Double killer

For fans of Killer Sudoku, Samurai Sudoku and other variants

Double killer

Postby Squirrel » Mon Oct 02, 2006 12:06 am

This is my first post on these forums, so hello everybody.:D


Rules: The same as a regular killer sudoku. Both grids yeild the same solution, but one cannot be solved without the use of the other.

Image

If I was to change one thing, it would be the empty space on the second grid, and maybe compensate by having larger regions. Comments would be greatly appreciated.

Enjoy.



If anybody would post a link to a site which helps create images of sudokus, that would be great. Using paint was a bit of a pain.


~Squirrel

EDIT: I made a typo on the puzzle: on the righthand grid, its a sum of 27 R3C1, not a 25. I fixed it in the link, but if you printed the origonal version, now you know.
Last edited by Squirrel on Tue Oct 03, 2006 4:53 pm, edited 1 time in total.
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Postby emm » Tue Oct 03, 2006 5:36 am

R3C1 says 27 not 25 or 17.
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Postby Jean-Christophe » Tue Oct 03, 2006 8:05 am

27 is the correct sum, like in the picture.

Squirrel made a typo while correcting his typo !
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Postby tarek » Tue Oct 03, 2006 8:15 am

I ike the idea of having two(or more) identical solution grids with identical starting clues.....but each puzzle has different constarints......One is a Killer the second a vanilla

Or even a samurai, with each sub-puzzle having different constarints(X+toridal+vanilla+killer with the centre being a killer toroidal x or something:(:idea: )

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Postby HATMAN » Tue Oct 03, 2006 9:06 pm

Tarek

This mixing is an interesting thought. There have been quite a few double killers posted on DJ's site, I enjoy them an will try this one.
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Postby djape » Wed Oct 04, 2006 8:38 am

tarek,

if you make a vanilla/killer Twin Sudoku, it's pretty much the same as if you had one Killer Sudoku with some givens, or if you like, with singleton cages (existing cages split into a singleton and whatever remains).

So, I don't think a vanilla/killer Twin hybrid would be interesting. The same applies to your suggestion for "X"+killer hybrid. If one solution is "diagonal" the other one must be too! They are the same, remember.

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Postby tarek » Wed Oct 04, 2006 8:58 am

djape wrote:So, I don't think a vanilla/killer Twin hybrid would be interesting. The same applies to your suggestion for "X"+killer hybrid. If one solution is "diagonal" the other one must be too! They are the same, remember.


Having thought about it for a second time:D , It looks that the twinning is just another way of mixing constraints.....it just one puzzle with all the constraints in place. A killer X or killer toridal X

The only twinning that may be viable are 2 killers with different cages....each can't be solved alone & start from a similar set of clues.....
Or different toroidals with the same concept. In any case... A vanilla should not be one of the twins & the ability of superimposing one on top of the other should be difficult.........

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Postby motris » Wed Oct 04, 2006 6:58 pm

djape's comments are quite right. I was thinking that something like a jigsaw/killer hybrid might be interesting. Have one jigsaw puzzle and a normal boxed puzzle where you put in some cages. In principle you could represent this all on one puzzle (different colors for the jigsaw versus box constraints), but it seems easiest to display as two linked puzzles. The jigsaw/vanilla interactions would put in some LoL-type steps and,
with the addition of cages, would add nice box/jigsaw piece constraints across the puzzle too. Again, jigsaw/killer might be a bit complex, but jigsaw/vanilla would work fine as another choice.
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Postby Ruud » Wed Oct 04, 2006 7:23 pm

motris wrote:I was thinking that something like a jigsaw/killer hybrid might be interesting.

On my website, I publish a regular Jigsaw Killer.

For killers with both 3x3 nonets and jigsaw groups, you can find several EG-Killers made by h3lix (and a recent one by hatman) on DJape's Killer forum. (EG=Extra Groups)

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Postby udosuk » Wed Oct 04, 2006 7:28 pm

One thing we haven't tried is a EG Killer (Jigsaw + normal boxes) with 2 duplicating sets of cages, i.e. a Double EG Killer... So as a double grid one side would be presented as a normal killer and the other side a jigsaw killer...
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Postby motris » Wed Oct 04, 2006 8:07 pm

The EG Killer fits exactly the description I had in mind, just presented as two linked puzzles instead of colored regions on one. I'll have to try these out.

udosuk's suggestion is an interesting one and I think having it presented in the "linked" puzzle form is the most reasonable way to combine having different sets of cages and different sets of shapes.
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Postby Squirrel » Fri Oct 06, 2006 1:55 am

There are some other plausable combinations for the two grids as well. One could be cages with sums, the other with products. A double GT killer (like the ones found at the bottom of www.killersudokuonline.com) would be VERY interesting. Having one a 00...7 killer sudoku and the other a regualr killer sudoku (changing the 8s and 9s into 0's ) would also be an interesting combination.


@ HATMAN: I wasn't aware that they had been done before. I'll make sure to try them sometime.


Heres another regular double killer for anybody who wants it. The first couple numbers are easy, but there is some very interesting logic needed later on in the puzzle.

Image


~Squirrel

P.S. that was an embarassing typo.
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Postby emm » Fri Oct 13, 2006 1:44 pm

Squirrel, thanks for the puzzles. I like the idea of a double killer.

Are they solvable without guessing? I had a go at the second two before I gave it to SumoCue and got 4 solutions. I was in a bit of a muddle by then and could easily have made a mistake.
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Postby Squirrel » Sun Oct 15, 2006 5:54 pm

Yes, both puzzles only have 1 solution and are solvable by logic.


Heres the solution to the second double killer puzzle. As you enter in the numbers, please place them in both grids 1 (left) and 2 (right). As I dont know all the shortforms for all the techniques, I just wrote it out in full. Hope you can follow it.


rule of 45 gives you
r6c3 = 5
r4c6 = 4
r2c7 = 4 (using regions from both grids)
r7c2 = 4 (using regions from both grids)

r6c9 = 3 (bottom four rows of second grid)
r4c1 = 7 (top 4 rows of 2nd grid)

unique sums in regions at r4c1 and r5c9:
r5c1 = 3
r5c9 = 6

This is where you get into the fun logic. The region at r6c4g1 and r6c5g2 share 3 cells in common (r6c5, r7c4, r8c4). The highest those cells can total is 17 (which would yeild a 1 in r6c4 (look at grid 1)), and the lowest they can equal is 15 (which would yeild a 9 in r7c5 (look at grid 2)). That means that r6c4 is either a 1 (the sum of the three square is 17), or a 2 (the sum ofthe three squares is 16). The sum of the three squares cant be 16 because of 3 in r6c9. (These possible sums also tell us that r7c5 = 7 or 9)

Directly to the left of the two grids we have been working with are two more which share 3 cells (located at r6c3g12). You can use the same logic using the possible sums of the three cells in common, or else you can say that the difference in the sums between the two regions is 3, so r6c4 (look at the specified region in grid 2) is three less than r7c2 (look at the specified region in grid 1). That means that because r6c4 = 1 or 2, r7c2 = 4 or 5. However, r7c2 cant be a 5 because r6c3 = 5, and is in the same region.

Therefore,
r7c2 = 4
r6c4 = 1
r7c5 = 7

You should be able to solve the rest of it fairly easily from here on. The rest of the sloution is still there if you need it.

looking at grid 2, we can see:
the region at r6c1 has a remainder of 8 which has to be a 2 and a 6.
r6c1 = 2
r6c2 = 6
the region at r6c3 has a remainder of 16 which has to be 7 and 9
r7c3 = 9
r8c3 = 7
using rule of 45 in bottom left nonet, r9c23 = 9, and has to contain a 3 (because of the 3 at r5c1).
r9c2 = 3
r9c3 = 6

looking at the grid 1:
rule of 45 reveals
r4c4 = 3

looking at grid 2:
the region at r3c4 has a remainder of 15
r4c3 = 8
r3c4 = 7
PENCILMARKS: looking at the region at r9c2, the remainder is 12 which has to be an 8 and a 5

looking at grid 1:
the region at r5c4 = 20, which has to be either 974 or 875 and therefore must contain a 7.
r5c6 = 7
the only place for a 4 in the left center nonet is r5c3
r5c3 = 4
the only spot for a 4 in the center center nonet is r6c6 (if r6c5 = 4 the the remainder in the region at r6c4 = 13, which has no possible sloutions)
r6c6 = 4
rule of 45 reveals
r4c9 = 9
you can now complete the center left nonet:
r4c2 = 1
r5c2 = 9
the only spot for a 9 in center center nonet is r6c5
r6c5 = 9

looking at both grids:
the regions at r2c6g1 and r2c6g2 share 3 cells and have a difference of 5. Therefore r4c6 is 5 greater than r3c5.
r4c6 = 6
r3c5 = 1
the regions at r2c7g1 and r2c7g2 share 3 cells and have a difference of 0. Therefore r3c7 = r4c6
r3c7 = 6

looking at grid 2:
r6 is missing a 7 and an 8, r6c7 <> 8 because of the region it is in
r6c7 = 7
r6c8 = 8
unique sum in region at r3c8
r4c8 = 5
unique sum in region at r6c6
r7c6 = 1
unique sum in region at r6c5: r78c4 = 8 and has to be 2 + 6
r7c4 = 2
r8c4 = 6
1 number left in r4
r4c5 = 2

looking at grid 1:
unique sum in region at r3c2: r3c23 = 11, and has to be 3 + 8
r3c2 = 8
r3c3 = 3
rule of 45 gives us
r1c2 = 5
1 number left in c2:
r2c2 = 7

looking at grid 2:
unique sum in region at r3c1
r3c1 = 4
only place for a 2 in top center nonet is r1c6 (cant be r23c6 because of region)
r1c6 = 2
unique sum in region at r2c6: r23c6 = 13, and has to be 8 + 5
r2c6 = 8
r3c6 = 5
2 numbers left in r3:
r3c7 = 9
r3c9 = 2
unique sum in region at r2c7
r2c7 = 5
only 1 place for a 6 in bottom right nonet:
r7c7 = 6
unique sum in region at r6c8: r7c89 = 8 and has to be 3 + 5
r7c8 = 3
r7c9 = 5
rule of 45 on r8 reveals r8c1 + r8c9 = 5 and has to be 1 + 4
r8c1 = 1
r8c9 = 4
only two numbers left in bottom left nonet:
r7c1 = 8
r9c1 = 5
unique sum in region at r1c9: r12c9 = 9 and has to be 1 + 8
r1c9 = 8
r2c9 = 1
two numbers left in top right box:
r1c7 = 3
r1c8 = 7
one number left in r9:
r9c9 = 7
unique sum in region at r1c5:
r1c5 = 6
two numbers remain in c1:
r1c1 = 9
r2c1 = 6
two numbers remain in r1:
r1c3 = 1
r1c4 = 4
one number remains in top left nonet:
r2c3 = 2
two numbers remain in top center nonet:
r2c4 = 9
r2c5 = 3
get rid of penclmarks:
r9c4 = 8
r9c5 = 4
two numbers remain in center center nonet:
r5c4 = 5
r5c5 = 8
one number remains in c5:
r8c5 = 5

looking at grid 1:
rule of 45 reveals:
r9c8 =2
two numbers remain in right center nonet:
r5c7 = 2
r5c8 = 1
two numbers remain in bottom center nonet:
r8c6 = 3
r9c6 = 9
three numbers remain in bottom right nonet:
r8c7 = 8
r8c8 = 9
r9c7 = 1



Hope that helps,

~Squirrel
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Postby emm » Mon Oct 16, 2006 7:35 am

Thanks for that squirrel... I see where I went wrong.
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