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[Marty R.]

I've learned a number of new techniques lately, such as W- and M-Wings, XY-Chains, pincer coloring, ERs, Medusa and others, but most of them require strong links and bivalue cells. What I am wondering is how you generally approach a puzzle that's loaded with polyvalue cells and won't budge? I've tried everything I know except the ocean dwellers.

Here is one example of such a puzzle. If you'd like to see the original, you're looking at it since not one cell has been solved.

[Steve R]

I rely heavily on conjugates, so much so that I’m usually stuck if I don’t see any.

Here r4c7 and r5c7 are conjugate with respect to 9, while r5c7 is also conjugate to r5c3 with respect to 1. Throw in the (19) in r4c5 and I think you can eliminate 1 from r4c1.

More formally, the conjugates suggest a chain:

r4c1 = 1 ==> r4c5 = 9 ==> r4c7 ≠ 9 ==> r5c7 = 9 ==> r5c3 = 1 ==> r4c1 ≠ 1

We all do puzzles in our own way but perhaps this approach will help occasionally.

Steve

PS And you’re right: there is a fish.

3 can be eliminated from r2c2 using the x-wing in columns 1 and 5.

[Marty R.]

Thank you Steve, I'll try that conjugate approach.

[Steve R]

We may be at cross purposes but I saw the x-wing as targeting rows 2 and 8 with the fin (r1c1) limiting the elimination to r2c2.

Steve

[Marty R.]

[Victor]

Marty, you've caused brief marital disharmony! - I got hooked on this instead of putting up a bookshelf, spending a happy 2 or 3 hours hunting and eventually finding one of my favourite animals, ALS.

Set A: R5 less C7: {1,3,4,6,7,9}
Set B : R4C5 on its own: {1,9}

I'm sure you're familiar with ALS, but in case someone's reading this who isn't, here goes with one way of looking at it.
Only one of A & B can have the 9 in box 5. If it's in A then B is just {1}. If it's in B, then A is {1,3,4,6,7} So there must be a 1 in either R5C3 or R4C5. So that kills the 1s in R4C13. (In general, the linking number, 9 here, is called x, the other one, 1 here, is z.)

(And, perhaps surprisingly, it's fairly easy thereafter.)

[Asellus]

Steve's chain also removes <1> from r4c3. So, it has the same result as Victor's ALS.

For those interested, Steve's chain in Eureka notation is:
(1)r4c13-(1=9)r4c5-(9)r4c7=(9-1)r5c7=(1)r5c3-(1)r4c13; r4c13<>1

[Asellus]

After those <1>s are removed, the puzzle only requires a c15 X-Wing on <3> and then, after considerable simplification, one XY-Wing.

[Asellus]

Getting back to Marty's original question about attacking tough puzzles, I posted another approach, using this puzzle as an example, in a new thread.

[cgordon]

I had a go at this one and couldn't get any further. Is this a Sudoku.com Very Hard puzzle? I don't recall any of these that couldn't be completed without a wing, UR or ER. I never got into chains - borderline guessing I say - in fact my reference guidelines suggest the ALS method could be considered a form of guessing. So there !!

Marty: One question: Using the Draw/Play and the Sweep functions - how did you remove the redundant 1's from Box 5 and the 6's from Box 7.

[Marty R.]

[cgordon]

[Marty R.]

[Victor]

Cgordon says:

[Asellus]

If ALS solutions in general are guessing, then so are XY Wings and XYZ Wings, since both of these are just forms of the "xz" ALS technique. If one were to forgo the former as "guessing," then consistency would demand that one forgo the latter. Me, I don't see that any of it is guessing in any way whatsoever.