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For a proper three-graph colouring $c$, associate the colors with the three roots of unity $d(i)=\exp (2\pi i(c(i)-1)/3)$, viewed as unit vectors in ${\mathbb{R}}^2$. Define $U_{ij}=2⟨d(i),d(j)⟩$ using the real inner product.

This is twice a Gram matrix, so it is a positive semidefinite matrix. Its diagonal is $2$. On every edge, the endpoint colors differ, so their vectors make angle $2\pi /3$ or $4\pi /3$ and $U_{ij}=2\cos (2\pi /3)=-1$. Thus $t=3$ and this $U$ are feasible in the second semidefinite program, provingThis is the regular simplex construction of a strict vector coloring.

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

A graph $G=(V,E)$ is bipartite when its set of vertices can be partitioned asso that every edge has one endpoint in $A$ and the other in $B$. Equivalently, $G$ has a proper graph colouring with at most two colours.

A $k$-semicoloring of a finite graph assigns one of $k$ colors to every vertex and is a proper graph colouring on an induced subgraph containing at least half the vertices. This is weaker than a proper coloring of the whole graph, and is useful as an intermediate step in a randomized algorithm.

This decision problem asks whether a finite undirected graph has a graph colouring with three colours. It is NP-complete: a colouring certifies membership in NP, while 3-SAT reduces using Boolean-pair colouring gadgets and three-colour clause gadgets.

Take auxiliary graph triangles $a,b,c$ and $d,t,e$, join $a$ to $d$, and add edges $bx,cy,ez$. When $t$ has colour $T$ and the inputs $x,y,z$ have colours in $\{T,F\}$, this graph admits an extension to a graph colouring with three colours exactly when at least one input is $T$. All-false inputs force $a=d=F$, contradicting their edge. Conversely, use $(a,b,c,d,e)=(T,F,B,F,B)$ if $x=T$, $(T,B,F,F,B)$ if $x=F,y=T$, and $(F,T,B,B,F)$ if $x=y=F,z=T$. The extension property is necessary for both directions of a polynomial-time many-one reduction.

A vector $k$-coloring, for $k>1$, assigns a unit vector $v_i$ to each vertex of a graph, such that $⟨v_i,v_j⟩\le -1/(k-1)$ on every edge. A graph colouring with $k$ colors gives a vector $k$-coloring by placing the colors at the vertices of a regular simplex. The Gram matrix of these vectors allows semidefinite programming to search for such a representation.