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Math 232: Quiz 4 Solutions - Hamilton Cycles and Paths, Quizzes of Discrete Mathematics

Solutions to quiz 4 questions related to finding hamilton cycles and paths in graphs. Step-by-step solutions for two problems, one involving counting the number of hamilton cycles in a graph and the other determining if a graph has a hamilton cycle or path.

Typology: Quizzes

Pre 2010

Uploaded on 07/29/2009

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koofers-user-5a7 🇺🇸

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02/10/09
Math 232: Quiz 4
Name:
Instructions
Show all your work in the space provided. No credit will be given to correct answers
with no supporting work.
If provided, write your final answer in the box.
1. [5] How many Hamilton cycles are there in the graph K3,3?
Solution: Consider the graph K3,3below
abc
def
Choosing alternating vertices from the two partition sets, we have 6×3×2×2×1×1 = 72
possible cycles where one point is fixed and the cycle has a direction. Since Hamilton
cycles use all vertices, we divide by 6 to remove the dependency of the starting vertex.
To get the total amount of undirected cycles, we also must divide by 2. Hence the total
number of Hamilton cycles is 72/12 = 6.
pf2

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02/10/

Math 232: Quiz 4

Name:

Instructions

  • Show all your work in the space provided. No credit will be given to correct answers with no supporting work.
  • If provided, write your final answer in the box.
  1. [5] How many Hamilton cycles are there in the graph K 3 , 3?

Solution: Consider the graph K 3 , 3 below

a (^) b c

d e^ f

Choosing alternating vertices from the two partition sets, we have 6× 3 × 2 × 2 × 1 ×1 = 72 possible cycles where one point is fixed and the cycle has a direction. Since Hamilton cycles use all vertices, we divide by 6 to remove the dependency of the starting vertex. To get the total amount of undirected cycles, we also must divide by 2. Hence the total number of Hamilton cycles is 72/12 = 6.

  1. [5] Find a Hamilton cycle, if one exists for the graph below. If the graph has no Hamilton cycle, determine whether it has a Hamilton path. Justify your answer.

a (^) b

c (^) d e (^) f

g (^) h

Solution: There is no Hamilton cycle since the degree of the vertices c and d is 2. Hence any Hamilton cycle must use edges {a,c}{c,g}{g,d}{d,a}. Since these 4 edges themselves form a cycle, no Hamilton cycle can exist.

The following is a Hamilton path:

a (^) b

c (^) d e (^) f

g (^) h