The New Sudoku Players' Forum

[yzfwsf]

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...4..3..3.......9..1.9..4.....8..1...87..9..6....2...2...5..7..1...6..2..3..84..

SE:8.9 \ Hodoku Rate: 24516 \YZF_SUDOKU Rate:12840

[totuan]

My path for this one – again, using uniqueness pattern to reduce steps.
After basic SSTS:

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 *-----------------------------------------------------------------------------*
 | 5789    256789  25679   | 4       126     157     | 3       2568    15678   |
 | 3       245678  24567   | 12568   126     157     | 125678  2568    9       |
 | 578     25678   1       | 23568   9       357     | 25678   4       5678    |
 |-------------------------+-------------------------+-------------------------|
 | 4579    234579  24579   | 3569    8       3459    | 2567    1       34567   |
 | 1       2345    8       | 7       346     345     | 9       2356    3456    |
 | 6       34579   4579    | 1359    134     2       | 578     358     34578   |
 |-------------------------+-------------------------+-------------------------|
 | 2       4689    469     | 139     5       1349    | 168     7       1368    |
 | 45789   1       4579    | 39      347     6       | 58      3589    2       |
 | 579     5679    3       | 129     127     8       | 4       569     156     |
 *-----------------------------------------------------------------------------*

01: (26=1)r12c5-r12c6=(1-4)r7c6=(4-7)r8c5=r9c5 => r9c5<>2, r9c4=2
02: (2)r3c7=r3c2-r5c2=r5c8 => r12c8<>2, r5c8=2
03: Present as diagram: => r6c5<>1, some singles

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(26-1)r23c7=r1c9-r9c9=r9c5*
 ||
(6)r7c7--(6)r9c8=r12c8-(6)r3c79
 ||    |                || 
 ||    |               (6)r3c4-(6=12)r12c5*
 ||    |                ||
 ||     -(6)r7c23=r9c2-(6)r3c2   
 ||               
(6)r4c7-r5c9=r5c5-(6=12)r12c5*

04: (1)r2c7=r1c9-r9c9=r9c5 => r2c5<>1
05: Present as diagram: => r2c23<>2

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(6)r2c7-(6=2)r2c5*
 ||
(6)r7c7--(6)r9c8=r12c8-(6)r3c79
 ||    |                || 
 ||    |               (6)r3c4-(6=2)r2c5*
 ||    |                ||
 ||     -(6)r7c23=r9c2-(6)r3c2   
 ||               
(6-2)r3c7=r2c7*
 ||
(6)r4c7-r5c9=r5c5-(6=2)r2c5*

06: Present as diagram: => r2c2<>6

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(6)r2c7*
 ||
(6)r4c7-(6)r5c9=r5c5-(3)r5c5=r6r5-r6c8=(3-9)r8c8=r9c8-(6)r9c8
 ||                 |                                  || 
 ||                  -----------(6=12)r12c5-r9c5=r9c9-(6)r9c9
 ||                                                    ||
(6-2)r3c7=r2c7-(2=6)r2c5*                             (6)r9c2*   
 ||
(6)r7c7-r7c23=r9c2*

07: Present as diagram: => r6c5<>4, some singles

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MUG(26)r13c2/r12c5/r23c7
 ||
(2-3)r4c2=(3-4)r4c9=r4c123-
 ||                        |
(6-8)r7c2=(8-4)r8c1=r4c1 ---(4)r5c2=[XY-wing:(345)r5c26/r6c5]-(3)r5c5=r6c5*
 ||
(6)r5c5---------------(6=12)r12c5-(1=47)r89c5*
 ||               |   
(6)r4c7-r5c9=r6c5-      (6)r3c4-(6=12)r12c5-(1=47)r89c5*
 ||                      ||
 ||      ---------------(6)r3c2-----   
 ||     |                ||         |
(6)r9c2----r8c9=r12c8---(6)r3c79    |
 ||     |                           |
(6)r7c7—---r89c9=r9c2---------------       

08: (6)r5c5=r5c9-r9c9=r9c2-r3c2=r3c4 => r12c5<>6, stte

totuan

[Ajò Dimonios]

Synthetic resolution after basics.
1)P(2r9c5)=>contradiction=>-2r9c5+2r9c4
2)P(2r5c2)=>cobtradiction=>-2r5c2+2r5c8
3)P(9r7c6)=>contradiction=>-9r7c6+9r4c6
4)P(1r7c7)=>contradiction=>-1r7c7+1r2c7
5)P(8r7c7)=>contradiction=>-8r7c7=>stte

or through two conjugated tracks. Every supposedly true track generator demonstrates the truth of the generator of its anti-track.

1)P(2r9c5)=>+7r8c5+4r7c6+5r5c6+3r3c6+1r6c4+1r9c5+2r9c4
2)P(2r5c2)=>+2r3c7 +2r5c8
3)P(9r7c6)=> +4r8c5+4r4c1+4r5c6+4r6c9+3r3c6+3r8c4+3r7c9+3r4c2+2r4c3+3r5c5+6r4c4+9r4c6

in the end

4) Trucks P (1r7c7) and P (1r2c7) are conjugated . If both are fully developed with the basic technique, they both lead to common eliminations (-25678r2c7; -1r2c45; -2r3c2; -758r3c7; -7r46c9) which are equivalent to + 1r2c7.
5) Solution for crossing of backdoor P (6r7c7) and antitrack P (8r7c7) by means of element common(+6r5c5) to the two tracks.
Paolo

[eleven]

1)P(2r9c5)=>+7r8c5+4r7c6+5r5c6+3r3c6+1r6c4+1r9c5+2r9c4
2)P(2r5c2)=>+2r3c7 +2r5c8

Before i continue reading, please explain your first 2 steps (in the result being the same as totuan's).
P(2r9c5)=>+7r8c5+4r7c6+5r5c6 why ???
P(2r5c2)=>+2r3c7 why ??

[Ajò Dimonios]

During the construction of tracks P (2r9c5), imposing that r9c5 is equal 2 and using the basic technique, 7 in C5 is the only candidate in r8c5, after having entered the 7, 4 in box 8 is the only candidate in r7c6, after having inserted the 4 you can eliminate the 3 in r5c6 which imposes the 5 in r5c6 as the only candidate, at this point the track is already invalid, (+ 7r8c5 + 4r7c6 + 5r5c6 + 3r3c6), true if r9c5 = 2, impose that 1 is single in r6c4 after elimination of 3 and 9 in the same cell for alignment with C5 and C6, at this point in box 8 candidate 1 is single in r9c5 (where the chain started from) and therefore r9c4 is equal to 2. Therefore r9c5 = 2 dictates that r9c4 is equal to 2, clearly r9c5 ≠ 2 => r9c4 = 2 since the two tracks are conjugated. For track P (2r5c2) the reasoning is similar.

Paolo

[eleven]

Thanks.
So you need a box-line elimination of 3 in b5 to get 5r5c6, which you did not mention.
For "P(2r5c2)=>+2r3c7" you indeed need a skyscraper, which you call "similar reasoning".

Fine, so i can spare my time to read any of your "solutions".

[Ajò Dimonios]

Hi
Eleven

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Eleven wrote:
So you need a box-line elimination of 3 in b5 to get 5r5c6, which you did not mention.
For "P(2r5c2)=>+2r3c7" you indeed need a skyscraper, which you call "similar reasoning".

Fine, so i can spare my time to read any of your "solutions".

In the resolution I simply reported the insertions obtained in the construction of a track. I took the eliminations that are obtained with the basic technique for granted. It is simply a matter of using the same technique that is used at the beginning of the resolution of any scheme. To prove that P (2r5c2) => + 2r3c7 you can certainly use a skyscraper but I only used the basic technique.
By "solution" I mean any logical method that leads you to resolution.

Paolo

[eleven]

To prove that P (2r5c2) => + 2r3c7 you can certainly use a skyscraper but I only used the basic technique.

At least i could not. So you should show, how you did that, and add all that to your solution path. Otherwise it is useless for others.

By "solution" I mean any logical method that leads you to resolution.

Ok, and the presentation of your logical method is as illuminating as the output of a backtracking solver.

[Ajò Dimonios]

Hi Eleven

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Eleven Wrote:
At least i could not. So you should show, how you did that, and add all that to your solution path. Otherwise it is useless for others.

During the construction of track P (2r5c2), by imposing that r5c2 is equal to 2, candidate 2 is eliminated in r3c2 which makes candidate 2 in r3c7 unique in r3 which in turn eliminates candidate 2 in r4c7 which makes unique in box 6 the candidate 2 in r5c8. So when r5c2 = 2 => r3c7 = 2 and then r5c8 = 2. Also in this case being the tracks P (2r5c2) and P (2r5c8) two tracks conjugated when r5c2 ≠ 2 => r5c8 = 2, in practice r5c8 = 2 both that r5c2 = 2 whether it is r5c2 ≠ 2 (as in a normal one AIC).

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Eleven Wrote:
Ok, and the presentation of your logical method is as illuminating as the output of a backtracking solver.

I just described the logic of an AIC.

[eleven]

Oh sorry, i mixed r2c5 with r5c2.

[denis_berthier]

Some puzzles in this section tend to become more interesting than the trivial ones in the past years, where a single step solution could be found (often with very artificial and unnecessarily complicated patterns).

Here is a solution in W7; no OR branching, no complicated nets, no hidden steps. Simplest-first strategy used.
I don't mean a manual solver should rigidly follow the same strategy. For whatever good or bad reason, a manual solver may concentrate on some part of the grid and he will take whatever comes. But the simplest-first strategy provides an idea of what's possible with a given puzzle and a given set of rules.
As the resolution path shows, it's a hard puzzle, with many eliminations by relatively long whips (length 6 or 7). As I showed in chapter 6 of PBCS, 98.54% of the Sudoku puzzles (in unbiased statistics, with only 0.065% relative error) can be solved with whips of length ≤ 5.

Hidden Text: Show

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(solve "...4..3..3.......9..1.9..4.....8..1...87..9..6....2...2...5..7..1...6..2..3..84..")
***********************************************************************************************
***  SudoRules 20.1.s based on CSP-Rules 2.1.s, config = W+SFin
***  using CLIPS 6.31-r761
***********************************************************************************************
hidden-single-in-a-column ==> r5c1 = 1
227 candidates, 1510 csp-links and 1510 links. Density = 5.88671006978285%
whip[1]: c6n7{r3 .} ==> r2c5 ≠ 7, r1c5 ≠ 7
finned-x-wing-in-columns: n4{c1 c5}{r8 r4} ==> r4c6 ≠ 4
finned-x-wing-in-columns: n1{c7 c6}{r7 r2} ==> r2c5 ≠ 1, r2c4 ≠ 1
biv-chain[3]: r3n3{c6 c4} - r8c4{n3 n9} - c6n9{r7 r4} ==> r4c6 ≠ 3
biv-chain[3]: c1n4{r4 r8} - b8n4{r8c5 r7c6} - c6n9{r7 r4} ==> r4c1 ≠ 9
whip[4]: r6n9{c3 c4} - r8c4{n9 n3} - r7n3{c6 c9} - r4n3{c9 .} ==> r4c2 ≠ 9
whip[4]: r8c4{n9 n3} - r7c4{n3 n1} - r6c4{n1 n5} - r4c6{n5 .} ==> r4c4 ≠ 9
whip[4]: r9n2{c4 c5} - r2c5{n2 n6} - r1c5{n6 n1} - r6n1{c5 .} ==> r9c4 ≠ 1
whip[5]: r2c5{n2 n6} - r1c5{n6 n1} - c6n1{r2 r7} - b8n4{r7c6 r8c5} - c5n7{r8 .} ==> r9c5 ≠ 2
hidden-single-in-a-block ==> r9c4 = 2
finned-x-wing-in-rows: n2{r3 r5}{c2 c7} ==> r4c7 ≠ 2
hidden-single-in-a-block ==> r5c8 = 2
finned-x-wing-in-columns: n3{c8 c5}{r8 r6} ==> r6c4 ≠ 3
whip[4]: r5n6{c9 c5} - r2c5{n6 n2} - r1c5{n2 n1} - r9n1{c5 .} ==> r9c9 ≠ 6
biv-chain[3]: c7n1{r2 r7} - r9c9{n1 n5} - r8c7{n5 n8} ==> r2c7 ≠ 8
biv-chain[4]: b3n2{r3c7 r2c7} - c7n1{r2 r7} - r9c9{n1 n5} - r8c7{n5 n8} ==> r3c7 ≠ 8
whip[5]: r6n9{c3 c4} - r8c4{n9 n3} - r7n3{c6 c9} - r4n3{c9 c2} - b4n2{r4c2 .} ==> r4c3 ≠ 9
hidden-single-in-a-row ==> r4c6 = 9
whip[5]: r3n2{c7 c2} - r1n2{c2 c5} - r2c5{n2 n6} - r5n6{c5 c9} - r3n6{c9 .} ==> r3c7 ≠ 7
whip[5]: r3n2{c7 c2} - r1n2{c2 c5} - r2c5{n2 n6} - r5n6{c5 c9} - r3n6{c9 .} ==> r3c7 ≠ 5
whip[5]: r5n6{c9 c5} - r2c5{n6 n2} - r1c5{n2 n1} - c6n1{r1 r7} - c6n4{r7 .} ==> r5c9 ≠ 4
whip[6]: c5n3{r6 r8} - r8c4{n3 n9} - r7c4{n9 n1} - r7c6{n1 n4} - r5c6{n4 n5} - r6c4{n5 .} ==> r4c4 ≠ 3
whip[6]: c8n6{r2 r9} - c8n9{r9 r8} - r8c4{n9 n3} - c8n3{r8 r6} - c5n3{r6 r5} - r5n6{c5 .} ==> r3c9 ≠ 6
whip[5]: r8c4{n3 n9} - c8n9{r8 r9} - r9n6{c8 c2} - r3n6{c2 c7} - c8n6{r1 .} ==> r3c4 ≠ 3
hidden-single-in-a-block ==> r3c6 = 3
whip[1]: b5n3{r6c5 .} ==> r8c5 ≠ 3
hidden-pairs-in-a-block: b8{r7c4 r8c4}{n3 n9} ==> r7c4 ≠ 1
hidden-single-in-a-column ==> r6c4 = 1
biv-chain[3]: r6c5{n3 n4} - c9n4{r6 r4} - r4n3{c9 c2} ==> r6c2 ≠ 3
biv-chain[4]: r9c9{n5 n1} - b8n1{r9c5 r7c6} - c6n4{r7 r5} - b5n5{r5c6 r4c4} ==> r4c9 ≠ 5
biv-chain[5]: b6n4{r6c9 r4c9} - c1n4{r4 r8} - b8n4{r8c5 r7c6} - b8n1{r7c6 r9c5} - r9c9{n1 n5} ==> r6c9 ≠ 5
biv-chain[5]: c8n3{r8 r6} - r6c5{n3 n4} - r8c5{n4 n7} - r9c5{n7 n1} - r9c9{n1 n5} ==> r8c8 ≠ 5
biv-chain[5]: c1n4{r8 r4} - b6n4{r4c9 r6c9} - r6c5{n4 n3} - c8n3{r6 r8} - r8c4{n3 n9} ==> r8c1 ≠ 9
whip[5]: r5n6{c9 c5} - c5n3{r5 r6} - c8n3{r6 r8} - c8n9{r8 r9} - c8n6{r9 .} ==> r1c9 ≠ 6
whip[4]: r9n6{c2 c8} - b3n6{r2c8 r3c7} - c7n2{r3 r2} - r2c5{n2 .} ==> r2c2 ≠ 6
whip[5]: r2c5{n6 n2} - b3n2{r2c7 r3c7} - r3n6{c7 c2} - r9n6{c2 c8} - b3n6{r2c8 .} ==> r2c4 ≠ 6
whip[5]: r2c4{n8 n5} - r2c8{n5 n6} - r9n6{c8 c2} - r3n6{c2 c4} - c4n8{r3 .} ==> r2c2 ≠ 8
biv-chain[6]: r9c9{n5 n1} - r9c5{n1 n7} - r8c5{n7 n4} - r6c5{n4 n3} - c8n3{r6 r8} - b9n9{r8c8 r9c8} ==> r9c8 ≠ 5
biv-chain[6]: r8c7{n8 n5} - r9c9{n5 n1} - b8n1{r9c5 r7c6} - c6n4{r7 r5} - r6c5{n4 n3} - c8n3{r6 r8} ==> r8c8 ≠ 8
naked-pairs-in-a-row: r8{c4 c8}{n3 n9} ==> r8c3 ≠ 9
biv-chain[4]: r8n8{c1 c7} - b9n5{r8c7 r9c9} - r9n1{c9 c5} - b8n7{r9c5 r8c5} ==> r8c1 ≠ 7
whip[6]: c1n4{r4 r8} - c5n4{r8 r5} - r5n6{c5 c9} - r5n3{c9 c2} - r4n3{c2 c9} - r4n4{c9 .} ==> r6c3 ≠ 4
whip[6]: c1n4{r4 r8} - c5n4{r8 r5} - r5n6{c5 c9} - r5n3{c9 c2} - r4n3{c2 c9} - r4n4{c9 .} ==> r6c2 ≠ 4
whip[5]: r6n4{c9 c5} - r6n3{c5 c8} - r6n8{c8 c7} - b9n8{r7c7 r7c9} - c9n3{r7 .} ==> r6c9 ≠ 7
whip[6]: c8n5{r2 r6} - c9n5{r5 r9} - r8c7{n5 n8} - r6c7{n8 n7} - r6c3{n7 n9} - r6c2{n9 .} ==> r2c7 ≠ 5
whip[6]: c7n1{r2 r7} - r7c6{n1 n4} - r5c6{n4 n5} - r4c4{n5 n6} - c7n6{r4 r3} - c7n2{r3 .} ==> r2c7 ≠ 7
whip[1]: c7n7{r6 .} ==> r4c9 ≠ 7
whip[6]: r5c6{n5 n4} - b8n4{r7c6 r8c5} - r8n7{c5 c3} - r2n7{c3 c2} - b4n7{r4c2 r4c1} - c1n4{r4 .} ==> r2c6 ≠ 5
biv-chain[4]: c6n5{r1 r5} - c6n4{r5 r7} - b8n1{r7c6 r9c5} - r9c9{n1 n5} ==> r1c9 ≠ 5
whip[6]: r2c5{n2 n6} - r2c7{n6 n1} - c7n2{r2 r3} - b3n6{r3c7 r1c8} - b1n6{r1c2 r3c2} - r9n6{c2 .} ==> r2c3 ≠ 2
whip[6]: r2c5{n2 n6} - r2c7{n6 n1} - c7n2{r2 r3} - b3n6{r3c7 r1c8} - b1n6{r1c2 r3c2} - r9n6{c2 .} ==> r2c2 ≠ 2
whip[6]: c6n4{r7 r5} - c6n5{r5 r1} - c6n7{r1 r2} - r2c2{n7 n5} - c8n5{r2 r6} - r5n5{c9 .} ==> r7c2 ≠ 4
biv-chain[7]: r9c9{n5 n1} - r9c5{n1 n7} - r8c5{n7 n4} - r6c5{n4 n3} - c8n3{r6 r8} - b9n9{r8c8 r9c8} - r9n6{c8 c2} ==> r9c2 ≠ 5
biv-chain[7]: b7n8{r7c2 r8c1} - c1n4{r8 r4} - b6n4{r4c9 r6c9} - r6c5{n4 n3} - c8n3{r6 r8} - b9n9{r8c8 r9c8} - r9n6{c8 c2} ==> r7c2 ≠ 6
whip[7]: c6n7{r1 r2} - r2n1{c6 c7} - r1c9{n1 n8} - c8n8{r1 r6} - c8n3{r6 r8} - c8n9{r8 r9} - c1n9{r9 .} ==> r1c1 ≠ 7
whip[7]: r2c4{n5 n8} - r2c8{n8 n6} - r9n6{c8 c2} - r3n6{c2 c4} - b2n5{r3c4 r1c6} - r5n5{c6 c9} - r3n5{c9 .} ==> r2c2 ≠ 5
whip[6]: r9n6{c2 c8} - c8n9{r9 r8} - c8n3{r8 r6} - r6c5{n3 n4} - r5n4{c5 c2} - r2c2{n4 .} ==> r9c2 ≠ 7
naked-pairs-in-a-row: r9{c2 c8}{n6 n9} ==> r9c1 ≠ 9
hidden-single-in-a-column ==> r1c1 = 9
biv-chain[3]: r9n7{c1 c5} - r8c5{n7 n4} - c1n4{r8 r4} ==> r4c1 ≠ 7
biv-chain[3]: c1n8{r3 r8} - r8c7{n8 n5} - r9n5{c9 c1} ==> r3c1 ≠ 5
whip[4]: c1n7{r3 r9} - r9n5{c1 c9} - r3c9{n5 n8} - r3c1{n8 .} ==> r3c2 ≠ 7
whip[5]: r3c1{n8 n7} - r9n7{c1 c5} - c5n1{r9 r1} - c5n2{r1 r2} - b2n6{r2c5 .} ==> r3c4 ≠ 8
hidden-single-in-a-block ==> r2c4 = 8
whip[4]: r2c8{n5 n6} - r9n6{c8 c2} - r3n6{c2 c4} - b2n5{r3c4 .} ==> r1c8 ≠ 5
whip[5]: c4n5{r3 r4} - b4n5{r4c1 r6c3} - c3n9{r6 r7} - r7n4{c3 c6} - r5c6{n4 .} ==> r3c2 ≠ 5
biv-chain[2]: r3n5{c9 c4} - b5n5{r4c4 r5c6} ==> r5c9 ≠ 5
biv-chain[3]: c9n5{r3 r9} - r8c7{n5 n8} - c1n8{r8 r3} ==> r3c9 ≠ 8
whip[1]: r3n8{c2 .} ==> r1c2 ≠ 8
whip[4]: r4c4{n6 n5} - r4c1{n5 n4} - r4c9{n4 n3} - r5c9{n3 .} ==> r4c7 ≠ 6
whip[1]: b6n6{r5c9 .} ==> r7c9 ≠ 6
naked-triplets-in-a-column: c7{r4 r6 r8}{n5 n7 n8} ==> r7c7 ≠ 8
whip[4]: r7n6{c3 c7} - r3n6{c7 c4} - r3n5{c4 c9} - r2c8{n5 .} ==> r2c3 ≠ 6
finned-x-wing-in-columns: n6{c3 c7}{r7 r1} ==> r1c8 ≠ 6
naked-single ==> r1c8 = 8
naked-triplets-in-a-column: c9{r1 r3 r9}{n1 n7 n5} ==> r7c9 ≠ 1
x-wing-in-columns: n1{c5 c9}{r1 r9} ==> r1c6 ≠ 1
naked-triplets-in-a-row: r7{c2 c4 c9}{n8 n9 n3} ==> r7c3 ≠ 9
hidden-single-in-a-column ==> r6c3 = 9
biv-chain[3]: r4c1{n4 n5} - r6c2{n5 n7} - r2c2{n7 n4} ==> r4c2 ≠ 4
biv-chain[3]: r4c1{n4 n5} - r6c2{n5 n7} - r2c2{n7 n4} ==> r5c2 ≠ 4
hidden-single-in-a-column ==> r2c2 = 4
whip[1]: r5n4{c6 .} ==> r6c5 ≠ 4
stte

[Mauriès Robert]

Hi Denis,
I agree with you, we can solve this with short, easy to spot channels. By hand, I'm at 25 successive steps with anti-tracks of 7 sequences at the most,but these chains only produce one or two eliminations at a time and I was about to give up..., but seeing that your resolution has more than 40, I'm going to continue my work and publish it.
Robert

[Mauriès Robert]

Hi all,
My step-by-step resolution by hand, using short anti-tracks (7 sequences at most). Some of the steps should be able to be removed, but I didn't do this optimization work, the task of solving this by hand is tedious enough as it is.
If you take longer sequences, up to 10 for example, the number of steps will be significantly reduced.
This resolution should be compared to Denis Berthier's without being compared, because here the subsets are used (TB) and the notion of target is not used.

resolution: Show

1) P'(2r12c5) : (-2r12c5) => 16r1c2c5->(75r12c3->3r3c3)->4r5c6->4r8c5->7r9c5 => -2r9c5 => r9c4=2.
2) P'(2r3c2) : (-2r3c2) => 2r3c7->2r5c8 => -2r5c2 => r5c8=2 (Skyscraper)
3) P'(1r12c6) : (-1r12c6) => 1r7c6->1r2c7 => -1r2c45.
4) P'(4r7c6) : (-4r7c6) => 4r8c5->4r4c1 => -4r4c6.
5) P'(4r5c6) : (-4r5c6) => 4r7c6->1r12c6->26r12c5->6r5c9 => -4r5c9.
6) P'(6r5c9) : (-6r5c9) => (6r5c5->2r2c5)->1r1c5->1r9c9 => -6r9c9.
7) P'(3r7c9) : (-3r7c9) => 3r8c8->9r8c8->(6r9c2 & 6r12c8)->6r3c4->3r3c6 => -3r7c6.
8) P'(3r7c4) : (-3r7c4) => 3r7c9->3r6c8 => -3r6c4.
9) P'(8r8c7) : (-8r8c7) => 5r8c7->1r9c9->1r2c7->2r3c7 => -8r23c7.
10 P'(1r2c7) : (-1r2c7) => [(1r7c1 & 1r1c9)->26r12c5->6r6c9]->(6r9c8->26r23c7) => -57r2c7.
11) P'(3r8c4) : (-3r8c4) => 9r8c4->9r9c8->(6r9c2 & 6r12c8) ->6r3c4 => -3r3c4 => r3c6=3.
12) P'(5r23c4) : (-5r23c4) => 57r12c6->1r1c5->1r6c4 => -5r6c4 =>139r678c4 => -39r4c4, -3r8c5.
13) P'(3r4c2) : (-3r4c2) => 3r4c9->[(4r6c9 & 3r7c6)->1r6c4]->3r6c5 & 9r4c6 => -3r6c2, -9r4c2.
14) P'(5r9c9) : (-5r9c9) => 1r9c9->7r9c5-4r8c5->4r4c1->4r6c9 = -5r6c9.
15) P'(9r4c6) : (-9r4c6) => 5r4c6->4r5c6->4r8c5->4r4c1 => -9r4c1.
16) P'(9r4c6) : (-9r4c6) => 9r4c3->2r4c2->3r4c9->3r7c4->1r6c4 => -9r6c4 => r6c4=1, r4c6=9.
17) P'(4r6c9) : (-4r6c9) => 4r4c9->[(3r4c2 & 4r8c1)->4r7c6->5r5c6]->4r5c2 => -4r6c23.
18) P'(6r12c8) : (-6r12c8) => 6r9c8->9r8c8->(3r6c8->4r6c5->5r5c6)->6r6c9 => -6r12c9.
19) P'(2r3c7) : (-2r3c7) => [(2r3c2 & 2r2c7)->1r2c6->4r7c6->5r5c5->6r4c4]->6r3c7 => -57r3c7 => -7r46c9.
20) P'(5r12c8) : (-5r12c8) => [(86r12C8->6r9c2)->6r3c4->6r5c5->3r6c5]->5r8c8 => -5r89c8.
21) P'(8r12c8) : (-8r12c8) => [(86r12C8->6r9c2)->6r3c4->6r5c5->3r6c5]->8r8c8 => -8r8c8 => -9r8c13
22) P'(8r7c2) : (-8r7c2) => 8r8c1->4r4c1->4r6c9->3r6c5->3r8c8->9r9c8->6r9c2 = > -6r7c2.
23) P'(5r8c3) : (-5r8c3) => 47r8c35->4r4c1->4r6c9->3r6c5->3r8c8->9r9c8->6r9c2 => -5r9c2.
24) P'(1r1c9) : (-1r1c9) => 1r2c7->2r3c7->(6r12c8->6r9c2)->(6r3c4->2r2c5)->1r1c5 => -1r1c6.
25) P'(6r12c5) : (-6r12c5) => 2r2c5->1r1c5->1r2c7->(2r3c7->6r12c8->6r9c2)->6r3c4 => 6r2c4.
26) P'(9r1c1) : (-9r1c1) => 9r9c1->5r9c9->1r9c5->1r2c6->7r1c6 => -7r1c1.
27) P'(2r2c5) : (-2r2c5) => [(6r2c5 & 2r1c5)->1r1c9]->2r2c7 => -2r2c34.
28) P'(6r2c57) : (-2r2c5) => 2r2c5->2r3c7->6r12c8->6r9c2 => -6r2c2.
29) P'(5r2c468) : (-5r2c468) => [[[8r2c4->(6r2c8->6r9c2)->6r3c4]->5r1c6->]5r3c9]->5r5c2 => -5r2c2.
30) P'(8r2c4) : (-8r2c4) => 5r2c4->7r1c6->7r3c9->47r2c34 => -8r2c2.
31) P'(6r9c2) : (-6r9c2) => 6r9c8->9r8c8->3r6c8->4r6c5->4r5c2->7r2c2 => -7r9c2 => -9r9c9 => r1c1=9.
32) P'(7r9c1) : (-7r9c1) => 7r9c5->4r8c5->4r4c1 => -7r4c1.
33) P'(5r89c1) : (-5r89c1) => 5r8c3->8r8c7->8r3c1 => -5r3c1.
34) P'(8r3c1) : (-8r3c1) => 7r3c1->7r1c9->5r1c6->8r2c4 => -8r3c4 => r2c4=8.
35) P'(4r2c2) : (-4r2c2) => 7r2c2->8r3c1->8r7c2 => -4r7c2.
36) P'(9r9c2) : (-9r9c2) => (6r9c2->6r12c8)->6r3c4->6r5c5->3r6c3->6r8c5->4r7c3 => -9r7c3 => r6c3=9.
37) P'(4r4c1) : (-4r4c1) => 5r4c1->7r6c2->4r2c2 => -4r45c2 => 4r5c56 => r6c5=3, stte.

Robert

[eleven]

Denis and Robert,

what in your opinion is the nicest step in your solution paths ?

[denis_berthier]

Denis and Robert,
what in your opinion is the nicest step in your solution paths ?

A good question. I'll be honest. There's no nicest step in my solution (or in any solution I've seen in this thread). In such puzzles requiring many hard eliminations, each step is hard to find but it has nothing remarkable.
I tried alternative solutions with braids, with g-whips and even with g-braids. If an elimination by a g-whip or a braid or a g-braid had led to a solution with related rating 6 instead of 7, I might have said it was the nicest step. g-whips and braids do appear in the solutions, but the ratings remain 7.
Another type of nicest step would've been a large Subset (or another pattern) that lowers the W rating, but there's none.
One other way there might have been a nicest step is if a different resolution path with whips had had a single whip[7] (or say two)
and all the rest would have been easier. But Robert failed to find one and I didn't try.

Most of the puzzles one encounters in reality require only very few eliminations (other than SSTS). It's easier to find one or two key steps.

Now, if you look at most of the puzzles proposed in this section of the forum in the past years, they are totally uninteresting for an opposed reason: they can be solved by bivalue-chains of length 3 or 4 (rarely more). People spend much time trying to find one-step solutions (I have no objection if they have fun with this - but the result is artificially complicated patterns). Compared to such puzzles, the more difficult recent ones are more interesting, because they require a little more neuron firings, but this is only relative.

Finding interesting puzzles for a human player is hard. This is one thing I've never been involved in and I don't have much to say about it. I would say a good criterion is, it requires using a variety of different patterns, though I would exclude artificial pattern combinations.

Let me ask you the same question: did you find any interesting step in this puzzle?