## Challenge: New set of 11 'Unsolvables'

Advanced methods and approaches for solving Sudoku puzzles
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`134x7k8 2      6      |34z5a8 1348       145A8  |9    137       1378      137K8   9      5      |2m38   6          1É8é   |4    1237      12378    1348    1e38   34l8   |9      12M34x8    7      |5    1236b     1236B8    ------------------------------------------------------------------------ 9       3678   2c-378 |1      2C4d78     4D6f8  |3G6g 3Ê7ê      5        1e36f   4      3ë7Ë   |5A7a   7Ì9ì       5a6F9n |2    8         1E3s7Â9h 5       1E6g78 2C78   |2M78   2789h      3      |1g6G 179H      4        ------------------------------------------------------------------------ 348     5      34L7i8 |6      1p347I8ä9È 2      |1G3g 134o9h    139      2J3468  368    9      |348    5          1P-48  |7    12R34O6B  1236      2j34y6  367I   1      |347á   3479       9N-4n  |8    5         2J36À9È`

Here is my translation of your chains with "/" mapping to "-"; "_" to "="; and "|*" to either an AHS, AAHS, or an ALS:

2r4c5C - 2r6c4 = 2r2c4 - AHS:(2r2c8 = 26r38c8) - 4r8c8 = 4r7c8 - AAHS:(4r7c13 = 24r89c1) - 6r89c1F = 6r4c6f - 4r4c6 = 4r4c5 - 2r4c5C => r4c3c=2

9r5c6n - 9r5c5 = 7r5c5 - 7r7c5 = 7r7c3 - ALS:(2378r456c3) - AHS:(8r4c2 = 248r4c356) - 6r4c6 = 6r5c6 - 9r5c6n => r9c6N=9

1r7c5p - ALS:(1349r7c789) - AAHS:(4r7c13 = 4r89c1) - 6r89c1 = 6r5c1 - 1r5c1 = 1r5c9 - 1r6c7 = 1r7c7 - 1r7c5p => r8c6P=1

Nice examples of AHS and AAHS which my solver doesn't have implemented. The second chain with a common cell between the ALS and AHS is especially interesting.

I just saw your edits. Its late and I'll look at your other chains later.
Mike Barker

Posts: 458
Joined: 22 January 2006

Hi Mike,

1) Typo : I'll work on my US version to meat your rules. I find nevertheless the "French" version easier to read for a poor old man with bad eyes. '-' and '=' are very close on my screen.

2) In my chains, you have no ALS and no AAHS. All are AHS;AC
|*r38c8 (6+?) |*r89c1 (2+?) |*r4c56 (4+?) |*r37c3 (4+?)
|*r4c356 (2;4+?) |*r7c135 (7;8+?) |*r589c1 (2;6+?)

For sure any AHS has an ALS as counterpart, but in the logic I am applying, I stick to AHS providing directly weak links.
ALS are providing 'OR' condition. Mix of 'OR' with strong and weak link is possible, but somehow misleading.

3) U:31 like that one can be solved in a kind of "One shot" set of AICs.

4) I will make a trial to switch from your interpretation to my own translation of the chains.

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`134x7k8 2      6      |34z5a8 1348       145A8  |9    137       1378      137K8   9      5      |2m38   6          1É8é   |4    1237      12378    1348    1e38   34l8   |9      12M34x8    7      |5    1236b     1236B8    ------------------------------------------------------------------------ 9       3678   2c-378 |1      2C4d78     4D6f8  |3G6g 3Ê7ê      5        1e36f   4      3ë7Ë   |5A7a   7Ì9ì       5a6F9n |2    8         1E3s7Â9h 5       1E6g78 2C78   |2M78   2789h      3      |1g6G 179H      4        ------------------------------------------------------------------------ 348     5      34L7i8 |6      1p347I8ä9È 2      |1G3g 134o9h    139      2J3468  368    9      |348    5          1P-48  |7    12R34O6B  1236      2j34y6  367I   1      |347á   3479       9N-4n  |8    5         2J36À9È `

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`2r4c5C - 2r6c4 = 2r2c4 - AHS:(2r2c8 = 26r38c8) - 4r8c8 = 4r7c8 - AAHS:(4r7c13 = 24r89c1) - 6r89c1F = 6r4c6f - 4r4c6 = 4r4c5 - 2r4c5C => r4c3c=2 []2r4c5-C/2r6c4_2r2c4/2r2c8_2r38c8/4r8c8|*r38c8_4r7c8/4r7c13_4r89c1/6r89c1|*r89c1_6r4c6/2r4c5-C|*r4c56 r4c3=2 `

I could write in my logic
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`2r4c5.C - 2r6c4 = 2r2c4 - 2r2c8 = AHS:(2r38c8-4r8c8) = 4r7c8 - 4r7c13 = AHS:(4r89c1 - 6r89c1) = 6r4c6.f - 4r4c6 = 4r4c5 - 2r4c5.C`

1) No problem to change the end of the chain. The solver looked for the shortest chain.
AHS(6r4c6f - 2r4c5C) is a shortcut for 6r4c6f - 4r4c6 = 4r4c5 - 2r4c5C
2) Using ALS, I could also write the chain in that way (is it accepted?)
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`2r4c5C - 2r6c4 = 2r2c4 - ALS:(2r2c8 | 4r7c8) - 4r7c13 = AHS:(4r89c1 - 6r89c1) = 6r4c6f - 4r4c6 = 4r4c5 - 2r4c5C`
The second AHS can not be switched to ALS because we are using in fact two different counterparts of the same AHS:
4r7c13 belongs to ALS: r7c13r8c2r9c2 6r4c6f belongs to ALS:r1235C1 (in fact 6r5c1==6r4c6)

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`9r5c6n - 9r5c5 = 7r5c5 - 7r7c5 = 7r7c3 - ALS:(2378r456c3) - AHS:(8r4c2 = 248r4c356) - 6r4c6 = 6r5c6 - 9r5c6n => r9c6N=9 []9r5c6-n/9r5c5_7r5c5/7r7c5_7r7c3/8r37c3|*r37c3_8r46c3/8r4c2_8r4c356/6r4c6|*r4c356_6r5c6/9r5c6-n r9c6=9 `

again in my logic (no shortcuts seen)

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`9r5c6.n - 9r5c5 = 7r5c5 - 7r7c5 = AHS:(7r7c3 -  8r37c3) = 8r46c3 - 8r4c2 = AHS:(8r4c356 - 6r4c6) = 6r5c6 - 9r5c6.n `

Could you comment your sequence ALS AHS, I not sure I got it.

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`1r7c5p - ALS:(1349r7c789) - AAHS:(4r7c13 = 4r89c1) - 6r89c1 = 6r5c1 - 1r5c1 = 1r5c9 - 1r6c7 = 1r7c7 - 1r7c5p => r8c6P=1 []1r7c5-p/4r7c135|*r7c135_4r7c8/4r7c13_4r89c1/1r5c1|*r589c1_1r5c9/1r6c7_1r7c7/1r7c5-p r8c6=1 `

my logic (no shortcut)

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`AHS:(1r7c5.p - 4r7c135) = 4r7c8 - 4r7c13  = AHS:(4r89c1 - 1r5c1) = 1r5c9 ....`
champagne
2017 Supporter

Posts: 6610
Joined: 02 August 2007
Location: France Brittany

Thanks much for translating the chains. The "-" and "=" are pretty standard here and elsewhere. The key is understanding your choices (AHS or AC) for which there is not a "standard" nomenclature so I pretty much winged it. To me an AHS ("m" candidates in "m+1" cells) in the first chain is r238c8=26 which is why I showed it the way I did. Thus if r2c4=2 then r38c8=26 which is a locked set.

I'm not sure I understand your ALS which is defined as "n+1" candidates in "n" cells. r237c8=123467 is not an ALS. On the other hand r7c789=1349 is. In this latter case if r7c5=1 then r7c789=349 which is a locked set. Nothing which sees these cells can contain 349. Specifically r7c13<>4. I see r7c13|r89c1=24 (there was a typo in my original post) as an AAHS ("m" candidates in "m+2" cells). I couldn't resolve this into an AHS.

I knew I had modified the start of your third chain somewhat. I saw the ALS r7c789=1349 before the AHS r7c135=78 which are complements. Understanding that you don't use ALS will help in the future. Interestingly I see ALS and AHS as both providing strong inferences (in an ALS if one candidate is false/not included in the set then the rest are true/form a locked set; in an AHS, if one cell is false/doesn't contain a hidden candidate then the rest are true/form a locked set).

You are correct that your ending for the first chain is shorter. I guess I'm not use to thinking in terms of one candidate AHS which are, of course, perfectly valid.

Your solution is great stuff - exactly what I believe Andrew is looking for with the "Unsolvables".
Mike Barker

Posts: 458
Joined: 22 January 2006

Hi Mike,

Reading your post, I got in trouble. we certainly have some definition problems.

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`134x7k8 2      6      |34z5a8 1348       145A8  |9    137       1378      137K8   9      5      |2m38   6          1É8é   |4    1237      12378    1348    1e38   34l8   |9      12M34x8    7      |5    1236b     1236B8    ------------------------------------------------------------------------ 9       3678   2c-378 |1      2C4d78     4D6f8  |3G6g 3Ê7ê      5        1e36f   4      3ë7Ë   |5A7a   7Ì9ì       5a6F9n |2    8         1E3s7Â9h 5       1E6g78 2C78   |2M78   2789h      3      |1g6G 179H      4        ------------------------------------------------------------------------ 348     5      34L7i8 |6      1p347I8ä9È 2      |1G3g 134o9h    139      2J3468  368    9      |348    5          1P-48  |7    12R34O6B  1236      2j34y6  367I   1      |347á   3479       9N-4n  |8    5         2J36À9È `

Let me first take my point 2.

2) In my chains, all are AHS;AC
|*r38c8 (6+?) |*r89c1 (2+?) |*r4c56 (4+?) |*r37c3 (4+?)
|*r4c356 (2;4+?) |*r7c135 (7;8+?) |*r589c1 (2;6+?)

The common point to all these sets is that one digit is unknown.

r38c8 has digits 12346. 6 is compulsory, one among the four others is valid.
r4c356 has digits 234678. 2 and 4 are compulsory, one among 3678 is valid.

These are set I called Almost cells (AC) in full tagging presentation.
1,2,3,4 in the first case 3,6,7,8 in the second cases are Choices.
Each couple in these sets is a weak link.

I understood this was the same as AHS but this does not match with your sentence:

To me an AHS ("m" candidates in "m+1" cells)

As long as we will not have clarified that point, I'll use AC and not AHS.

Complementary set to an AC in any object containing the AC is an ALS. ("m+1" candidates in "m" cells).
Unknown digits are the same in both entities,
The digit in excess in the ALS is the digit valid in the AC.

you wrote

I'm not sure I understand your ALS which is defined as "n+1" candidates in "n" cells. r237c8=123467 is not an ALS.

The three forms for the same AIC

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`2r4c5-C/2r6c4_2r2c4/2r2c8_2r38c8/4r8c8|*r38c8_4r7c8/4r7c13_4r89c1/6r89c1|*r89c1_6r4c6/2r4c5-C|*r4c56 2r4c5.C - 2r6c4 = 2r2c4 - 2r2c8 = AHS:(2r38c8-4r8c8) = 4r7c8 - 4r7c13 = AHS:(4r89c1 - 6r89c1) = 6r4c6.f - 4r4c6 = 4r4c5 - 2r4c5.C2r4c5C - 2r6c4 = 2r2c4 - ALS:(2r2c8 | 4r7c8) - 4r7c13 = AHS:(4r89c1 - 6r89c1) = 6r4c6f - 4r4c6 = 4r4c5 - 2r4c5C`

The first one indicates clearly that the AC is r38c8. This can be guessed in the second one, with a risk to be wrong.
ALS complementary to r38c8 is r12467c8. 2r2c8 and 4r7c8 belong to that ALS.

I could suggest as final writing

2r4c5.C - 2r6c4 = 2r2c4 - 2r2c8 = AC r38c8:(2r38c8-4r8c8) = 4r7c8 - 4r7c13 = AC r89c1:(4r89c1 - 6r89c1) = 6r4c6.f - 4r4c6 = 4r4c5 - 2r4c5.C
or
2r4c5C - 2r6c4 = 2r2c4 - ALS r12467c8:(2r2c8 | 4r7c8) - 4r7c13 = AC r89c1:(4r89c1 - 6r89c1) = 6r4c6f - 4r4c6 = 4r4c5 - 2r4c5C

I'll prepare in parallel a post on U31.
champagne
2017 Supporter

Posts: 6610
Joined: 02 August 2007
Location: France Brittany

U31 seen by my solver. (sorry, still with my writing rules)
again, a selection of the best.
....5.....4.2.8.9...7.6.4.2.8.....7.7.6...2.5.5.....4.3.8.2.1...7.9.1.......3.... 31

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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       -------------------------------------------------------------------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   -------------------------------------------------------------------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  `

#1r2c5=>r2c5=7
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`[]1r2c3/1r2c5 []3r2c3/3r13c2_3r5c2/9r5c2_9r5c5/9r4c5_1r4c5/1r2c5  []5r2c3==5r89c1/5r3c1_5r3c8/5r7c8_6r7c8/6r7c2_9r7c2/9r5c2_9r5c5/9r4c5_1r4c5/1r2c5 `

#3r8c9
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`[]3r8c7/3r8c9 []6r8c7/6r789c8_6r1c8/3r15c8|*r15c8_3r8c8/3r8c9    []5r8c7/5r7c8_6r7c8/6r7c2_9r7c2/9r5c2_9r5c5/8r58c5|*r58c5_8r6c5/8r5c45_8r5c8/8r389c8_8r15c8/3r15c8|*r15c8_3r8c8/3r8c9       []8r8c7/8r8c5_4r8c5/4r7c46_4r7c9/9r7c9_9r7c2/.....`

#3r8c7
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`[]3r5c8/3r5c6_4r5c6/4r5c5_4r8c5/....._3r8c8/3r8c7 []3r8c8/3r8c7    []3r1c8/3r2c79_3r2c3/3r13c2_3r5c2/9r5c2_....._3r8c8/3r8c7`

r8c8=3

#2r8c3
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`[]5r8c3/2r8c3 []5r8c78/5r7c8_6r7c8/6r8c789_6r8c1/6r12c1_6r1c2/2r1c2_2r9c2/2r8c3 []5r8c1/5r3c1_5r3c8/5r7c8_.....`

#4r8c13
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`[]8r9c4==4r8c5-J/4r8c13 []8r9c8/8r3c8_5r3c8/5r7c8_6r7c8/6r7c2_9r7c2/9r5c2_9r5c5/4r5c5_4r8c5/4r8c13    []8r9c7/8r8c789r9c8_8r9c79/4r79c9|*r7c9r9c79_4r8c9/4r8c13  []8r9c9/8r8c789r9c8_....`

r8c3=5

still available in the same form at that point

#8r15c8
#1r2c9
#1r4c3
#1r4c9
#6r78c9r9c79
#6r89c8
#9r9c3

using all the stuff, the game is over. I did'nt check what happens after the three fix.
champagne
2017 Supporter

Posts: 6610
Joined: 02 August 2007
Location: France Brittany

A couple translations from your U#31 solution. The first is a Kraken Blossom (in this case a trivalue cell). The translation is straight forward except that I used AIC short hand for the bivalue cells and strong links sharing a cell. I would establish the complementary tag, 5r2c3==5r89c1 via 5r2c3 - 5r89c3 = 5r89c1 and 5r2c3 = 5r89c3 - 5r89c1, but I assume you used your complementary tagging rules.

Kraken Blossom (r2c3=135) => r2c5<>1
[]1r2c3 - 1r2c5
[]3r2c3 - 3r13c2 = 3r5c2 - 9r5c2 = 9r5c5 - (9=1)r4c5 - 1r2c5
[]5r2c3==5r89c1 - 5r3c1 = 5r3c8 - (5=6)r7c8 - (6=9)r7c2 - 9r5c2 = 9r5c5 - (9=1)r4c5 - 1r2c5

I don't see how your second elimination is valid as shown, specifically 9r5c5/8r58c5|*r58c5. If it is it would be pretty amazing because later on you establish 8r5c45_8r5c8.
[]5r8c7/5r7c8_6r7c8/6r7c2_9r7c2/9r5c2_9r5c5/8r58c5|*r58c5_8r6c5/8r5c45_8r5c8/8r389c8_8r15c8/3r15c8|*r15c8_3r8c8/3r8c9

I think I may see the difference in our terminology and what I refer to as an AHS is related to, but different from, your AC. Again I see an AHS as "m" digits which only occur in "m+1" cells resulting in a strong inference. Your AC seems to refer to a similar set, but based on the weak inference - different perspectives on the same thing.

Kraken Unit (r9c4789=8) => r8c13<>4
[]8r9c4==4r8c5J - 4r8c13
[]8r9c8 - (8=5)r3c8 - (5=6)r7c8 - (6=9)r7c2 - 9r5c2 = (9-4)r5c5 = 4r8c5 - 4r8c13
[]8r9c7 - 8r8c789|r9c8 = AC:(8r9c79 - 4r79c9) = 4r8c9 - 4r8c13
[]8r9c9 - 8r8c789|r9c8 = ....

I would tend to see the third chain as:
[]8r9c7 - AHS:(79r9c7 = 79r79c9) - 4r79c9 = 4r8c9 - 4r8c13
Mike Barker

Posts: 458
Joined: 22 January 2006

[Withdrawn]
Last edited by daj95376 on Tue Jan 29, 2008 9:32 pm, edited 1 time in total.
daj95376
2014 Supporter

Posts: 2624
Joined: 15 May 2006

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        ------------------------------------------------------------------- 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    ------------------------------------------------------------------- 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

[]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"
champagne
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Thanks for your translations and don't feel I'm suggesting you change for these posts. I'm just used to seeing things a different way (like a bivalue only showing up once to cover both candidates) so its easier for me to see it this way. I see what you are doing now. Because I focus on the strong inferences, I see the chain as:

[]5r8c7 - 5r7c8 = 6r7c8 - 6r7c2 = 9r7c2 - 9r5c2 = 9r5c5 - AHS:(4r5c5 = 4r8c5) - 8r58c5 = 8r6c5 ...

but the extra link isn't necessary, so I could get used to:
[]5r8c7 - 5r7c8 = 6r7c8 - 6r7c2 = 9r7c2 - 9r5c2 = AC:r58c5(9r5c5 - 8r58c5) = 8r6c5 ...
Mike Barker

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Joined: 22 January 2006

Mike wrote

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`[]5r8c7 - 5r7c8 = 6r7c8 - 6r7c2 = 9r7c2 - 9r5c2 = 9r5c5 - AHS:(4r5c5 = 4r8c5) - 8r58c5 = 8r6c5 ... but the extra link isn't necessary, so I could get used to: []5r8c7 - 5r7c8 = 6r7c8 - 6r7c2 = 9r7c2 - 9r5c2 = AC:r58c5(9r5c5 - 8r58c5) = 8r6c5 ...`

The second form stick to the solver logic, so I'll use it. Next prints will be done using the rules we have applied here. (subject to
full debugging).

Similar adjustments will have to take place at level 4 of full tagging process. To avoid any trouble, I'll call AC2 the sets I introduced at that level:

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`m cells having m+2+x digits with m-2 digits known. (for the time being, m cells having m+2 digits with m-2 known, equal to four unknown digits among them two are valid).`

ps: if of any interest, I can give the start proposed by the solver for other puzzles of the lot, but they are not "one shot puzzles".
Last edited by champagne on Tue Jan 29, 2008 6:44 pm, edited 1 time in total.
champagne
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Its up to you if you'd like to post more. I would say definitely if the solution shows some different aspect. For example, you really haven't shown complementary tagging here or an AC2 example (might be worthwhile even if it isn't strictly necessary). Otherwise having other examples of AC may be helpful if someone is trying to understand the technique. I know I've learned a lot from the two you have posted.

At this point, I'm of the opinion that AC are a different way of looking at AHS (at this point I couldn't say better or worse - though for you better since that's what you've programmed ). Do you agree? Does that make an AC2 a different way of looking at AAHS (m-2 candidates only appearing in m cells)?

A few more questions: Under what circumstances would the cells identified inside of the parens of "AC:rXcY(...)" be different than rXcY? When you select a subset of the eliminations, what criteria do you use? TIA
Mike Barker

Posts: 458
Joined: 22 January 2006

Hi Mike,

A quick answer on easy points and I’ll search to-morrow appropriate examples of other aspects of the process.

The file of “unsolvable” has a limited number of puzzles. No chance to use AC2 for example. None of these puzzles request level 4 of my process. I’ll select in that file those corresponding to some of your expectations.

You wrote

At this point, I'm of the opinion that AC are a different way of looking at AHS (at this point I couldn't say better or worse - though for you better since that's what you've programmed ). Do you agree?

I am not familiar with what has been done elsewhere, so you can better answer that question.
What is for sure is that my use of ALS is quite different from what you have shown here. I have studied and included in my program about two years ago analysis of couples of ALS. When I implemented ALS/AC analysis, this became redundant. I discarded the code.

I understood you are comparing several methods. This is a huge task. I preferred to limit the code to essential and to look for cracking of hardest puzzles.

You wrote
Does that make an AC2 a different way of looking at AAHS (m-2 candidates only appearing in m cells)?

I only know what has done Steve Kurzhals on Easter Monster. As you could see, for me, AC2 are looked as another dimension of candidates, the couple tags, or to stay out of tagging, couples of digits. In the small part of the AC2 I am considering (2 valid out of four digits), each AC2 generates six “super candidates couple digits”. Nobody up to now indicated me references of similar approaches.

You wrote
A few more questions: Under what circumstances would the cells identified inside of the parens of "AC:rXcY(...)" be different than rXcY? When you select a subset of the eliminations, what criteria do you use? TIA

I hope I got your point.

The rule I apply is a very general and trivial one.

If you have a weak link between groups eg 5r123c4 - 6r5c7r6c8r7c9. Any subset of the left side is a weak link to any subset of the right side eg: 5r1c4 – 6r6c8r7c9.

Reversely, if 2r1c1 – 6r7c5 and 2r2c1 – 6r7c5 then 2r12c1 – 6r7c5

Handling in optimized way these properties is not so easy. As the solver do it thru additional pieces in AICs, (as you could see), it was not a true problem. At level 4, with AC2, you can simplify the path with appropriate handling of “vicinity problems”, so I am working on it.

Tarek 191 and may be the new version of Easter Monster I am working on can give us good examples of what you would see at level 4. It should be much shorter than the set of examples I already gave.
champagne
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In the list of unsolvables, other puzzles where the first tagging leads to new fix.
All of them thru AICs.

6...8...5.4...128..8.....6...7..23.....5.8.....17......6.....4...43...2.3...9...6 24

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`6       1a2Î7ä9 2Ê9ê   |2S49    8       479     |1A4W79 3     5      579     4       3b59   |6c9C    3B5Z6C7 1       |2      8     7Ë9ë   1A2579  8       23B59  |249     2i34x57 3j45ã79 |1479   6     1R4Y79 --------------------------------------------------------------------458d9   5Á9á    7      |1q46C9n 146c    2       |3      1ç5f9 148D   249     23e9    6      |5       1P3E4   8       |14Y7   17l9ç 12m47  245f8D9 23E9    1      |7       3Ì4ì    3J49N   |6      5F9f  2M48d  --------------------------------------------------------------------12579   6       258g9è |128G    12I57   5Í7í    |15k79o 4     3      1579    1579    4      |3       15à7    6       |8      2     179O   3       1257    258G   |12t4h8g 9       4H57    |15K7   1l7L  6   `

[] AC: r2c35(6r2c5.C - 5r2c35) = 5r2c1 - 5r23c3 = 5r79c3 - 5r8c12 = 5r8c5 - 5r23c5 = 5r3c6 - 3r3c6 = AC: r2378c5(3r23c5 - 6r2c5.C)
[]5r6c1.f - 5r4c2 = 9r4c2 - 9r4c4 = 9r6c6 - 9r6c8.f

7 r2c4=6 B
8 r4c5=6 B
9 r6c8=5 B

..5...8...7.3...1.8..1.5..96......74...2.6...29......83....4..1.1...9.4...6...3.. 28

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`1      23x4À6l 5      |4m6Ê79a 24679A   27     |8     236    23r67e 4D9d   7       249D   |3       246È8c   2c8C   |4p5b6 1      25B6   8      2346L   23Z4   |1       2467e    5      |4P67E 23x6   9      -------------------------------------------------------------------6      35Ã8    1s38   |589A    1589a    13n8   |2     7      4      45Å7Í  3458    13478  |2       14á57ì8Î 6      |1o59f 359F   3R5r   2      9       134á7ì |4M57    1457     13N7   |1O56g 3y56G  8      -------------------------------------------------------------------3      258     2789d  |5678    2w56Ê78  4      |5679F 2568q9 1      57     1       2h78i  |56é78I  3        9      |567   4      2H56ë7 4k579D 24K58   6      |578     1j2578   1J2u78 |3     258Q9d 257   `

[]2r2c56 - 2r1c6 = 7r1c6 - 7r3c5 = 7r3c7 - 4r3c7 = AC: r2c79(4r2c7 - 6r2c79) = AC: r2c56(6r2c5 - 2r2c56)
[]5r5c9.r - 5r2c9.B = AC: r1c9r23c7(5r2c7.b - 3r1c9.r)

4 r2c6=8
5 r5c9=3 B

.....8.6.19.....7...5.2.4..93.....4....2.7....6.....89..3.7.8...5.....26.1.3..... 29

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`23j4   2l47a 2Ê4ê     |1457A9b 1345  8      |1r259B 6     1235   1      9     2T4Y6m8c |456     3456  3456   |2Ë5ë   7     23X58C 3J6M8  7A8a  5        |167a9   2     1369é  |4      13E9b 138c   ----------------------------------------------------------------9      3     12u7Å    |1568d   1568D 156    |1257   4     12v57i 45k8   4a8A  1K48     |2       9     7      |6      13e5Â 13E5   2w5K7H 6     127      |145     13f45 13F45  |12u57Å 8     9      ----------------------------------------------------------------246Ä   2L4l  3        |4569    7     2o4569 |8      1g59è 1G4p5  47h8   5     47H89n   |148D9   148   149    |3      2     6      2468   1     246M89N  |3       4568Æ 2O4569 |57i9b  5Ì9ì  4P57I  `

[]9r9c3.N - 9r9c78 = AC: r9c13569(9r9c36 - 7r9c9) = 7r4c9 - 2r4c9 = 2r46c7 - 2r2c7 = AC: r2c39(2r2c39 - 6r2c3) = 6r9c3 - 9r9c3.N
[] AC: r7c4689(2r7c6.o - 6r7c46) = 6r7c1 - 6r9c3 = AC: r2c39(6r2c3 - 2r2c39) = 2r2c7 - 2r46c7 = AC: r49c9(2r4c9 - 4r9c9) = 4r7c9 - 4r7c2 = 2r7c2 - 2r7c6.o

4 r8c3=9 B
5 r9c6=2 B
champagne
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I'm somewhat less convinced that AC and AHS are two views of the same thing. Here are AHS views of your AC eliminations/placements for the last puzzle (plus some clean up which is only for aesthetic purposes). The second AC in the first chain only contains two cells while the AHS contains three. On the other hand it may be more appropriate to consider all nodes included in the AC cells as part of the AHS (otherwise there are hidden t-links in the chain). This increases the similarity between AC and AHS. There ends up being two flavors of AHS, those that link with a strong link and those that link with a weak link. Something to think about. Thanks again for the solutions. FWIW, it would be nice to keep this thread focused on only the unsolvable puzzles.

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`[] 9r9c3.N - 9r9c78 = AC: r9c13569(9r9c36 - 7r9c9) = 7r4c9 - 2r4c9 = 2r46c7   - 2r2c7 = AC: r2c39(2r2c39 - 6r2c3) = 6r9c3 - 9r9c3.N [] AHS:(9r9c3.N - 2468r9c3 = 2468r9c1569 - 7r9c9) = (7-2)r4c9 = 2r46c7    - AHS:(2r2c7 = 28r2c39 - 6r2c3) = 6r9c3 - 9r9c3.N [] AC: r7c4689(2r7c6.o - 6r7c46) = 6r7c1 - 6r9c3 = AC: r2c39(6r2c3 - 2r2c39)   = 2r2c7 - 2r46c7 = AC: r49c9(2r4c9 - 4r9c9) = 4r7c9 - 4r7c2 = 2r7c2 - 2r7c6.o [] AHS:(2r7c6.o - 459r7c6 = 459r7c489 - 6r7c46) = 6r7c1 - 6r9c3 = AHS:(6r2c3 - 8r2c3 = 8r2c9 - 2r2c39)   = 2r2c7 - 2r46c7 = AHS:(2r4c9 - 7r4c9 - r9c9 - 4r9c9) = 4r7c9 - (4=2)r7c2 - 2r7c6.o`
Mike Barker

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Joined: 22 January 2006

Hi mike,

1) could you give me a link to a detailed definition of AHS and it's uses. It could help me.

2) Ok to keep that thread clean. I'll post answers to your questions in "ull tagging method". The first example should come this afternoon.
champagne
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