MGF 1107 Section 15.2 Euler Paths and Euler Circuits Puzzle books ...

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Objective 2: The learner will use Euler’s Theorem. Euler’s Theorem: Number of Odd Vertices in a Connected Graph (Must be an even number) Exactly two •At least one Euler path but no Euler circuit •Each Euler path must start at one of the odd vertices and end at the other one Zero •At least one Euler circuit (which is also an Euler path) •It can start and end at any vertex. More than two •No Euler paths and No Euler circuits Example 2: Given the number of odd and even vertices, determine if the graph has an Euler path (but no Euler circuits), an Euler circuit, or neither. a) The graph has 24 even vertices and no odd vertices. b) The graph has 84 even vertices and two odd vertices. c) The graph has 55 even vertices and six odd vertices. Example 3: Explain why the graph in problem 7 on page 837 has at least one Euler path. Use trial and error to find one such path. Solution: A (3) – odd, B (3) – odd, C (2) – even, D (4) – even, E (2) even. The graph has exactly two odd vertices so it has at least one Euler path. One path : A, C, D, A, B, D, E, B
Example 4: Explain why the graph in problem 10 on page 837 has an Euler circuit. Find one possible circuit. A B C D E Solution: A (4) – even, B (2) – even, C (4) – even, D(4) – even, E(2) – even The graph has no odd vertices so it has at least one Euler circuit. One circuit: A, B, C, E, D, C, A, D, A Example 5: Work problem 12 on page 837. Solution: A (2) – even, B(3) – odd, C(3) – odd, D(2) – even, E(3) – odd We’ve found more than two odd vertices so the graph has no Euler paths and no Euler circuits.
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MGF 1107 Section 15.2 Euler Paths and Euler Circuits Puzzle books ...

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