If a bipartite graph $H$ has $k$ vertices and <maximum degree of a graph>[maximum degree] $d$, then
$$
R(H)=O(d^{2d}k).
$$
A <dependent random choice> argument finds, in one colour, enough vertices whose every subset of at most $d$ vertices has a large common neighbourhood; a greedy embedding then places $H$.

Bounded-degree bipartite Ramsey bound

Let $N=C d^{2d}k$ and red-blue colour $K_N$. Split its vertices into two classes of comparable size. At least half the cross-edges have one colour, say red. The <bounded-degree bipartite Ramsey bound>, proved by <dependent random choice>, gives a set $U$ in one class such that every subset of at most $d$ vertices of $U$ has at least $k$ common red neighbours in the other class.

Let $H=X\sqcup Y$ be a bipartition. Embed $X$ injectively into $U$. List the vertices of $Y$ and embed them one at a time. Each vertex of $Y$ has at most $d$ already embedded neighbours, whose common red neighbourhood has at least $k$ vertices; fewer than $k$ host vertices have yet been used, so a fresh choice is available. This greedily constructs a red copy of $H$. Thus
$$
R(H)\leq C d^{2d}k=O(d^{2d}k).
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 122 / 4 / b / Solution

{wiki}

Dependent random choice finds a large vertex set whose small subsets have large common neighbourhoods. In a bipartite graph with parts $A,B$, choose random vertices of $B$ with repetition and take their common neighbourhood $X\subseteq A$; convexity gives a lower bound for $\mathbb E|X|$, while counting subsets of $X$ with small common neighbourhood allows their deletion.


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