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

Advanced methods and approaches for solving Sudoku puzzles
My solver doesn't see a Turbot Fish in the 2s here, which I believe means that it can't be possible to perform a Broken Wing elimination. Instead, it suggests the chain:
r9c7=8 => r9c3=7 => r8c2=4 => r8c9=8, a contradiction that gives us r9c7=7.
After that, things proceed smoothly until the following point:
Code: Select all
. . . | . 1 . | . . .
. . . | . 7 . | . . 1
6 1 . | 8 . 9 | . . 3
-------+-------+------
4 . 5 | . 3 . | 6 1 9
1 . . | 9 6 . | . . 7
7 6 9 | 5 . 1 | 3 8 .
-------+-------+------
. . 6 | 1 9 . | . . 5
. . 2 | . 5 . | 1 9 .
9 5 1 | 4 8 2 | 7 3 6

25  389  3478 |  236   1  3456 |  2489  2467  248
25  389   348 |  236   7  3456 |  2489   246    1
6    1    47 |    8  24     9 |   245  2457    3
----------------+----------------+-----------------
4   28     5 |   27   3    78 |     6     1    9
1  238    38 |    9   6    48 |   245   245    7
7    6     9 |    5  24     1 |     3     8   24
----------------+----------------+-----------------
38   47     6 |    1   9    37 |   248    24    5
38   47     2 |  367   5   367 |     1     9   48
9    5     1 |    4   8     2 |     7     3    6

Consider the chain r1c4~2~r1c9-2-r6c9-2-r6c5-2-r3c5.
When the cell r1c4 contains the value 2, so does the cell r3c5 - a contradiction.
Therefore, the cell r1c4 cannot contain the value 2.
- The move r1c4:=2 has been eliminated.
Consider the chain r2c4-2-r4c4-2-r6c5-2-r6c9-2-r1c9~2~r1c1-2-r2c1.
When the cell r2c4 contains the value 2, so does the cell r2c1 - a contradiction.
Therefore, the cell r2c4 cannot contain the value 2.
- The move r2c4:=2 has been eliminated.
The cell r3c5 is the only candidate for the value 2 in Box 2.

I wonder whether these Turbot Fishes correspond to your Broken Wing. Both Fishes involve cells you've highlighted in green.
rubylips

Posts: 149
Joined: 01 November 2005

rubylips wrote:My solver doesn't see a Turbot Fish in the 2s here, which I believe means that it can't be possible to perform a Broken Wing elimination.

I am inclined to agree with that. I've stumbled into enough "happy accidents" to believe it happens to others as well.

rubylips wrote:
Code: Select all
Consider the chain r1c4~2~r1c9-2-r6c9-2-r6c5-2-r3c5.
When the cell r1c4 contains the value 2, so does the cell r3c5 - a contradiction.
Therefore, the cell r1c4 cannot contain the value 2.

Consider the chain r2c4-2-r4c4-2-r6c5-2-r6c9-2-r1c9~2~r1c1-2-r2c1.
When the cell r2c4 contains the value 2, so does the cell r2c1 - a contradiction.
Therefore, the cell r2c4 cannot contain the value 2.

I wonder whether these Turbot Fishes correspond to your Broken Wing. Both Fishes involve cells you've highlighted in green.

The first chain is definitely for a turbot fish, but the second has seven sides (instead of five) so it's some other X-cycle pattern. However, another true turbot fish may be represented by the chain ...

r3c7~2~r1c9-2-r6c9-2-r6c5-2-r3c5~2~r3c7 implying r3c7<>2

After that elimination in box 3, a coloring map in 2s looks like ...
Code: Select all
C . . | . . . | 2 2 a
c . . | a . . | 2 2 .
. . . | . A . | . a .
- - - + - - - + - - -
. a . | A . . | . . .
. A . | . . . | 2 2 .
. . . | . a . | . . A
- - - + - - - + - - -
. . . | . . . | B b .
. . . | . . . | . . .
. . . | . . . | . . .

Since either 'C' or 'c' must be true, and 'C' true excludes 'a' true, and 'c' true excludes 'a' true ... all locations colored 'a' must be false.

So r1c9<>2, r2c4<>2, r3c8<>2, r4c2<>2, and r6c5<>2 in one swell foop.
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Jeff wrote:
I checked the deductions using x-cycle and found that they are not valid. Can you check them again?

I checked again and can see no flaw in the logic. (Of course the answer is right: that’s not the issue.) Perhaps there’s something here that x-cycle doesn’t go at. The logic of the Broken Wing is of course different in that it concerns not the cells in the loop but the cells outside the loop that keep the loop from being perfectly conjugate. Just to develop the argument from first principles for this case:
1. The cells marked in amber are all the cells that preclude the five cells marked in green from being a perfect conjugate loop.
2. Arguing negatively, if one of the cells marked in amber is not a 2, then the loop is a perfect conjugate loop of five cells and the contradiction ensues.

Perhaps it’s the fact that several of the guardians in the first pass are doing “double duty” and reducing the conjugate pairings to only two that makes the x-cycle logic balk at a positive conclusion. Of course it would be nice if I could point to a case where the answer couldn’t be found by an alternate route!

rubylips wrote:
My solver doesn't see a Turbot Fish in the 2s here, which I believe means that it can't be possible to perform a Broken Wing elimination.

Of course I can’t answer for your solver but the only thing I’ve claimed is that the Broken Wing uses the Turbot Fish shape, which is visibly there in the diagram. Since the logics do proceed on different paths, there’s no reason to be surprised that the Broken Wing can apply in some circumstances where the x-cycle logic may not, and vice versa. I’ll stand by the logic above and invite you to consider it for flaws.
rubylips wrote:
I wonder whether these Turbot Fishes correspond to your Broken Wing. Both Fishes involve cells you've highlighted in green.

They aren’t all the same cells, but in fact the Broken Wing works quite nicely with both loops you describe and gives the same answers by inspection. The loop cells are marked in green and the guardians in amber.

In this case the guardians all see r1c4 so it cannot contain a 2.

This loop of seven is a much rarer animal than the five, but the guardians again behave quite nicely. They eliminate r1c9 and would do so even if r1c4 (shown here as still holding a 2) had not been eliminated. With r1c4 gone, they are in fact the same guardians as for the second pass at the loop I diagrammed earlier, giving a nice “second opinion” on the logic I followed.
ronk wrote
However, another true turbot fish may be represented by the chain ...
r3c7~2~r1c9-2-r6c9-2-r6c5-2-r3c5~2~r3c7

Your loop is below with the Broken Wing guardians highlighted (sorry for the clipoff at the bottom).

Yes, that’s one of the four ways my first diagram above (or the second Broken Wing that applies immediately thereafter) could have been drawn. It’s just that in this case the Broken Wing gets there at an earlier stage.
Cheers,
Rod
RodHagglund

Posts: 9
Joined: 19 December 2005

In the first grid of this puzzle (in page 1) we have

[r3c7|r3c8]-2-[r3c5]=2=[r6c5]-2-[r6c9]=2=[r1c9|r2c9]-2-[r3c7|r3c8]

which implies r3c7, r3c8<>2 => r3c5=2.

Carcul
Carcul

Posts: 724
Joined: 04 November 2005

Post cancelled.
Last edited by Jeff on Wed Jan 04, 2006 1:41 am, edited 1 time in total.
Jeff

Posts: 708
Joined: 01 August 2005

RodHagglund wrote:The logic of the Broken Wing is of course different in that it concerns not the cells in the loop but the cells outside the loop that keep the loop from being perfectly conjugate. Just to develop the argument from first principles for this case:
1. The cells marked in amber are all the cells that preclude the five cells marked in green from being a perfect conjugate loop.
2. Arguing negatively, if one of the cells marked in amber is not a 2, then the loop is a perfect conjugate loop of five cells and the contradiction ensues.

Statement #2 has pretty much convinced me.

However, in your illustration for rubylips' chain of length 7, I think r2c7 and r2c8 also need to be highlighted as "guardians". If not, one might arrive at the false conclusion that r1c1#2 can be eliminated too.

Ron
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Jeff wrote:Every elimination can be expressed as a forcing chain or forcing net. Since that cannot be found, could you provide a reasonable proof to show that the deductions are valid.

Hi Rod, I apologise for my oversight. Your deductions are totally correct as proven by Carcul's grouped x-cycle (see diagram below).

Carcul wrote:[r3c7|r3c8]-2-[r3c5]=2=[r6c5]-2-[r6c9]=2=[r1c9|r2c9]-2-[r3c7|r3c8]
which implies r3c7, r3c8<>2 => r3c5=2.

Jeff

Posts: 708
Joined: 01 August 2005

Hi RodHagglund,

I thought your Broken Wing technique would be applicable to this puzzle, but I can't detect any candidate 8 elimination for the situation where multi-coloring makes the elimination r5c7<>8.

The potential (but contradictory) 7-sided x-cyle is colored green, and (what I think are) the guardians are colored amber. Colored as guardians are those cells which would prevent conjugate pairs in rows 4 and 5 and col 7. The dilemma is that no cell of the x-cycle 'sees' all the guardians.

Is that the correct to way identify guardians? If the guardians are correct, is it possible there just isn't a Broken Wing elimination even though multi-coloring can make one? Something else?

TIA, Ron
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

ronk wrote:The potential (but contradictory) 7-sided x-cyle is colored green, and (what I think are) the guardians are colored amber. Colored as guardians are those cells which would prevent conjugate pairs in rows 4 and 5 and col 7. The dilemma is that no cell of the x-cycle 'sees' all the guardians.

Is that the correct to way identify guardians? If the guardians are correct, is it possible there just isn't a Broken Wing elimination even though multi-coloring can make one? Something else?

To the best of my knowledge that is indeed the correct way to identify the guardians, but unfortunately they don't intersect anywhere, so no elimination can be made.

I want to share few thoughts, though...

As far as I have gathered (which isn't very far, admittedly), Broken Wing is perfect for application in conjunction with regular Turbot Fish and/or X-Cycle detection. In the example above, though, the 8s form a typical discontinuous X-Cycle, with the discontinuity in R5C7.

The reason I think the two strategies could be attempted in conjunction with eachother is that they both require a loop of strong and/or weak edges; X-Cycles in an alternating fashion, while Broken Wing can potentially have them in any order. (Long loops with a lot of scattered "guardians" are unlikely to lead anywhere, though, I think, as one can only eliminate candidates from the intersection of all the guardian cells (regardless of its participation in the loop)... But of course, if they *do* intersect, all is good.

Here's an artificially created sample of a loop (marked with "+") in 2s that is neither a Turbot Fish, nor an X-Cycle, due to the three adjacent weak edges;
Code: Select all
.  .  .  |  .  .  .  |  . +2  .
. +2  .  |  . g2  .  |  .  . +2
.  .  .  |  .  .  .  |  .  .  .
---------+-----------+---------
. g2  .  |  . #2# .  |  .  .  .
.  .  .  |  .  .  .  |  .  .  .
.  .  .  |  .  .  .  |  .  .  .
---------+-----------+---------
.  .  .  |  .  .  .  |  .  .  .
. +2  .  |  . g2  .  |  . +2  .
.  .  .  |  .  .  .  |  .  .  .

One of the guardians (marked with "g") have to be a 2 to avoid a 5-sided conjugate loop (impossible), and all the guardians "see" the center cell, from which 2 can be eliminated (marked with "#").

I'll try to construct a real puzzle that resembles this situation, but I'm not too good at that, so don't hold your breath... (Nor am I sure it's possible at all, but I guess I'll find out...) ;)
vidarino

Posts: 295
Joined: 02 January 2006

vidarino wrote:Here's an artificially created sample of a loop (marked with "+") in 2s that is neither a Turbot Fish, nor an X-Cycle, due to the three adjacent weak edges;
Code: Select all
.  .  .  |  .  .  .  |  . +2  .
. +2  .  |  . g2  .  |  .  . +2
.  .  .  |  .  .  .  |  .  .  .
---------+-----------+---------
. g2  .  |  . #2# .  |  .  .  .
.  .  .  |  .  .  .  |  .  .  .
.  .  .  |  .  .  .  |  .  .  .
---------+-----------+---------
.  .  .  |  .  .  .  |  .  .  .
. +2  .  |  . g2  .  |  . +2  .
.  .  .  |  .  .  .  |  .  .  .

One of the guardians (marked with "g") have to be a 2 to avoid a 5-sided conjugate loop (impossible), and all the guardians "see" the center cell, from which 2 can be eliminated (marked with "#").

Hi vidarino, The broken wing is indeed very powerful for this instance. An indirect way to deduce r4c5=2 is via an almost swordfish nice loop (fillet O fish) or an almost x-wing nice loop.

Almost swordfish x-cycle (fillet O fish):
[r2c9]=2=[almost swordfish:r2c25|r4c25|r8c258]-2-[r1c8]=2=[r2c9] => r2c9=2

Almost x-wing x-cycle:
[r1c8]-2-[r8c8]=2=[almost x-wing:r4c25|r8c25]-2-[group:r2c2|c2c5]=2=[r2c9]-2-[r1c8] => r1c8<>2

Carcul, could you see if a pure grouped x-cycle is applicable here? It seems to me that there is nothing much that nice loops can't do nowadays.
Jeff

Posts: 708
Joined: 01 August 2005

It is odd, since broken wing and filet-o-fish were derived from completely different logic, but the colored cells denoting the broken wing often are the very same cells which are part of a sashimi swordfish, its fin, and/or its deleted cells. I could use the broken wing logic almost as a signpost for the filet-o-fish (which seems to have a slight advantage when it comes to getting rid of candidates)

By the way, Ronk's grid...

Contains a filet-o-fish x-wing in r16c17 with a fin in r6c8 which deletes the 8 in r5c7.
Myth Jellies

Posts: 593
Joined: 19 September 2005

Hi Jeff and Vidarino.

Vidarino wrote:Here's an artificially created sample of a loop (marked with "+") in 2s that is neither a Turbot Fish, nor an X-Cycle, due to the three adjacent weak edges;

Jeff wrote:Carcul, could you see if a pure grouped x-cycle is applicable here? It seems to me that there is nothing much that nice loops can't do nowadays.

Vidarino, your example above is very nice, but how can we possibly conclude that no other argument can be applied if we have just one "2" in boxes 1,2,4,5,7,8,9? Jeff, for me (or other) to see if grouped nice loops can be applied here, I would obviously need the complete grid on "2s". Making artificial examples for proving the usefulness of a technique is very nice, but how about to find a real example?

Regards, Carcul
Carcul

Posts: 724
Joined: 04 November 2005

Carcul wrote:Vidarino, your example above is very nice, but how can we possibly conclude that no other argument can be applied if we have just one "2" in boxes 1,2,4,5,7,8,9? Jeff, for me (or other) to see if grouped nice loops can be applied here, I would obviously need the complete grid on "2s". Making artificial examples for proving the usefulness of a technique is very nice, but how about to find a real example?

I think I managed to make one, actually! (I had to disable X- and Y-Cycles and Turbot Fish to make my solver stop here, but the 2s end up close to what I outlined in my draft above.

Puzzle:
Code: Select all
.4...5..7
5.7..3.4.
.8.....19
.3.15....
...4.6.8.
.....92..
.7.2...5.
8.4...9.6
.9.....3.

Pencil marks:
Code: Select all
1269      4   1269  |     68   1268      5  |      3     26      7
5    126      7  |      9    126      3  |     68      4     28
236      8    236  |     67   2467    247  |      5      1      9
-----------------------+-----------------------+-----------------------
267      3    268  |      1      5    278  |     67      9      4
279     25    259  |      4    237      6  |      1      8     35
4    156   1568  |    378    378      9  |      2     67     35
-----------------------+-----------------------+-----------------------
36      7     36  |      2      9    148  |     48      5     18
8    125      4  |    357     37     17  |      9     27      6
12      9    125  |   5678   4678   1478  |    478      3    128

And finally, the 2s:
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

The Broken Wing loop as outlined above (guardians at R5C2 and R2C5) will eliminate the 2 in R5C5, leaving a single 2 in that box, which takes the puzzle a bit onwards. (Not too far, though, it seems...)

(Unfortunately, I also accidentally made this a Turbot Fish as well, since there are now 3 adjacent strong edges (box3, C8 and R8), but hopefully it proves my point anyway. I'll keep looking for a better example, though.)

Vidar
vidarino

Posts: 295
Joined: 02 January 2006

OK, nevermind, you may disregard my previously posted sample. Here's one that appears to be a better demonstration.

It actually defeats my solver (which admittedly isn't that clever), despite me having re-enabled X- and Y-Cycle and Turbot Fish detection. (It even uses a Y-Cycle and a Turbot Fish to reach the shown scenario.)

Initial puzzle:
Code: Select all
.4...5..7
5.79.3.4.
.8.....19
.3.59....
...7.6.8.
......2..
.7.....5.
8.43.19.6
.9.....3.

Stops here:
Code: Select all
1269      4   1269  |   1268   1268      5  |      3     26      7
5     12      7  |      9    126      3  |     68      4     28
236      8    236  |    246   2467    247  |      5      1      9
-----------------------+-----------------------+-----------------------
124      3    128  |      5      9     28  |     67     67     14
1249    125   1259  |      7    123      6  |     14      8     35
7      6    158  |    148   1348     48  |      2      9     35
-----------------------+-----------------------+-----------------------
1236      7   1236  |   2468   2468      9  |    148      5   1248
8     25      4  |      3    257      1  |      9     27      6
126      9   1256  |   2468 245678   2478  |   1478      3   1248

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.

(Edit: A word of warning, though. I haven't verified it 100% yet, but this puzzle may have more than one solution.)

(Edit 2: To make the puzzle unique, see if you can force R1C1 to become 1. (If not, just set it to 1. ;) That, along with application of some Uniqueness Rectangles later makes it solvable.)

Vidar
Last edited by vidarino on Wed Jan 18, 2006 8:17 am, edited 1 time in total.
vidarino

Posts: 295
Joined: 02 January 2006

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.
Jeff

Posts: 708
Joined: 01 August 2005

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