is not in the Hamilton circuit, Rule 1 at h and f is applied, forcing b c , b d , c e , d e to be used. But they form a subcircuit and violates Rule 2. So c d is in the Hamilton circuit. In the same way, you can show that f h and j l should be in the circuit. Observe vertex d . Either b d or d e should be in the circuit and the two cases are symmetric. Observe vertex h . Either g h or h i should be in the circuit and the two cases are symmetric. Without loss of generality, suppose b d and g h are in the Hamilton circuit, then you should include a b , c d , e n ; and a g , f i , i n in the circuit by using Rule 1. Now you get a subcircuit, which violates Rule 2. So there does NOT exist a Hamilton circuit. Method 2: There are 3 regions bounded by 8 edges (including the outside region), and 6 bounded by 3 edges in the planar graph. So we know that r 8 + r 8 = 3 , and r 3 + r 3 = 6 . By using Thm 3 in class, if there exists a Hamilton circuit, then 6( r 8 r 8 ) + 1( r 3 r 3 ) = 0 ....
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This note was uploaded on 08/30/2010 for the course AMS 301 taught by Professor Arkin during the Spring '08 term at SUNY Stony Brook.
 Spring '08
 ARKIN

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