12
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Place two crosses on two cells of each row and column of this 9×9 board, and circles elsewhere, so that the number on the right of each row indicates the number of circles between its two crosses, while the number below each column indicates the number of circles between its two crosses.

enter image description here

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  • $\begingroup$ This is almost like a nonogram! $\endgroup$ – JGibbers Nov 30 '18 at 15:31
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    $\begingroup$ Nitpick: the number on the right of each row indicates the number of circles between its two crosses, while the number below each column indicates the number of circles between its two crosses. Both sentences have the same content ("indicates the number of circles between its two crosses") so it's confusing: it implies the row and column numbers mean different things but they actually have the same meaning. $\endgroup$ – Voile Dec 1 '18 at 16:05
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Answer is probably this (confessing that this is trial and error):

enter image description here

Approach:

I have started where Row:Col differences are 0:0 (which is either 6,6/9,9 or 6,9/9,6)
Using the walls as an aid, I have used it to identify the adjacent X
i.e. using 6,9 to identify 6,8 as gap for Row 6 is 0.
Similarly using 9,6 to identify 8,6
Next confirmed Xs in-order are 8,1 ; 4,1 ; 4,8


From from 6,9 the next guesses are 5,9 or 7,9 which has a lengthy iteration process ...... (to-be-completed when the time allows)

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5
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Starting from top row

(3rd, 7th) column, (2, 4), (3, 5), (1, 8), (7, 9), (8, 9), (2, 4), (1, 6), (5, 6)

| | |X| | | |X| | | 3
| |X| |X| | | | | | 1
| | |X| |X| | | | | 1
|X| | | | | | |X| | 6
| | | | | | |X| |X| 1
| | | | | | | |X|X| 0
| |X| |X| | | | | | 1
|X| | | | |X| | | | 4
| | | | |X|X| | | | 0
3 4 1 4 5 0 3 1 0

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4
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At the risk of running over my lunch break at work:

Solution: (verified by other answers)

SolutionPic

This is a pure logic solution, similar to Sudoku and others. I forget the name of the exact type of puzzle, but the trick is to fill the entire grid with crosses (+) and only fill in circles (O) where they MUST be.

For example:

In row 4, there must be 6 O between +. Therefore, there will be 2 possible sets that will satisfy the condition. LogicExplanation As you can see, the O in yellow MUST be as such.

Using this trick, we can fill out rows 4 and 8 as well as columns 2, 4, and 5.

Step1

A-ha! We have our first set of + in column 5. But what next?

Looking back at column 4 and applying the trick (since we already have 2 O), the bottom entry in column 4 must be O... Then following the logic condition, row 9 and column 6 are immediately solved, as follows. Step2

The rest is an exercise of repetition. Apologies for the bad Excel snaps and some poor formatting, and hope there is no confusion (or worse, a mistake!).

Cheers, KF

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  • $\begingroup$ Could you explain this a bit more: "A-ha! We have our first set of + in column 5." I follow your logic that in column 5 we have O in rows 4, 5, 6, 8. I can see how logically you'd have a O in row 2 as well, but how do you eliminate rows 1 & 7 as possible +? $\endgroup$ – Dean Dec 1 '18 at 5:41
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Solution

o o x o o o x o o
o x o x o o o o o
o o x o x o o o o
x o o o o o o x o
o o o o o o x o x
o o o o o o o x x
o x o x o o o o o
x o o o o x o o o
o o o o x x o o o

Approach:

Starting with the 4th row, I placed an X on the left, then 6 o's, then another x, then a final o. In the two columns with x's in, I then placed o's in the cells which couldn't have x's in. From this, I was then able to place o's in other cells which couldn't have x's in, and x's in cells which had to have x's in. Eventually the grid was filled and I checked to ensure each row and column met their criteria. The question did ask, after all, to put o's where there weren't x's.

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