How Many Ways Are There To Draw This Graph?
Вставка
- Опубліковано 11 чер 2024
- We count the number of ways of drawing the K_5 graph, without lifting a pen off the page. This is equivalent to counting the number of labelled Euler paths in K_5, the complete graph with 5 vertices.
00:00 Intro
00:26 Paths vs cycles
01:27 Pairs of cycles
04:13 3 & 7
11:12 4 & 6
15:11 5 & 5
18:58 Conclusion
I always find these combinatorial problems overwhelming. I look for a short cut but often it just needs to be slogged out like you did. Well done explaining all the steps. You are a patient man!
I love your hard cuts at the end. They were a bit jarring at first like “oh it’s over” but you do it every time and it kinda feels like your thing I like it.
Forget the maths.. impressive magic trick with the colour change at 2:03
the symmetry that you get from a fully connected graph makes it really intuitive to cover all cases just by drawing specific ones like that
Yes, and unfortunately this sort of problem becomes much harder without the nice symmetry!
Cool. I would have liked a discussion of symmetries afterwards. (I think) if you can rotate the graph then the starting point can just be the top vertex and with mirror image the second vertex can always be to the right. This gives 2640 / (5*2) ways
Or just 264, since 5*2 = 10. :P
If you take total point permutation symmetry instead then the first cycle hits N points clockwise and the secind cycle is either 5: 2 choices, 6: 4 choices, 7: 12 choices for a total of 34 permutationally distinct paths
I may be overcounting the 3x7 and 4x6 paths for the ones that contain sub cycles of 3 or 4
How good of an upper bound would 5 * 9!/((2!)^4) be? "We sample one vertex and a sequence of 9 vertices from 5 options such that 4 vertices must repeat 2 times" should be a good estimate for an Euler cycle but not sure.
I make this 113400, so a lot bigger than our final answer. For the first cycle from the starting point to itself, I feel like a random sequence of vertices could often create a path of viable edges, but when we get to the next cycle, it's less likely that the edges are still possible, or haven't been used already.
I suppose it's also unlikely that all 10 edges would be used exactly once, with no repetitions or edges unused.
This hit me at about 17:30 in.. Is this a Halloween video?
It's some fortuitous timing! 😂