The New Sudoku Players' Forum

by **Mauriès Robert** » Tue Sep 22, 2020 6:38 pm

Hi all,
I propose you this puzzle.
Good resolution
Robert
.2.5.9.7.....4.9.2..93.....93...1..62.......71..8...94.....576.6...1.....4.6.3.5.

puzzle: Show

resolution: Show

by **SpAce** » Wed Sep 23, 2020 2:13 am

Hi Robert,

Disclaimer: I didn't have time to start solving a multi-stepper from scratch, so I cheated and looked at your eliminations (not the actual moves). They seemed pretty efficient, so I built my own moves around them. I haven't checked how close they are to yours, but I guess they can't be very far off.

Step 1. AIC: Show

- 4x4 PM: Show

Step 2. AIC (with nesting): Show

- 6x6 TM: Show

Step 3. AIC (with nesting): Show

- 5x5 TM: Show

--
I actually found steps 2 and 3 together, but it gets pretty ugly if they're written as a single move.

Step 2. Memory chain (with a subchain): Show

- 12x12 TM: Show

--
PS. To avoid boring copycatting like this, you probably shouldn't post your own resolution right away

Website · by **denis_berthier** » Wed Sep 23, 2020 4:51 am

Even though I activated whips, only simple short reversible chains (bivalue-chains and z-chains) are needed.

Code: Select all: (solve ".2.5.9.7.....4.9.2..93.....93...1..62.......71..8...94.....576.6...1.....4.6.3.5.") *********************************************************************************************** *** SudoRules 20.1.s based on CSP-Rules 2.1.s, config = W+SFin *** Using CLIPS 6.32-r770 *********************************************************************************************** hidden-single-in-a-row ==> r7c4 = 4 naked-single ==> r5c4 = 9 hidden-single-in-a-block ==> r5c6 = 4 hidden-single-in-a-block ==> r4c3 = 4 hidden-single-in-a-column ==> r3c9 = 5 hidden-single-in-a-column ==> r2c1 = 5 hidden-single-in-a-column ==> r2c4 = 1 158 candidates, 788 csp-links and 788 links. Density = 6.35% whip[1]: r4n8{c8 .} ==> r5c8 ≠ 8, r5c7 ≠ 8 whip[1]: r4n7{c5 .} ==> r6c6 ≠ 7, r6c5 ≠ 7 x-wing-in-columns: n2{c4 c8}{r4 r8} ==> r8c7 ≠ 2, r8c6 ≠ 2, r8c3 ≠ 2, r4c7 ≠ 2, r4c5 ≠ 2 biv-chain[3]: r4n8{c7 c8} - c8n2{r4 r8} - b9n4{r8c8 r8c7} ==> r8c7 ≠ 8 biv-chain[3]: r8c7{n3 n4} - r1n4{c7 c1} - c1n3{r1 r7} ==> r7c9 ≠ 3, r8c3 ≠ 3 biv-chain[3]: c9n3{r1 r8} - r8c7{n3 n4} - r1n4{c7 c1} ==> r1c1 ≠ 3 hidden-single-in-a-column ==> r7c1 = 3 z-chain-bn[3]: b7n9{r8c2 r7c2} - b8n9{r7c5 r9c5} - b8n7{r9c5 .} ==> r8c2 ≠ 7 z-chain-rc[3]: r8c6{n8 n7} - r2c6{n7 n6} - r1c5{n6 .} ==> r3c6 ≠ 8 biv-chain[3]: c6n8{r8 r2} - r2c8{n8 n3} - c9n3{r1 r8} ==> r8c9 ≠ 8 biv-chain[4]: r4n5{c5 c7} - b6n8{r4c7 r4c8} - b6n2{r4c8 r6c7} - r6n3{c7 c5} ==> r6c5 ≠ 5 biv-chain[4]: r2c8{n8 n3} - c9n3{r1 r8} - r8c7{n3 n4} - c8n4{r8 r3} ==> r3c8 ≠ 8 biv-chain[4]: r8n9{c2 c9} - c9n3{r8 r1} - r2c8{n3 n8} - c6n8{r2 r8} ==> r8c2 ≠ 8 z-chain-rn[4]: r8n7{c6 c3} - r8n5{c3 c2} - r8n9{c2 c9} - r9n9{c9 .} ==> r9c5 ≠ 7 whip[1]: r9n7{c3 .} ==> r8c3 ≠ 7 biv-chain[3]: r8n7{c6 c4} - r4c4{n7 n2} - c6n2{r6 r3} ==> r3c6 ≠ 7 naked-pairs-in-a-column: c6{r3 r6}{n2 n6} ==> r2c6 ≠ 6 whip[1]: r2n6{c3 .} ==> r1c3 ≠ 6, r3c2 ≠ 6 biv-chain[3]: r3n2{c6 c5} - c5n7{r3 r4} - r4c4{n7 n2} ==> r6c6 ≠ 2 naked-single ==> r6c6 = 6 naked-single ==> r3c6 = 2 naked-pairs-in-a-block: b4{r6c2 r6c3}{n5 n7} ==> r5c3 ≠ 5, r5c2 ≠ 5 whip[1]: b4n5{r6c3 .} ==> r6c7 ≠ 5 biv-chain-rc[3]: r6c3{n7 n5} - r8c3{n5 n8} - r9c1{n8 n7} ==> r9c3 ≠ 7 hidden-single-in-a-block ==> r9c1 = 7 whip[1]: c1n8{r3 .} ==> r1c3 ≠ 8, r2c2 ≠ 8, r2c3 ≠ 8, r3c2 ≠ 8 biv-chain-rc[3]: r6c7{n3 n2} - r4c8{n2 n8} - r2c8{n8 n3} ==> r5c8 ≠ 3, r1c7 ≠ 3 naked-single ==> r5c8 = 1 naked-single ==> r3c8 = 4 naked-single ==> r3c1 = 8 naked-single ==> r1c1 = 4 hidden-single-in-a-column ==> r8c7 = 4 biv-chain-rc[4]: r4c7{n8 n5} - r4c5{n5 n7} - r3c5{n7 n6} - r1c5{n6 n8} ==> r1c7 ≠ 8 naked-pairs-in-a-block: b3{r1c7 r3c7}{n1 n6} ==> r1c9 ≠ 1 whip[1]: c9n1{r9 .} ==> r9c7 ≠ 1 biv-chain[3]: r9n1{c9 c3} - r1c3{n1 n3} - r1c9{n3 n8} ==> r9c9 ≠ 8 biv-chain[4]: r4c8{n8 n2} - c4n2{r4 r8} - r8n7{c4 c6} - c6n8{r8 r2} ==> r2c8 ≠ 8 stte

by **Mauriès Robert** » Wed Sep 23, 2020 6:43 am

Hi Space,

SpAce wrote:Disclaimer: I didn't have time to start solving a multi-stepper from scratch, so I cheated and looked at your eliminations (not the actual moves). They seemed pretty efficient, so I built my own moves around them. I haven't checked how close they are to yours, but I guess they can't be very far off.

Your resolution is equivalent to mine, and its interest lies in the fact that it is expressed with correctly written AICs and TMs, which I do not know how to do.
Cordialy
Robert

by **Mauriès Robert** » Wed Sep 23, 2020 6:57 am

Hi Denis,
Your SudoRules solver is set to find a resolution with the shortest strings and in this case the number of strings used is free. Is it configurable to find a resolution with a minimum number of strings whose length is free or limited?
Cordialement
Robert

by **Cenoman** » Wed Sep 23, 2020 12:03 pm

Two steps:

Code: Select all: +-----------------------+-------------------------+------------------------+ | 348 2 1368 | 5 68 9 | 13468 7 M138 | | 5 678 3678 | 1 4 G678 | 9 M3-8 2 | | 478 1678 9 | 3 2678E* 2678D* | 1468 14-8 5 | +-----------------------+-------------------------+------------------------+ | 9 3 4 | A27 257F* 1 | 258 A28 6 | | 2 568 568 | 9 356 4 | 135 13 7 | | 1 567 567 | 8 2356 26 | 235 9 4 | +-----------------------+-------------------------+------------------------+ | 38 189 1238 | 4 289 5 | 7 6 L1389 | | 6 e589-7 e2358-7 | Ha27B* 1 Ha278C* | J2348 J2348 d389 | | 78 4 1278 | 6 b2789 3 | K128 5 Lc189 | +-----------------------+-------------------------+------------------------+

1. (7)r8c46 = (7-9)r9c5 = r9c9 - r8c9 = (95)r8c23 => -7 r8c23
2. (8=27)r4c48 - [(7)r8c4 = (7-8)r8c6 = (8-2)r3c6 = (2-7)r3c5 = (7)r4c5] = (8)r2c6^ - (8=72)r8c46 - r8c78 = (2-1)r9c7 = r79c9 - (1=38)b3p35 => -8 r2c8^, -8 r3c8; ste
(Note: the embedded chain is tagged from B* to F*, behind the pencilmarks, the spoiler chain is tagged from G to M, skiping I)

Website · by **denis_berthier** » Wed Sep 23, 2020 12:15 pm

Mauriès Robert wrote:Hi Denis,
Your SudoRules solver is set to find a resolution with the shortest strings and in this case the number of strings used is free. Is it configurable to find a resolution with a minimum number of strings whose length is free or limited?
Cordialement
Robert

No, because such a condition is not expressible in purely logical terms.

by **SpAce** » Wed Sep 23, 2020 6:36 pm

Mauriès Robert wrote:Your resolution is equivalent to mine, and its interest lies in the fact that it is expressed with correctly written AICs and TMs, which I do not know how to do.

Indeed. Now that I took a look at your actual moves, it seems that our first steps are identical. The second and third take a bit different routes for the same effect:

Your step 2:

8x8 BTM: Show

AIC (with nesting): Show

That would probably translate to a braid, while mine to a whip.

Your step 3:

7x7 TM: Show

AIC (with nesting): Show

That would probably translate to a braid as well, just like my third step.

(Denis is welcome to correct my assumptions, but if I've understood anything, the difference between whips and braids is trivial to see in the matrix form. For the same reason it's easy to see why braids are confluent and whips are not.)

by **Ajò Dimonios** » Wed Sep 23, 2020 7:26 pm

Hi Cenoman

Cenoman wrote:
2. (8=27)r4c48 - [(7)r8c4 = (7-8)r8c6 = (8-2)r3c6 = (2-7)r3c5 = (7)r4c5] = (8)r2c6^ - (8=72)r8c46 - r8c78 = (2-1)r9c7 = r79c9 - (1=38)b3p35 => -8 r2c8^, -8 r3c8; ste

The second step is not clear to me. My problem is to understand what is the logical motivation that allows us to discard the conclusions of the first logical chain, the one with (= (8) r3c6), which is a discontinuous nice loop and to accept vice versa the conclusions of the second, the one with ( = (8) r2c6) ^?

Paolo

by **Cenoman** » Wed Sep 23, 2020 10:50 pm

Ajò Dimonios wrote:The second step is not clear to me. My problem is to understand what is the logical motivation that allows us to discard the conclusions of the first logical chain, the one with (= (8) r3c6), which is a discontinuous nice loop and to accept vice versa the conclusions of the second, the one with ( = (8) r2c6) ^?

My step 2 is actually a kraken column written on a single line:
Kraken column (8)r238c6 => -8 r2c8^, -8 r3c8
(8)r2c6^ - (8=72)r8c46 - r8c78 = (2-1)r9c7 = r79c9 - (1=38)b3p35
(8-2)r3c6 = (2-7)r3c5 = r4c5 - (7=28)r4c48
(8-7)r8c6 = r8c4 - (7=28)r4c48

Only two AIC's are considered in my step 2:
(keeping only the endpoints) (8)r4c8 == (38)b3p35 and the sub-chain (8)r4c8 == (8)r2c6
I see no chain having 8r3c6 as an endpoint. The chain inside the brackets can't be split. It is an almost chain, as the strong link 8r8c6 = 8r3c6 is spoiled by 8r2c6, treated separately.

by **SpAce** » Wed Sep 23, 2020 11:10 pm

Ajò Dimonios wrote:
Cenoman wrote:
2. (8=27)r4c48 - [(7)r8c4 = (7-8)r8c6 = (8-2)r3c6 = (2-7)r3c5 = (7)r4c5] = (8)r2c6^ - (8=72)r8c46 - r8c78 = (2-1)r9c7 = r79c9 - (1=38)b3p35 => -8 r2c8^, -8 r3c8; ste

The second step is not clear to me. My problem is to understand what is the logical motivation that allows us to discard the conclusions of the first logical chain, the one with (= (8) r3c6), which is a discontinuous nice loop and to accept vice versa the conclusions of the second, the one with ( = (8) r2c6) ^?

You need to read it like this:

(8=2)r4c8 - (2=7)r4c4 - [nested AIC] = (8)r2c6The whole nested chain is a single node in the outer AIC. It has a weak link with 7r4c4 and a strong link with 8r2c6. If you read it from left to right, 7r4c4 is assumed true and thus the nested AIC is false and 8r2c6 is true. If you read it from right to left, it's vice versa. The chain in the nested part could be written in either orientation. This would work just the same:

(8=2)r4c8 - (2=7)r4c4 - [(7)r4c5 = (7-2)r3c5 = (2-8)r3c6 = (8-7)r8c6 = (7)r8c4] = (8)r2c6If one wants to go nuts with nested chains, adding another would get both eliminations without needing the subchain:

(8=27)r4c84 - [(7)r4c5 = (72-8)r3c56 = (87)r8c64] = 8r2c6 - [(8=72)r8c64 - r8c78 = (21)b9p739 - (1=3,8)b3p35] = (8)r1c9 => -8 r23c8

uncompressed: Show

@Cenoman: Nice job! Much better than my horrible memory chain

(I see you also replied while I was typing this.)

by **Cenoman** » Thu Sep 24, 2020 8:48 am

Hi Paolo and SpAce,

It was "round midnight" when I posted yesterday! I felt that some more information about embedded AIC's would be useful.
SpAce did it. Thanks !

BTW is 'embedded' synonym of 'nested' ? (English isn't my native language)

I like the short writing

(8=2)r4c8 - (2=7)r4c4 - [nested AIC] = (8)r2c6

I'd just add a hint:
- the weak link to the nested AIC is a valid weak link to both endpoints,
- the strong link to the nested AIC complements a strong link inside, between a subset of a 3-SIS (or more).
No matter which link is on the left side or the right side.
It could be written as well: (8)r2c6 = [nested AIC] - (7=2)r4c4 - (2=8)r4c8 without flipping the nested AIC left-to-right (or with flipping it too)
In the example above:
7r4c4 is weakly linked to 7r8c4 AND to 7r4c5
8r2c6 is the candidate missing to the strong link 8r8c6=8r3c6 (SIS: 8c6)
My definition of a SIS: a set of candidates, one out of which, at least, must be true.

by **SpAce** » Thu Sep 24, 2020 10:55 am

Hi Cenoman,

I'd just add a hint:
- the weak link to the nested AIC is a valid weak link to both endpoints,
- the strong link to the nested AIC complements a strong link inside, between a subset of a 3-SIS (or more).
7r4c4 is weakly linked to 7r8c4 AND to 7r4c5
8r2c6 is the candidate missing to the strong link 8r8c6=8r3c6 (SIS: 8c6)

Very good clarifications!

No matter which link is on the left side or the right side.
It could be written as well: (8)r2c6 = [nested AIC] - (7=2)r4c4 - (2=8)r4c8

Indeed. To me that orientation has always been the more natural way to read it, just like with any almost-patterns. Approaching from the weak-link side is harder. Then again, it's important to learn to see it both ways.

Cenoman wrote:BTW is 'embedded' synonym of 'nested' ?

Good question. They're very similar, and I'm sure either one gets understood.

Perhaps 'nested' implies more clearly a hierarchical containment structure of objects of the same type. For example, in programming there are nested loops and nested classes etc, which in turn may contain deeper nesting of those same types. Similarly nested AICs may contain other nested AICs, and nested T&E may contain several levels of T&E branches.

On the other hand, 'embedded' might more naturally imply a flatter hierarchy (just one containment level) and different types, like 'embedded SQL' within a programming language. I guess an analogy in sudoku might be an AIC having nodes written in another language, such as 'embedded' UFG fishes.

To add to the confusion there's also the term 'subchain' which in my vocabulary means a shorter part of a top level AIC that provides additional eliminations with different end-points (but could be written as a separate AIC too). Looks like you're using it in the same meaning. In other contexts 'subchain' could also mean a branch, so it's a bit ambiguous, but I don't think that's a problem as we can simply use 'branch' for that purpose.

I don't even remember where I originally picked the terms 'nested' and 'subchain' from, so I don't know how "official" they are. They just made sense to me so I've been using them.

by **Ajò Dimonios** » Thu Sep 24, 2020 12:22 pm

Hi Cenoman

Thanks Cenoman your explanation is clear and comprehensive.

Paolo

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