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[Mogulmeister]

This another extreme from Andrew Stuart - the current one.

Enjoy.

[Marty R.]

Stuart replaced Mepham, is that correct? How does the Extreme rating compare to the Diabolical?

[Mogulmeister]

Andrew is little short of being a Sudoku god (alongside Myth Jellies, Ruud, Mike Barker, Jeff et al) and worked for Michael Mepham as the main brains and compiler for the puzzles.

The Extreme is (usually) tougher than the Diabolical.

[Asellus]

I'd say that this one was tougher than most Diabolicals. I eventually solved it but the route wasn't pretty! Besides sashimi fish, Medusa coloring, and ALS chains, I resorted to an implication chain based (somewhat loosely!) on an AB, BC, CA type of UR pattern before it succumbed. Don't ask me to recreate it, though!

[Johan]

This one was very tough, after x-chain or AIC(I don't know the precise name for this approach) on <4>. Starting in R8C3 using a strong link I ended up in R1C3,R2C3, with both cells having two weak links on <4>, which erases <4> in both cells.
After that there was some kind of a finned swordfish on <5>, with two fin cells in R1C8 and R1C9, so I eliminated the <5> in R2C9, because the common number <5> lies in the same box as the two finned cells and the puzzle opens up, but I'm not sure it was a finned swordfish and the elimination on <5> was correct.

[Asellus]

Johan,

I believe that what you have shown is not a Finned Swordfish. For a finned fish based on columns, the fin must share one of the columns with the fish (and similarly with rows). Also, the eliminations only occur in cells which are not part of either the fish or the fin.

Your candidate Swordfish is based on columns (C189). Either R1C8 could be a fin or R1C9 could be a fin, but not both. In either case, it does not help you because there are no eligible candidates for elimination. Your elimination is in a cell which is part of your swordfish, which is not valid.

Let's suppose that there was no <5> in R1C9 but there was, instead, a <5> in R5C1. Then, you would have a Finned Swordfish with the Swordfish part as you've marked it and R5C1 as the Fin. The <5>s in R2C46 would be eliminated since they are both eliminated by both the Swordfish and the Fin.

I don't see any way to make a fish-related elimination of a <5> in the grid that you've posted. Maybe someone can see something that I can't.
[/i]

[Asellus]

Given the grid as posted by Johan after eliminating those <4>s, the next step isn't immediately obvious. So, I followed some good advice I received elsewhere and cleared my mind. That took away all of the predetermined tools and opened me to new insights.

We already know that we have some strong links on 4. The other thing that stood out to me were all the 36 relationships. The 36 bivalue in R6C7 means that Box 4 R6 cannot be both 3 and 6. R78C1 threatens to become a 36 pair if R9C1 is 1. And, that depends on the 4 in R9C5.

So, let's follow this 36 trail and let R9C5, etc., be 4. That means that R4C7 is also 4, which means that R6C7 is 6 and R6 C9 is 3 and that 36 is eliminated from R6 in Box 4.

But, if R9C5 is 4, then R9C1 is 1, creating that R78C1 36 pair and making R46C1 a 59 pair. At this point, the only candidate left for R6C2 is 8. But, the 36 pair in C1 of Box 7 means that R7C2 must also be 8, which is impossible.

So R9C5 cannot be 4. And, by strong links, several other cells cannot be 4, and the puzzle falls apart.

I have no idea how to notate this. But, if you clear your mind, it's really quite obvious.

[Johan]

Asellus,

Thanks for the explanation,I think I'd just made a lucky guess on <5> to solve this one.

[Mogulmeister]

A forcing chain is one way.......I was looking for some other things......

[nataraj]

Actually, three ALS eliminations do the trick.

1) r4c3 (6,8 ) and r6c2,7,8,9 (3,5,6,7,8 ) have restricted common 8 in box 4. The other common candidate 6 can be removed from r6c1 and r6c3.

2) r9c4 (1,9) and r2,3,5,7,8 c6 (1,3,4,5,7,9) restricted common 1, remove 9 from r1c4 (sees both '9's)

3) r1c4 (7,5) and r1c8,r1c9,r2c7 (3,4,5,7), 5 is restricted common, remove 7 from r1c5

How one is supposed to find those? No idea (Andrew Stuart's solver suggested the three steps mentioned above)

On the other hand, it didn't take too long to find the solution using a coloring scheme basically the same way Asellus described it:
- start with '4's because there are so many strong links.
Set A(g): r1c5,r2c7,r3c3,r4c8,r8c6,r9c2
Set B(r): r1c8,r4c7,r8c4,r9c5
- color some more candidates, arrive at the contradiction in red.
Accept the green candidates. Done.

It seems to me, that with those very tough puzzles, "looking for some other things" can be an (impossible?) challenge...

[nataraj]

In fact, the "elegant solution" is not done yet. After doing the ALS (see previous post) there are a few simple steps and then - this position:

[Mogulmeister]

nat,

You don't need the forcing chain. !

Regarding the ALS: