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Can you put 14 crosses in a 6 by 6 grid so that there are an even number of crosses on each row and column?

An elementary school teacher asked me for help with this. Below is a photo of the book they are using. The text is in Swedish and it says "Draw 14 crosses in the grid so that there is an even number of crosses in each horizontal and vertical row". I already gave my answer to the teacher, so why don't you try to solve it!

6 by 6 grid

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  • 3
    $\begingroup$ Here is a quick html grid you can play with instead of paper: dlabs.me/14grid.html $\endgroup$ – Asleepace Jan 17 '18 at 20:58
  • $\begingroup$ The way you define the problem "Draw 14 crosses in the grid so that there is an even number of crosses in each horizontal and vertical row" is more akin to the 8 queens problem than what people are doing here, this is being solved as "Draw 14 crosses in the grid so that there are the same number of rows with each number of crosses horizontally as there are vertically" As a side note it's not possible to solve with the way the question is actually worded. $\endgroup$ – Andrew Jan 17 '18 at 21:53
  • $\begingroup$ I am now realizing i was reading even as equal and the question may have meant even (not odd) $\endgroup$ – Andrew Jan 17 '18 at 22:14

11 Answers 11

12
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I believe there are many solutions. One of them is this:

X X X X _ _
X X _ _ _ _
X _ X _ _ _
X _ _ X _ _
_ _ _ _ X X
_ _ _ _ X X

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    $\begingroup$ Once we have a valid solution, swapping any two rows or any two cols gives another valid solution. I think we should mark solutions which can be related by this kind of permutation equivalent, and if we do so, I don't think there are many solutions that are not equivalent. Your solution has the bonus that there are even numbers on the diagonal as well. $\endgroup$ – Weijun Zhou Jan 17 '18 at 8:47
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    $\begingroup$ Btw, I just updated the solution because the new one is nicer lol. If you say the main diagonal, then it's true they are also even. Let see if we can do that for all diagonals. $\endgroup$ – athin Jan 17 '18 at 8:50
  • $\begingroup$ It looks like it's even for all the diagonals with an even number of items. $\endgroup$ – Jakob Lovern Jan 17 '18 at 22:12
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I tried to make the solution as compact as possible

 X X X _ X _
 X X _ X X _
 X X X X _ _
 X X _ _ _ _
 _ _ _ _ _ _
 _ _ _ _ _ _

This one is 4x5. It is impossible to fit in 4x4, 3x5, or 3x6

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  • $\begingroup$ 0 isn't an even number $\endgroup$ – Strawberry Jan 17 '18 at 11:59
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    $\begingroup$ @Strawberry yes it is $\endgroup$ – Kruga Jan 17 '18 at 12:05
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    $\begingroup$ OK. Fair enough! $\endgroup$ – Strawberry Jan 17 '18 at 12:51
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    $\begingroup$ @JPhi1618 why ever not? 0 is a perfectly cromulent even number. $\endgroup$ – Geoffrey Brent Jan 17 '18 at 21:05
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    $\begingroup$ @JPhi1618 - I was taught that $0$ is even when I was first introduced to the concept in early elementary school (can't remember which grade after more than 45 years). It wasn't until I was an adult that I discovered that there were people in the world who somehow had trouble with this obvious fact. $\endgroup$ – Paul Sinclair Jan 18 '18 at 3:30
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I think this solution is independant from the ones already published.

Consider those blocks:

A

X X _
X _ X
_ X X

B

X X
X X

C

X X
X X

Place them anywere on your 6x6 board, and your get dozens of solutions.

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  • $\begingroup$ Looks like B and C are equivalent. $\endgroup$ – Jakob Lovern Jan 17 '18 at 22:14
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    $\begingroup$ They are. But you have to place both to reach the 14 crosses on the board that are asked for. $\endgroup$ – Evargalo Jan 17 '18 at 22:16
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Two symmetrical 5x5 solutions:

X _ X X X
_ _ _ X X
X _ _ _ X
X X _ _ _
X X X _ X

and

_ _ _ X X
_ X X X X
_ X _ X _
X X X X _
X X _ _ _

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    $\begingroup$ ... which become 6x6 solutions by leaving the last row and column empty? (since 0 is an even number) $\endgroup$ – Rand al'Thor Jan 17 '18 at 12:18
  • $\begingroup$ @Randal'Thor Sure, but we'd loose the symmetry $\endgroup$ – Vincent Jan 17 '18 at 14:07
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    $\begingroup$ Well, the question is about a 6x6 square, so you'd need that extra row and column for this to be a valid answer :-) $\endgroup$ – Rand al'Thor Jan 17 '18 at 14:08
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Another solution, which I believe is non-equivalent to any existing ones:

X X X X X X
X X X X _ _
X _ _ _ X _
X _ _ _ _ X
_ _ _ _ _ _
_ _ _ _ _ _

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2
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X X _ _ X X
_ X X _ _ _
_ _ X X _ _
_ _ _ X X _
_ _ _ _ X X
X _ _ _ X _

Even number of "X" in each row and column...

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  • $\begingroup$ Please format your proposed answer as a spoiler so that it doesn't spoil the puzzle for others. $\endgroup$ – JeffC Jan 17 '18 at 16:20
  • $\begingroup$ Sorry. It was my first time answering a question here. Will keep in mind. $\endgroup$ – Sid Jan 18 '18 at 5:03
2
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Another Solution is this one. Symmetric and Pleasing to the eyes.

enter image description here

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1
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This is, in my opinion, the simplest solution. Not "beautiful", just simply valid, but unique compared to others.

X X _ X X _
_ _ X _ X _
X X X X _ _
_ _ _ _ X X
_ _ _ _ _ _
_ _ _ _ X X

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0
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Following the "Air Bud" rules of lateral thinking...

X _ _ _ _ X
_ X _ _ X _
_ _ X X _ _
XX _ X X _ _
_ X _ _ X _
X _ _ _ _ X Where there are two crosses in the cell in the first column.

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    $\begingroup$ Or all 14 could go to a single cell while we are at it :) $\endgroup$ – Yunus Nedim Mehel Jan 17 '18 at 18:19
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There are plenty of solutions, a piece of code maybe able to find them all. All you have to have is:

5 columns with 2 crosses and
1 column with 4 crosses

and therefore (1 row with 4 crosses and 5 rows with 2 crosses)

An example:

x - - x - -
x x - x - x
- x x - - -
- - x - - x
- - x - x -
- - x - x -

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0
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Kept making mistakes at this, so now the 14grid.html will check for valid solutions!

enter image description here

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