The New Sudoku Players' Forum

[denis_berthier]

The full solving path (with the complete forcing nets):
[...]
Forcing Net |
r1c1=3 --> r1c1=3 r3c1=5 r5c1=4 r3c6=3 r8c5=3 r7c3=3 r6c4=3 r4c7=3 r6c6=4 r3c7=1 r6c5=9 r4c3=2 r6c9=1 r4c1=1 r4c2=9 r9c3=1 r8c6=1 r9c5=6 r1c5=7 r2c4=9 r2c9=7 r3c8=6 r4c5=5 r4c8=7 r5c6=7 r6c7=5 r7c6=9 r7c9=2 r8c9=9 r9c4=4 r9c8=5 r1c6=6 r2c5=1 r2c6=5 r2c7=4 r3c2=7 r5c3=5 r6c3=7 r7c4=7 r7c7=6 r8c1=2 r8c7=7 r8c8=4 r9c2=void => (-3)r1c1.
[...]
r1c8=9 --> r1c8=9 r2c9=7 r2c4=9 r9c4=4 r9c8=5 r4c8=7 r6c7=5 r7c4=7 r7c7=6 r8c7=7 r8c8=4 r9c2=6 r9c5=9 r4c5=5 r6c5=7 r8c2=void => (-9)r1c8.
[...]

I tried to eliminate the same two candidates as you.
Starting non trivial part of solution with the following RESOLUTION STATE:

Code: Select all

   34        1         3467      8         3679      3679      2         4679      5         
   8         2         4567      79        15679     15679     1467      3         179       
   35        567       9         2         4         13567     167       67        8         
   1259      579       1257      6         3579      8         1357      57        4         
   45        3         457       1         2         457       9         8         6         
   6         8         1457      3479      3579      34579     1357      2         17       
   23459     4569      23456     3479      8         34679     4567      1         279       
   12349     469       8         5         13679     134679    467       4679      279       
   7         4569      1456      49        169       2         8         4569      3

For the first elimination, I find a long whip, but much shorter than your T&E procedure:
whip[23]: r3n3{c1 c6} - r8n3{c6 c5} - r4n3{c5 c7} - r6n3{c7 c4} - r7n3{c4 c3} - c3n2{r7 r4} - r4n1{c3 c1} - r4n9{c1 c2} - c2n7{r4 r3} - r3c8{n7 n6} - r3c7{n6 n1} - c9n1{r2 r6} - c3n1{r6 r9} - c5n1{r9 r2} - b2n5{r2c5 r2c6} - r3n5{c6 c1} - r5c1{n5 n4} - r5c6{n4 n7} - c5n7{r6 r1} - r2c4{n7 n9} - r1n9{c6 c8} - r9n9{c8 c5} - c5n6{r9 .} ==> r1c1 ≠ 3
leading (after more Singles) to:

Code: Select all

   4         1         6         8         379       379       2         79        5         
   8         2         5         79        1679      1679      4         3         79       
   3         7         9         2         4         5         1         6         8         
   1         9         2         6         57        8         3         57        4         
   5         3         47        1         2         47        9         8         6         
   6         8         47        3         579       479       57        2         1         
   29        456       3         479       8         679       567       1         279       
   29        46        8         5         13679     13679     67        479       279       
   7         456       1         49        69        2         8         459       3

For the second, it's much shorter also:
whip[6]: r2c9{n9 n7} - c4n7{r2 r7} - r7n4{c4 c2} - r8n4{c2 c8} - r9c8{n4 n5} - c2n5{r9 .} ==> r1c8 ≠ 9
stte

There are generally lots of anti-backdoor pairs. I didn't try different pairs; but it is very likely that there are some that allow a solution with only two shorter whips. If you want to play with the idea, here are the 56 anti-backdoor pairs (with stte), among the 16836 candidate pairs.

Code: Select all

n4r9c4, n5r3c1          n4r8c8, n5r3c1          n4r5c1, n5r9c8          n4r5c1, n4r9c4          n4r5c1, n4r8c8          n4r5c1, n7r8c7          n4r5c1, n6r8c2          n4r5c1, n6r7c7          n4r5c1, n7r7c4          n4r5c1, n5r7c2          n4r5c1, n5r6c7          n4r5c1, n7r4c8          n4r5c1, n5r4c5          n4r5c1, n7r2c9          n4r5c1, n9r2c4          n4r5c1, n9r1c8          n3r3c6, n5r9c8          n3r3c6, n4r9c4          n3r3c6, n4r8c8          n3r3c6, n7r8c7          n3r3c6, n6r8c2          n3r3c6, n6r7c7          n3r3c6, n7r7c4          n3r3c6, n5r7c2          n3r3c6, n5r6c7          n3r3c6, n7r4c8          n3r3c6, n5r4c5          n3r3c6, n7r2c9          n3r3c6, n9r2c4          n3r3c6, n9r1c8          n5r3c1, n5r9c8          n5r3c1, n7r8c7          n5r3c1, n6r8c2          n5r3c1, n6r7c7          n5r3c1, n7r7c4          n5r3c1, n5r7c2          n5r3c1, n5r6c7          n5r3c1, n7r4c8          n5r3c1, n5r4c5          n5r3c1, n7r2c9          n5r3c1, n9r2c4          n5r3c1, n9r1c8          n3r1c1, n5r9c8          n3r1c1, n4r9c4          n3r1c1, n4r8c8          n3r1c1, n7r8c7          n3r1c1, n6r8c2          n3r1c1, n6r7c7          n3r1c1, n7r7c4          n3r1c1, n5r7c2          n3r1c1, n5r6c7          n3r1c1, n7r4c8          n3r1c1, n5r4c5          n3r1c1, n7r2c9          n3r1c1, n9r2c4          n3r1c1, n9r1c8         

Notice that the last one is your pair.

Remarks:
1) I don't think the name Forcing-net is correct for your eliminations. As I see them, they are mere T&E. A forcing net starts with a bivalue (or trivalue) pair; you start from a single candidate. I know names are very fluctuating.
2) this puzzle illustrates what I said earlier: the 1-step or 2-step constraint can lead to totally absurd solutions, such as using a whip[26] when a solution with short bivalue-chains is available.

Helàs, no one-step solution

I checked: there are no BRT-, W1- or S- anti-backdoors.

[denis_berthier]

.
ALL the 56 anti-backdoor pairs give rise to a two-step solution !!!!!

I found a pair with shorter whips than above:

whip[19]: r1c1{n4 n3} - r3n3{c1 c6} - r8n3{c6 c5} - r4n3{c5 c7} - r6n3{c7 c4} - r7n3{c4 c3} - c3n2{r7 r4} - r4n1{c3 c1} - c3n1{r6 r9} - c5n1{r9 r2} - b2n5{r2c5 r2c6} - r5c6{n5 n7} - c5n7{r6 r1} - c5n6{r1 r9} - c6n6{r8 r1} - r1n9{c6 c8} - r2n9{c9 c4} - r9n9{c4 c2} - r4n9{c2 .} ==> r5c1 ≠ 4
singles and one whip[1]
whip[4]: r7n4{c4 c2} - r8c2{n4 n6} - c7n6{r8 r7} - r7n5{c7 .} ==> r9c4 ≠ 4
stte

[denis_berthier]

.
Questions to yzfwsf

What is the "one trick" this puzzle was supposed to have ??????
How does your solver solve it?

[Hajime]

1) I don't think the name Forcing-net is correct for your eliminations. As I see them, they are mere T&E.

I agree again. This is the way I implemented it in SiSeSuSo.
But is is conform the definition in http://sudopedia.enjoysudoku.com/Forcing_Net.html

[DEFISE]

W-rating = 3 and here is a solution with 2 whips[3]:

Singles: 8r5c8, 8r6c2, 8r9c7, 8r1c4, 2r1c7, 2r3c4, 8r3c9
Alignment: 9r6b5 => -9r4c5
Alignment: 4c2b7 => -4r7c1 -4r7c3 -4r8c1 -4r9c3
Triplet: 345c1r135 => -5r4c1 -3r7c1 -5r7c1 -3r8c1
Singles: 3r7c3, 2r4c3, 3r6c4, 3r4c7, 1r4c1, 9r4c2, 7r3c2, 6r3c8, 1r3c7, 1r6c9, 1r9c3
Alignment: 4c4b8 => -4r7c6 -4r8c6
Naked pair: 79r2c49 => -7r2c5 -9r2c5 -7r2c6 -9r2c6 -7r2c7
Single: 4r2c7
Hidden pair: 13r8c56 => -6r8c5 -7r8c5 -9r8c5 -6r8c6 -7r8c6 -9r8c6
Alignment: 7r8b9 => -7r7c7 -7r7c9
Naked pair: 29r7c19 => -9r7c4 -9r7c6
Alignment: 9b8r9 => -9r9c8

whip[3]: r8c2{n6 n4}- c8n4{r8 r9}- r9n5{c8 .} => -6r9c2
Singles: 6r9c5, 7r7c6, 4r7c4, 9r9c4, 7r2c4, 9r2c9, 7r1c8, 5r4c8, 7r4c5, 7r6c7, 2r7c9, 9r7c1, 2r8c1, 6r8c7, 5r7c7, 6r7c2, 4r8c2, 9r8c8, 7r8c9, 5r9c2, 4r9c8, 7r5c3
Xwing: 5r35c16 => -5r2c6 -5r6c6

whip[3]: r1c5{n3 n9}- r6c5{n9 n5}- b2n5{r2c5 .} => -3r3c6
STTE

[DEFISE]

Sorry,
I must not use the triplet as a basic technique, if we want to limit ourselves to 2 steps of level >= 3.
So I can't find better than Denis: a whip [19] + a whip [4]

[AnotherLife]

.
What is the "one trick" this puzzle was supposed to have ??????

I do not understand what this long discussion is all about.
1. If we allow 'lcls' steps, as Cenoman said above, then there is a one-step solution with a forcing net, as I showed above.
2. If we restrict ourselves to two steps except for singles, then this puzzle is solvable with a naked triple and a forcing net.
3. If we want a one-step solution after Singles Only then we might never find it.

EDIT
4. If we allow 'lcls' steps and two other steps, then we can solve it in a very simple way with a UR and a XY-Wing, as I showed in my first post.

[denis_berthier]

1) I don't think the name Forcing-net is correct for your eliminations. As I see them, they are mere T&E.

I agree again. This is the way I implemented it in SiSeSuSo.
But is is conform the definition in http://sudopedia.enjoysudoku.com/Forcing_Net.html

In Sudopedia, the only difference between Forcing-net and T&E is, a Forcing-net is restricted to start from a candidate in a bivalue cell - a totally absurd restriction if the branch starting from the other candidate is not used.
Sudopedia has not been updated for a long time and it's totally outdated on some topics.