The <Lopsided Lovász local lemma> states that, for bad events $E_i$ with a lopsidependency graph, numbers $x_i\in[0,1)$ satisfying
$$
\mathbb P(E_i)\leq x_i\prod_{j\sim i}(1-x_j)
$$
imply $\mathbb P(\bigcap_iE_i^c)>0$.

Fix any $0<\eta<1/6$, put $N=\lfloor k^{4/3+\eta}\rfloor$, and sample $G(N,p)$ with
$$
p=aN^{-2/3},
$$
where $a>0$ is a sufficiently small absolute constant. Let $A_Q$ be the event that a specified copy $Q$ of $C_4$ is present, and let $B_S$ be the event that a specified $k$-set $S$ is independent. Then
$$
\mathbb P(A_Q)=p^4,
\qquad
\mathbb P(B_S)=(1-p)^{\binom k2}leq e^{-pk^2/3}=:q.
$$
For product measures, the standard monotone-event lopsidependency graph joins an increasing event $A_Q$ to a decreasing event $B_S$ only when they use a common edge. Events of the same monotonicity need no lopsidependency edge. Thus each $A_Q$ has at most $4\binom{N-2}{k-2}$ neighbours of type $B$, and each $B_S$ has at most $Ck^2N^2$ neighbours of type $A$.

Set $x_A=2p^4$ and $x_B=2q$. Since
$$
pk=k^{1/9-2\eta/3+o(1)}\gg\log k,
$$
we have
$$
\binom{N-2}{k-2}x_B=o(1).
$$
On the other side,
$$
Ck^2N^2x_A=O(a^4k^2N^{-2/3}),
$$
whereas $-\log q\geq(a/3)k^2N^{-2/3}$. Choosing $a$ small makes both local-lemma inequalities hold. There is therefore a graph on $N$ vertices containing no $C_4$ and no independent $k$-set. Taking, for example, $\eta=1/12$ proves
$$
R(C_4,K_k)\geq\Omega(k^{4/3+1/12}).
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 122 / 2 / b / Solution

{c}
{wiki=Lovász_local_lemma#Lopsided_Lovász_local_lemma}

Let $A_1,\ldots,A_m$ be bad events with a lopsidependency graph. If numbers $x_i\in[0,1)$ satisfy
$$
\mathbb P(A_i)\leq x_i\prod_{j\sim i}(1-x_j)
$$
for every $i$, then $\mathbb P(\bigcap_iA_i^c)>0$. Unlike the ordinary <Lovász local lemma>, a lopsidependency graph may omit pairs whose interaction can only make their simultaneous avoidance easier.

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