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Chapter 4 sec. 2 A famous and difficult problem to solve in graph theory. 1. Hamiltonian path- A path that passes through all the vertices of a graph exactly once. 2. Hamiltonian circuit- If a Hamilton path begins and ends at the same vertex. (If a graph has a Hamilton circuit we will say it is Hamiltonian) Although the definitions of Hamilton path and Euler path sound similar, they are not the same. Why? In producing a Hamilton path, you do not have to trace every edge, as with an Euler path. A A B B C C D D Graph (a) E E Graph (b) A) If we start from pt. A, the path is AEDCB is the Hamilton path. The path AEDCBA is the Hamilton circuit. B) There are no Hamilton path or Hamilton circuits. A complete graph- Is one in which every pair of vertices is joined by an edge. completed graph with n vertices is denoted by Kn A A B C K3 A B C D K4 A C B D ABCDA, ABDCA, ACBDA, ACDBA, ADBCA, ADCBA Q: What if I ask you to find all the Hamilton circuits for a K6 or K10? A: I hope that you will not list all the combinations, but tell me how many Hamilton circuits there are. Kn has (n-1)(n-2)(n-3)(n-4)…..3x2x1 Hamilton circuits. When there is a pattern as shown, it is called a factorial. Which is written as (n-1)!. Therefore, K6 = 5x4x3x2x1 = 5! = 120 Hamilton circuits. K10 = 9! = 362,880 Hamilton circuits. 3 Methods to solve the Travel Salesperson. 1. Brute Force 2. Nearest Neighbor 3. Best Edge Weighted Graph- we assign numbers to the edges of a graph. Weights- edges Weight the numbers on the of a path- is the sum of the weights of the edges of the path. You want to conference in 5 different cities across the U.S. The weights represents money. (In other problems it could represent distance or time.) The five cities are: ◦A = Albuquerque ◦L = Las Vegas ◦S = Seattle ◦B = Boston ◦D = Dallas S 350 230 240 310 B 210 290 A 110 250 180 L 170 D Find the Hamilton circuit that has the smallest weight. Solution: ◦ ALDSBA = $1,120 ◦ ASBDLA = $1,110 ◦ ADBLSA = $1,210 Adv. Or Disadv.? 1. Begin at A (Albq) 2. View the weights around A. ◦ AB = 210 ◦ AS = 230 ◦ AD = 180 ◦ AL = 110 3. Since AL has the lowest weight, that is our next vertex. 4. View the weights around L. ◦ LD = 170 ◦ LS = 240 ◦ LB = 310 5. D is the lowest and next vertex. 6. Keep doing this til you reach A. Solution: ALDBSA = $1,100 1. Begin by choosing any edge with the smallest weight. 2. Choose any remaining edge in the graph with the smallest weight. 3. Keep repeating step 2; however, do not allow a circuit to form until all vertices have been used. Also, because the final hamilton circuit cannot have three edges joined to the same vertex, new allow this to happen during the construction of the circuit. 1. Smallest edge to largest: AL(110), LD(170), AD(180), AB(210), AS(230), LS(240), BD(250), SD(290), LB(310), SB(350) 2. AL, LD, (cannot use AD for it closes the circuit.) DB, BS, SA