The New Sudoku Players' Forum

[Leren]

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*-----------*
|..2|..6|7..|
|...|...|.2.|
|6..|4..|5.8|
|---+---+---|
|.3.|.1.|..5|
|..4|..9|..6|
|...|...|.8.|
|---+---+---|
|59.|34.|..2|
|...|...|1.3|
|18.|56.|4..|
*-----------*
..2..67.........2.6..4..5.8.3..1...5..4..9..6.......8.59.34...2......1.318.56.4..

[pjb]

Code: Select all

 389    a145     2      | 189    3589   6      | 7     f139-4  149   
 389    b1457    1589   | 1789   35789  3578   | 6      2      149   
 6      c17      19     | 4      2     d37     | 5     e139    8     
------------------------+----------------------+---------------------
 2789    3       89     | 6      1      478    | 29     47     5     
 278     15      4      | 278    3578   9      | 23     17     6     
 279     6       15     | 27     357    3457   | 239    8      147   
------------------------+----------------------+---------------------
 5       9       7      | 3      4      1      | 8      6      2     
 4       2       6      | 789    789    78     | 1      5      3     
 1       8       3      | 5      6      2      | 4      79     79     


(4)r1c2 = (4-7)r2c2 = (7)r3c2 - (7=3)r3c6 - (3)r3c8 = (3-4)r1c8 => -4 r1c8 ; stte

Phil

[P.O.]

Code: Select all

3489   145    2      189    3589   6      7      1349   149             
34789  1457   1589   1789   35789  3578   6      2      149             
6      17     19     4      2      37     5      139    8               
2789   3      89     6      1      478    29     479    5               
278    1257   4      278    3578   9      23     137    6               
279    6      159    27     357    3457   239    8      1479           
5      9      7      3      4      1      8      6      2               
24     24     6      789    789    78     1      5      3               
1      8      3      5      6      2      4      79     79       

after singles.

depth: 5  candidate: 1  from cell(s)
((2 2 1) (1 4 5 7)) ((3 2 1) (1 7))

((1 0) (5 2 4) (1 2 5 7))
((1 0) (6 3 4) (1 5 9))
((5 1) (2 3 1) (1 5 8 9))
((5 2) (6 6 5) (3 4 5 7))
((4 3) (4 6 5) (4 7 8))
((4 4) (1 8 3) (1 3 4 9))
((1 5) (1 2 1) (1 4 5))

ste.


[yzfwsf]

Code: Select all

3489   145    2      189    3589   6      7      1349   149             
34789  1457   1589   1789   35789  3578   6      2      149             
6      17     19     4      2      37     5      139    8               
2789   3      89     6      1      478    29     479    5               
278    1257   4      278    3578   9      23     137    6               
279    6      159    27     357    3457   239    8      1479           
5      9      7      3      4      1      8      6      2               
24     24     6      789    789    78     1      5      3               
1      8      3      5      6      2      4      79     79       

after singles.

depth: 5  candidate: 1  from cell(s)
((2 2 1) (1 4 5 7)) ((3 2 1) (1 7))

((1 0) (5 2 4) (1 2 5 7))
((1 0) (6 3 4) (1 5 9))
((5 1) (2 3 1) (1 5 8 9))
((5 2) (6 6 5) (3 4 5 7))
((4 3) (4 6 5) (4 7 8))
((4 4) (1 8 3) (1 3 4 9))
((1 5) (1 2 1) (1 4 5))

ste.


I don’t understand the logic you expressed. Can you explain it further?

[P.O.]

hi yzfwsf i will try

The start of the chain are the cells:
(5 2 4) (1 2 5 7))
(6 3 4) (1 5 9))

a bilocation on candidate 1.

so if (5 2 4)(1 2 5 7) is not 1 (6 3 4)(1 5 9) is 1
if (6 3 4)(1 5 9) is 1 it is not 5
so (2 3 1)(1 5 8 9) is 5 by bilocation
this eliminate 5 from (2 6 2)(3 5 7 8)
so (6 6 5)(3 4 5 7) is 5 by bilocation thus it is not 4
so (4 6 5)(4 7 8) is 4 by bilocation
this eliminate 4 from (4 8 6)(4 7 9)
so (1 8 3)(1 3 4 9) is 4 by bilocation
cell (2 3 1)(1 5 8 9) which is 5
and cell (1 8 3)(1 3 4 9) which is 4
eliminate 4 and 5 from cell (1 2 1)(1 4 5) which thus is 1

in the end:
when (5 2 4)(1 2 5 7) is 1 cells (2 2 1)(1 4 5 7) (3 2 1) (1 7) cant be 1
when it is not cell (1 2 1)(1 4 5) is 1 so those 2 cells cant ever be one
these eliminations solve the puzzle.

[jco]

Code: Select all

.-------------------------------------------------------.
| 389   145   2    | 189   3589   6    | 7    1349  149 |
| 389  a145-7 1589 | 1789  35789  3578 | 6    2    a149 |
| 6    d17    19   | 4     2     c37   | 5    139   8   |
|------------------+-------------------+----------------|
| 2789  3     89   | 6     1     b478  | 29   47    5   |
| 278   15    4    | 278   3578   9    | 23   17    6   |
| 279   6     15   | 27    357   a3457 | 239  8    a147 |
|------------------+-------------------+----------------|
| 5     9     7    | 3     4      1    | 8    6     2   |
| 4     2     6    | 789   789   b78   | 1    5     3   |
| 1     8     3    | 5     6      2    | 4    79    79  |
'-------------------------------------------------------'

SS(4): r6c6,r2c2 connected by r6c9,r2c9

(4)r2c2 == (4)r6c6 - (4=87)r48c6 - (7)r3c6 = (7)r3c2 => -7 r2c2; ste

[yzfwsf]

hi PO, please explain the coordinate system in your logical expression.

I guess I have guessed that the first bracket is coordinate RC and count of the candidates for cell , the second bracket is the candidates for this cell.

[P.O.]

Well, in the full notation there are some codes that help to see the relationship between the links:

((1 0) (5 2 4) (1 2 5 7))
((1 0) (6 3 4) (1 5 9))
((5 1 10) (2 3 1) (1 5 8 9))
((5 2 1) (6 6 5) (3 4 5 7))
((4 3 10) (4 6 5) (4 7 8))
((4 4 1) (1 8 3) (1 3 4 9))
((1 5 77) (1 2 1) (1 4 5))

I've removed them because some explanation is needed to understand them.
Also this system of codes is not stable yet.

so the notation is made of 3 lists of numbers
the second list is the coordinates of the cell in Row Col Box
the third list is the content of the cell: its candidates
in the first list:
the first number is the candidate set in the cell
the second number is the depth in the Breadth First Search algorithm
the third number is the code.

for this example:
the first 2 lists don't have any code: they are the start bilocation

the 10 in the third and fifth list means the link is obtained by bilocation following the elimination in the preceding link of the candidate set in this one

the 1 in the fourth and sixth list means the link is obtained by bilocation following the elimination of a candidate of the same sort of the preceding link

the 77 in the seventh list means some elimination is made in the cell and the remaining candidate is set

there are as many codes as there are relationships between the links.

Also you can see i program in (Lots of ((Irritatingly Spurious) (Parentheses)))

[jco]

Hello P.O.,

It would be very nice if you could mark in the grid (i.e., the board that you have displayed with the candidates shown in each cell) with letters (a,b,c,...) the sequence of cells being visited from the first one to the last one, for your chosen move. In case the move is a pattern without order for cells, you can mark the cells relevant to the pattern. Of course, words explaining the move and the marked grid are good combination when the move has an elaborate notation.
Marking the grid would allow us to understand the Sudoku move that is being made, helping to understand quickly your notation.

Regards,
JCO

Edit: improved the text.

[eleven]

P.O., so you have to "remember" the 5r2c3 to get the 1r1c2 from 4r1c8.

In AIC notation:
1r5c2 = (1-5)r6c3 = *5r2c3 - r2c6 = (5-4)r6c6 = r4c6 - r4c8 = 4r1c8 - (4|*5 = 1)r1c2 => -1r23c2

I guess, it looks simpler as a whip ...

[P.O.]

Code: Select all

3489  i+1-4-5   2        189    3589    6        7     h13+49   149             
34789  ×1457   c1+589    1789   35789  d3-578    6      2       149             
6      ×17      19       4      2       37       5      139     8               
2789   3        89       6      1      f+478     29    g-479    5               
278   a-1257    4        278    3578    9        23     137     6               
279    6       b+1-59    27     357    e3-4+57   239    8       1479           
5      9        7        3      4       1        8      6       2               
24     24       6        789    789     78       1      5       3               
1      8        3        5      6       2        4      79      79     


JCO:
i tried to mark the board as you suggested:
+ is to set the following candidate
- is to eliminate the following candidate
× is for the candidate(s) eliminated by the relationship between the start and the end of the chain
there is a ambiguity with the first candidate as it is eliminated to start the chain and set to make the finals eliminations.

ELEVEN:
in the chains the algorithm builds a link is added to the chain on the consequences of all the precedings links not only the last one: in R1C2 the 5 has been eliminated by the link R2C3, it is no longer in the cell when R1C8 is reached: is this remembering?

speaking of notation I find some of them here very hard to understand, the ones i post are the data structure and the data used by the algorithm to build the chains: a list of lists of numbers: the coordinates of the cell, its candidates, the candidate set in the cell, and a code to see the relationship between two links.

[jco]

Hello P.O.,

Code: Select all

3489  i+1-4-5   2        189    3589    6        7     h13+49   149             
34789  ×1457   c1+589    1789   35789  d3-578    6      2       149             
6      ×17      19       4      2       37       5      139     8               
2789   3        89       6      1      f+478     29    g-479    5               
278   a-1257    4        278    3578    9        23     137     6               
279    6       b+1-59    27     357    e3-4+57   239    8       1479           
5      9        7        3      4       1        8      6       2               
24     24       6        789    789     78       1      5       3               
1      8        3        5      6       2        4      79      79     


Many thanks! Really appreciated!
I understood the move with your explanation (on notation used in the grid) and by looking at the board.
Best regards,
JCO

[jco]

Hi again P.O.,

Code: Select all

3489  i+1-4-5   2        189    3589    6        7     h13+49   149             
34789  ×1457   c1+589    1789   35789  d3-578    6      2       149             
6      ×17      19       4      2       37       5      139     8               
2789   3        89       6      1      f+478     29    g-479    5               
278   a-1257    4        278    3578    9        23     137     6               
279    6       b+1-59    27     357    e3-4+57   239    8       1479           
5      9        7        3      4       1        8      6       2               
24     24       6        789    789     78       1      5       3               
1      8        3        5      6       2        4      79      79     


Your solution is distinct (different logic) but related to the following move

Code: Select all

.------------------------------------------------------------.
| 389 uaA145    2    | 189   3589    6    | 7    b1349  C149 |
| 389    457-1  g1589| 1789  35789  f3578 | 6     2     C149 |
| 6      7-1     19  | 4     2       37   | 5     139    8   |
|--------------------+--------------------+------------------|
| 2789   3      89   | 6     1      d478  | 29   c47     5   |
| 278  iv15     4    | 278   3578    9    | 23    17     6   |
| 279    6     h15   | 27    357    e3457 | 239   8      147 |
|--------------------+--------------------+------------------|
| 5      9      7    | 3     4       1    | 8     6      2   |
| 4      2      6    | 789   789     78   | 1     5      3   |
| 1      8      3    | 5     6       2    | 4     79     79  |
'------------------------------------------------------------'

Kraken Cell (145)r1c2
(1)r1c2
||
(4)r1c2 - r1c8 = r4c8 - r4c6 = (4-5)r6c6 = r2c6 - r2c3 = (5-1)r6c3 = (1)r5c2
||
(5)r1c2-(5=1)r5c2
------------------
=> -1 r12c2; ste

The logic for this other move is that for each possible values in the cell r1c2 we have a chain that implies that if the start digit is true in r1c2, then 1 is false in r12c2 (all chains start at r1c2). The chain that starts with (4)r1c2 visits the same cells (but in the other direction) of your marks in the board.
Anyway, I noticed this relation after I understood your Sudoku move.

Edit: improved the text and removed unnecessary almost chain.

[eleven]

in the chains the algorithm builds a link is added to the chain on the consequences of all the precedings links not only the last one: in R1C2 the 5 has been eliminated by the link R2C3, it is no longer in the cell when R1C8 is reached: is this remembering?

Yes, while (pure) AIC-links are independent from preceding ones.
Your's is similar to (some of) Denis' SudoRules chains. Finding such chains manually is harder, because there are many things to remember, when you follow one (you don't know in forward, which are useful).

speaking of notation I find some of them here very hard to understand, the ones i post are the data structure and the data used by the algorithm to build the chains: a list of lists of numbers: the coordinates of the cell, its candidates, the candidate set in the cell, and a code to see the relationship between two links.

Once you explained it, i can read them, and if you mark the cells in the grid, it should be easy to understand.

[denis_berthier]

.

Code: Select all

Resolution state after Singles and whips[1]:
   +-------------------+-------------------+-------------------+
   ! 3489  145   2     ! 189   3589  6     ! 7     1349  149   !
   ! 34789 1457  1589  ! 1789  35789 3578  ! 6     2     149   !
   ! 6     17    19    ! 4     2     37    ! 5     139   8     !
   +-------------------+-------------------+-------------------+
   ! 2789  3     89    ! 6     1     478   ! 29    47    5     !
   ! 278   1257  4     ! 278   3578  9     ! 23    17    6     !
   ! 279   6     159   ! 27    357   3457  ! 239   8     147   !
   +-------------------+-------------------+-------------------+
   ! 5     9     7     ! 3     4     1     ! 8     6     2     !
   ! 24    24    6     ! 789   789   78    ! 1     5     3     !
   ! 1     8     3     ! 5     6     2     ! 4     79    79    !
   +-------------------+-------------------+-------------------+

1) There is an easy solution with bivalue-chains[≤4]:

Code: Select all

hidden-pairs-in-a-block: b4{n1 n5}{r5c2 r6c3} ==> r6c3 ≠ 9, r5c2 ≠ 7, r5c2 ≠ 2
hidden-single-in-a-column ==> r8c2 = 2
naked-single ==> r8c1 = 4
whip[1]: b4n7{r6c1 .} ==> r2c1 ≠ 7
x-wing-in-columns: n5{c3 c6}{r2 r6} ==> r6c5 ≠ 5, r2c5 ≠ 5, r2c2 ≠ 5
finned-x-wing-in-rows: n1{r5 r3}{c8 c2} ==> r2c2 ≠ 1, r1c2 ≠ 1
hidden-triplets-in-a-row: r6{n1 n4 n5}{c3 c9 c6} ==> r6c9 ≠ 7, r6c6 ≠ 7, r6c6 ≠ 3
singles ==> r9c9 = 7, r9c8 = 9, r3c3 = 9, r4c3 = 8
whip[1]: c6n3{r3 .} ==> r1c5 ≠ 3, r2c5 ≠ 3
naked-pairs-in-a-row: r4{c6 c8}{n4 n7} ==> r4c1 ≠ 7
x-wing-in-rows: n1{r3 r5}{c2 c8} ==> r1c8 ≠ 1
biv-chain[3]: r2n4{c9 c2} - c2n7{r2 r3} - b1n1{r3c2 r2c3} ==> r2c9 ≠ 1
biv-chain[3]: r2n5{c6 c3} - r1c2{n5 n4} - r2c2{n4 n7} ==> r2c6 ≠ 7
biv-chain[3]: r1c1{n8 n3} - r2n3{c1 c6} - b2n5{r2c6 r1c5} ==> r1c5 ≠ 8
biv-chain[3]: r1c5{n9 n5} - r2n5{c6 c3} - r2n1{c3 c4} ==> r2c4 ≠ 9
biv-chain[4]: r3c2{n7 n1} - b4n1{r5c2 r6c3} - r6c9{n1 n4} - r2n4{c9 c2} ==> r2c2 ≠ 7
stte

2) 1-step solutions:

Code: Select all

***********************************************************************************************
***  SudoRules 20.1.s based on CSP-Rules 2.1.s, config = TyW+W
***  Using CLIPS 6.32-r779
***  Running on MacBookPro Retina Mid-2012 i7 2.7GHz, 16GB 1600MHz DDR3, MacOS 10.15.7
***  Download from: https://github.com/denis-berthier/CSP-Rules-V2.1
***********************************************************************************************
(find-sudoku-1-steppers-wrt-W1 "..2..67.........2.6..4..5.8.3..1...5..4..9..6.......8.59.34...2......1.318.56.4..")

===> There are 13 W1-anti-backdoors:
n5r1c5 n4r1c8 n5r2c3 n1r3c2 n7r3c6 n3r3c8 n4r4c6 n7r4c8 n5r5c2 n1r5c8 n1r6c3 n5r6c6 n4r6c9

all of which lead to a 1-step solution with whips[≤8].
The simplest require a whip[6], e.g.:

Code: Select all

whip[6]: c6n5{r2 r6} - r6n4{c6 c9} - c8n4{r4 r1} - r1c2{n4 n1} - r3n1{c3 c8} - c8n3{r3 .} ==> r1c5 ≠ 5
stte
OR:
whip[6]: c9n4{r2 r6} - r6n1{c9 c3} - r5n1{c2 c8} - r3n1{c8 c2} - r1c2{n1 n5} - b4n5{r5c2 .} ==> r1c8 ≠ 4
stte
OR:
whip[6]: c6n5{r2 r6} - r6n4{c6 c9} - c8n4{r4 r1} - r1c2{n4 n1} - r3n1{c3 c8} - c8n3{r3 .} ==> r2c3 ≠ 5
stte
OR:
whip[6]: c3n1{r3 r6} - c3n5{r6 r2} - r1c2{n5 n4} - r2n4{c2 c9} - r6n4{c9 c6} - c6n5{r6 .} ==> r3c2 ≠ 1
stte
OR:
whip[6]: r6n1{c9 c3} - b4n5{r6c3 r5c2} - r5n1{c2 c8} - r3n1{c8 c2} - r1c2{n1 n4} - c8n4{r1 .} ==> r6c9 ≠ 4
stte

(Note: SudoRules total computation time for this result = 11.45s on an old MacBookPro).
My conclusion: for this puzzle, the requirement for a 1-step solution leads to totally absurd solutions, compared to the simplest-first one.


3) 2-step-solutions:
Considering there's no reasonable 1-step solution, it's natural to try the 2-step ones.

Code: Select all

***********************************************************************************************
***  SudoRules 20.1.s based on CSP-Rules 2.1.s, config = TyW+W
***  Using CLIPS 6.32-r779
***  Running on MacBookPro Retina Mid-2012 i7 2.7GHz, 16GB 1600MHz DDR3, MacOS 10.15.7
***  Download from: https://github.com/denis-berthier/CSP-Rules-V2.1
***********************************************************************************************
(find-sudoku-2-steppers-wrt-W1 "..2..67.........2.6..4..5.8.3..1...5..4..9..6.......8.59.34...2......1.318.56.4..")

There remains 127 candidates after Singles and whips[1] have been applied.
===> 62 of them can be eliminated by rules in W8:
n4r1c1 n9r1c1 n1r1c2 n4r1c2 n3r1c5 n5r1c5 n1r1c8 n4r1c8 n9r1c8 n1r1c9 n4r2c1 n7r2c1 n8r2c1 n9r2c1 n1r2c2 n5r2c2 n7r2c2 n5r2c3 n8r2c3 n9r2c3 n8r2c4 n3r2c5 n5r2c5 n3r2c6 n7r2c6 n8r2c6 n1r2c9 n4r2c9 n1r3c2 n1r3c3 n7r3c6 n3r3c8 n9r3c8 n7r4c1 n8r4c1 n9r4c3 n4r4c6 n8r4c6 n7r4c8 n7r5c1 n8r5c1 n2r5c2 n5r5c2 n7r5c2 n7r5c4 n3r5c5 n7r5c5 n8r5c5 n1r5c8 n1r6c3 n9r6c3 n5r6c5 n3r6c6 n5r6c6 n7r6c6 n4r6c9 n7r6c9 n2r8c1 n4r8c2 n7r8c6 n7r9c8 n9r9c9

This leaves 5921 candidate pairs as possible anti-backdoor-pairs
After checking each of them, there are 1028 W1-anti-backdoor-pairs for the rules in W8.
1023 of them lead to a 2-step solution in W8. Here are the simplest 8:

Code: Select all

z-chain-rn[2]: r5n1{c2 c8} - r3n1{c8 .} ==> r1c2 ≠ 1
biv-chain[4]: r1n5{c2 c5} - c6n5{r2 r6} - b5n4{r6c6 r4c6} - c8n4{r4 r1} ==> r1c2 ≠ 4
stte
OR:
z-chain-rn[2]: r5n1{c2 c8} - r3n1{c8 .} ==> r1c2 ≠ 1
biv-chain[4]: c6n5{r2 r6} - b5n4{r6c6 r4c6} - c8n4{r4 r1} - r1c2{n4 n5} ==> r1c5 ≠ 5, r2c2 ≠ 5, r2c3 ≠ 5
stte
OR:
z-chain-rn[2]: r5n1{c2 c8} - r3n1{c8 .} ==> r1c2 ≠ 1
biv-chain[4]: b6n4{r4c8 r6c9} - r6n1{c9 c3} - b4n5{r6c3 r5c2} - r1c2{n5 n4} ==> r1c8 ≠ 4
stte
OR:
z-chain-rn[2]: r5n1{c2 c8} - r3n1{c8 .} ==> r1c2 ≠ 1
biv-chain[4]: c6n5{r2 r6} - b5n4{r6c6 r4c6} - c8n4{r4 r1} - r1c2{n4 n5} ==> r2c3 ≠ 5, r1c5 ≠ 5, r2c2 ≠ 5
stte
OR:
z-chain-rn[2]: r5n1{c2 c8} - r3n1{c8 .} ==> r1c2 ≠ 1
biv-chain[4]: r6n4{c6 c9} - c8n4{r4 r1} - r1c2{n4 n5} - b4n5{r5c2 r6c3} ==> r6c6 ≠ 5
stte
OR:
z-chain-rn[2]: r5n1{c2 c8} - r3n1{c8 .} ==> r1c2 ≠ 1
biv-chain[4]: r6n1{c9 c3} - b4n5{r6c3 r5c2} - r1c2{n5 n4} - c8n4{r1 r4} ==> r6c9 ≠ 4
stte
OR:
biv-chain[4]: r1n5{c2 c5} - c6n5{r2 r6} - b5n4{r6c6 r4c6} - c8n4{r4 r1} ==> r1c2 ≠ 4
biv-chain[3]: r1c2{n1 n5} - c3n5{r2 r6} - b4n1{r6c3 r5c2} ==> r3c2 ≠ 1, r2c2 ≠ 1
stte
OR:
biv-chain-bn[2]: b4n1{r5c2 r6c3} - b4n5{r6c3 r5c2} ==> r5c2 ≠ 2, r5c2 ≠ 7
hidden-single-in-a-column ==> r8c2 = 2
naked-single ==> r8c1 = 4
whip[1]: b4n7{r6c1 .} ==> r2c1 ≠ 7
biv-chain[4]: c8n3{r1 r3} - r3c6{n3 n7} - c2n7{r3 r2} - c2n4{r2 r1} ==> r1c8 ≠ 4
stte

(Note: SudoRules total computation time for this result = 15m 44s on the same old MacBookPro; juts what I needed to take my second morning tea.)

Notice that some of them require only bivalue-chains[≤4] and no z-chain. You can consider the computation time as very long, but:
- I activated rules much longer than reasonable (max-length 8, when 4 or 5 was the upper limit of reasonable)
- SudoRules had to consider 5921 candidate-pairs before finding only a few interesting ones.
Needless to say, a human solver would've to do the same job and has 0 chance to find these 2-step solutions.


[Edit]: I tried to restrict max chain length to 4, but still keeping all the possible chains, typed or not:
There remains 127 candidates after Singles and whips[1] have been applied.
===> 21 of them can be eliminated by rules in W4:
n4r1c1 n1r1c2 n4r1c2 n1r1c8 n4r2c1 n7r2c1 n9r2c1 n1r2c2 n5r2c2 n5r2c5 n1r3c3 n9r3c8 n2r5c2 n7r5c2 n9r6c3 n5r6c5 n3r6c6 n7r6c6 n7r6c9 n2r8c1 n4r8c2
This leaves 2436 candidate pairs as possible anti-backdoor-pairs
After checking each of them, there are 280 W1-anti-backdoor-pairs for the rules in W4.
19 of them lead to a 2-step solution in W4. Of course, the simplest of them are as before.
Total computation time is reduced to 3m 11s
[end edit]