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The Johnson graph $J(5,2)$ is the line graph of $K_5$ and is a $6$-regular graph on $10$ vertices and $30$ edges. As $K_5$ has chromatic index $5$ so the chromatic number of $J(5,2)$ is $\chi(J(5,2))=5$.

By Vizing's theorem the chromatic index of $J(5,2)$ is $6$ or $7$. I want to know whether it is known.

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    By Vizing's theorem the chromatic index of $J(5,2)$ is $4$ or $5$, because $4$ is the maximal degree of vertices of $K_5$. – 2017-02-19
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    @Wolfram: Vizing's theorem has to be applied to $J(5,2)$ which is $6$-regular. – 2017-02-19
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    "The edges of every simple undirected graph may be colored using a number of colors that is at most one larger than the maximum degree $\Delta$ of the graph." If we color the vertices of $J(5,2)$, then we apply the theorem to $K_5$, if we color the edges of $J(5,2)$, then, yes, we need $6$ or $7$ colors. – 2017-02-19

1 Answers 1

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My computer found a 6-edge-coloring of Johnson(5,2) using a backtracking algorithm:

A 6-edge-coloring of Johnson(5,2)

I acknowledge the use of GAP and the graph package Grape. Here's my code:

EdgeSet:=function(G)
  return Concatenation(List(Vertices(G),v1->List(Adjacency(G,v1),v2->[v1,v2])));
end;;

RequirePackage("grape");
G:=JohnsonGraph(5,2);
F:=Filtered(EdgeSet(G),i->i[2]>i[1]);
Cols:=List([1..30],i->0);

NewColor:=function(i)
  local c,n;

  for c in [1..6] do
    if(ForAny([1..i-1],j->Intersection(F[i],F[j])<>[] and Cols[j]=c)) then continue; fi;
    Cols[i]:=c;

    if(i=30) then
      return Cols;
    else
      n:=NewColor(i+1);
      if(n<>fail) then return n; fi;
    fi;

    Cols[i]:=0;
  od;
  return fail;
end;;