Opposites of 1

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Opposites of 1

Postby evert » Tue Nov 04, 2008 3:01 pm

I thought about the following constraint:
Classical sudoku, and also the rule that the cells opposite to a cell containing a 1 should contain each of the numbers 1 - 9 once.

Example of such a grid:

Code: Select all
617|435|829
328|976|514
495|128|736
---+---+---
531|297|468
746|813|952
982|654|371
---+---+---
159|782|643
274|369|185
863|541|297


Puzzle - based on another grid:

Code: Select all
.86|...|...
25.|..6|4..
...|...|.3.
---+---+---
...|...|...
.4.|.1.|6..
...|4..|3..
---+---+---
3..|9..|..1
9..|...|...
...|1..|2..


In this case - the additional constraint is only implemented by singles.:(
I wonder what kind of solving techniques may be aimed at here.
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Postby coloin » Wed Nov 05, 2008 9:56 am

In you grid.....which cell is "opposite to a cell containing a 1" in box 3 ?
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Postby tarek » Wed Nov 05, 2008 10:45 am

does r5c5 have to be 1 ?
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Postby udosuk » Wed Nov 05, 2008 11:38 am

coloin wrote:In you grid.....which cell is "opposite to a cell containing a 1" in box 3 ?
In the example grid the 6 in r3c9 is opposite to the 1 in r7c1.

tarek wrote:does r5c5 have to be 1 ?
Yes because if you have two 1s opposite to each other then you're left with seven 1s to map to 2..9 which is not enough.:idea:

Triple click to view the solution for the puzzle I wrote:186345972
253796418
794821536
539672184
847513629
612489357
328954761
961237845
475168293

Nice puzzle, thanks!:)
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Postby Glyn » Wed Nov 05, 2008 3:20 pm

evert Very interesting it did solve with singles + the opposites rule, although once I had all the 1's it seemed easier to treat the opposites as an extra group.
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Postby evert » Wed Nov 05, 2008 11:12 pm

Potentiallay: we have not only possible placements of candidates - but also possible participations in a constraint ... which may or may not have interaction with the 'clear' constraints.
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Postby Smythe Dakota » Fri Nov 07, 2008 2:15 am

A vanilla sudoku has 27 constraints -- 9 each for rows, columns, and boxes. This opposites business adds just 1 more constraint.

You could, instead, add 9 new constraints, by doing the same for each digit, not just 1.

You could also add 18 more, by using horizontal and vertical opposites as well as 180-degree rotational opposites.

Then you'd be up to 27, allowing all 27 of the original constraints to be dropped.

Would this make a feasible puzzle? Apparently not, because then each digit would have to appear in r5c5.

But maybe there could be some kind of variation on this idea ...

Bill Smythe
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Postby HATMAN » Sun Nov 16, 2008 6:03 am

I've got a solution based on opposites with one nine opposites group and two more with a set of eight opposites. so the 5's have a 45 group the 9's have a 36 group and the 1's have a 44 group. You can probably find a solution with another 8 group but I doubt that you can fit two more in.

Now all I've got to do is create an interesting puzzle.
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Postby evert » Sun Nov 16, 2008 10:20 pm

I suppose you mean that - for example - the opposite cells of 5 contain each of the numbers 1-4&6-8 only once?
Or else what do you mean by "the 5's have a 45 group the 9's have a 36 group"?
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Postby udosuk » Mon Nov 17, 2008 5:17 am

evert I think what Maurice meant was this:

HATMAN wrote:I've got a solution based on opposites with one "nine opposites group" and two more with a set of "eight opposites", so the 5's have a 45 group, the 9's have a 36 group, and the 1's have a 44 group...

All nine of the 5-cells in the solution grid are opposite to nine different digits - one each from 1 to 9. Of course the one opposite to a 5 must be in r5c5.

Eight of the nine of the 9-cells in the solution grid are opposite to eight different digits - one each from 1 to 8. The remaining 9-cell must also be opposite to one of these eight digits but we don't know which or where. We can't have two 9s opposite to each other as it will leave us only seven 9s to map to 1..8.

Eight of the nine of the 1-cells in the solution grid are opposite to eight different digits - one each from 2 to 9. The remaining 1-cell must also be opposite to one of these eight digits but we don't know which or where. We can't have two 1s opposite to each other as it will leave us only seven 1s to map to 2..9.

:idea:
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Postby evert » Mon Nov 17, 2008 10:54 am

In that case there will be a digit which occurs twice in one of the cells opposite to a 1-cell.
Also there will be a digit which occurs twice in one of the cells opposite to a 9-cell.
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Postby udosuk » Mon Nov 17, 2008 1:09 pm

evert wrote:In that case there will be a digit which occurs twice in one of the cells opposite to a 1-cell.
Also there will be a digit which occurs twice in one of the cells opposite to a 9-cell.

Yep that's what I think. Perhaps Maurice has other elaborations.
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Postby HATMAN » Mon Nov 17, 2008 2:20 pm

Twice and only twice i.e not tripletets or two pairs.

I'm still getting nowhere with even the structural puzzle design - perhaps carefully chosen jigsaws? (not nine of them of course).
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Re: Opposites of 1

Postby dyitto » Sun Nov 07, 2010 12:09 am

I've made some puzzles that should be a little bit more advanced with the opposites-group:

Code: Select all
.73|1..|..4
5..|..9|.1.
.4.|...|...
---+---+---
..2|..7|1.8
...|...|...
.16|..2|9..
---+---+---
.3.|...|2..
..7|...|..5
9..|...|...

...|...|.1.
...|...|...
...|.7.|39.
---+---+---
...|9.5|6..
.5.|2.3|...
29.|...|..1
---+---+---
...|14.|.8.
..6|..9|...
..1|..6|...

...|...|3..
...|269|...
..6|..5|.9.
---+---+---
9..|...|7.1
..8|.1.|...
...|.2.|...
---+---+---
1..|3..|85.
...|1..|...
.8.|.4.|.3.


Also, I've made one that should be much more difficult then the ones above:

Code: Select all
...|1..|6..
.3.|2.8|...
.9.|.5.|...
---+---+---
1.3|...|..9
6..|...|8..
...|.8.|...
---+---+---
..2|.7.|...
...|..4|...
3.9|.6.|..1


I'm not sure if any guessing is needed here, I didn't try it myself yet.
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Re: Opposites of 1

Postby dyitto » Sun Nov 07, 2010 6:55 am

And another one:
Code: Select all
6..|5..|...
.8.|4..|...
.9.|6..|...
---+---+---
1..|8..|5.3
...|..5|..2
...|...|...
---+---+---
..3|.6.|..7
..6|...|8..
...|...|15.

My solver needs 3 guessings but maybe a human player can come up with better ideas.
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