Hi Mike

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`1268c9 12i36â9 1239 |134a7 5 34A79d |3678 1m36Ä8 13678 `

156Z 4 13q5Y |2 1K7k 8 |35c67 9 1O367

5c8C 1b3r9d 7 |1B3b 6 3d9D |4 5C8c 2

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124e9 8 1234E9 |135f6 1í9Í 2l35F6 |369 7 1369

7 13s9g 6 |1348 14j89G 34Î |2 1M38ê 5

129 5 1239 |13678 17K89 2L367 |3689 4 13689

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3 6H9h 8 |4567 2 4567 |1 5W6w 4U67ç9H

2456Á 7 245 |9 4J8j 1 |3568 2n3568 3468

124569 12I69 12459 |45678J 3 4567 |567è89ë 2N568 46789

Definitions to start. For me an AC is m cells containing m+x digits with m-1 digit known.

if m=1, we have a cell. The smallest AC is a cell.

if m=2, any number of digits in candidates from three to nine and one bivalue located in the set.

if m=3....

eg: r58c5 is a two cells AC. Bivalue 4j=4J; other digits 189.

For the solver, this is equivalent to weak links 1r5c5 - 8r58c5 1r5c5 - 9r5c5 8r58c5 - 9r5c5

It is exactly what is done in a cell.

eg2: r15c8 is a two cells AC. im=iM bivalue;

other digits 368=> 3r15c8 - 6r1c8 3r15c8 - 8r15c8 6r1c8 - 8r15c8

These weak links and the Choices AC:r58c5(1;8;9) AC:r15c8(3;6;8) are the only stuff extracted from ALS/ACs and entering in the main process.

In AIC chains it will always be one of these weak links. A kraken Blossom can be based on a Choice coming out of an AC.

Now the chains

If the first set is a Kraken Blossom, all others are as well.

BTW, any T chain with a single node as an equivalent Kraken Blossom form as I explained in my last post in "full tagging method".

First kraken Blossom : ok. Your shortcuts are easy to catch. I am not sure I'll do it.

I explained above 8r58c5-9r5c5 and 8r15c8 - 3r15c8, so the second kraken blossom based on cell r8c7 shoul be easy to understand. I translate it

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`[]3r8c7 - 3r8c9 `

[]6r8c7 - 6r789c8 = AC: r15c8(6r1c8 - 3r15c8) = 3r8c8 -3r8c9

[]5r8c7 - 5r7c8 = 6r7c8 - 6r7c2 = 9r7c2 - 9r5c2 = AC:r58c5(9r5c5 - 8r58c5) = 8r6c5 - 8r5c45 = 8r5c8 - 8r389c8 = AC: r15c8(8r15c8 - 3r15c8) = 3r8c8 - 3r8c9

[]8r8c7 - 8r8c5 = 4r8c5 - 4r7c46 = 4r7c9 - 9r7c9 - 9r7c2/..... same as above

you wrote

I don't see how your second elimination is valid as shown, specifically 9r5c5/8r58c5|*r58c5.

I hope this is now clear. Again, this is the only effect of ALS/AC stuff searching for AICs in my levels 2 and 3.

you wrote

If it is it would be pretty amazing because later on you establish 8r5c45_8r5c8.

8r5c45 = 8r5c8 is a group bivalue. Nothing to establish. But may-be your point is different.

There is a (relative) weakness in my process. Here for example, the solver has in memory AC: r15c8(8r15c8 - 3r15c8). He has not 8r5c8 - 3r15c8.

Consequently, he writes

= 8r5c8 - 8r389c8 = AC: r15c8(8r15c8 - 3r15c8) =

which could has been as well AC: r15c8(8r5c8 - 3r15c8)

Often, when I prepare a post, I add this shortcuts, but I do it mainly at start and ending point of the chain.

I will likely improve later the print. It does not affect the logic, so there is for me no urgency to do it.

I could have done it here

[]8r9c7/8r8c789r9c8_8r9c79/4r79c9|*r7c9r9c79_4r8c9/4r8c13

your idea

[]8r9c7 - AHS:(79r9c7 = 79r79c9) - 4r79c9 = 4r8c9 - 4r8c13

my own way to shortcut

[]AC:r7c9r9c79(8r9c7 - 4r79c9) = 4r8c9 - 4r8c13

You can find several examples of such shortcuts in the thread "full tagging examples"