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storm_norm · Joined: 18 Oct 2007 Posts: 1741

convoluted one stepper Rolling Eyes

keith · Posted: Fri Oct 10, 2008 11:33 pm Post subject:

Norm,

I have a W-wing <13> (link on <1> in R9) plus coloring (in C9) that says R2C1 is not <3>.

EDIT: This is incorrect. I did not write down my original solution, and I will try to find it again.

Keith

storm_norm · Joined: 18 Oct 2007 Posts: 1741

Keith,
your w-wing is this??

(3=1)r8c5-(1)r9c6=(1)r9c8-(1=3)r6c8... so you would extend it... -(3)r6c9=(3)r2c9...

like that??

but where does r2c1 see both 3's??

the 3 in r2c5 is weakly linked to the 3 in r8c5.
the chain would need one more strong link from the 3 in r2c5 that sees the 3 in r2c1.

daj95376 · Joined: 23 Aug 2008 Posts: 3854

storm_norm · Joined: 18 Oct 2007 Posts: 1741

keith · Posted: Sat Oct 11, 2008 12:38 pm Post subject:

Norm and Danny,

I have acknowledged my error above, in the original message.

Keith

arkietech · Posted: Sat Oct 11, 2008 1:55 pm Post subject:

I got it down to needing 2 xy-chains but still looking for the pony Shocked

daj95376 · Joined: 23 Aug 2008 Posts: 3854

arkietech · Posted: Sat Oct 11, 2008 5:58 pm Post subject:

Asellus · Posted: Sat Oct 11, 2008 6:42 pm Post subject:

There are some other interesting things present, though none of them alone solve it.

A Kite (or Skyscraper) removes <4> from r1c2.
A Sue de Coq in r2c1 and r3c1236 removes <6> from r2c2 and <4> from r3c4.
A 47 UR in r19c46 removes <4> from r1c4 and <7> from r9c6.
(After the Kite and the Sue de Coq, there is a Type 6 49 UR.)

cgordon · Joined: 04 May 2007 Posts: 769 Location: ontario, canada

I used a double whammy ER on <2> - another on <4> - then the Type 6 UR which was also a double whammy. Then I got stuck. I looked at the Sue de Coq but for me all it does is confirm that R3C123 cannot contain a 3 and a 6 or a 4 and a 9 which means there has to be a <7> which I already knew cos they are the only two <7> in R3.
Strange eh!

Asellus · Posted: Sat Oct 11, 2008 9:41 pm Post subject:

Craig,

You're not quite seeing the Sue de Coq. A Sue de Coq is two or more overlapping Almost Locked Sets (ALS) where the total number of digits involved equal the total number of cells involved and where the Almost Locked Sets outside the overlap cells do not have any digits in common.

This might sound bewildering, but the concept isn't quite as hard to grasp as the cumbersome language of describing it.

The 5 cells r2c1 and r3c1236 contain the 5 digits {34679}, satisfying the first requirement. The two (in this case) overlapping ALS are r2c1|r3c123 and r3c1236 (each of which are 4 cells containing the 5 digits {34679}). The overlapping cells are r3c123. One external (outside the overlap) ALS is the {36} in r2c1; the other is the {49} in r3c6. These two external ALS do not have any digits in common, satisfying the second requirement.

Whether r2c1 is <3> or <6>, r3c1236 will become a Locked Set containing <4> and <9>. Thus, there cannot be either of these two digits anywhere else in r3. That is how <4> is eliminated from r3c4.

Whether r3c6 is <4> or <9>, r2c1|r3c123 will become a Locked Set containing <3> and <6>. Thus, there cannot be either of these two digits anywhere else in box 1, eliminating <6> from r2c2.

It should now be evident why those non-overlap Almost Locked Sets (in this case, the bivalues {36} and {49}) cannot have any digits in common. If they did, they wouldn't be able to induce those Locked Sets.

While there were only two eliminations in this case, Sue de Coq sometimes produces many eliminations and can be very helpful.

cgordon · Joined: 04 May 2007 Posts: 769 Location: ontario, canada

Asellus: Got to go out now - but I'm gonna have to study that. When I first read an explanation for Sue de Coqs it was along the lines if you have three cells containing a 34679 seeing a 36 and a 49, the three cells cannot contain a 3 as well as a 6. or a 4 as well as a 9, because that would negate the two pairs. So the choices are 347, 379 and 467. So the threee cells have to contain a 7. In fact I'd bet the farm the explanation came from the lady herself. I read it a long time ago and always wanted to find one to impress the Forum. Never did though.

Cheers

daj95376 · Joined: 23 Aug 2008 Posts: 3854

Asellus · Posted: Sat Oct 11, 2008 11:00 pm Post subject:

Craig,

I, too, recall Sue's original example containing a conjugate pair within the overlap cells, as is the case with the <7>s in this puzzle. However, this is not necessary for a Sue de Coq and there are many examples that are otherwise, including earlier ones posted on this board. I'll look for an example or two.

Asellus · Posted: Sat Oct 11, 2008 11:51 pm Post subject:

Okay... here are some good examples of various Sue de Coq situations that may be helpful for understanding them.

First, a fairly simple example with no conjugate digit in the overlap (in the second grid of the post).

Next, a nice example with 3 external ALS, 2 binaries sharing a row and 1 binary sharing a box.

Here is one with locked candidates in the overlap cells rather than a conjugate pair. It also has a 2-cell external {136} ALS sharing the box with the overlap cells. (The grid is shown a few posts above the linked post.)

And, finally, here is one that shows that a conjugate pair (or locked candidates) can be FORMED within the overlap cells of a Sue de Coq, leading to additional eliminations. (I believe this was the case with Sue's original example.)