Math Puzzle Thread

[Ramo]

You just have to draw it out. Unless I confused myself, the answer is no, since as soon as three adjacent faces (of the cube) are filled, you have to end up in the middle to fill that particular face. So you can't fill out the last face without hitting the center.

[LordShiva]

Originally posted by Lul Thyme
Here's one I heard at the same party.
It's not as fun though.


"You have a 3x3x3 cube, you start in one corner and want to end in the middle, traversing each 1x1x1 cube exactly once, going from a 1x1x1 cube to an adjacent one (sharing a face).
Can you?"



(is there a Hamiltonian path in the graph starting in the corner and ending in the middle).

No. Label the adjacent cubes as either "odd" or "even," such that you need to move from odd to even or even to odd in each move. Since 27 is an odd number, you'll need to end on the same type of cube as the one you started in, but the centre cube is not of the same label as any of the corners.

[Ramo]

That's a much better way of thinking about it...

[Kuciwalker]

Elegant

[Lul Thyme]

Originally posted by Ramo
You just have to draw it out. Unless I confused myself, the answer is no, since as soon as three adjacent faces (of the cube) are filled, you have to end up in the middle to fill that particular face. So you can't fill out the last face without hitting the center.

'
I'm not sure I understand your argument.
LordShiva got it.

[LordShiva]

Place 4 knights of each colour on a chess board, so that each knight attacks three opposing knights and no friendly knights.

[Lul Thyme]

Originally posted by LordShiva
Place 4 knights of each colour on a chess board, so that each knight attacks three opposing knights and no friendly knights.

found it
fits on a 5x6
White: a2, d3, d5, e2
black: b4, c1, c3, f4

pretty cool, the graph is the cube and it is drawn in isometric perspective or whatever.

EDIT:
ill try to think of something soon, but if somebody has a good one, feel free to post.

[KrazyHorse]

Originally posted by Kuciwalker
Elegant

It's a classic solution to that type of pathing problem...

[LordShiva]

Originally posted by Lul Thyme

found it
fits on a 5x6
White: a2, d3, d5, e2
black: b4, c1, c3, f4

Yup

[KrazyHorse]

Yeah, that one's a cool problem.

[Lul Thyme]

Here's one until I can find something better.
Show that some of the requirements in Shiva's problem are redundant, so the problem is not aesthetically pleasing

(I checked this quickly, so may be wrong)

[Kuciwalker]

I don't see it.

[Lul Thyme]

As far as I can tell, the requirement that knights of the same color don't attack each other is redundant, and just asking that each knight attacks 3 opposite knights gives a unique solution, up to rotation\reflection\changing the colors.

[Kuciwalker]

I meant I didn't see what the problem was. I didn't see the "show" in your post, and thought you'd linked to a [broken] image.