The New Sudoku Players' Forum

[Luke]

Some examples from this puzzle just to illustrate hidden set logic :
3. 28r1c13=6r1c3-(6=4)r7c3-(4=2)r8c3-(24=1)r8c1-(124=5)r9c1-5r9c2=(5-2)r2c2 :=><2>r2c2

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 *--------------------------------------------------------------------*
 | 28     236    268    | 2369   4      5      | 7      1      289    |
 | 2458   23456  1      | 2369   239    7      | 3469   3489   2489   |
 | 9      23467  2467   | 1      8      36     | 346    5      24     |
 |----------------------+----------------------+----------------------|
 | 6      1      478    | 5      379    3489   | 2      3489   489    |
 | 3      247    9      | 24     127    148    | 5      48     6      |
 | 248    24     5      | 23469  239    34689  | 349    7      1      |
 |----------------------+----------------------+----------------------|
 | 7      8      46     | 49     5      2      | 1      469    3      |
 | 124    2469   246    | 7      139    1349   | 8      2469   5      |
 | 1245   2459   3      | 8      6      149    | 49     249    7      |
 *--------------------------------------------------------------------*

Also a pretty impressive display of a "Multiple Inference Net/Secondary Implication Stream" in action. Too bad that doesn't work out as an acronym for "memory."
28r1c13=6r1c3-(6=4)r7c3-(4=2)r8c3-(24=1)r8c1-(124=5)r9c1-5r9c2=(5-2)r2c2 :=><2>r2c2

[Steve K]

I always am partial to continuous loops, when they present themselves:
I found 3 in this path:
1) ss
2) HubSpokesRim: hsr (3)r6c4=(3)r12c4-(3=6)r3c6-(6)r12c4=(6)r6c4 => r2c5<>3, r6c4<>249
3) (3)r8c6=(3-1)r8c5=(nt148)r5c568-(4)r5c4=(4)r7c4 => r8c6<>4
3) alt (3)r8c6=(3-1)r8c5=(hp17-hp27)r5c25=(2-4)r5c4=(4)r7c4 => r8c6<>4
4) hsr (4)r8c8=(4)r8c123-(4=6)r7c3-(6)r8c23=(6)r8c8 => r9c12<>4, r8c8<>29 one single
5) using aur 46 r78c38, (2)r8c3=(9-6)r7c8=(6)r7c3 => r8c3<>6
6) (np24=1)r8c13-(1)r9c1=(1-4)r9c6=(4)r9c7 => r8c8<>4 2 singles,
(np28)r1c13, 2 singles, (np36)r13c46, (np48)r45c89, fsf(9)r247
7) (3)r4c56=(3-9)r4c8=(9-4)r7c8=(4)r7c4-(4=2)r5c4-(2=3)r6c5 loop => r6c46<>3,r5c5<>2 5 singles
8) (4)r7c4=(4-2)r5c4=(2-7)r5c2=(7-3)r3c2=(3-4)r3c7=(4)r9c7 => r9c6,r7c8<>4 ste

[ronk]

Some examples from this puzzle just to illustrate hidden set logic :
3. 28r1c13=6r1c3-(6=4)r7c3-(4=2)r8c3-(24=1)r8c1-(124=5)r9c1-5r9c2=(5-2)r2c2 :=><2>r2c2

...
Also a pretty impressive display of a "Multiple Inference Net/Secondary Implication Stream" in action. Too bad that doesn't work out as an acronym for "memory."
28r1c13=6r1c3-(6=4)r7c3-(4=2)r8c3-(24=1)r8c1-(124=5)r9c1-5r9c2=(5-2)r2c2 :=><2>r2c2

Maybe this will help. If you consider ALS and AALS nodes, there actually is no "memory".

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 NL: r2c2 -2- als:r1c13 -6- als:r78c3 -24- aals:r89c1 -5- r9c2 =5= r2c2 ==> r2c2<>2

AIC: (2)r2c2 - (2=6)als:r1c13 - (6=24)als:r78c3 - (24=5)aals:r89c1 - (5)r9c2 = (5)r2c2 ==> r2c2<>2


Note that use of an AALS [ed: requires two weak inferences] on at least one side of the AALS.

[DonM]

1: (9=4)r7c4-(4=6)r7c3-als(6=28)r1c13-(28=9)r1c9 => r1c4<>9 -> r1c9=9 -> np(48)r4c9/r5c8:r4c8<>4,8, r6c7<>4
2: (2=4)r3c9-(4=8)r4c9-r4c3=r6c1-(8=2)r1c1 => r3c23<>2 -> r3c9=2
3: (9)r2c5=r2c4-r7c4=r7c8-(9=4)r9c7-als(4=36)r23c7-(3)r2c8=r4c8-(3=9)r6c7 => r6c45<>9
4: (1)r8c5=r5c5-als(1=48)r5c68-(4=2)r5c4-(2=3)r6c5 => r8c5<>3 -> r8c6=3 -> r3c6=6 -> r2c7=6 -> r6c4=6
5: (9=4)r9c7-r9c6=r7c4-(4=2)r5c4-(2=3)r6c5-(3=9)r6c7-loop => r5c5<>2, r9c128<>4
6: (3=4)r3c7-(4=7)r3c3-r4c3=(7-2)r5c2=r5c4-(2=3)r6c5 => r6c7<>3=9

STTE

Point of interest after step 5, but doesn't help solution: Sue-de-Coq(12469)r8c123/(19)r3c5/(46)r7c3: r8c8<>9 (also would have duplicated step 5's r9c12<>4)

[Allan Barker]

7 Sided Broken Loop with 1 target that sees 4 guardians, with guardian appendix

Congratulations on a very nice find. Most of us think of the "broken wing" in terms of a single digit, but the principle certainly holds for multi-digit broken wings as well. To your knowledge, has anyone ever posted a multi-digit BW before

Apologies, this is a mistake I often make when multi-digit logic works by the same logic as a single digit counterpart. More correctly, this is a loop of strong links + guardians that eliminates a candidate analogously to Broken Wings.

I do have one question Allan -- how is a 7-sided loop different than a network?

Draco, I'm using the term loop in a general sense. The set logic I use is not based on the same kind of implications as nice loops and networks, so I'm not a good one to speculate on their differences. However, I think most all methods like chains, discontinuous nice loops, and related networks rely on odd length of something. They also require an odd number of base + cover sets.

[Luke]

If you consider ALS and AALS nodes, there actually is no "memory".

That was very good of you to take the time to write up the als/aals perspective, and throw in AIC notation to boot. Thank you! I'm sure you've got better things to do, but after studying this I still need some help figuring it out.

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*--------------------------------------------------------------------*
 |A28     236   A268    | 2369   4      5      | 7      1      289    |
 | 2458  -23456  1      | 2369   239    7      | 3469   3489   2489   |
 | 9      23467  2467   | 1      8      36     | 346    5      24     |
 |----------------------+----------------------+----------------------|
 | 6      1      478    | 5      379    3489   | 2      3489   489    |
 | 3      247    9      | 24     127    148    | 5      48     6      |
 | 248    24     5      | 23469  239    34689  | 349    7      1      |
 |----------------------+----------------------+----------------------|
 | 7      8     B46     | 49     5      2      | 1      469    3      |
 |C124    2469  B246    | 7      139    1349   | 8      2469   5      |
 |C1245  *2459   3      | 8      6      149    | 49     249    7      |
 *--------------------------------------------------------------------*

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NL: r2c2 -2- als:r1c13 -6- als:r78c3 -24- aals:r89c1 -5- r9c2 =5= r2c2 ==> r2c2<>2

AIC: (2)r2c2 - (2=6)als:r1c13 - (6=24)als:r78c3 - (24=5)aals:r89c1 - (5)r9c2 = (5)r2c2 ==> r2c2<>2


I get lost right off the bat, sorry. The first als (A) is (268)r1c13. I can understand (28=6)r1c13, so I was surmising that the notation means (2)r2c2-(28=6)als:r1c13. But throwing the (8) in there like that can't be right.

The second als (B) is (246)r78c3. The restricted common between the two is (6), so the weak link makes sense. So far it's an als-xy chain, right?

I go off the tracks again with the aals (C). I understand the aals concept a little bit (OK, not much beyond N+2.) Now it looks as though there's a grouped weak inference between the B and C nodes, but I get lost trying to get to (24=5.) Does it mean "(some-two-candidate-combination-of-124)=5?" There can't be an RC involved like before, since the nodes are both in the same house. The rest is straight forward, even to me.

Thanks for any help.

[aran]

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*--------------------------------------------------------------------*
 |A28     236   A268    | 2369   4      5      | 7      1      289    |
 | 2458  -23456  1      | 2369   239    7      | 3469   3489   2489   |
 | 9      23467  2467   | 1      8      36     | 346    5      24     |
 |----------------------+----------------------+----------------------|
 | 6      1      478    | 5      379    3489   | 2      3489   489    |
 | 3      247    9      | 24     127    148    | 5      48     6      |
 | 248    24     5      | 23469  239    34689  | 349    7      1      |
 |----------------------+----------------------+----------------------|
 | 7      8     B46     | 49     5      2      | 1      469    3      |
 |C124    2469  B246    | 7      139    1349   | 8      2469   5      |
 |C1245  *2459   3      | 8      6      149    | 49     249    7      |
 *--------------------------------------------------------------------*

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NL: r2c2 -2- als:r1c13 -6- als:r78c3 -24- aals:r89c1 -5- r9c2 =5= r2c2 ==> r2c2<>2

AIC: (2)r2c2 - (2=6)als:r1c13 - (6=24)als:r78c3 - (24=5)aals:r89c1 - (5)r9c2 = (5)r2c2 ==> r2c2<>2


I get lost right off the bat, sorry. The first als (A) is (268)r1c13. I can understand (28=6)r1c13, so I was surmising that the notation means (2)r2c2-(28=6)als:r1c13. But throwing the (8) in there like that can't be right.

In addition to what Ronk may respond just a couple of points.
1. The ALS presentation is in my view more succinct that I personally prefer eg in that first move you have to make the mental jump r2c2=2 =>r1c1=8 and hence r1c3=6.
In my own chain formulation I don't require such effort
2. Making a discontinuous loop from the elimination candidate so far as I can tell seems to be quite common in NL (as in the above with 2r2c2). I assume it may mean that after finding an elimination by whatever method, it is reformulated to that style (remembering that any elimination x can always be presented as a discontinuous loop starting "x-" ). I may well be wrong but that is how it strikes me.
I prefer to write the chain as I found it.

[Luke]

I prefer to write the chain as I found it.

I have the advantage of already understanding your approach. It seems so simple and productive. I can't say it's any better than the als approach because I still don't quite get the als approach. When I do, I'll be better qualified to make judgements about what works for me. (Hmm...that sounds snarky but I don't intend it that way! )

Continuing my woodsheddin', I found a link to the original discussion on Eureka of "ALS Chains - A Tutorial With Graphics." (I think I should go back over Don's reposting of that.) Reading some of the unhacked posts makes me re-think my notion that the relation btwn B and C can't be based on a RC. Now I'm thinking that it might be based on a "double restricted common." Once again, just guessing.

Also interesting was a reference to "remembering." It was described as "doing a Berthier." Was "remembering" anything like the m-word (that doesn't exist)? It appears that way given the context but I can't find the referenced discussion.

[aran]

Also interesting was a reference to "remembering." It was described (derided?) as "doing a Berthier." Was "remembering" anything like the m-word (that doesn't exist)?

Subject to rescrutiny I think that Denis Berthier's 't' as in his xyt, xyzt chains is "memory" or "remembering" which I think he used in restrictive or specific contexts...unlike me...

[ronk]

Reading some of the unhacked posts makes me re-think my notion that the relation btwn B and C can't be based on a RC. Now I'm thinking that it might be based on a "double restricted common."

Yes, sets B & C are doubly-linked, as indicated by the appearance of the <24> weak inferences between sets B & C. Based on your prior post, it seems you are expecting all the candidates to appear in the AIC expression. If so, IMO that would be ...

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                 A                 B                   C
(2)r2c2 - (2=86)als:r1c13 - (6=24)als:r78c3 - (24=15)aals:r89c1 - (5)r9c2 = (5)r2c2 ==> r2c2<>2

Now the above reads OK left-to-right (L2R), but not right-to-left (R2L). The biggest problem with R2L is ... if <5> is removed from C, then pairs <12>, <14> and <24> all appear to be possibilities. However, pair <24> is clearly impossible. This dilemma is fixed by considering sets B & C together as one ALS, which I should have done in the first place. The AIC expressions then might look like ...

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With ALS RCCs only:
(2)r2c2 - (2=6)als:r1c13 - (6=5)als:[r78c3,r89c1] - (5)r9c2 = (5)r2c2 ==> r2c2<>2

With all ALS digits:
(2)r2c2 - (28=86)als:r1c13 - (6124=1245)als:[r78c3,r89c1] - (5)r9c2 = (5)r2c2 ==> r2c2<>2

The symmetry of the ALS notation is a strong indication that the AIC has both L2R and R2L validity. In my preferred NL notation:

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r2c2 -2- als:r1c13 -6- als:[r78c3,r89c1] -5- r9c2 =5= r2c2 ==> r2c2<>2

[ttt]

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 *--------------------------------------------------------------------*
 | 28     236    268    | 2369   4      5      | 7      1      289    |
 | 2458   23456  1      | 2369   239    7      | 3469   3489   2489   |
 | 9      23467  2467   | 1      8      36     | 346    5      24     |
 |----------------------+----------------------+----------------------|
 | 6      1      478    | 5      379    3489   | 2      3489   489    |
 | 3      247    9      | 24     127    148    | 5      48     6      |
 | 248    24     5      | 23469  239    34689  | 349    7      1      |
 |----------------------+----------------------+----------------------|
 | 7      8      46     | 49     5      2      | 1      469    3      |
 | 124    2469   246    | 7      139    1349   | 8      2469   5      |
 | 1245   2459   3      | 8      6      149    | 49     249    7      |
 *--------------------------------------------------------------------*

My path for this one:
01: (hp13)r8c56=(1)r9c6-(1)r5c6=(1-hp27)r5c25=(2)r5c4-(2)r12c4=(2-9)r2c5=(9)r12c4-(9=4)r7c4 => r8c6<>4

02: r8c2-6-r7c3-4-r7c4=4=r9c6-4-r9c7-9-r9c2=9=r8c2-6-r8c2
(6=4)r7c3-(4)r7c4=(4)r9c6-(4=9)r9c7-(9)r9c2=(9)r8c2 => r8c2<>6, some singles

03: r6c5-9-r2c5=9=r2c4=9=r7c4=9=r7c8-9-r4c8=9=r6c7-9-r6c5
9’s r2c5=r2c4-r7c4=r7c8-r4c8=r6c7 => r6c5<>9

04: -2-r6c5-3-r6c7=9=r9c7-4-r9c6=4=r7c4=4=r5c4-2-
(2=3)r6c5-(3=9)r6c7-(9=4)r9c7-(4)r9c6=(4)r7c4-(4=2)r5c4 => Loop: r6c46<>3, r5c5<>2, r9c128<>4, some singles

05: r3c2=3=r3c7=3=r6c7-3-r6c5-2=r5c4=2=r5c2=7=r3c2
(3)r3c7=(3)r6c7-(3=2)r6c5-(2)r5c4=(2-7)r5c2=(7)r3c2 => r3c2<>3, singles to the end

I’m not sure for NL notation, please correct me. Thanks
At move 01, we can start from (3)r8c6=(3-1)r8c5=(1)r5c5… but I found this move based on the considering almost hidden pair (13)r8c56 first so I keep it as original.

PS. I try to translate my solution to NL notation for someone who can’t follow Eureka/AIC notation then help them understand our solutions.

ttt