Combinatorics Through Guided Discovery, 2004a

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175 79. We didn’t explicitly say to use induction here, but, especially in this context, induction is a natural tool to try here. But we don’t have a variable n to induct on. That means you have to choose one. So what do you think is most useful. The number of blocks in the partition? The size of the first block of the partition? The size of the set we are partitioning? Or something else? 80. Think about how you might have gone from the number of double decker cones to the number of triple decker cones in Problem 6. 81. Perhaps the first thing one needs to ask is why proving that if there are ( m+n−2 )people m−1 in a room, then there are either at least m mutual acquaintances or at least n mutual strangers proves that R(m, n) exists. Can you see why this tells us that there is some number R of people such that if R people are in a room, then there are m mutual acquaintances or n mutual strangers? And why does that mean the Ramsey Number exists? Additional Hint: Naturally it should come as no surprise that you will use double induction, and you can use either form. As you think about how to use induction, the Pascal relation will come to mind. This suggests that you want to make assumptions involving ( m+n−3 ) m−1 people in a room, or (m+n−3 m−2 ) people in a room. Now you have to figure out what these assumptions are and how they help you prove the result! Recall that we have made progress before by choosing one person and asking whether this person is acquainted with at least some number of people or unacquainted with at least some other number of people. 82. One expects to need double induction again here. But only because of the location of the problem and because the sum looks like double induction. And those reasons aren’t enough to mean you have to use double induction. If you had this result in hand already, then you could us it with double induction to give a second proof that Ramsey Numbers exist. Additional Hint: What you do need to show is that if there are R(m − 1, n)+ R(m, n −1) people in a room, then there are either m mutual acquaintances or n mutual strangers. As with earlier problems, it helps to start with a person and think about the number of people with whom this person is acquainted or nonacquainted. The generalized pigeonhole principle tells you something about these numbers. 83.b. If you could find four mutual acquaintances, you could assume person 1 is among them. And by the generalized pigeonhole principle and symmetry, so are two of the people to the first, second, fourth and eighth to the right. Now there are lots of possibilities for that fourth person. You now have the hard work of using symmetry and the definition of who is acquainted with whom to eliminate all possible combinations of four people. Then you have to think about nonacquaintances. 86.a. What is the definition of R(n, n)?
176 D. Hints to Selected Problems 86.b. If you average a bunch of numbers and each one is bigger than one, what can you say about the average? 86.c. Note that there are 2 (n 2 ) graphs on a set of n vertices. 86.d. A notation for the sum over all colorings c of K m is ∑ c:c is a coloring of K m , and a notation for the sum over all subsets N of M that have size n is ∑ N:N⊆M, |N|=n 86.e. If you interchange the order of summation so that you sum over subsets first and colorings second, you can take advantage of the fact that for a fixed subset N , you can count count the number of colorings in which it is monochromatic. 86.f. You have an inequality involving m and n that tells you that R(n, n) > m. Suppose you could work with that inequality in order to show that if the inequality holds, then m is bigger than something. What could you conclude about R(n, n)? 87. Remember, a subset of [n] either does or doesn’t contain n. 90.b. A first order recurrence for a n gives us a n as a function of a n−1 . 91. Suppose you already knew the number of moves needed to solve the puzzle with n − 1rings. 92. Ifwehaven − 1circles drawn in such a way that they define r n−1 regions, and we draw a new circle, each time it crosses another circle, except for the last time, it finishes dividing one region into two parts and starts dividing a new region into two parts. Additional Hint: Compare r n with the number of subsets of an n-element set. 98. You might try working out the cases n =2, 3, 4 and then look for a pattern. Alternately, you could write a n−1 = ba n−2 + d, substitute the right hand side of this expression into a n = ba n−1 + d to get a recurrence involving only a n−2 , and then repeat a similar process with a n−2 and perhaps a n−3 and see a pattern that is developing. 102.a. There are several ways to see how to do this problem. One is to draw pictures of graphs with one edge, two edges, three edges, perhaps four edges and figure out the sum of the degrees. Another is to ask what deleting an edge does
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vi a great deal of time to helping
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viii because they use an important
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xii Discovery had been placed on th
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xiv Contents 2.3.1 Undirected graph
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xvi Contents C Exponential Generati
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2 1. What is Combinatorics? learned
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24 1. What is Combinatorics? (vii)
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26 1. What is Combinatorics? · Pro
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28 1. What is Combinatorics? 1.4 Su
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32 2. Applications of Induction and
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68 3. Distribution Problems number
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74 4. Generating Functions • Prob
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78 4. Generating Functions x k in t
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100 5. The Principle of Inclusion a
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104 6. Groups acting on sets on. We
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108 6. Groups acting on sets Let us
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110 6. Groups acting on sets Proble
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112 6. Groups acting on sets 6.1.7
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114 6. Groups acting on sets ⇒ ·
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116 6. Groups acting on sets • Pr
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118 6. Groups acting on sets 3, and
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120 6. Groups acting on sets Proble
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Page 151 and 152: 134 A. Relations Problem 329. Here
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