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Transcript
MT5821 Advanced Combinatorics
Problems 2
2.1. Show that R2 (2, l) = l for all l. More generally, Rk (k, k, . . . , k, l) = l.
2.2. Prove the Handshaking Lemma:
Lemma 2.1 In any group of people at a meeting, the number of people who have
shaken hands an odd number of times is even.
Now use this lemma to show that R2 (3, 4) ≤ 9, as follows. Examine carefully
the proof in the notes to show that, if we have coloured the 2-subsets of {1, . . . , 9}
so as to avoid both 3-sets with all edges red and 4-sets with all edges blue, then
necessarily each point lies on three red and five blue edges. Now interpreting the
red edges as handshakes, show that the Handshaking lemma is contradicted.
2.3. Show that R2 (3, 4) ≥ 9 by finding a configuration of eight points with the
edges coloured red and blue in such a way as to avoid sets of 3 with all edges red
and sets of 4 with all edges blue. (Hint: Start with a red octagon.)
2.4. Dirichlet’s principle is used to prove the following result.
Theorem 2.2 Let α be an irrational number. Then there exist infinitely many
rational numbers p/q for which
α − p < 1 .
q q2
Prove this theorem, by showing that for every natural number n we can find
a rational number p/q with q < n + 1 and |α − p/q| < 1/nq. For this, let cn =
nα −bnαc for any n. Now the n+1 numbers c1 , c2 , . . . , cn+1 lie in the unit interval.
Divide the interval into n equal parts. By the doocot principle, two of the numbers
ci and c j must lie in the same part, and so differ by at most 1/n.
2.5. A finite set of points in the plane is said to be in general position if no three
of the points are collinear. A convex n-gon is a set of n points in general position,
none of which is in the interior of the polygon formed by the others.
1
(a) Prove that, given five points in general position in the plane, some four of
the points form a convex 4-gon.
(b) Use Ramsey’s Theorem to prove that there is a function F with the property
that, given F(n) points in the plane, some n of them form a convex n-gon.
(Hint: F(n) = R4 (5, n). You need to show that given n points, if any four of
them form a convex 4-gon, then all the points form a convex n-gon.)
Remark This result is called the Happy End Theorem. The picture below shows
George and Esther Szekeres. Esther proved (a), and then George proved (b) (some
years before the photograph was taken).
2.6. Suppose that t + 2 is a prime number. Show that the divisibility conditions of
Proposition 2.9(b) are satisfied for k = t + 1 and n = 2t + 2. [This does not mean
that such Steiner systems exist. For t = 9, Mendelsohn and Hung showed by
computer that S(4, 5, 15) does not exist; hence by Proposition 2.9(a), S(9, 10, 20)
does not exist either.]
2.7. This exercise gives a product construction for Steiner triple systems. Let
(Y, T ) and (Z,U) be Steiner triple systems. Now let X be the Cartesian product
Y × Z, and let S consist of all the 3-sets {(y1 , z1 ), (y2 , z2 ), (y3 , z3 )} for which
• either y1 = y2 = y3 , or {y1 , y2 , y3 } is a block of T ;
• either z1 = z2 = z3 , or {z1 , z2 , z3 } is a block of U.
2
Show that (X, S) is a Steiner triple system. [Note that we cannot have the first
option in both clauses of the definition, since if y1 = y2 = y3 and z1 = z2 = z3 then
the putative 3-set would have only one element.]
2.8. This exercise gives a “doubling” construction for Steiner triple systems. Let
(Y, T ) be a Steiner triple system on n points. Take X = Y × {0, 1} ∪ {∞}, and let
T consist of the 3-sets of one of the following types:
• {∞, (y, 0), (y, 1)}, for all y ∈ Y ;
• {(y1 , e1 ), (y2 , e2 ), (y3 , e3 )}, where {y1 , y2 , y3 } is a block of (Y, T ), e1 , e2 , e3 ∈
{0, 1}, and e1 + e2 + e3 is even.
Prove that (X, S) is a Steiner triple system on 2n + 1 points.
3