dailysudoku.com

[Marty R.]

I had a tough time with this one, using two Medusa wraps sandwiched between two Hidden URs. There must be a better way.

[Asellus]

This one was tough. I assume Marty means "extended" Medusa, which I am avoiding. My path was not short, but contains some interesting loops:

[1] 15UR r23c58: r2c5<>5
[2] Loop: (5)r1c4=(5)r9c4 - ALS[(5)r9c235=(8)r9c2] - (8)r1c2=(8-5)r1c4; r9c1<>5, r2c2<>8, r1c4<>69
[3] Loop: (9-5)r9c4=(5-8)r1c4=(8)r1c2 - (8)r9c2=(8-7)r9c1=(7)r8c1 - (7=9)r8c8 - (9)r8c4=(9-5)r9c4; r9c4<>13, r9c1<>34
[4] Loop: (1-9)r8c4=(9-5)r9c4=(5)r1c4 - (5=1)r3c5 - (1)r89c5=(1-9)r8c4; r8c4<>36
[5] (3)r4c4=(3)r4c1 - (3=7)r8c1 - (7=9)r8c8 - (9=1)r8c4 - (1=3)r4c4; r4c4=3
(There is a 16UR at this point, but I ignored it.)
[6] Skyscraper c34: r1c2<>5
[7] (6)r2c5 - (6)r2c8=(6-5)r1c8=(5-8)r1c4=(8-6)r2c5; r2c5<>6

Medusa multi-coloring combined with examining ALS inferences aided in uncovering the AICs.

Pretty loopy solution, no?

[Marty R.]

[daj95376]

Asellus: An impressive solution!!! Someday, I'm going to learn how to use/understand an ALS in a chain.

[Asellus]

[daj95376]

[Asellus]

I awoke this morning realizing something: in a continuous loop, because the strong inference within an ALS becomes conjugate (i.e., one of those digits must be false), all other digits (those not involved in the loop inference) must be true within the ALS.

This means that in my Step [2] above, we can add some more eliminations as a result of the loop:

[2] Loop: (5)r1c4=(5)r9c4 - ALS[(5)r9c235=(8)r9c2] - (8)r1c2=(8-5)r1c4; r9c1<>5, r2c2<>8, r1c4<>69, r9c14<>3, r9c46<>1

Since the <1> and <3> in the r9c235 ALS must both be true, <1> and <3> cannot exist anywhere else in r9.

This change eliminates the Step [3] loop and the solution is shorter by one step.

(Why I suddenly thought of this, I have no idea. Sudoku wasn't even on my mind... that I was aware of.)

[keith]

[storm_norm]

[Asellus]

Norm,

Interesting that your first loop is the same as mine... or at least I think it is, since you don't show the complete loop. The only way that I can see to connect the "=(8)r9c1" to the "(5)r9c4=" is via the 3-cell ALS in r9 that I used. It's curious that you don't show how you did it. (I hope you don't think that because the <8> and <5> at the ends of the AIC you show are both in r9, the loop is complete! That's not enough if the digits are dissimilar. You still need to find an AIC connection between them.)

And, assuming you've read my followup post, you will see that, as I discovered, the second loop (which also matches mine in effect, if not in the particulars) isn't necessary because the first loop actually performs the same eliminations (and more). I'm sort of excited about that "ALS in a continuous loop" insight because it is powerful (a sort of "leveraging"!)... even if it is something that probably doesn't come up all that often.

I didn't retrace the rest of your (shorter) steps. But, because I took more steps than you, I'm too tired to do so.

Keith,

Thanks for the diversion! I've often had solutions to difficult or obscure problems pop into my head out of the blue, usually either after waking up in the morning or else while in the shower any time of day. It's like the synapses keep trying to fine new pathways long after you've stopped concentrating on the problem.

[daj95376]

I find it interesting that all of the chains/loops seem to rely on extending the 2-String Kite on <8> in r1c1.

[storm_norm]