The New Sudoku Players' Forum

[denis_berthier]

Now I can guarantee to generate puzzles of SE9.1 ~ 9.3 in 1 minute, generally no more than 30 seconds.

That's great !

[yzfwsf]

If you have similar collections for skfr = 9.2 and skfr = 9.4, I'm curious to see how the proportion of puzzles in T&E(1) varies.

Please download 3 attachments, remove '. 7z' and unzip from '9+.7z.001'

9+.7z.001.7z

(256 KiB) Downloaded 138 times

9+.7z.002.7z

(256 KiB) Downloaded 119 times

9+.7z.003.7z

(145.87 KiB) Downloaded 132 times

[denis_berthier]

If you have similar collections for skfr = 9.2 and skfr = 9.4, I'm curious to see how the proportion of puzzles in T&E(1) varies.

Please download 3 attachments, remove '. 7z' and unzip from '9+.7z.001'

9+.7z.001.7z

9+.7z.002.7z

9+.7z.003.7z

Hi yzfwsf
Thanks for this new collection of 30,000 puzzles.
The analysis shows that
- 29918 are in T&E(1),
- 73 in gT&E(1) = T&E(B1, 1)
- the remaining 9 in B2B = T&R(B2, 1)

This tends to confirm the impression I had after analysing the first collection: a much larger proportion of them are in T&E(1) than usually appears with SER > 9.1 puzzles.
We have the puzzling result that your generator finds puzzles with higher SER than the top-down ones, but most of them remain in T&E(1) even though their SER would suggest they are beyond.

There appears some obscure point: the relationship between skrf and SER. I don't use skfr, but if I remember well, it was supposed to be a faster version of SER. But I see that many puzzles in both collections with the same skfr happen to have largely different SER.

[yzfwsf]

Hi denis_berthier:

After the test run, skfr should very often rate in that area .1 or .2 below SE.
The reason is that SE fails very often in looking for the shortest path toward it's own rule.

I have however the following problem that could lead sometimes to a +0.1 in skfr rating.

In the dynamic process, you have many ways to establish a false for a candidate.
Depending on the order in which you develop the chains, it can be that the best possibility toward SE final count is not seen when I start the reverse search.

The test run in the "dynamic" area has moreover 4 goals:

a) check that the process works properly
b) check that the count is done strictly following SE rules
c) check that lower rating in skfr are justified
d) explain higher rating in skfr

I think a) is ok now up to level 2 and I start test for level 3.
All other items are still requiring comparative tests that need a lot of hand work.
At the end, d) should be the exception.
champagne

I know this may open a can of worms but I thought that with the work on the Sukaku explainer (which doubles up as sudoku explainer as well), there could be a scope for SE ratings / Resolution rule order / Resolution rule addition/deletion.

The current rules & ratings (which occasionally) overlap are:

Code: Select all

1.0: Last value in block, row or column
1.2: Hidden Single in block
1.5: Hidden Single in row or column
1.7: Direct Pointing
1.9: Direct Claiming
2.0: Direct Hidden Pair
2.3: Naked Single
2.5: Direct Hidden Triplet
2.6: Pointing
2.8: Claiming
3.0, 3.2, 3.4: Naked Pair, X-Wing, Hidden Pair
3.6, 3.8, 4.0: Naked Triplet, Swordfish, Hidden Triplet
4.2, 4.4: XY-Wing, XYZ-Wing
4.5 - 5.0: Unique rectangles and loops
5.0, 5.2, 5.4: Naked Quad, Jellyfish, Hidden Quad
5.6 - 6.0: Bivalue Universal Graves
6.2: Aligned Pair Exclusion
6.5 - 7.5: Bidirectioal X-Cycles and Y-Cycles
6.6 - 7.6: Forcing X-Chains
7.0 - 8.0: Forcing Chains, Bidirectional Cycles
7.5 - 8.5: Nishio
8.0 - 9.0: Cell/Region Forcing Chains
8.5 - 9.5: Dynamic Forcing Chains
9.0 - 10.0: Dynamic Forcing Chains (+)
> 9.5: Nested Forcing Chains

I'll start by 1st suggesting a change towards the lower end of the ratings as follows:

Code: Select all

1.2: Hidden single (1.5 if not in block)
1.6: Naked single//2.3 ---> 1.6
1.7: Direct Pointing
1.9: Direct Claiming
2.0: Direct Hidden Pair
2.6: Pointing
2.8: Claiming
2.9: Hidden pair//3.4 ---> 2.9
3.0: Direct Hidden Triplet//2.5 ---> 3.0
3.1, 3.2: Naked pair, X-Wing//3.0 ---> 3.1
3.6, 3.8, 4.0: Naked triplet, Hidden triplet, Swordfish//3.8 ---> 4.0 4.0 ---> 3.8
4.2, 4.4: XY-Wing, XYZ-Wing
4.5 - 5.0: Unique Rectangles and Loops
5.0, 5.2, 5.4: Naked quad, Hidden quad, Jellyfish//5.2 ---> 5.4 5.4 ---> 5.2
5.6 - 6.0: Bivalue Universal Graves
6.2: Aligned Pair Exclusion
6.5 - 7.5: X-Cycles, Y-Cycles
6.6 - 7.6: Forcing X-Chains
7.0 - 8.0: Forcing Chains, XY-Cycles
7.5: Aligned Triplet Exclusion
7.5 - 8.5: Nishio
8.0 - 9.0: Multiple chains
8.5 - 9.5: Dynamic chains
9.0 - 10.0: Dynamic chains (+)
>9.5: Nested Forcing Chains

One thing I was considering is increasing the space between the current ratings through multiplying all the current ratings by 10 … But then: should we allow overlapping as it is happening now or not? The increase in space can allow adding more techniques in the future if needed …
tarek

I think ser9.1 ~ 9.3 fall into T & E (1), which should be a normal phenomenon, because as mentioned above,T & E (1) and T & E (2) overlap in 9.0 ~ 9.5 in SudokuExplainer

BTW：Few 10+ puzzle for you

Hidden Text: Show

2.4...8...5...8.6....2....96..3...1..4..5.6....7..4..54...2......31...9..6...37.. 104/12/12
.2...76....52....78..36....7..9..8..9.2....7..8...1..9.1..5..4.4.8..31....6.....8 102/12/12
4..6.......7..94...9..7.....3..2...4..87....22....5.6.5..8..1....1.4..3..8...1.4. 102/102/96
..2....4.59....7...3..6...5.4...1..3...6....7....4.9..1...9.5...6...5..1..4.3..9. 105/12/12
.2..9......57.......9..6.3.....8.2.6.1.4....7..8..1....6...9.4...1.4...58..5..9.. 104/12/12
.....6..1..5.7....9..4...2..2..8.3..1.......8..6..5.4.....9...4..2..7.6..3.8..7.. 104/104/92
6....92....9.5...654.6...1.2..5..6....1.4..9..6.......1.....9...7.8...4...4.6...7 103/12/12
..5..6.7.47.9.....9...7.1..1..4...2..2...3.....8.1...32...3..9...68..3.......4..5 102/12/12
8....9....7.2.8.....1.6..8..4.3....91......7...3.1.2...5...2.6...2...4..6...9...3 101/12/12
.1...9..86..4..3....5.7..2.1....6.8..6..9.1....83....6..1..4..3.2.9..8..7...8..5. 101/12/12
4....95...6.3....9..9..4.2.1...4.....4...1..7..28...1..3.1....25...2.7....8..3.6. 104/104/78
..29...3..3..6.4..5....2..93..8....7..1.7..8..7....6..4..6......2...3........9.15 105/105/26
.1.....7.5...2...8.....93....4..5..72..8..4...8.....9..2..7...5..73...1.6....4... 104/12/12
.....74......1...21..2...8..5..2.9..3....4..56.87.....5...9...3.3.6..2....4....9. 104/104/84
.2......1..86...5.9.....8....79...8..4..3.9..8....1...6..3...7..5...42....3.5.... 104/104/34
....6...95....34...7.8...2...3.5.2...2......11....4.6.6...7.1...1.9....2..2..6.7. 103/12/12
..5.3...8.....65..3..8..21......8.7.9..5....1.4..7.6...6.41..2....6....9..1..7... 102/12/12
.2...7..3..82..6..4...6.....1...2.5.8..1.......2.4.3....48....2..7.5..9.9.....4.. 102/12/12
6.....8....3.2..5..9...6..71..7....9.....45....2.8..7..3.5..7....6.3..4.5....1..3 102/12/12
.1...9.8...58....4....3.7..1..5...6...8.2.....7...15..4..........3.8...2.6.7...9. 102/12/12
21.3....7.....26....6.7..1..6..8.7....57....37....4.6.9.....8....3.....4.7.5...9. 101/12/12
.3.6....2....31.9..8..9....7..9....3.1...62...45.7....1...2.6.......3..7..27...19 101/12/12
..8..2..3..3.1..4..5.4..8.14.5..9.6..1.6.......6...3..3....64....192.....8......6 101/12/12
.....6...4..5.2.7..5......4....9.2..5.4.....97...2..6.2..7..9.......4..6.83....5. 101/12/12

[denis_berthier]

I think ser9.1 ~ 9.3 fall into T & E (1), which should be a normal phenomenon, because as mentioned above,T & E (1) and T & E (2) overlap in 9.0 ~ 9.5 in SudokuExplainer

And I have some doubts about any equivalence such as forcing-chains <=> T&E(1) or nested-forcing-chains <=> T&E(2)

[m_b_metcalf]

AFAIK, there's been no study on any correlation between maximum number of clues per house and difficulty.

The 3 clue per house makes it more likely that boxes would display a "Diagonal pattern of clues": Which essentially can be mapped on an "X". Puzzles with boxes displaying these diagonal patterns are associated with higher SE difficulty rating.

Out of curiosity, I generated random grids and then took out all clues not on mini-diagonals, thus:

Code: Select all

 7 . 3 2 . 8 4 . 1
 . 2 . . 6 . . 3 .
 4 . 1 7 . 3 2 . 8
 2 . 6 5 . 7 1 . 9
 . 7 . . 1 . . 2 .
 8 . 4 9 . 2 6 . 5
 3 . 2 1 . 4 9 . 6
 . 9 . . 2 . . 4 .
 6 . 5 3 . 9 7 . 2

I then removed further clues at random until the result was minimal and kept only those passing an internal filter that corresponds to ~SE8.4. This I repeated for 10 minutes and applied SE to the 52 puzzles generated. Of these, 27 were >=SE 9.0, examples being

Code: Select all

 . . 1 2 . 3 4 . 5
 . . . . 1 . . 3 .
 . . . 6 . . . . .
 6 . 5 7 . . . . 8
 . 1 . . 8 . . 2 .
 . . 2 . . . 9 . .
 . . . 3 . . . . .
 . . . . . . . 9 .
 3 . 7 . . 4 1 . 6   ED=9.0/2.3/2.3

 . . 1 2 . 3 . . .
 . 4 . . 5 . . 3 .
 . . 6 . . 7 . . .
 1 . . . . . 8 . .
 . 8 . . 2 . . 4 .
 . . 9 . . 8 3 . .
 2 . . 6 . . 4 . 5
 . . . . . . . 2 .
 6 . 5 . . . . . 7   ED=9.1/1.2/1.2

 1 . . . . 2 3 . .
 . 4 . . 5 . . . .
 . . 6 7 . 1 4 . .
 3 . . . . 5 2 . .
 . . . . 7 . . 3 .
 7 . 8 2 . . . . 6
 . . 2 . . 9 . . 8
 . . . . 2 . . 9 .
 . . 5 1 . . 6 . .   ED=9.2/9.2/6.7

Using biased patterns does seem to be an interesting approach.

Regards,

Mike

[yzfwsf]

Out of curiosity, I generated random grids and then took out all clues not on mini-diagonals, thus:

Code: Select all

 7 . 3 2 . 8 4 . 1
 . 2 . . 6 . . 3 .
 4 . 1 7 . 3 2 . 8
 2 . 6 5 . 7 1 . 9
 . 7 . . 1 . . 2 .
 8 . 4 9 . 2 6 . 5
 3 . 2 1 . 4 9 . 6
 . 9 . . 2 . . 4 .
 6 . 5 3 . 9 7 . 2

I'm curious how you got this non minimal puzzle and how much time it took?

[denis_berthier]

Out of curiosity, I generated random grids and then took out all clues not on mini-diagonals, thus:

Code: Select all

 7 . 3 2 . 8 4 . 1
 . 2 . . 6 . . 3 .
 4 . 1 7 . 3 2 . 8
 2 . 6 5 . 7 1 . 9
 . 7 . . 1 . . 2 .
 8 . 4 9 . 2 6 . 5
 3 . 2 1 . 4 9 . 6
 . 9 . . 2 . . 4 .
 6 . 5 3 . 9 7 . 2

I'm curious how you got this non minimal puzzle and how much time it took?

When you take any complete grid and you randomly delete any set of 33 values as here, that leaves 48 givens. The probability that the resulting puzzle has more than one solution (Ie. any solution other than the initial complete grid) is so close to 0 that it's impossible to compute. As a result, the time it took is equal to the time for generating a full grid.
[Edit]: this is true only for random deletions. In the present case, the deleted clues are in fixed cells. I would bet the result is true in the present case, but there are probably counter-examples with other specific deletion templates (such as if you delete a full band).

[m_b_metcalf]

Out of curiosity, I generated random grids and then took out all clues not on mini-diagonals, thus:

Code: Select all

 7 . 3 2 . 8 4 . 1
 . 2 . . 6 . . 3 .
 4 . 1 7 . 3 2 . 8
 2 . 6 5 . 7 1 . 9
 . 7 . . 1 . . 2 .
 8 . 4 9 . 2 6 . 5
 3 . 2 1 . 4 9 . 6
 . 9 . . 2 . . 4 .
 6 . 5 3 . 9 7 . 2

I'm curious how you got this non minimal puzzle and how much time it took?

When you take any complete grid and you randomly delete any set of 33 values as here, that leaves 48 givens. The probability that the resulting puzzle has more than one solution (Ie. any solution other than the initial complete grid) is so close to 0 that it's impossible to compute. As a result, the time it took is equal to the time for generating a full grid.
[Edit]: this is true only for random deletions. In the present case, the deleted clues are in fixed cells. I would bet the result is true in the present case, but there are probably counter-examples with other specific deletion templates (such as if you delete a full band).

This is rather curious. In order to answer the questions, I made two tests. The first was to measure how long it takes to generate a random grid. This turns out to be 28 microseconds. To delete 36 clues leaving 45 is trivial. But then I checked how many of the resulting puzzles have more than one solution and the answer was not, as Denis and I expected, close to 0, but 31%! I was very surprised. The reason is due to the regularity of the deletions. For the grid above, one of the two solutions is

Code: Select all

  7  6* 3  2  5* 8  4  9* 1
  9  2  8  4  6  1  5  3  7
  4  5* 1  7  9* 3  2  6* 8
  2  3  6  5  4  7  1  8  9
  5  7  9  8  1  6  3  2  4
  8  1  4  9  3  2  6  7  5
  3  8  2  1  7  4  9  5  6
  1  9  7  6  2  5  8  4  3
  6  4  5  3  8  9  7  1  2

where an uncovered UA6 is marked.

Regards,

Mike

[denis_berthier]

But then I checked how many of the resulting puzzles have more than one solution and the answer was not, as Denis and I expected, close to 0, but 31%! I was very surprised. The reason is due to the regularity of the deletions.

That's very interesting. I was almost certain that this pattern had nothing special wrt to uniqueness.

It may be interesting to understand if / how this is related to puzzles being harder.

[denis_berthier]

I have a pure pattern-based resolution path. I will give it later, after more people give this puzzle a try. It is worth it.

I have already made a resolution by hand, I hope to give it soon.

I'm still curious to see your findings

[Mauriès Robert]

Hi Denis,
I'm moving on, I'm moving on... but as I'm very busy , I'm not finished yet.
I'm at 26 steps from 6 to 20 sequences, because I don't want to use extension (OR) or direct contradictions.
As soon as I'm done, I'll put it online.
I can send you these steps by private message if you want to see what I've done already.
Thank you for your interest.
Robert

[denis_berthier]

I can send you these steps by private message if you want to see what I've done already.

I can wait for the complete answer. I was just wondering about how TDP would help you deal with this.

[Mauriès Robert]

Hi all,
Here is my resolution (by hand) with TDP.
In this resolution I have imposed on myself not to use the direct contradiction, nor the OR (bifurcation) condition.
Some sequences can be avoided and thus reduce the number of steps, but working by hand, I did not rethink the resolution to remove them.
With the OR (bifurcation) condition, solving this puzzle is obviously faster.
Robert

resolution: Show

Note that the different nested brackets "memorize" the elements of the sub-strings.

With two conjugated tracks:
1) P(9r2c3) : 9r2c3->2r7c3->(2r5c7 & 2r9c9)->34r7c79 and P(9r8c3) : [9r8c3->9r9c8->(8r8c8 & 9r9c7)]->9r2c1
=> -9r3c12, -9r2c8, -34r8c8, -4r9c8.
With successive anti-tracks:
2) P'(3r8c4) : (-3r8c4) => (3r8c7->89r89c8->5r9c9->2r9c1->2r2c3->9r8c3)->8r8c8 => -8r8c4.
3) E=7b1p238
P'(E) : (-E) => { 8r1c3->36r13c2->(3r6c1->9r6c2)->[(7r6c8->45r4c78) & (7r9c2->7r7c6->6r4c6)] }->8r9c6->8r7c1->7r4c1
=> -7r23c1.
4) E={1r4c3, 1r8c5, 9r8c3}
P'(E) : (-E) =>{ [9r2c3 & (1r4c4->1r9c5->7r79c6)]->7r2c8->7r4c7 }->48r4c13->1r5c3
=> -1r8c3 =>-1r56c2.
5) E={7r1c3, 1r4c3}
P'(E) : (-E) =>{ [(8r1c3 & 1r5c3)->(1r4c4->6r4c6)->8r5c5]->8r4c1 }->4r4c3 => -7r4c3.
6) E={9r3c7, 7r4c7}
P'(E) : (-E) => 9r8c7->9r2c3->[(2r7c3 & 7r1c3)->2r9c9->(5r89c8->37r26c8)->4r4c8->7r6c8->7r4c1]->7r7c6->7r3c5
=> -7r3c7.
7) E={8r1c3, 2r3c1}
P'(E) : (-E) => [(29r2c13 & 7r1c3)->7r4c7->79r6c12->3r5c2]->3r3c1 => -8r3c1.
8) P'(3r5c2) : (-3r5c2) =>8r5c2->(8r1c3->14r45c3)->9r8c3->9r2c1->9r6c2 => -3r6c2.
9) E={9r9c8, 5r4c8}
P'(E) : (-E) => { [(47r4c8 & 9r8c78)->(9r2c3->2r7c3)->7r1c3->7r4c7]->4r4c8->4r5c3 }->8r8c3->8r9c8 => -5r9c8
10) P'(7r1c37) : (-7r1c37) => { [(8r1c3->14r4c3->9r8c3->9r2c1) & (7r4c7->79r6c12->3r5c2)]->[(8r4c1->8r5c5->8r3c4) & 3r3c1]->2r3c5 }->4r1c5
=> -7r1c5.
11) P'(3r8c4) : (-3r8c4) => 3r8c7->5r9c9->2r9c1->4r9c45 => -4r8c4.
12) E={3r8c4, 3r2c4}
P'(E) : (-E) => [(3r8c7->5r9c9->2r9c1->2r2c3->5r2c4)->5r8c5->5r4c6->6r7c6]->3r7c4 => -3r3c4.
13) E={8r8c8, 5r26c8}
P'(E) : (-E) => { {5r2c46->48r1c5} & { 8r9c8->9r8c78->9r2c3->2r7c3->[ (2r9c9->5r8c78) & (7r1c3->7r4c7->79r6c12->3r5c2)->68r13c2 ] }->1r8c2->48r8c5 }->86r4c46->5r4c8
=> -5r8c8 => r3c7=9.
14) P'(8r8c8) : (-8r8c8) => 9r8c8->(9r2c3->2r7c2)->7r1c2->8r13c2 => -8r8c2.
15) E={7r1c3, 7r2c68}
P'(E) : (-E) => { [(2r2c4->2r3c1) & (8r1c3->14r45c3)->9r8c3->9r2c1->9r6c2)]->3r6c1->(8r5c2 & 7r4c1->7r1c7->7r3c5) }->8r3c4->8r4c6->6r7c6->7r9c6->7r7c3
=> -7r2c3.
16) P'(3r5c3) : (-3r5c2) => [(8r5c2->8r1c3->7r7c3->2r2c3->9r8c3)->1r9c2]->6r8c2->6r1c9->6r5c8 => -3r5c8.
17) P'(3r5c3) : (-3r5c2) => { [3r6c1->9r6c2->7r4c1->7r1c7] & [(8r5c2->8r1c3->7r7c3->2r2c3->9r8c3)->1r9c2]->6r8c2 }->3r1c2 => -3r3c2.
18) E={3r1c2, 7r1c3}
P'(E) : (-E) =>{ { { (3r5c2 & 8r1c3)->[(8r4c1->8r5c5) & (14r45c3->9r8c3->2r2c3) & (8r9c2->1r8c2) & 45r1c5] }->8r3c4->8r7c6->(6r4c6 & 45r8c5) }->5r46c4->3r2c4->3r8c7->5r8c5->4r1c5 }->75r1c67
=> -7r1c2.
19) E={3r8c4, 6r4c4}
P'(E) : (-E) => { [(3r8c7->5r9c9->2r9c1->2r2c3->9r8c3->9r2c1->9r6c2->7r46c1) & 6r4c6]->5r12c6->3r2c4->3r6c6->7r7c3->7r3c2->7r12c6->7r9c5 }->8r9c6->1r9c2->6r8c2
=> -6r8c4.
20) P'(6r8c1) : [(6r8c2->1r9c2)->9r6c2->7r46c1]->7r7c3->2r2c3->9r2c1 => -9r8c1.
21) E={6r8c1, 8r8c8}
P'(E) : (-E) => { [(9r8c8->9r2c3->2r7c3->34r7c79) & (6r8c2->6r3c1)]->(2r2c1->3r6c1)->9r6c2->7r4c1 }->8r7c1 => -8r8c1.
22) E={1r8c2, 8r13c2}
P'(E) : (-E) => { [(1r8c2->6r8c1) & (8r1c3->7r7c3->2r2c3->9r8c3->8r8c8)]->[(3r3c1 & 35r28c4 & 45r18c5)->5r4c6->47r4c78->6r5c8]->47r34c8->7r2c6 ->8r9c6
=> -8r9c2.
23) E={2r7c3, 8r1c3}
P'(E) : (-E) => { [(2r2c3->9r8c3->9r2c1->9r6c2) & 7r1c3]->[(7r9c2 & 1r8c2)->(7r7c6 & 6r4c6)->7r3c5] }->2r3c4 & 35r28c4->8r4c4->8r5c3
=> -8r7c3.
24) E={7r7c3, 4r7c79}
P'(E) : (-E) => { [7r1c3 & 45b9p49->(3r8c4 & 2r7c79->2r2c3->9r8c3->9r2c1->9r6c2) ]->7r9c2->7r7c6->6r4c6 }->[(5r2c4->5r6c5->5r8c7->4r9c9) & (7r3c4->2r3c4->4r1c5)]->4r7c4
=> -4r7c3.
25) P'(3r5c2) : (-3r5c2) => { { 8r5c2->(3r6c1 & 8r1c3)->[(7r3c2->9r6c2)->1r9c2]->6r8c2->(4r8c1 & 6r3c1) }->[(7r6c8 & 34r3c89)->5r2c8]->4r4c8->5r4c7 }->3r8c7
=> -3r5c7.
26) P'(2r5c7) : (-2r5c7) => 2r7c7->7r7c3->[8r1c3 & (7r3c2->9r6c2)->1r9c2->6r8c2-4r8c1]->7r4c1->45r4c78 => -4r5c7 => r5c7=2.
27) P'(5r8c7) : (-5r8c7) => (5r9c9->2r9c1->7r7c3->[(7r3c2->9r6c2)->1r9c2]->6r8c2->4r8c1 => -4r8c7.
28) P'(2r9c9) : { (-2r9c9) => 2r7c9->7r7c3->{ [(8r1c3 & 7r3c2)->9r6c2]->1r9c2->6r8c2->(4r8c1 & 6r1c9) }->7r4c1->3r6c1->3r5c9 }->4r3c9
=> -4r9c9 => -4r7c14.
29) E={8r1c3, 8r5c5}
P'(E) : (-E) => { { [7r1c3->(7r4c7 & 2r7c3->2r9c9->5r8c7)] & (1r5c5->8r4c46) }->4r4c1->4r8c3 }->8r8c5 => -8r1c5.
30) P'(4r9c4) : (-4r9c4) => 4r3c4->5r1c5->5r2c8->7r2c6->7r9c5 => -4r9c5.
31) P'(7r4c7) : (-7r4c7) => { 7r1c7->[(7r3c2 & 8r1c3->8r5c2)->9r6c2->1r9c2]->6r8c2->4r8c1 }->7r4c1 => -7r4c8.
32) P'(7r1c3) : (-7r1c3) => 8r1c3->8r5c2->(3r6c1->9r6c2)->7r6c8->7r1c7 => -7r1c6.
33) P'(7r1c7) : (-7r1c7) => 7r1c3->2r7c3->2r9c9->5r8c7 => -5r1c7.
34) P'(4r4c8) : (-4r4c8) => 5r4c8->5r1c9->6r1c2->3r5c2->79r6C12->7r4c7 => -4r4c7.
35) P'(2r7c3) : (-2r7c3) => 7r7c3->7r1c7->5r4c7->5r9c9->2r7c9 => -2r7c1
36) E={7r3c2, 46r3c8}
P'(E) : (-E) => { [ 7r1c3->7r4c7->( (79r6c12->3r5c2) & (5r8c5->3r8c4) ) ] & [6r5c8->4r4c8->4r5c3->1r4c3] }->{ 1r5c9 & [(8r8c3 & 68r13c2->6r8c1)->4r8c5->5r1c5->2r2c4->2r6c5] }->1r9c5->7r3c5
=> -7r3c8.
37) P'(1r6c9) : (-1r6c9) => 1r5c9->3r5c2->79r6c12->7r4c7->7r2c8->5r1c9 => -5r6c9
With two conjugated tracks:
38) P(6r1c2) : 6r1c2->3r5c2->79r6c12->7r4c7->7r2c8->5r1c9 and P(6r1c9) : 6r1c9->5r2c8->7r1c7->8r1c3->7r3c2
=> -34r1c9, -3r2c8, -6r3c2, -8r1c2
With successive anti-tracks:
39) P'(4r8c1) : (-4r8c1) => 6r8c1->6r1c2->5r1c9->4r1c5->4r9c4 => -4r9c1 => r9c4=4.
40) P'(7r3c5) : (-7r3c5) => { [7r3c2->(8r1c3 & 9r6c2)]->1r9c2 }->6r8c2->3r1c2->5r1c6->4r1c5 => -4r3c5 => r1c5=4 & r7c7=4.
With two conjugated tracks:
41) P(5r1c6) : 5r1c6->8r1c3->7r1c7->5r4c7 and P(5r1c9) : 5r1c9->6r1c2->6r8c1->4r4c1->5r4c8->7r4c7->3r1c7
=> -5r6c8 & -5r4c46 => r6c9=1 and -3r1c6 => -3r2c1, -2r2c4, -2r3c1.
With anti-track:
42) P'(6r7c4) : (-6r7c4) => 6r7c6->3r2c6->5r2c4->5r1c9->5r8c7->3r8c4 => -3r7c4.
With two conjugated tracks and end:
43) P(3r5c2) : 3r5c2->8r3c2->2r3c4->2r6c5 and P(8r5c2) : 8r5c2->1r5c5->(1r8c4->3r2c4)->5r6c4->2r6c5
=> r6c5=2 => solution with singles.

[Ajò Dimonios]

Hi Robert.

Beautiful resolution of a very difficult scheme.
Paolo