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A hypergraph is linear when two distinct hypergraph edges intersect in at most one vertex of a hypergraph. Equivalently, every pair of distinct vertices of a hypergraph has hypergraph codegree at most one. This restricts local overlap without bounding the hypergraph chromatic number: the random alteration method gives linear three-uniform hypergraphs of arbitrarily large hypergraph chromatic number.

Use the random alteration method. Take arbitrarily large $n$ divisible by $2k$, and sample a binomial random graph with edge probability $p=n^{-1+1/(2g)}$. Let $Z$ count cycles of lengths $3,\dots ,g-1$. For each length $\ell$, its expected count is at most $(np{)}^{\ell }/(2\ell )$, soThe sum is empty if $g=3$. By the Markov inequality, $Z\le n/2$ with probability tending to one.

Set $s=n/(2k)$. The expected number of independent $s$-vertex sets isbecause the negative exponent has order $n^{1+1/(2g)}$, dominating the term of order $n$. Thus with probability tending to one there is no independent set of size $s$.

Choose a realization having both properties. Delete one vertex from each of its short cycles, using at most $Z$ deletions. The induced remaining graph $H$ has at least $n/2$ vertices and girth at least $g$. Deletion cannot create a new independent set, so $\alpha (H)\le s-1$. Every colour class is independent, givingIn particular a graph of the required girth and chromatic number exists. This proves graphs of arbitrarily high girth and chromatic number rather than invoking that existence theorem.

We use the random alteration method, controlling large hypergraph independent sets and then removing the few forbidden overlaps. Fix $k=2005$, let $n$ tend to infinity, and independently include each three-element subset of an $n$-vertex set as an hypergraph edge with probability $p=n^{-7/4}$. Call the resulting three-uniform hypergraph $H_0$, and let $s=\lceil n/(2k)\rceil$.

For any prescribed $s$-vertex set, the probability that it is a hypergraph independent set is $(1-p{)}^{\left(\genfrac{}{}{0ex}{}{s}{3}\right)}$. Hence the union bound givesbecause $p\left(\genfrac{}{}{0ex}{}{s}{3}\right)$ has order $n^{5/4}$, with a positive constant depending only on $k$.

Let $Y$ count unordered pairs of distinct hypergraph edges meeting in two vertices. Each potential pair consists of its common pair and two distinct additional vertices, soThe Markov inequality implies $\mathbb{P}(Y\ge n/2)=O(n^{-1/2})$. We may therefore fix one realization for which $Y<n/2$ and no $s$-vertex hypergraph independent set exists. For each offending hypergraph edge pair choose one vertex from its union, and let $D$ be the set of all chosen vertices. Delete $D$, together with all incident hypergraph edges, and call the induced hypergraph $H$. Then $|D|\le Y<n/2$, so $N=|V(H)|>n/2$. The next two parts verify both desired properties of this same $H$.

The binomial upper bound for a Ramsey number gives $R(3,t)\le \left(\genfrac{}{}{0ex}{}{t+1}{2}\right)=O(t^2)$. For the triangle-free Ramsey lower bound by alteration, a binomial random graph $G(n,p)$ has expected triangle count $\left(\genfrac{}{}{0ex}{}{n}{3}\right)p^3$ and expected independent $t$-set count $\left(\genfrac{}{}{0ex}{}{n}{t}\right)(1-p{)}^{\left(\genfrac{}{}{0ex}{}{t}{2}\right)}$. Choose a realization whose sum of these counts is at most its expectation. Delete one vertex from each surviving forbidden set, iterating through the originally listed sets. The remaining graph has neither a triangle nor an independent $t$-set. This random alteration method provesFor large $t$, set $n=\lfloor (t/(4\log t){)}^{3/2}\rfloor$ and $p=n^{-2/3}$. The triangle term is at most $n/6$. Using $\left(\genfrac{}{}{0ex}{}{n}{t}\right)\le (en/t{)}^t$ and $1-p\le e^{-p}$, the logarithm of the independent-set term divided by $t$ is at mostwhich tends to $-\infty$. That term is eventually below $1$. Thus $R(3,t)>5n/6-1$ and

If $m=0$, use a single color. Otherwise chooseThe previous bound gives $\mathbb{E}B\le n/4$, including the case $r=0$. By Markov inequality, $\mathbb{P}(B>n/2)\le 1/2$. Sample the hyperplanes, count the monochromatic edges, and repeat if $B>n/2$. The success probability is at least $1/2$, so at most two draws are needed on average.

For a successful draw, select one endpoint of each monochromatic edge and let $D$ be the union of the selected vertices. Then $|D|\le B\le n/2$. Every monochromatic edge meets $D$, so the original colors give a proper graph colouring on the induced subgraph with vertex set $W=V\setminus D$. Since $|W|\ge n/2$, the coloring of all vertices is a semicoloring. This is a random alteration method; no additional colors for $D$ are required.

The number of available colors satisfies $2^r\le 2\max \{1,(4m/n{)}^{{\log }_{3}2}\}$. A simple graph has $m\le n(n-1)/2$, hence semicoloring by independent hyperplanes gives

Letwhere the positive constant $c$ will be chosen small, and sample the binomial random graph $G(N,p)$. If $X$ counts its copies of $C_4$, thenIf $Y$ counts its independent $k$-sets, thenHere $pk\sim c^{-3/4}\log k$. Choosing $c$ sufficiently small makes the negative exponential term dominate $k\log (eN/k)$, so $\mathbb{E}Y=o(1)$. By Markov inequality, with positive probability $X<N/2$ and $Y=0$.

Starting from such a graph, delete one vertex from each remaining $C_4$. This random alteration method removes fewer than $N/2$ vertices, destroys every $C_4$, and cannot create an independent set of order $k$. The resulting graph has at least $N/2$ vertices, so its edges and nonedges give a red-blue colouring with neither a red $C_4$ nor a blue $K_k$. Consequently

Suppose a graph with $n$ vertices and $m>0$ edges has a vector coloring with adjacent inner products at most $-1/2$. Take $r$ independent copies of random hyperplane rounding and use the $r$ signs as a color. Each edge remains monochromatic with probability at most $3^{-r}$, so the expected number $B$ of monochromatic edges is at most $m{3}^{-r}$ by linearity of expectation.

Choose $r=\max \{0,\lceil {\log }_3(4m/n)\rceil \}$. Then Markov inequality gives $\mathbb{P}(B>n/2)\le 1/2$. On success, delete one endpoint of every monochromatic edge. This random alteration method leaves at least $n/2$ vertices properly colored withThe removed vertices can retain their original colors, as the semicoloring only requires the retained induced subgraph to be proper. A failed draw can be detected and repeated, with at most two trials on average. The case $m=0$ needs just one color. This is the elementary independent-hyperplane construction in Karger, Motwani and Sudan's paper on approximate graph coloring.