TITLE: Is there more than 1 way to make a 17-node graph such that there are no 4-cycles and each node has at least four edges?
QUESTION [6 upvotes]: I'm working on the 17x17 challenge, and this sub-problem has come up. I have one solution to this problem that you can see here:

For some complex reasons (that I can elaborate on if needed) I know that this graph is unique if I add the following constraint: There has to exist a subset of at least 6 5 nodes that are not connected to each other and are not individually part of a 3-cycle. In the graph above, any 5-set of pink nodes satisfies this constraint. What I do not know is if it is still unique when I remove this extra constraint.
I'm coming from a CS background so I may be missing something basic from a graph theory perspective. Any references that may help me either prove that this is the case, disprove it by producing one or more different graphs or (even better) by producing all possible alternative solutions would be deeply appreciated.
EDIT: On closer examination, it seems the graph contains a double Hamiltonian circuit that uses every edge. No idea if this is relevant. [not interesting]

REPLY [6 votes]: Yes, here are some
Graph 1, order 17.
  0 : 5 8 12 13;
  1 : 6 9 13 16;
  2 : 7 10 13 15;
  3 : 8 9 14 15;
  4 : 9 10 11 12;
  5 : 0 11 15 16;
  6 : 1 10 14 16;
  7 : 2 11 13 14;
  8 : 0 3 12 14;
  9 : 1 3 4 15;
 10 : 2 4 6 12;
 11 : 4 5 7 16;
 12 : 0 4 8 10;
 13 : 0 1 2 7;
 14 : 3 6 7 8;
 15 : 2 3 5 9;
 16 : 1 5 6 11;
Graph 2, order 17.
  0 : 5 8 14 15;
  1 : 6 9 15 16;
  2 : 7 11 13 16;
  3 : 8 9 10 13;
  4 : 9 11 12 14;
  5 : 0 10 11 15;
  6 : 1 10 12 16;
  7 : 2 12 13 15;
  8 : 0 3 14 16;
  9 : 1 3 4 15;
 10 : 3 5 6 13;
 11 : 2 4 5 16;
 12 : 4 6 7 14;
 13 : 2 3 7 10;
 14 : 0 4 8 12;
 15 : 0 1 5 7 9;
 16 : 1 2 6 8 11;
Graph 3, order 17.
  0 : 5 7 11 13;
  1 : 6 8 13 15;
  2 : 7 9 10 15;
  3 : 8 10 14 16;
  4 : 9 11 12 14;
  5 : 0 11 15 16;
  6 : 1 12 13 16;
  7 : 0 2 10 13;
  8 : 1 3 14 15;
  9 : 2 4 12 15;
 10 : 2 3 7 16;
 11 : 0 4 5 14;
 12 : 4 6 9 16;
 13 : 0 1 6 7;
 14 : 3 4 8 11;
 15 : 1 2 5 8 9;
 16 : 3 5 6 10 12;
Graph 4, order 17.
  0 : 5 7 14 16;
  1 : 6 8 12 13;
  2 : 7 9 10 12;
  3 : 8 10 11 15;
  4 : 9 11 13 14;
  5 : 0 13 15 16;
  6 : 1 11 12 16;
  7 : 0 2 10 14;
  8 : 1 3 14 15;
  9 : 2 4 12 15;
 10 : 2 3 7 13;
 11 : 3 4 6 16;
 12 : 1 2 6 9;
 13 : 1 4 5 10;
 14 : 0 4 7 8;
 15 : 3 5 8 9;
 16 : 0 5 6 11;
Graph 5, order 17.
  0 : 5 7 11 13;
  1 : 6 8 12 15;
  2 : 7 9 10 15;
  3 : 8 10 14 16;
  4 : 9 12 13 16;
  5 : 0 11 15 16;
  6 : 1 11 12 14;
  7 : 0 2 10 13;
  8 : 1 3 13 15;
  9 : 2 4 14 15;
 10 : 2 3 7 16;
 11 : 0 5 6 14;
 12 : 1 4 6 16;
 13 : 0 4 7 8;
 14 : 3 6 9 11;
 15 : 1 2 5 8 9;
 16 : 3 4 5 10 12;
Graph 6, order 17.
  0 : 5 7 12 13;
  1 : 6 10 13 16;
  2 : 7 8 14 15;
  3 : 8 9 10 12;
  4 : 9 11 13 14;
  5 : 0 10 11 15;
  6 : 1 12 15 16;
  7 : 0 2 14 16;
  8 : 2 3 11 12;
  9 : 3 4 13 15;
 10 : 1 3 5 14;
 11 : 4 5 8 16;
 12 : 0 3 6 8;
 13 : 0 1 4 9;
 14 : 2 4 7 10;
 15 : 2 5 6 9;
 16 : 1 6 7 11;
Graph 7, order 17.
  0 : 5 7 11 14;
  1 : 6 11 13 15;
  2 : 7 8 12 13;
  3 : 8 9 14 15;
  4 : 9 10 11 12;
  5 : 0 10 13 16;
  6 : 1 12 14 16;
  7 : 0 2 12 15;
  8 : 2 3 11 16;
  9 : 3 4 13 14;
 10 : 4 5 15 16;
 11 : 0 1 4 8;
 12 : 2 4 6 7;
 13 : 1 2 5 9;
 14 : 0 3 6 9;
 15 : 1 3 7 10;
 16 : 5 6 8 10;
Graph 8, order 17.
  0 : 5 7 13 14;
  1 : 6 11 13 15;
  2 : 7 8 10 12;
  3 : 8 9 15 16;
  4 : 9 10 11 14;
  5 : 0 11 12 16;
  6 : 1 12 14 15;
  7 : 0 2 10 13;
  8 : 2 3 14 16;
  9 : 3 4 12 13;
 10 : 2 4 7 15;
 11 : 1 4 5 16;
 12 : 2 5 6 9;
 13 : 0 1 7 9;
 14 : 0 4 6 8;
 15 : 1 3 6 10;
 16 : 3 5 8 11;