In this graph, each vertex has just three edges, all of which would be needed in a K4, but the three vertices on the other ends of those edges only connect in the complement.

In the complement, each vertex is connected to four others; measured from the center, two vertices are 90 degrees away and the other two are 135 degrees away. But however we choose three of these four vertices, two will be 45 degrees from each other and only share an edge in the original graph. ]]>

I’m not sure whether this applies usefully to designing polyedge tiling puzzles, though.

]]>did you hear this problem?

we have a graph with 6 vertices,we can prove that there is a k3 in this graph or there is a k3 in complement of this graph.

what can we say about a graph with 8 vertices? is this sentence true?:

we have a graph with 8 vertices,we can prove that there is a k4 in this graph or there is a k4 in complement of this graph.

]]>One embedding with a little less symmetry that might work is a pair of concentric equilateral triangles with the same orientation. (Equivalently, a triangular prism including the diagonals of the rectangles.) This only has three crossings! That might be too easy though, so you might be able to add an extra constraint, like having all three pairs of pentaedges represented in the three crossings.

]]>*Lemma: The five edges making up the pentagram must be occupied by all three pentaedges, with consecutive groups of 1, 2, and 2 edges in some order.*

A check of which of a pentagram’s edges cross is enough to show this.

Now, the second K6 contains both a red loop and a blue snake (a list of standardized names would be helpful). Neither of these can take up two edges of the pentagram, so they must appear separately, Q.E.D.

Not sure whether it’s possible. The asterisk (first K6, red) is one pentaedge that must take two legs of the pentagram, but my mind’s just going to wander off now.

]]>Keep in mind that there are 24 vertices, all of degree 5. The following are all of the partitions of 5 with 3 or fewer parts:

(5)

(4,1)

(3,2)

(3,1,1)

(2,2,1)

The pentaedges contain the following number of vertices of each degree:

5: 1

4: 2

3: 10

2: 25

1: 27

The 5 and 4 degree vertices have only one possible partition where they can fit, so we can pull them out, leaving:

3: 10

2: 25

1: 25

and 21 total vertices.

Now we can set up some equations:

a) |3,2| + |3,1,1| = 10

b) |3,2| + 2·|2,2,1| = 25

c) 2·|3,1,1| + |2,2,1| = 25

d) |3,2| + |3,1,1| + |2,2,1| = 21

Then |2,1,1| = 11 (by subtracting a from d,)

|3,2| = 3 (by substituting in b,)

and |3,1,1| = 7 (by substituting in a or c.)