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November 26

Can there be more than 25 sources in a 36 square square?

Or >26 if conveyer belt squares are available so that effluent can be merged and/or turned left or right and drain squares no longer have to border their source squares?

Diagonal drainage not allowed.

In the first case perfection would need only 1 overhead square per five (=28.8 desired squares per 36) cause Greek crosses can tesselate no gaps but edge gaps prevent this for finite squares of course. Sagittarian Milky Way (talk) 03:56, 26 November 2019 (UTC)[reply]

I can't make any sense of this question or the near-duplicate immediately above it. Could you explain what you are asking in math terms? Most of us here are unlikely to understand industrial plumbing or whatever the application area is. 67.164.113.165 (talk) 07:36, 26 November 2019 (UTC)[reply]

[1] Sagittarian Milky Way (talk) 13:55, 26 November 2019 (UTC)[reply]

My interpretation is: Is it possible to color the 36 squares in a 6 by 6 grid to make 10 black and 26 white squares, so that each of the white squares is adjacent to a black square? In the diagram shown in the link, the white squares are represented by arrows and the black squares are represented by dollar signs or blanks. That diagram has 25 white and 11 black squares. If that interpretation is correct then Black = {(1, 2), (2, 1), (1, 4), (4, 1), (2, 6), (6, 2), (3, 3), (4, 5), (5, 4), (6, 6)} is a possibility with 10 black, 26 white. It seems unlikely that there is a 9 black 27 white coloring but I don't have a proof. It's not too hard to show, as I think the OP did if I'm interpreting correctly, that 7 black, 29 white is impossible. --RDBury (talk) 15:09, 26 November 2019 (UTC)[reply]

Wow I'm dumb, the two blank squares in the upper left can become arrows if the arrow in between became their dollar sign. Which would give 26. Sagittarian Milky Way (talk) 16:23, 26 November 2019 (UTC)[reply]

@RDBury: If I made my sketch correctly, I found a 9 black 26 white configuration (and one being empty or unused):

	$				$
			$
$			$
					$
	$	$
×				$

--CiaPan (talk) 21:16, 26 November 2019 (UTC)[reply]

I think the remaining open question is now whether 27 white squares is possible. The problem can be transformed into an integer programming program, and with something of this size it would take an industrial IP solver less time to solve it than it would take to input the data. The relaxed problem is small enough to be solved by hand (6 variables, 6 inequalities) if you apply the symmetries of the square, but the solution only proves that 28 white squares is impossible. One could ask similar questions about different sized rectangles, and I believe the corresponding problem for general graphs is equivalent to the minimum vertex cover problem. --RDBury (talk) 15:43, 27 November 2019 (UTC)[reply]

I was wondering what the general rectangle problem is called or classified as, yes it does seem to be a minimum vertex cover problem. Sagittarian Milky Way (talk) 19:14, 27 November 2019 (UTC)[reply]