Homework |
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Solution: By viewing the points on the circle and the intersection of two chords as vertices, we obtain a plane graph. Each intersection of the |
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chords is determined by four points on the circle, and hence our graph has v = |
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4 and each vertex on the circumference of the circle has degree n + 1, the Handshake Lemma (Theorem 363) we have a total of |
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edges. Discounting the outside face, our graph has |
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faces or regions.
Homework
382 Problem Determine whether there is a simple graph with eight vertices having degree sequence 6, 5, 4, 3, 2, 2, 2, 2.
383 Problem Determine whether the sequence 7, 6, 5, 4, 4, 3, 2, 1 is graphic.
384 Problem (IMO 1964) Seventeen people correspond by mail with one another—each o ne with all the rest. In their letters only three different topics are discussed. Each pair of correspondents deals with only one of these topics. Prove that there at least three people who write to each other about the same topic.
385 Problem If a given convex polyhedron has six vertices and twelve edges, prove that every face is a triangle.
386 Problem Prove, using induction, that the sequence
n, n, n −1, n −1, . . . , 4, 4, 3, 3, 2, 2, 1, 1
is always graphic.
387 Problem Seven friends go on holidays. They decide that each will send a postcard to three of the others. Is it possible that every student receives postcards from precisely the three to whom he sent postcards? Prove your answer!
Answers
383 Using the Havel-Hakimi Theorem, we have
7, 6, 5, 4, 4, 3, 2, 1 →
5, 4, 3, 3, 2, 1, 0 →
3, 2, 2, 1, 0, 0 →
1, 1, 0, 0 →
This last sequence is graphic. Hence the original sequence is graphic.
384 Choose a particular person of the group, say Charlie. He corresponds with sixteen others. By the Pigeonhole Principle, Charlie must write to at least six of the people of one topic, say topic I. If any pair of these six people corresponds on topic I, then Charlie and this pair do the trick, and we are done. Otherwise, these six correspond amongst themselves only on topics II or III. Choose a particular person from this group of six, say Eric. By the Pigeonhole Principle, there must be three of the five remaining that correspond with Eric in o ne of the topics, say topic II. If amongst these three there is a pair that corresponds with each other on topic II, then Eric and this pair correspond on topic II, and we are done. Otherwise, these three people only correspond with one another on topic III, and we are done again.
385 Let x be the average number of edges per face. Then we must have x f = 2e. Hence x = 2e = 24 = 3. Since no face can have fewer
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than three edges, every face must have exactly three edges.
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98 |
Chapter 8 |
386 The sequence 1, 1 is clearly graphic. Assume that the sequence
n −1, n −1, . . . , 4, 4, 3, 3, 2, 2, 1, 1
is graphic and add two vertices, u, v. Join v to one vertex of degree n −1, one of degree of n −2,, etc., one vertex of degree 1. Since v is joined to n −1 vertices, and u so far is not joined to any vertex, we have a sequence
n, n −1, n −1, n −1, n −2, n −2, . . . , 4, 4, 3, 3, 2, 2, 1, 0.
Finally, join u to v to obtain the sequence
n, n, n −1, n −1, . . . , 4, 4, 3, 3, 2, 2, 1, 1.
387 The sequence 3, 3, 3, 3, 3, 3, 3 is not graphic, as the number of vertices of odd degree is odd. Thus the given condition is not realisable.
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