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SudoQ
Joined: 02 Aug 2011
Posts: 127
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 Posted: Wed Mar 14, 2012 10:45 pm Post subject: HoDoKu puzzle (2) |
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Just one more!
The following puzzle is a bit interesting because it doesn't seem to have any nice bi cells.
..1......34.........9..1.627..15.24.9.42.38.7.12.48..912.3..9.........23......4..
After basics: | Code: | |----------------|----------------|---------------|
| 2 6 1 | 589 3789 57 | 357 3589 4 |
| 3 4 578 | 5689 2789 26 | 1 589 58 |
| 58 78 9 | 4 378 1 | 357 6 2 |
|----------------|----------------|---------------|
| 7 38 38 | 1 5 9 | 2 4 6 |
| 9 5 4 | 2 6 3 | 8 1 7 |
| 6 1 2 | 7 4 8 | 35 35 9 |
|----------------|----------------|---------------|
| 1 2 6 | 3 78 4 | 9 578 58 |
| 4 789 578 | 589 1 57 | 6 2 3 |
| 58 3789 3578 | 5689 2789 26 | 4 78 1 |
|----------------|----------------|---------------| |
/SudoQ |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Thu Mar 15, 2012 4:52 am Post subject: |
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| (8=7)r7c5-[(7=5)r8c6 AND 7r2c5=7r2c3]-(57=8)r8c3 => 8r7c5=8r8c3 => -8r8c4; +8r8c23 => -8r9c1; stte |
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SudoQ
Joined: 02 Aug 2011
Posts: 127
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| Posted: Thu Mar 15, 2012 8:24 am Post subject: |
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Hi JC,
I knew you'd find something!
There is also a short tri-cell solution...
By the way, what is 'stte'?
/SudoQ |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Thu Mar 15, 2012 11:23 am Post subject: |
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| SudoQ wrote: | | There is also a short tri-cell solution... |
Hi SudoQ,
Maybe my initial solution ... 
Colouring from (58)r3c1 yields -5r8c6 and stte. To wit :
5-SIS Chain : Kite(7r8c6=7r1c6-7r2c5=7r2c3)-5r2c3=5r3c1-(5=8)r9c1-(87#2=5)r8c3 => (-7r8c3,) -5r8c6; stte
A simpler solution is, of course, available. But this was not the purpose of my post 
| SudoQ wrote: | | ... what is 'stte'? |
stte = singles (only) to the end.
Best Regards, JC. |
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SudoQ
Joined: 02 Aug 2011
Posts: 127
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| Posted: Thu Mar 15, 2012 12:41 pm Post subject: |
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Hi JC,
the tri-cell solution I've found is this:
| Code: | r8c3=5 -> r8c6=7 -> r7c5=8 ->
=7 -> r2c5=7 -> r7c5=8 ->
=8 -> r8c4<>8 |
What do you write after that, where it's not 'singles (only) to the end'?
/SudoQ |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Thu Mar 15, 2012 1:29 pm Post subject: |
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Hi again SudoQ,
Your tri-cell solution is based on the 4-SIS : (578)r8c3, (57)r8c6, (78)r7c5, 7r2c35 => -8r8c4.
You've written the solution as a "Kraken Cell" R8C3. One could also write an Almost XY Wing( 57-8 ) or an Almost W Wing( 78 ).
In my first post, I presented a Chain based on the same SIS . Just to avoid revisiting a SIS in the other presentations.
| SudoQ wrote: | | What do you write after that, where it's not 'singles (only) to the end'? |
sstte = simple sudoku techniques (= basics) to end. However, I now often give the details of the basics used before stte.
Best Regards, JC. |
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SudoQ
Joined: 02 Aug 2011
Posts: 127
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| Posted: Thu Mar 15, 2012 3:07 pm Post subject: |
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Hi JC,
and thanks for the information.
I realize (once again  ) that it's time to learn a more established syntax.
But for me it seems more difficult to understand than 'Kraken Cells'...
/SudoQ |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Thu Mar 15, 2012 3:58 pm Post subject: |
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| JC Van Hay wrote: | sstte = simple sudoku techniques (= basics) to end. However, I now often give the details of the basics used before stte.
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I recall SSTS (Simple Sudoku Technique Set) being used -- often just for basics. If it was the last step listed, then "to the end" was implied.
However, since it's now common practice to list grids with the prefix after basics, I skip all basics in my solution -- unless they're somehow relevant.
BTW: an alternate way to present JC's (and SudoQ's) solution.
| Code: | (8)r8c3=[(8-7)r7c5-r2c5=r2c3-(7=5)r8c3-(5=7)r8c6-(7=8)r7c5] => r8c4<>8
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Thu Mar 15, 2012 5:23 pm Post subject: |
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| daj95376 wrote: | BTW: an alternate way to present JC's (and SudoQ's) solution.
| Code: | (8)r8c3=[(8=7)r7c5-r2c5=r2c3-(7=5)r8c3-(5=7)r8c6-(7=8)r7c5] => r8c4<>8
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Mea culpa. I forgot this one.
However, to avoid the "looping" in the "lasso" and borrowing a notation of abi on the French Forum, I would write ...
(8=7)r7c5-7r2c5=7r2c3-([8=]7=5)r8c3-(5=7)r8c6 => -8r8c4 because 8r8c3=[8r7c5=7r8c6] => 8r8c3=8r7c5
Best Regards, JC. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Thu Mar 15, 2012 11:07 pm Post subject: |
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| JC Van Hay wrote: | ... to avoid the "looping" in the "lasso" ...
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[Withdrawn]
RonK did a much better job of describing my (muttled) thoughts!!!
Last edited by daj95376 on Thu Mar 15, 2012 11:30 pm; edited 2 times in total |
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ronk
Joined: 07 May 2006
Posts: 398
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| Posted: Thu Mar 15, 2012 11:26 pm Post subject: |
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| daj95376 wrote: | | I guess that every discontinuous loop could be considered a "lasso". However, I reserve the use of lasso for cases where more than one cell is used twice in the endpoints. |
I'm away from my computer, so I'm having to trust my untrustworthy memory, but ... I think berthier's lasso actually looks like a lasso. Assert an ultimately false candidate, follow the "xyzt-chain", and the contradiction occurs at some location on the chain other than the starting point. |
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