22/7 4 (SER 6.6)

Post puzzles for others to solve here.

22/7 4 (SER 6.6)

Postby mith » Mon Feb 22, 2021 5:34 pm

Code: Select all
+-------+-------+-------+
| . 3 . | . . . | 1 4 . |
| . . . | . 1 5 | . 9 . |
| . . 2 | 6 . . | . . . |
+-------+-------+-------+
| . . . | 5 . . | . 3 . |
| . . 5 | . . . | 8 . . |
| . 9 . | . . 7 | . . . |
+-------+-------+-------+
| . . . | . . 9 | 3 . . |
| . 2 . | 3 8 . | . . . |
| . 4 6 | . . . | . 2 . |
+-------+-------+-------+
.3....14.....15.9...26........5...3...5...8...9...7........93...2.38.....46....2.


(Not a hard puzzle, but I'm curious if anyone can find a nice one-step solution.)
mith
 
Posts: 950
Joined: 14 July 2020

Re: 22/7 4 (SER 6.6)

Postby pjb » Mon Feb 22, 2021 10:09 pm

No single digit elimination produces stte result, therefore needs a multidigit approach which I can't see for now. In mean time, simple 2 AIC solution:
(2=8)r1c6 - (8=4)r2c4 - (4=2)r7c4 => -2 r1c4
(4=6)r5c6 - (6=4)r6c5 - (4)r6c7 = (4)r8c7 => -4 r8c6; stte

Phil
pjb
2014 Supporter
 
Posts: 2574
Joined: 11 September 2011
Location: Sydney, Australia

Re: 22/7 4 (SER 6.6)

Postby Cenoman » Mon Feb 22, 2021 11:12 pm

mith wrote:I'm curious if anyone can find a nice one-step solution

Me too ! I'm even curious if anyone can find an ugly one-step solution...

In the meantime, let's have fun with two-step solutions:
Code: Select all
 +------------------------+------------------------+------------------------+
 |  5        3     89     |  289    7       28     |  1      4      6       |
 |  4678     678*  478    |  4-8    1       5      |  2      9      3       |
 |  49       1     2      |  6      349     34     |  57     8      57      |
 +------------------------+------------------------+------------------------+
 |  124678   678*  1478   |  5      2469    2468   |  467    3      12479   |
 |  12467    67    5      |  1249   23469   2346   |  8      167    12479   |
 |  12468*   9     3      |  1248*  246     7      |  456    156    1245    |
 +------------------------+------------------------+------------------------+
 |  178      5     178    |  24     246     9      |  3      167    147     |
 |  179      2     179    |  3      8       46     |  4567   1567   1457    |
 |  3        4     6      |  7      5       1      |  9      2      8       |
 +------------------------+------------------------+------------------------+

1. Kite (8)r6c4 = r6c1 - r5c2 = r2c2 => -8 r2c4; 26 placements and basics

Code: Select all
 +-----------------+-----------------+---------------------+
 |  5    3    9    |  8    7    2    |  1      4     6     |
 |  6    8    7    |  4    1    5    |  2      9     3     |
 |  4    1    2    |  6    9    3    |  57*    8     57*   |
 +-----------------+-----------------+---------------------+
 |  1    67*  4    |  5    2    8    |  67*    3     9     |
 |  2    67*  5    |  9    3    46   |  8     a167*  147   |
 |  8    9    3    |  1    46   7    |  456    56    2     |
 +-----------------+-----------------+---------------------+
 |  7    5    8    |  2    46   9    |  3      6-1  c14    |
 |  9    2    1    |  3    8    46   | b4567*  567* b457*  |
 |  3    4    6    |  7    5    1    |  9      2     8     |
 +-----------------+-----------------+---------------------+

2. MUG (567)r3c79, r45c2, b6p15, r8c789 using internals
(1)r5c8 == (4)r8c79 - (4=1)r7c9 => -1 r7c8; ste
Cenoman
Cenoman
 
Posts: 2760
Joined: 21 November 2016
Location: France

Re: 22/7 4 (SER 6.6)

Postby mith » Mon Feb 22, 2021 11:16 pm

Cenoman wrote:
mith wrote:I'm curious if anyone can find a nice one-step solution

Me too ! I'm even curious if anyone can find an ugly one-step solution...


Right? There are plenty of simple two step solutions, but the two steps are so far apart...
mith
 
Posts: 950
Joined: 14 July 2020

Re: 22/7 4 (SER 6.6)

Postby RSW » Tue Feb 23, 2021 1:59 am

Cenoman wrote:
mith wrote:I'm curious if anyone can find a nice one-step solution

Me too ! I'm even curious if anyone can find an ugly one-step solution...

Here's an ugly one step solution:
Code: Select all
 +-----------------+--------------------+-----------------+
 | 5      3   89   |T8-29   7     2-8   | 1    4    6     |
 | 4678   678 478  | 4-8    1     5     | 2    9    3     |
 | 49     1   2    | 6      39-4  3-4   | 57   8    57    |
 +-----------------+--------------------+-----------------+
 | 124678 678 1478 | 5     269-4 T8-246 | 467  3    12479 |
 | 12467  67  5    | 129-4 2369-4 4-236 | 8    167  12479 |
 | 12468  9   3    | 12-48 26-4   7     | 456  156  1245  |
 +-----------------+--------------------+-----------------+
 | 178    5   178  | 2-4  B246    9     | 3    167  147   |
 | 179    2   179  | 3    B8      6-4   | 4567 1567 1457  |
 | 3      4   6    | 7     5      1     | 9    2    8     |
 +-----------------+--------------------+-----------------+

Generic Exocet: (2468)r78c5, r2c4 r4c6 => -236r5c6 -246r4c6 -29r1c4 -4r7c4 -4r5c45 -4r46c5 -4r3c56 -4r8c6 -48r6c4 -8r2c4 -8r1c6; stte
Proviso: I'm not entirely convinced that the proof logic of my solver is correct. It's a work in progress.
Proof(?):: Show
Start of GE proof:
6 eliminated from r7c5 by singles chain
2 eliminated from r7c5 by singles chain
Base cell r7c5 has a single candidate 4 and is the true base digit
Base cell r8c5 has a single candidate 8 and is the true base digit
No contradiction for 4r7c5 8r8c5 4r2c4 8r4c6
Contradiction for -4r7c5 8r8c5 4r2c4 8r4c6
Contradiction for 4r7c5 -8r8c5 4r2c4 8r4c6
Contradiction for 4r7c5 8r8c5 -4r2c4 8r4c6
Contradiction for 4r7c5 8r8c5 4r2c4 -8r4c6
Confirm 4r7c5 & 8r8c5 & 4r2c4 & 8r4c6
** -4r5c4 -4r6c4 2r7c4 -2r1c4 -2r5c4 -2r6c4 -4r2c1 -4r2c3 -4r3c5 3r3c6 -3r5c6 9r3c5 -9r4c5 -9r5c5 4r3c1 -4r4c1 -4r5c1 -4r6c1 8r1c4 1r6c4 9r5c4 -9r5c9 -1r6c1 -1r6c8 -1r6c9 9r1c3 -9r8c3 2r1c6 -2r5c6 -8r4c1 -8r4c2 -8r4c3 -4r4c5 -4r5c5 -4r6c5 -4r7c9 6r8c6 4r5c6 -4r5c9 -6r8c7 -6r8c8 3r5c5 4r4c3 -4r4c7 -4r4c9 6r7c8 -6r5c8 5r6c8 -5r8c8 -5r6c7 -5r6c9 8r6c1 -8r2c1 -8r7c1 8r7c3 7r2c3 1r8c3 -1r8c1 7r8c8 1r5c8 -1r5c1 -1r5c9 -1r4c9 9r8c1 -7r8c7 -7r8c9 1r7c9 -1r8c9 7r7c1 6r2c1 -6r4c1 -6r5c1 -6r2c2 -7r4c1 2r5c1 1r4c1 7r5c9 5r3c9 4r8c9 2r6c9 -2r4c9 6r6c5 2r4c5 4r6c7 5r8c7 7r3c7 6r4c7 7r4c2 8r2c2 6r5c2 9r4c9 ?r7c5 ?r8c5 -8r1c4 -8r6c4 -8r2c2 -8r2c3 9r1c4 -9r5c4 8r1c3 -8r7c3 -9r3c5 3r3c5 -3r5c5 4r3c6 -4r5c6 9r3c1 -9r8c1 -6r6c8 -8row_7 -2r4c6 4r4c6 6r5c6 ?r8c6
4r7c5 & -4r2c4 leads to contradiction using singles chains
-8r1c4 -8r6c4 -8r2c1 -8r2c2 -8r2c3 2r1c6 -2r4c6 -2r5c6 9r1c4 -9r5c4 8r1c3 -8r4c3 -8r7c3 -9r3c5 3r3c5 -3r5c5 4r3c6 -4r4c6 -4r5c6 6r8c6 8r4c6 -8r4c1 -8r4c2 3r5c6 -6r8c7 -6r8c8 9r3c1 -9r8c1 -4r4c5 -4r5c5 -4r6c5 2r7c4 -2r5c4 -2r6c4 -4r7c9 6r7c8 -6r5c8 -6r6c8 8r6c1 -8r7c1 -8row_7
?r7c5
8r8c5 & -8r4c6 leads to contradiction using singles chains
-4r3c5 -4r4c5 -4r5c5 -4r6c5 2r7c4 -2r1c4 -2r5c4 -2r6c4 -4r7c9 6r8c6 -6r4c6 -6r5c6 -6r8c7 -6r8c8 2r1c6 4r4c6 3r3c6 -3r5c6 9r3c5 -9r4c5 -9r5c5 4r3c1 -4r2c1 -4r4c1 -4r5c1 -4r6c1 -4r2c3 8r1c4 4r2c4 -4r5c4 -4r6c4 1r6c4 9r5c4 -9r5c9 -1r6c1 -1r6c8 -1r6c9 9r1c3 -9r8c3 2r5c6 ?r1c6
?r8c5
All target combination complements for this base pair lead to contradiction. Exocet is proven.
Furthermore true base values are determined to be 4r7c5 & 8r8c5

48 are true base digits.

Rule 3: Nonbase candidate 26 is invalid in target cells.
Eliminating 2 6 from target r4c6.
Rule 6: Base candidates that don't appear in a mirror node can be eliminated from the corresponding target cell.
Eliminating 8 from target r2c4.
Eliminating 4 from target r4c6.
Rule 7: If one cell of a mirror node has only nonbase candidates, then all nonbase candidates can be removed from the other cell of the mirror node.
Eliminating 2 3 6 from mirror r5c6.
Eliminating 2 9 from mirror r1c4.
Rule 10a: True base digit 4 can be removed from all cells in sight of both base cells.
Eliminating 4 from r7c4.
Eliminating 4 from r3c5.
Eliminating 4 from r4c5.
Eliminating 4 from r5c5.
Eliminating 4 from r8c6.
Eliminating 4 from r6c5.
Rule 10b: True base digit 4 can be removed from cells in sight of both target cells.
Eliminating 4 from r5c4.
Eliminating 4 from r6c4.
Eliminating 4 from r3c6.
Eliminating 8 from r6c4.
Eliminating 8 from r1c6.
RSW
 
Posts: 619
Joined: 01 December 2018
Location: Western Canada

Re: 22/7 4 (SER 6.6)

Postby denis_berthier » Tue Feb 23, 2021 5:05 am

mith wrote:
Code: Select all
+-------+-------+-------+
| . 3 . | . . . | 1 4 . |
| . . . | . 1 5 | . 9 . |
| . . 2 | 6 . . | . . . |
+-------+-------+-------+
| . . . | 5 . . | . 3 . |
| . . 5 | . . . | 8 . . |
| . 9 . | . . 7 | . . . |
+-------+-------+-------+
| . . . | . . 9 | 3 . . |
| . 2 . | 3 8 . | . . . |
| . 4 6 | . . . | . 2 . |
+-------+-------+-------+
.3....14.....15.9...26........5...3...5...8...9...7........93...2.38.....46....2.


(Not a hard puzzle, but I'm curious if anyone can find a nice one-step solution.)


Mith, congrats for this noticeable puzzle.
It has no BRT-anti-backdoor, no W1-anti-backdoor and no S-anti-backdoor. As a result, there can't be any 1-elimination solution.
There could still be a 1-step solution, but it would have to eliminate at least two candidates.
I tried Forcing-T&E and Forcing{3}-T&E but they still need two steps.


SudoRules normal solution: Show
(solve ".3....14.....15.9...26........5...3...5...8...9...7........93...2.38.....46....2.")
***********************************************************************************************
*** SudoRules 20.1.s based on CSP-Rules 2.1.s, config = W+SFin
*** Using CLIPS 6.32-r779
*** Download from: https://github.com/denis-berthier/CSP-Rules-V2.1
***********************************************************************************************
9 singles
Starting non trivial part of solution with the following RESOLUTION STATE:
Code: Select all
   56789     3         789       289       279       28        1         4         2567     
   4678      678       478       248       1         5         267       9         3         
   14579     157       2         6         3479      34        57        8         57       
   124678    1678      1478      5         2469      2468      2467      3         124679   
   12467     167       5         1249      23469     2346      8         167       124679   
   12468     9         3         1248      246       7         2456      156       12456     
   1578      1578      178       24        246       9         3         1567      14567     
   1579      2         179       3         8         46        4567      1567      14567     
   3         4         6         7         5         1         9         2         8         
196 candidates, 1186 csp-links and 1186 links. Density = 6.21%

naked-pairs-in-a-row: r3{c7 c9}{n5 n7} ==> r3c5 ≠ 7, r3c2 ≠ 7, r3c2 ≠ 5, r3c1 ≠ 7, r3c1 ≠ 5
6 singles
finned-x-wing-in-columns: n8{c6 c2}{r4 r1} ==> r1c3 ≠ 8
26 singles
finned-x-wing-in-columns: n4{c7 c5}{r6 r8} ==> r8c6 ≠ 4
stte


I computed all the anti-backdoor pairs (there are 105), in case anyone wanted to try some of them. The format is: n1r1c1, n2r2c2:
BRT-anti-backdoor-pairs: Show
Code: Select all
589, 983          589, 864          589, 931          589, 816          589, 914          487, 983          487, 864          487, 931          487, 816          487, 914          486, 983          486, 864          486, 931          486, 816          486, 914          479, 983          479, 864          479, 931          479, 816          479, 914          178, 983          178, 864          178, 931          178, 816          178, 914          675, 983          675, 864          675, 931          675, 816          675, 914          275, 739          275, 537          474, 739          474, 537          465, 983          465, 864          465, 931          465, 816          465, 914          459, 983          459, 864          459, 931          459, 816          459, 914          159, 983          159, 864          159, 931          159, 816          159, 914          658, 758          656, 983          656, 864          656, 931          656, 816          656, 914          154, 589          154, 487          154, 486          154, 479          154, 178          154, 675          154, 465          154, 459          154, 159          154, 656          154, 752          154, 747          154, 642          154, 739          154, 537          752, 983          752, 864          752, 931          752, 816          752, 914          747, 983          747, 864          747, 931          747, 816          747, 914          642, 983          642, 864          642, 931          642, 816          642, 914          739, 983          739, 871          739, 864          739, 842          739, 931          739, 824          739, 816          739, 914          739, 813          537, 983          537, 871          537, 864          537, 842          537, 931          537, 824          537, 816          537, 914          537, 813          214, 739          214, 537         
denis_berthier
2010 Supporter
 
Posts: 3974
Joined: 19 June 2007
Location: Paris

Re: 22/7 4 (SER 6.6)

Postby denis_berthier » Sun Feb 28, 2021 10:43 am

.
I finally decided to try some of the anti-backdoor pairs I found in my previous post.
To be clear about what I did:
1) the resolution path starts from the givens, with:
Code: Select all
singles ==> r9c6 = 1, r9c4 = 7, r9c5 = 5, r9c7 = 9, r9c9 = 8, r9c1 = 3, r6c3 = 3, r3c8 = 8, r2c9 = 3
Starting non trivial part of solution with the following RESOLUTION STATE:
   56789     3         789       289       279       28        1         4         2567     
   4678      678       478       248       1         5         267       9         3         
   14579     157       2         6         3479      34        57        8         57       
   124678    1678      1478      5         2469      2468      2467      3         124679   
   12467     167       5         1249      23469     2346      8         167       124679   
   12468     9         3         1248      246       7         2456      156       12456     
   1578      1578      178       24        246       9         3         1567      14567     
   1579      2         179       3         8         46        4567      1567      14567     
   3         4         6         7         5         1         9         2         8         

and these trivial Singles are not considered as steps.

2) Non counted steps between two eliminations can consist of Singles
3) Final steps (after the second whip[≥1] elimination is found) can only be Singles.
4) What I find is not only 2-step but also 2-elimination before stte.


Code: Select all
whip[7]: c4n1{r6 r5} - c4n9{r5 r1} - b2n8{r1c4 r1c6} - r1c3{n8 n7} - r2n7{c3 c7} - r3c9{n7 n5} - r3c7{n5 .} ==> r6c4 ≠ 8
singles ==> r4c6 = 8, r1c6 = 2, r2c7 = 2, r1c9 = 6, r1c1 = 5, r7c2 = 5, r2c2 = 8, r2c4 = 4, r2c3 = 7, r1c3 = 9, r1c4 = 8, r1c5 = 7, r8c3 = 1, r4c3 = 4, r7c3 = 8, r7c1 = 7, r8c1 = 9, r2c1 = 6, r3c2 = 1, r3c1 = 4, r3c6 = 3, r3c5 = 9, r7c4 = 2, r6c4 = 1, r5c4 = 9, r4c9 = 9, r4c1 = 1, r5c1 = 2, r6c1 = 8, r6c9 = 2, r4c5 = 2, r5c5 = 3
whip[2]: r7n4{c9 c5} - r6n4{c5 .} ==> r8c7 ≠ 4
stte


There are other possibilities, but all involve at least a whip[7].
denis_berthier
2010 Supporter
 
Posts: 3974
Joined: 19 June 2007
Location: Paris


Return to Puzzles