The Ramsey Number for 3-Uniform Tight Hypergraph Cycles Article

The Ramsey Number for 3-Uniform Tight Hypergraph Cycles 189
√
By symmetry, we may assume that |Vred| 5s + c0 s.Wefirstprove
that the azure component A contains a matching M (3)
s whose vertex set is in Vblue. Again,
this is true even for t =6s− 2.
Observation 7.5. There exists a matching MA = M (3)
s ⊂ A with V (MA) ⊂ Vblue.
Proof. Let Vblue = V ′ ∪ V ′′ be a partition of Vblue such that |V ′ | = s. Since |V ′′ | � 6s − 2 −
(s − 1) − s � 4s − 1, Theorem 6.1 applied to the induced 2-colouring Kred[V ′′ ] ∪ Kblue[V ′′ ]
of K[V ′′ ] implies that there exists a matching M = M (3)
s in a monochromatic component
of K[V ′′ ] (which is contained in some monochromatic component C in K).
If C is blue, then it must be A, because V ′′ is a subset of Vblue, andwearedone.
Hence assume C = Cred is red. By (7.1), we have C (3)
4 �⊂ Cred. To avoid C (3)
4 in Cred, for
each edge xyz ∈ M and any vertex a ∈ V ′ , at least one of the edges xya, xza, yza must
be a blue edge, and, consequently, also in A, because a ∈ V ′ ⊂ Vblue. Thus, using all s
vertices a ∈ V ′ and s edges of M, we greedily find a matching MA of size s in A. Clearly,
V (MA) ⊂ V ′ ∪ V ′′ = Vblue.
In view of Observation 7.5 and the assumption (7.1), we know that
C (3)
4 �⊂ A. (7.3)
We distinguish two subcases. In the first one we assume that almost all pairs of vertices
from Vblue are contained in the shadows of at most two red components.
Subcase 2a. There exist two red components C1 red and C2 red
��
�
�
� Vblue
�
such that
�
� \(∂C
2
1 red ∪ ∂C 2 red) �
� < 6t. (7.4)
We now prove a series of observations. Recall that by Observation 6.3 the scarlet
component S exists whenever Vred �= ∅. We now show that in that case one of C1 red and
C2 red equals S or can be replaced by S.
Observation 7.6. If Vred �= ∅, then there exists a red component Cred such that
��
�
� ��
�
�
� Vblue
� � V
�
�
�
Proof. Note that � V
2
2
\(∂Cred ∪ ∂S) �
� = �
�
2
\(∂Cred ∪ ∂S) �
� < 24t. (7.5)
� � � Vblue
1 \ 2 ⊂ ∂S. If|∂Cred | � 18t holds, then with Cred = C2 red
��
�
�
�
�
V
�
\(∂Cred �
∪ ∂S) �
2
�
(7.4)
< 6t + |∂C 1 red| � 24t.
we have
Hence, suppose that |∂C 1 red | > 18t and |∂C2 red | > 18t. We claim that there exist vertices
u, v, w ∈ Vblue such that uv ∈ ∂C 1 red , uw ∈ ∂C2 red ,andvw ∈ ∂C1 red ∪ ∂C2 red .
This follows from a simple graph-theoretic fact.