HW2 - Solution to HW 2 Ning SUN July 28, 2009 2.2#4 (f) One...

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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.

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HW2 - Solution to HW 2 Ning SUN July 28, 2009 2.2#4 (f) One...

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