Find an example of a regular triangle-free $4$-chromatic graph Find an example of a regular triangle-free $4$-chromatic graph
I know that for every $k \geq 3$ there exists a triangle-free $k$-chromatic graph.
So if I can find a triangle-free graph $H$ such that $\chi(H)=3$, then I can use the Mycielski construction to obtain a triangle-free graph $G$ such that $\chi(G)=4$. However, the regular part keep getting me stuck. I try some odd cycle, I also tried the Petersen graph but still can't get a regular triangle-free $4$-chromatic graph.
I wonder if anyone can give me a hint, please.
 A: If you're lazy you can check all the graphs on Wikipedia. The Hoffman–Singleton graph, for example, is strongly regular, has girth 5, and has chromatic number 4.
A: Another example is given by Kneser graphs $K(n,k)$ with suitable parameters.
By Lovasz' theorem, the chromatic number of $K(n,k)$ is given by $n-2k+2$.
Moreover, if $n<3k$ we have that $K(n,k)$ is triangle-free, so:

$\color{red}{K(8,3)}$ is a triangle-free, $10$-regular graph on $56$ vertices with chromatic number $\chi=4$.

However, the minimal example of a triangle-free, regular graph with $\chi=4$ is given by another "topological graph", the Clebsch graph, with $16$ vertices and degree $5$.
A: Discussed this question with my guide after getting a specific answer by joining two Mycielskian of $C_5$ appropriately (so $22$ vertices). He told me the following technique, which can be generalized to handle such questions.
Construct the Mycielskian of $C_5$ (any triangle free $4$-chromatic graph will do). We have five vertices with degree $3$, five with degree $4$ and one with degree $5$. Make five copies of this. Add a new vertex and join to the corresponding degree $4$ vertex in each of the five copies. Do this for the rest degree $4$ vertices. For degree $3$ vertices take two vertices and join both of them to the corresponding degree $3$ vertex in each of the five copies. Now the final graph is regular, triangle free and $4$-chromatic.  
This technique is already published (unable to find the paper).

ADDENDUM: 
For this general method, you use the exact same coloring in all five copies ($a_i$,$b_i$,...,$e_i$ for $i = 1$ to $11$) of the Mycielskians (color 1 to 4). Every corresponding degree 4 vertex of each is connected to a new vertex (i.e. $a_1$,$b_1$,...,$e_1$, having same color say 1, is connected to a new vertex whose degree is $5$ which can be assigned a different color than 1). Do this for the rest degree 4 vertices ($a_i$,$b_i$,...,$e_i$ for $i = 2$ to $5$). 
Every corresponding degree 3 vertex of each is connected to two new vertices (i.e. $a_6$,$b_6$,...,$e_6$, having same color say 1, is connected to two new vertices whose degree is $5$ which can be assigned a different color than 1). Do this for the rest degree 4 vertices ($a_i$,$b_i$,...,$e_i$ for $i = 7$ to $10$).
This results in a regular $\Delta$-free $4$-chromatic graph.
PARTICULAR ANSWER: (not using the general method described above which would result in a graph of order $70$)
The following is a particular answer to this question using Mycielskian resulting in a graph of order $22$. These are two copies of Mycielskian of $C_5$. I have not drawn all edges within each Mycielskian to improve the clarity. The left 5 vertices of Mycielskian were degree 4 vertices hence have only one matching, where as the middle five vertices of Mycielskian were degree 3 vertices hence have two matchings to make them regular.

A: The Gewirtz graph, a strongly regular graph with parameters $(56,10,0,2)$, is another triangle-free graph with chromatic number $\chi=4$.
It is worth noting that Gewirtz graph is not isomorphic to Kneser graph $K(8,3)$. The Gewirtz graph has independence number $\alpha(G) = 16$ (i.e. the cardinality of a largest set of vertices pairwise nonadjacent, e.g. see Brouwer and Van Maldeghem' survey) whereas $\alpha(K(8,3)) = 21$ (the Erdős-Ko-Rado theorem  states that $\alpha(K(n,k)) = \binom{n-1}{k-1}, n\geq 2k $).
A: A search at House of Graphs returns 112 such graphs.

