## "broken wwing" - a new use for the "turbot fi

Advanced methods and approaches for solving Sudoku puzzles
Jeff wrote:
vidarino wrote:......I had to disable X- and Y-Cycles and Turbot Fish to make my solver stop here......

Hi vidarino, What are you talking about? Isn't turbot fish a subset of x-cycle? If you had disabled x-cycle, you would have disabled turbot fish at the same time.

Ah, indeed, sorry for not making that clear. My solver identifies the simpler cases of X-cycles (X-Wing, 222-Swordfish and Turbot Fish) separately, to make the solution path clearer for a human follower. Also, the cases have different "difficulty ratings", making up the puzzles' final "difficulty score". That value is highly experimental currently, though, and fluctuates wildly. ;)
vidarino

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Joined: 02 January 2006

Vidarino wrote:
Code: Select all
` 2  .  2   | 2  2  .   | . +2  .  . +2  .   | .  2  .   | .  . +2  2  .  2   | 2  2  2   | .  .  . -----------+-----------+----------  2  .  2   | .  .  2   | .  .  .  2  2  2   | . #2# .   | .  .  .  .  .  .   | .  .  .   | 2  .  . -----------+-----------+----------  2  .  2   | 2  2  .   | .  .  2  . +2  .   | .  2  .   | . +2  .  2  .  2   | 2  2  2   | .  .  2 `

Here the Broken Wing pattern eliminates the 2 in R5C5.

Even considering r1c1=1 the puzzle have 5 solutions.

Regards, Carcul
Last edited by Carcul on Wed Jan 18, 2006 10:14 am, edited 1 time in total.
Carcul

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Carcul wrote:BTW, even considering r1c1=1 the puzzle have 5 solutions.

[11]=6 yields 1 solution but also eliminates X-cycle logic
gsf
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Carcul wrote:Making r5c5=2 doesn't lead to any contradiction. In fact, from r5c5=2 a turbot fish emerges that imply r2c2=2. Am I missing something?

Making R5C5=2 will make the outlined loop (+) consist solely of conjugate pairs of 2, and a 5-sided strongly linked loop is impossible. That's the whole essence of the Broken Wings technique, I believe.

BTW, even considering r1c1=1 the puzzle have 5 solutions.

Hmm, OK, I'll investigate a bit. I'm just implementing Uniqueness Rectangles in my solver now, and as far as I could see, it managed to solve the rest if make R5C5<>2 and R1C1=1. I'll re-check, though.

Edit: Hmm, I just re-checked, and my solver *does* manage to solve it if I set R1C1=1 after having done the Broken Wing elimination. It's not impossible that I have an error in my program, though. It would certainly not be the first time. But things do look okay at first glance... I can try to scribble down a solution path if necessary. (Oops, one thing just struck me: My solution *does* rely on Unique Rectangles, so perhaps the grid *does* have multiple solutions, but a happy accident makes the UR do a correct elimination... Will investigate further.)

Edit 2: Oh crap, that was exactly the situation. I disabled the UR test, and found several solutions by using trial and error. Oh well, R1C1=6 does, as gsf pointed out, lead to a unique solution. The point was nevertheless to demonstrate a Broken Wing which wasn't an X-Cycle, which I hope I did.

Regards,
Vidar-who-won't-take-lightly-on-hand-building-demo-puzzles-from-now-on
vidarino

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Hi Vidarino.

Carcul wrote:Making r5c5=2 doesn't lead to any contradiction. In fact, from r5c5=2 a turbot fish emerges that imply r2c2=2.

This is obviously not correct, and the elimination r5c5<>2 is correct.

But with r1c1=1 the puzzle have 5 solutions, as stated by Simple Sudoku.

Regards, Carcul
Carcul

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Joined: 04 November 2005

vidarino wrote:....... The point was nevertheless to demonstrate a Broken Wing which wasn't an X-Cycle, which I hope I did.

This has been proven some time ago. At that time, the particular broken wing could not be expressed as an x-cycle, but a grouped x-xycle.
Jeff

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Carcul wrote:
Carcul wrote:Making r5c5=2 doesn't lead to any contradiction. In fact, from r5c5=2 a turbot fish emerges that imply r2c2=2.

This is obviously not correct, and the elimination r5c5<>2 is correct

When a puzzle has multiple solutions -- 62 in this case, unless I missed some changes after the original post -- how do you determine whether a placement or elimination is correct?

[edit: I guess that was a stupid question, because you can just make an assertion (T&E) and see if it leads to a contradiction. What I was thinking was ... if there are many solutions, aren't there likely to be quite a few placements that are common to all the solutions?]

Ron
Last edited by ronk on Wed Jan 18, 2006 11:08 am, edited 1 time in total.
ronk
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Jeff wrote:
vidarino wrote:....... The point was nevertheless to demonstrate a Broken Wing which wasn't an X-Cycle, which I hope I did.

This has been proven some time ago. At that time, the particular broken wing could not be expressed as an x-cycle, but a grouped x-xycle.

Well, my involvement just started as a reply to ronk's question about guardians, and I answered that his guardians were correctly placed, but didn't intersect anywhere, after which I created an artificial scenario where they *did* intersect, after which Carcul asked for a real example, so I made one. (whew) ;)

Anyway, I stand humbled when it comes to the complexity of building demo puzzles like this one. I hope gsf releases the source to his sudoku program soon, so that I can add what I'm looking for and do an exhaustive search on a large database of puzzles instead. Hint hint.

Vidar
vidarino

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ronk wrote:When a puzzle has multiple solutions -- 62 in this case, unless I missed some changes after the original post -- how do you determine whether a placement or elimination is correct?

Ron

In this particular case, setting R5C5=2 would yield a contradiction, so that elimination must be correct. Most other techniques will also do their stuff without hiccups, but the ones that rely on uniqueness (Uniqueness Rectangles, Almost UR, BUG+n) might very well give bogus results. (And indeed, I erroneously thought my latest posted grid had a unique solution, thanks to a UR pattern with a "happy accident" outcome.)

Vidar
vidarino

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