Combinatorics Through Guided Discovery, 2004a

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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)?