dailysudoku.com

[Marty R.]

This is my classic nightmare of a puzzle: few bivalue cells, loads of four- and five-candidate cells and I can't eliminate a thing.

[nataraj]

Neither coloring nor Medusa seem to help ...

Where'd you find this baby?


Edit: I let Andrew Stuart's Sudoku Solver take a shot at it.

"Death Blossom" and - get this - "Bowman Bingo"
I'd say just a wee bit above my head

[Marty R.]

[storm_norm]

S.E. ===> 8.9 !!!

[Marty R.]

[keith]

This is the rating from Sudoku Explainer.

Keith

[Marty R.]

[storm_norm]

[ravel]

You can find the SE rating scale here. So VH's would get a rating of 4.2 or 4.4 (like the one by gsf i just posted), but these puzzles also might require a hidden triple (naked something) or a swordfish.

Only a few puzzles with rating above say 7.5 can be solved in an elegant way (for my taste), most of them are pure chaining work.

[Marty R.]

Thanks for that information.

[Asellus]

Well... after staring at this thing for quite a while, my attention was drawn to some ALSs and grouped links that looked as if they might have potential. Some complex multi-branched AICs later, I managed to eliminated some <1>s. This comes mighty close to forcing... so close many won't see any difference. But, I was led there by those patterns and not just a wild stab in the dark. Still... elegant, it is not!

Because of all the branching, I will break the AICs up into separate fragments and label them with letters in brackets.

[A] (1=2)r3c1
[B] from A: (2)r3c1-(2)r3c7=(2)r1c9-(2)r6c9=(2)r5c7
[C] from A: (2)r3c1-(2=3)r2c1-(3)r2c5=(3)r1c4
[D] from C: (3)r1c4-(3=2)r2c5
[E] from B & D: (2)r2c5|r5c7-(2)r5c456|r9c4=(2)r6c4
[F] from D & E: (2)r2c5|r6c4-(2)r8c45|r9c4=(2)r9c6
[G] from A & F: (2)r3c1|r9c6-(2)r89c1|r9c2=(2-7)r8c2=(7)r7c2-(7)r7c46=(7)r8c4
[H] from C, G & E: (2)r6c4|(3)r1c4|(7)r8c4-({237}=4)r5c4-(4)r5c2=(4)r6c2
[I] from E & H: (2)r6c4|(4)r6c2-({24}={358})r6c589-({38}=9)r6c3-(9=1)r8c3

Thus, we have the following strong inference: (1)r3c1=(1)r8c3
and
r1c3|r89c1<>1

It doesn't appear to be of all that much help in moving forward. But, at least it's something.

[keith]

[nataraj]

[ravel]

Its easy to see it for r4c1=1 and r4c1=9:
r4c1=1 => r3c1=2 => r2c1=3 => r2c5<>3 => r1c4=3 => r1c4<>9
r4c1=9 => r4c46<>9 => r6c4=9 => r1c4<>9

But i had some troubles to see, why r4c1=3 should imply r1c4<>9. My first attempt ended in an empty cell
r4c1=3 => [r4c7=7 and (r8c1=9 => r8c8<>9 => r7c7=9)] => r2c7=8 => r2c6=9

And the elimination is not very useful ...

[nataraj]

Thanks, ravel! At least now I understand what susser is trying to say.

I can only guess at the method susser uses to arrive at these conclusions.

In comparison, "Bowman Bingo" at least has a cute name.

[Asellus]

It didn't surprise me that those <1> eliminations are related to <2>s: the AICs I listed are filled with <2>s, including all those of r6. (When pointed out after the fact, I can see how the 3 <2>s of r6 all lead to the <1> eliminations. But, I can't see any way to find such things other than explicit trial and error.)

After much more staring at this thing, I have found a rather simpler branched AIC that is useful:

[A]: ALS[(8)r8c9=(1)r78c9]
[B] from A: (1)r78c9-(1=2)r1c9-(2)r1c2=(2)r23c1
[C] from A & B: (1)r78c9|(2)r23c1-ALS[({12}={348})r9c178]-(8)r9c46=(8)r8c4; r8c8<>8

The notation may look formidable, but the AIC is actually not so hard to follow. Here's the current grid so you can follow along:

[keith]

[Asellus]

It is also possible to eliminate a couple of <5>s by exploiting that same r9 locked set:

[A]: (5=1)r7c9
[B] from A: (1)r7c9-(1=2)r1c9-(2)r1c2=(2)r23c1
[C] from A & B: (1)r7c9|(2)r23c1-({12}={348})r9c178-(8)r9c46=(8)r8c4
[D] from A & C: (1)r7c9|(8)r8c4-({18}=5)r8c9; r6c9|r8c8<>5

It may or may not be interesting that these <5>s and the <8> don't turn up in the Bingo.
[Edit to remove inappropriate ALS notation.]

[Asellus]

Hmmm... I keep being drawn back to this thing!