The New Sudoku Players' Forum

[Draco]

It's no Easter Monster, but something to perhaps pique your interest. Here's the base puzzle with PM's after SSTS:

Code: Select all

.1..27..5..7....3.4.85..6.7..23....6....8....1....29..7.6..32.1.5....7..2..74..6.

369 1     39   | 4689  2    7     | 48   489   5   
5   269   7    | 14689 169  14689 | 148  3     2489
4   29    8    | 5     3    19    | 6    129   7   
---------------+------------------+----------------
89  4789  2    | 3     1579 1459  | 1458 14578 6   
369 34679 3459 | 1469  8    14569 | 1345 2457  234
1   34678 345  | 46    567  2     | 9    4578  348
---------------+------------------+----------------
7   489   6    | 89    59   3     | 2    4589  1   
389 5     1349 | 2     169  1689  | 7    489   3489
2   389   139  | 7     4    1589  | 358  6     389

I've found a 6-step contradiction net that cracks the puzzle to SSTS to solve. I'll post that (and, if people really want to see them, a list of SSTS backdoors) if things stall. Past experience tells me that there will be plenty of solutions without my doing that .

Cheers...

- drac

my ref: Extreme 29 (fc>=5) #1575000049 (5 sec)

[DonM]

Draco- thanks for the puzzle. Am also glad to see that the backdoors won't make an appearance, or at least early appearance. Fwiw: I don't know how the other manual solvers feel about it, but for me, if I work on a manual solution and then happen to see a backdoor cell mentioned, it takes away absolutely any incentive to continue my solution, not to mention the fact that I probably won't even start one if I think a backdoor is going to be presented while solving is going on. IMO, computer solver derived backdoors and manual solving don't mix very well at all. Maybe it's just me though.

[ttt]

Code: Select all

 *--------------------------------------------------------------------*
 | 369    1      39     | 4689   2      7      | 48     489    5      |
 | 5      269    7      | 14689  169    14689  | 148    3      2489   |
 | 4      29     8      | 5      3      19     | 6      129    7      |
 |----------------------+----------------------+----------------------|
 | 89     4789   2      | 3      1579   1459   | 1458   14578  6      |
 | 369    34679  3459   | 1469   8      14569  | 1345   12457  234    |
 | 1      34678  345    | 46     567    2      | 9      4578   348    |
 |----------------------+----------------------+----------------------|
 | 7      489    6      | 89     59     3      | 2      4589   1      |
 | 389    5      1349   | 2      169    1689   | 7      489    3489   |
 | 2      389    139    | 7      4      1589   | 358    6      389    |
 *--------------------------------------------------------------------*

My path start from:
r7c2-9-r7c4-8=r12c4=8=r2c6=4=r12c4=4-r6c4-6=r1c4=6=als:r1c13=9=r23c2-9-r7c2 => r7c2<>9
(9=8)r7c4-(8)r12c4=(8-4)r2c6=(4)r12c4-(4=6)r6c4-(6)r1c4=(6-np39)r1c13=(9)r23c2
After above move my path require one dual Kraken to downgrade this puzzle.

Other finding: Look AUR(48)r12c47 => (48)r2c6=(1)r2c7 => r2c6<>1
r2c6-6-AUR(48)r12c47:48-r2c6|1=r2c7=1=r3c8=1=r3c6-1=r89c6=1=r8c5=6=r8c6-6-r2c6 => r2c6<>6
(48)r2c6=(1)r2c7-(1)r3c8=(1)r3c6-(1)r89c6=(1-6)r8c5=(6)r8c6

Please correct me for NL notation, thanks.

ttt

[Steve K]

1) ss
2) (ht127)r345c8=(7-6)r5c2 =[fxw(6)r1c41,r5c146] -(6=4)r6c4-(4)r12c4=(4-8)r2c6=(8)r89c6-(8=9)r7c4-(9=5)r7c5-(5)r7c8=(489)r178c8 =>r345c8<>489
3) (9=2)r3c2-(2)r2c2=(2-9)r2c9=(9)r1c8 => r1c13<>9 simple sudoku to end

[storm_norm]

(48)r2c6=(1)r2c7-(1)r3c8=(1)r3c6-(1)r89c6=(1-6)r8c5=(6)r8c6

Please correct me for NL notation, thanks.

UR48[(48)r2c6=(1)r2c7]...

perhaps?

[storm_norm]

ttt said:

Other finding: Look AUR(48)r12c47 => (48)r2c6=(1)r2c7 => r2c6<>1
r2c6-6-AUR(48)r12c47:48-r2c6|1=r2c7=1=r3c8=1=r3c6-1=r89c6=1=r8c5=6=r8c6-6-r2c6 => r2c6<>6
(48)r2c6=(1)r2c7-(1)r3c8=(1)r3c6-(1)r89c6=(1-6)r8c5=(6)r8c6

ttt,
I have an alternative view of your elimination of 6 in r2c6

Code: Select all

*--------------------------------------------------------------------*
 | 369    1      39     | 4689   2      7      | 48     489    5      |
 | 5      269    7      | 14689  169   K14689  |D148    3      2489   |
 | 4      29     8      | 5      3     I19     | 6     H129    7      |
 |----------------------+----------------------+----------------------|
 | 89     4789   2      | 3      1579  G1459   | 1458   14578  6      |
 | 369    34679  3459   |A1469   8     L14569  | 1345   12457  234    |
 | 1      34678  345    |B46     567    2      | 9      4578   348    |
 |----------------------+----------------------+----------------------|
 | 7      489    6      |C89     59     3      | 2      4589   1      |
 | 389    5      1349   | 2     J169   E1689   | 7      489    3489   |
 | 2      389    139    | 7      4     F1589   | 358    6      389    |
 *--------------------------------------------------------------------*

the UR that you are using is in r12c47 on {4,8}
notice that if all these three are false, then the deadly pattern remains...
8 at C
4's at AB
1 at D
now you can form this net.

(8)C - (8)EF = (8)K
||
(4)AB - (4)GL = (4)K
||
(1)D- (1)H = (1)I -(1)EF = (1-6)J = (6)E; r2c6 <> 6

[ronk]

Please correct me for NL notation, thanks.
...
My path start from:
r7c2-9-r7c4-8=r12c4=8=r2c6=4=r12c4=4-r6c4-6=r1c4=6=als:r1c13=9=r23c2-9-r7c2 => r7c2<>9
(9=8)r7c4-(8)r12c4=(8-4)r2c6=(4)r12c4-(4=6)r6c4-(6)r1c4=(6-np39)r1c13=(9)r23c2
After above move my path require one dual Kraken to downgrade this puzzle.

Other finding: Look AUR(48)r12c47 => (48)r2c6=(1)r2c7 => r2c6<>1
r2c6-6-AUR(48)r12c47:48-r2c6|1=r2c7=1=r3c8=1=r3c6-1=r89c6=1=r8c5=6=r8c6-6-r2c6 => r2c6<>6
(48)r2c6=(1)r2c7-(1)r3c8=(1)r3c6-(1)r89c6=(1-6)r8c5=(6)r8c6

You're definitely getting closer. Not sure why, but I had to move the ALS in the first.
(9=8)r7c4-(8)r12c4=(8-4)r2c6=(4)r12c4-(4=6)r6c4-(6)r1c4=(6)r1c1-(6=np29)r23c2
r7c2-9-r7c4-8-r12c4=8=r2c6=4=r12c4-4-r6c4-6-r1c4=6=r1c1-6-als:r23c2-9-r7c2

In the second, if you're going to drop the AUR details from the AIC, then doing so for the NL seems fair.
(48)r2c6=(1)r2c7-(1)r3c8=(1)r3c6-(1)r89c6=(1-6)r8c5=(6)r8c6
r2c6=48|1=r2c7-1-r3c8=1=r3c6-1-r89c6=1=r8c5=6=r8c6-6-r2c6

There are only three valid inference propagations:
-x- A =x= (or reversed for x-chains)
=x= A =y= (implicit weak inference)
-x- A -y- (implicit strong inference because A is bivalued)

No such thing as "-x=" or "=x-"

[ttt]

2) (ht127)r345c8=(7-6)r5c2 =[fxw(6)r1c41,r5c146] -(6=4)r6c4-(4)r12c4=(4-8)r2c6=(8)r89c6-(8=9)r7c4-(9=5)r7c5-(5)r7c8=(489)r178c8 =>r345c8<>489

Very… very nice path! But for more clearly I think it should be:

2) (ht127)r345c8=(7)r6c8-[X-wing(7)r46c25]=(7-6)r5c2 =[fxw(6)r1c41,r5c146] -(6=4)r6c4-(4)r12c4=(4-8)r2c6=(8)r89c6-(8=9)r7c4-(9=5)r7c5-(5)r7c8=(489)r178c8 =>r345c8<>489

BTW, thanks ronk for correcting me. I’ll try to be better lately

ttt

[Steve K]

Thank-you ttt. IMO, the x wing on 7's is not required, as 7 is already bilocal in row 5. Perhaps I should have used the complimentary ALS:
(4589)r1678c8=(7)r6c8-r5c8=r5c2......

[DonM]

Okay, so I want to know what you two guys had for breakfast.

[ttt]

Code: Select all

*--------------------------------------------------------------------*
 | 369    1      39     | 4689   2      7      | 48     489    5      |
 | 5      269    7      | 14689  169    14689  | 148    3      2489   |
 | 4      29     8      | 5      3      19     | 6      129    7      |
 |----------------------+----------------------+----------------------|
 | 89     4789   2      | 3      1579   1459   | 1458   14578  6      |
 | 369    34679  3459   | 1469   8      14569  | 1345   12457  234    |
 | 1      34678  345    | 46     567    2      | 9      4578   348    |
 |----------------------+----------------------+----------------------|
 | 7      489    6      | 89     59     3      | 2      4589   1      |
 | 389    5      1349   | 2      169    1689   | 7      489    3489   |
 | 2      389    139    | 7      4      1589   | 358    6      389    |
 *--------------------------------------------------------------------*

My path for this one - not as elegant as Steve’s.
01: r7c2-9-r7c4-8-r12c4=8=r2c6=4=r12c4-4-r6c4-6-r1c4=6=r1c1-6-als:r23c2-9-r7c2
(9=8)r7c4-(8)r12c4=(8-4)r2c6=(4)r12c4-(4=6)r6c4-(6)r1c4=(6)r1c1-(6=np29)r23c2 => r7c2<>9

02: r5c13-3-r56c2=3=r9c2-3-r9c7=3=r3c7-3-r5c13
3’s r56c2=r9c2-r9c7=r5c7 => r5c13<>3

03: I don’t know how to present this by NL notation. Present as Dual Kraken: => r2c8<>48, some singles

Code: Select all

(3)r9c2-(3)r8c1=(3)r1c1-(3=9)r1c3-(9=np48)r1c78
 ||
(9)r9c2-(9)r23c2=(9)r1c13-(9=np48)r1c78
 ||
(8)r9c3---(8=4)r7c2-(4)r4c2
       |             || 
        ------------(8)r4c2
                     ||
                    (9)r4c2-(9)r23c2=(9)r1c13-(9=np48)r1c78
                     ||
                    (7)r4c2-(7)r5c2=(7-2)r5c8=(2-1)r3c8=(1)r2c8

04: r5c8-24-r5c8=7=r5c2=6=r2c2-6-r2c5-9-r7c5-5-r7c8=5=als:r56c8-24-r5c8
(7)r5c8=(7-6)r5c2=(6)r2c2-(6=9)r2c5-(9=5)r7c5-(5)r7c8=(hp57)r56c8 => r5c8<>24, singles to the end

Hope my NL notation is better , but please correct me if it’s wrong. Thanks

ttt

[Luke]

Okay, so I want to know what you two guys had for breakfast.

Whatever it was sure wasn't in my Wheaties.

This is pretty impressive. Guess I won't be posting my 25 move resolution path now....

2) (ht127)r345c8=(7-6)r5c2 =[fxw(6)r1c41,r5c146] -(6=4)r6c4-(4)r12c4=(4-8)r2c6=(8)r89c6-(8=9)r7c4-(9=5)r7c5-(5)r7c8=(489)r178c8 =>r345c8<>489

Perhaps I should have used the complimentary ALS:
(4589)r1678c8=(7)r6c8-r5c8=r5c2......

I see the latter around here much more often than the former.

I found the strong link into the finned x-wing very interesting. I have learned about x-wings used like this but never a finned x-wing. Very nice .

[Draco]

Okay, so I want to know what you two guys had for breakfast.

Whatever it was sure wasn't in my Wheaties.

Or in mine. Wow, nice solutions! I also enjoy seeing the back-and-forth solution tweaks. Not sure that I will ever reach this level, but seeing how these paths evolve is certainly a learning experience.

Cheers...

- drac

[Draco]

1) ss
2) (ht127)r345c8=(7-6)r5c2 =[fxw(6)r1c41,r5c146] -(6=4)r6c4-(4)r12c4=(4-8)r2c6=(8)r89c6-(8=9)r7c4-(9=5)r7c5-(5)r7c8=(489)r178c8 =>r345c8<>489
3) (9=2)r3c2-(2)r2c2=(2-9)r2c9=(9)r1c8 => r1c13<>9 simple sudoku to end

If (ht127)r345c8 is true then how can you force r5c2=7? I can see how if (ht458)r345c8, but then b6 has no 7's and that wouldn't work. I must be misreading something somewhere (maybe I need something different for breakfast )... ?

Cheers...

- drac

[Steve K]

Code: Select all

 
 
 *--------------------------------------------------------------------*
 | 369    1      39     | 4689   2      7      | 48     489    5      |
 | 5      269    7      | 14689  169    14689  | 148    3      2489   |
 | 4      29     8      | 5      3      19     | 6      129    7      |
 |----------------------+----------------------+----------------------|
 | 89     4789   2      | 3      1579   1459   | 1458   14578  6      |
 | 369    34679  3459   | 1469   8      14569  | 1345   2457   234    |
 | 1      34678  345    | 46     567    2      | 9      4578   348    |
 |----------------------+----------------------+----------------------|
 | 7      489    6      | 89     59     3      | 2      4589   1      |
 | 389    5      1349   | 2      169    1689   | 7      489    3489   |
 | 2      389    139    | 7      4      1589   | 358    6      389    |
 *--------------------------------------------------------------------*


--------------------------------------------------------------------------------

1) ss
2) (ht127)r345c8=(7-6)r5c2 =[fxw(6)r1c41,r5c146] -(6=4)r6c4-(4)r12c4=(4-8)r2c6=(8)r89c6-(8=9)r7c4-(9=5)r7c5-(5)r7c8=(489)r178c8 =>r345c8<>489
3) (9=2)r3c2-(2)r2c2=(2-9)r2c9=(9)r1c8 => r1c13<>9 simple sudoku to end

If (ht127)r345c8 is true then how can you force r5c2=7?

The short answer: I cannot. Rather, I use that at least one of [(hidden triple 127)r345c8, (7)r5c2] is true.

The longer answer:
a) (7)r5c2=(7)r5c8
b) There exists an almost hidden pair (2 candidates in three locations): (ahp12)r345c8.
Therefor, (7)r5c2=(ht127)r345c8.
Note: it is not required that (7)r5c2-(ht127)r345c8, thus this usage is atypical, but quite "legal".

Often, one sees something like (7)r6c8=(ht127)r345c8. But there is no reason not to go at right angles from a candidate overlayed (in this case 7) upon an almost hidden tuple such as the almost hidden pair 12 above. For this purpose, rows, columns, and boxes are perpendicular to each other. In the case I used above, it seemed most useful to use the bilocal 7's in row 5 intersecting with the 12's in column 8.

I found the strong link into the finned x-wing very interesting. I have learned about x-wings used like this but never a finned x-wing. Very nice .

Thanks. I have also found other almost fish often useful (finned swordfish and even an occasional finned jellyfish). Any almost technique is fair game, imo.