Download 1. The proof follows easily by induction on n. For

1. The proof follows easily by induction on n. For n = 3, both statements
can be easily verified. Assume it is true for any 2-colouring of K n−1
and consider an arbitrarily 2-coloured K n for n ≥ 4. Let v be any
vertex. By induction, Kn − v has a spanning tree T with all edges
of the same colour, say 1. If any edge incident with v has colour 1,
then adding this edge to T gives a spanning tree of K n of colour 1. If
there is no such edge, then the edges of colour 2 incident with v form
a spanning tree of Kn .
Similarly, if Kn − v has a Hamilton cycle with all edges of the same
colour, replace any edge xy in the cycle by the edges xv and vy. This
gives a Hamilton cycle in Kn that is either monochromatic or the union
of two monochromatic paths.
If Kn − v has a Hamilton cycle C that is the union of two monochromatic paths, let x be a vertex that is a common endpoint of the two
paths. Without loss of generality, assume the edge xv is coloured 1.
Let xy be the edge of colour 2 incident with x in C. Replace the edge
xy by edges xv and vy in the cycle. This gives a Hamilton cycle in
Kn that is either monochromatic or the union of two monochromatic
paths.
2. It was shown in class that R(3, 3) = 6, R(3, 4) = 9 and R(k, l) ≤ R(k −
1, l) + R(k, l − 1). This implies R(3, 5) ≤ 14 and R(4, 4) ≤ 18. To show
equality, it is required to construct 2-colourings of K 13 and K17 . It is
possible to systematically derive several properties of these colourings,
and use these properties to cut down the search. Alternatively, you
can read up the description of these graphs.
The colouring of K17 is constructed as follows:
The vertices are {0, 1, . . . , 16} and edge {i, j} is coloured 1 if and only
if i − j ≡ k mod 17 for some k ∈ {1, 2, 4, 8, 9, 13, 15, 16}. These are
numbers that have a square root modulo 17. To verify that this does
not have a K4 of colour 1, by symmetry, it can be argued that if there
is one, then there is one including vertices {0, 1}. The only vertices
adjacent to both are 2, 9 and 16 and they are not adjacent to each
other. Similarly, there is no monochromatic K 4 of colour 2.
The colouring of K13 is similar. The vertices are {0, 1, . . . , 12} and
edge {i, j} is coloured 1 iff i − j ≡ k mod 13 for some k ∈ {1, 5, 8, 12}.
Using symmetry, it is easy to verify that there is no triangle of colour
1 or a K5 of colour 2.
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3. R(3, 3, 3) ≤ 17 follows from the results done in class as R(2, 3, 3) =
R(3, 3) = 6 is even.
To show that it is equal to 17, a 3-colouring of K 16 should be found
such that there is no monochromatic triangle. To describe such a
colouring, use the fact that there exists a finite field with 16 elements.
Label the vertices of K16 by the elements of this field. Let α be an
element in the field that generates the multiplicative group of nonzero elements in the field. The edge {a, b} is coloured k mod 3 where
αk = a − b. Verify that this has no monochromatic triangle. Can you
give a formal proof using properties of finite fields?
If you are not familiar with finite fields, this is a simple construction. Consider the set of all polynomials of degree at most 3 whose
coefficients are 0 or 1. Addition/multiplication of coefficients is done
modulo 2. Addition of polynomials is the usual polynomial addition.
Multiplication of polynomials is done modulo the polynomial x 4 +x+1.
This defines the field and α = x generates all non-zero polynomials.
This is the only 3-colour Ramsey number known exactly.
4. Prove that every tournament of order at least 2 k−1 contains a transitive
subtournament of order k. This is trivial for k = 1 and prove by
induction. Suppose T is a tournament of order at least 2 k for some
k ≥ 1. Let v be any vertex. Then either the indegree or outdegree of
v is ≥ d(2k − 1)/2e = 2k−1 . Without loss of generality, the outdegree
is at least 2k−1 . By induction, the subtournament induced by the
out-neighbours of v contains a transitive subtournament of order k.
Together with v this gives a transitive subtournament of order k + 1
in the original tournament.
To prove the second part, show that there exists a tournament of order
2(k−1)/2 that does not contain a transitive subtournament of order k.
Let the vertices of the tournament be {1, 2, . . . , n}. For every pair of
vertices {i, j}, randomly add either the directed edge (i, j) or (j, i)
with probability 1/2. The probability that a subset of k vertices forms
a transitive tournament is
k
k!.2−( 2) .
Since there are nk possible subsets, the probability that there exists
a transitive subtournament of size k is at most
!
k
k
n
.k!.2−( 2) < n.2−(k−1)/2 .
k
2
If n ≤ 2(k−1)/2 this is strictly less than 1. This implies there exists a
tournament of order 2(k−1)/2 containing no transitive subtournament
of order k.
5. This is known as Schur’s theorem. A simple proof shows that N (r) ≤
N = R(3, 3, . . . , 3; r, 2), the smallest number n such that in any rcolouring of Kn there exists a monochromatic triangle. Consider the
set of numbers {1, 2, . . . , N }. Given any r-colouring of these numbers, define an r-colouring of the complete graph with vertex set
{1, 2, . . . , N } as follows: for i < j let the colour of edge {i, j} be
the colour of the number j − i. Now there exists a monochromatic
triangle in this colouring, say {i, j, k} with i < j < k. Then x = j − i,
y = k − j and z = k − i give the required triple.
This bound is probably weak. It is easy to check that N (2) = 5. The
bound for N (3) obtained above is 17. A lower bound of 14 can be seen
from the colouring {1, 4, 7, 10, 13}, {2, 3, 11, 12}, {5, 6, 8, 9}.
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