00:01
We're asked to find the crossing number of the peterson graph.
00:08
Now, we have, by the last example in this section, that pearson's graph is not planar.
00:19
So it follows that pearson's graph has a crossing number must be greater than or equal to 1.
00:40
Now, suppose you remove an edge from the graph.
01:05
The graph will not become planar because the graph will still contain a subgraph homeomorphic to k33.
01:42
So for example, if the edge that was removed was not between a vertex of the set fdj and a vertex of the set eih, then by the same reason in the textbook we have that the graph still contains a subgroup, subgraph homeomorphic to k -33, and otherwise we can just rename one of the other vertices so that again we can use.
02:18
Use the statement in the example.
02:24
So this means that the crossing number graph is non -planner still.
02:49
And so the new graph has a crossing number greater than or equal to 1.
03:02
And so it follows that peterson graph has a crossing number greater than or equal to 2, since we're adding back an edge.
03:29
And now, let's move the vertical of peterson's graph so we can obtain exactly two crossings.
03:40
So for example, i could move a and b inside as well as e inside.
04:06
So i'll draw the new pearson graph down here.
04:23
We'll keep c and d in the same place.
04:45
But now let's move f up above a.
04:58
Well, we have the f is going to be connected to a.
05:32
And let's put g in the middle of a, b, and e...