Homework Two Answers

- 1. vertices are labeled 0 – 16. Each vertex has arcs to next 3 (e.g. 8 -> 9, 10, 11). Second player wins by always getting to these states: 16,12,8, 4.
- 2. By pidgeonhole principle. N vertices can have at most degree n-1, but if it does, there can't be a vertex of degree 0. So only n-1 different degrees possible.
- 3. not isomorphic: cube has all even cycles, pinwheel has odd cycles.
- 4. a) LHS has 3 verts of degree 4; b) isomorphic
- 5. Number of vertices N must be odd, since number of odd vertices must be even. So in complete graph degree of each vertex is even. Thus if x has odd degree in G, x has odd degree in G-bar. So N-1.
- 6. LHS - yes; RHS- no: dkjic is an odd circuit
- 7. sure - but too hard to draw
- 8. First is impossible - reducible to (2 0 0 0 0); second can be reduced to (2 1 1 1 1), which is a real graph.
- 9. 11. K11 has 55 edges. K10 only has 45 edges, so it's not possible to get to 50 with just 10 vertices.

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Topic revision: r2 - 2012-02-28 - RobbieMoll

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