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: thats not the issue.) Perhaps theres something here that x-cycle doesnt 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 its 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 couldnt 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 cant answer for your solver but the only thing Ive 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, theres no reason to be surprised that the Broken Wing can apply in some circumstances where the x-cycle logic may not, and vice versa. Ill 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 arent 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 wroteHowever, 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, thats one of the four ways my first diagram above (or the second Broken Wing that applies immediately thereafter) could have been drawn. Its just that in this case the Broken Wing gets there at an earlier stage.
Cheers,
Rod