Is the Erdős–Faber–Lovász Conjecture open still? According to Wikipedia it is unsolved still, but I think this is not hard to solve this conjecture.

Conjecture: If $n$ complete graphs, each having exactly $n$ vertices, have the property that every pair of complete graphs has at most one shared vertex, then the union of the graphs can be colored with $n$ colors.

My attempt: Consider the, $n-1$ complete graphs $K_n$ which all has a common vertex, say $u$. then this graph can be colored with $n$ colors. by choosing $n-1$ vertices (one from each graph $K_n\backslash \{u\}$) by different color and construct a graph with this vertices and a new vertex, say $v$. we can assign to $v$ color of $u$. Why this is not correct or maybe I don't understand the problem ?

  • $\begingroup$ It is difficult to understand what you're doing. Are you trying to construct a family of counterexamples to the conjecture? $\endgroup$ – Henning Makholm Oct 11 '17 at 12:46
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    $\begingroup$ What I think you're saying (although the explanation is not really clear) is that for a particular configuration of $n$ complete graphs which satisfies the assumptions, a coloring is possible. But you have to prove that for every configuration of $n$ complete graphs which satisfies the assumptions, a coloring is possible. $\endgroup$ – Hans Lundmark Oct 11 '17 at 12:51
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    $\begingroup$ See the question and your answers: math.stackexchange.com/questions/275194/… $\endgroup$ – MathOverview Oct 11 '17 at 12:53

I don't think you understand the problem. It's not just $n-1$ complete graphs with one vertex common to all of them. Each pair of complete graphs can have at most one shared vertex, but different pairs can have different shared vertices. So for $n=4$ you could have the graph pictured: the $K_4$s are circled and the five central vertices are each shared by two of the complete graphs. As $n$ increases, the number of possible configurations increases rapidly.

enter image description here

  • $\begingroup$ For $n=4$ we must have $4$ $K_4$. $\endgroup$ – C.F.G Oct 11 '17 at 12:58
  • $\begingroup$ Ah, sorry, misread. Will update the diagram, thanks. $\endgroup$ – Especially Lime Oct 11 '17 at 13:08

Looks likes you missed the point regarding common vertices. At most ONE common vertex is permissible for each pair of k-graph. So each graph among the k, can have up to k-1 common vertices in common with one or another of the remaining k-1 egraphs.

Since no pair of graphs have more than one vertex in common, total number of edges remains the same as taken separately for each graph viz Kk(k-1)/2 edges in the Union of graphs.

Trust this will help you in making fresh efforts.


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