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mdm12Section1_5_OddSolutionsFinal

# mdm12Section1_5_OddSolutionsFinal - Q1-3 Q9 Q15-21 Q5-7...

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Q1-3 Q5-7 Q9 Q11-13 Q15-21 Q23 CHAPTER 1 Section 1.5, pp. 49–52 Solutions to Odd Number Problems 1. a) deg(A) = 2, deg(B) = 2, deg(C) = 3, deg(D) = 2, deg(E) = 3 Since there are exactly two vertices of odd degree, the network is traceable. b) deg(P) = 4, deg(Q) = 5, deg(R) = 3, deg(S) = 4, deg(T) = 5, deg(U) = 3. Since there are more than two odd vertices, the network is not traceable. 3. a) Since no more than three provinces/territories (British Columbia, Yukon, Northwest Territories) are adjacent at one time, three colours are needed. b) Since the U.S.A. creates a fourth adjacent area, four colors are needed. Top

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5. In the existing network 1: deg(A) = 3, deg(B) = 5, deg(C) = 3, deg(D) = 3. It is possible to make the network traceable by adding a bridge joining the two regions A and C. In this new network 2: deg(A) = 4, deg(B) = 5, deg(C) = 3, deg(D) =4. Hence there are only two vertices of odd degree, making the network traceable. 7. a) vertices A B C D E F G H I J K degree 3 6 2 4 3 3 4 4 4 5 6 Four vertices, A, E, F, and K are of odd degree.
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