The New Sudoku Players' Forum

[P.O.]

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

4.........2..9.5....3..5..1..........9..5.7....8..1..4.6..2.9....1..4..7...8.....

after basics including basic fish:

4       1578    5679    2367    13678   23678   2368    23678   23689           
1678    2       67      13467   9       3678    5       34678   368             
6789    78      3       267     4678    5       2468    26789   1               
23567   13457   2567    2367    34678   236789  12368   23568   235689           
1236    9       246     2346    5       2368    7       12368   2368             
23567   357     8       23679   367     1       236     23569   4               
3578    6       457     1357    2       37      9       14      358             
23589   358     1       3569    36      4       2368    23568   7               
2357    3457    2579    8       1367    3679    14      2356    2356     



[eleven]

This is a puzzle for coloring fans, there are exactly 2 ways to place all 149 (one with 5r1c3, the other with 5r8c4 and 5r1c2).
So the quickest way for manual solvers probably is to mark them and see, that the latter can't be true cause the single 8r3c5 kills the last 8's in box 1.

[Cenoman]

After basics only (without the basic swordfishes)

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 +---------------------------+------------------------------+-----------------------------+
 |  4        1578    5679    |    12367    13678   23678    |  2368    236789    23689    |
 |  1678     2     i'67      |    13467    9       3678     |  5       34678     368      |
 |i'6789   i'78      3       |    2467    a4-678   5        |  2468    246789    1        |
 +---------------------------+------------------------------+-----------------------------+
 |  123567  c13457   24567   |    234679  b34678   236789   |  12368   1235689   235689   |
 |  1236     9       246     |    2346     5       2368     |  7       12368     2368     |
 |  23567    357     8       |    23679   h367     1        |  236     23569     4        |
 +---------------------------+------------------------------+-----------------------------+
 |  3578     6      e457     | g'g1357     2       37       |  9      f13458     358      |
 |h'23589    358     1       |  g'3569    h36      4        |  2368    23568     7        |
 |  23579   d3457    24579   |    8       h1367    3679     |  12346   123456    2356     |
 +---------------------------+------------------------------+-----------------------------+

1. (4)r3c5 = r4c5 - r4c2 = r9c2 - r7c3 = (4-1)r7c8 = r7c4 - (1=367)r689c5 => -67 r3c5
2. (4)r3c5 = r4c5 - r4c2 = r9c2 - r7c3 = (4-1)r7c8 = (15-9)r78c4 = r8c1 - (9=678)b1p678 => -8 r3c5; lcls, 21 placements

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 +-------------------+--------------------+---------------------+
 |  4     1    5     |  367    678   2    |  368   367    9     |
 |  67    2    67    |  1      9     38   |  5     4      38    |
 |  9     8    3     |  67     4     5    |  26    267    1     |
 +-------------------+--------------------+---------------------+
 |  367   4    267   |  237    78    9    |  1     2358*  256   |
 |  1     9    26    |  4      5     38   |  7     38*    26    |
 |  5     37*  8     |  2367   67    1    |  23*   9      4     |
 +-------------------+--------------------+---------------------+
 |  38    6    4     |  5      2     7    |  9     1      38    |
 |  28    5    1     |  9      3     4    |  268   26     7     |
 |  237   37*  9     |  8      1     6    |  4     25-3   25    |
 +-------------------+--------------------+---------------------+

3. X-chain: (3)r9c2 = r6c2 - r6c7 = r45c8 => -3 r9c8; ste

[DEFISE]

Swordfish in rows: 1r257c148 => -1r1c4 -1r4c1 -1r4c8 -1r9c8
Swordfish in rows: 4r257c348 => -4r3c4 -4r3c8 -4r4c3 -4r4c4 -4r9c3 -4r9c8
Hidden pairs: 14b9p27 => -3r7c8 -5r7c8 -8r7c8 -2r9c7 -3r9c7 -6r9c7
Swordfish in rows: 9r368c148 => -9r1c8 -9r4c4 -9r4c8 -9r9c1

whip[6]: r1n1{c2 c5}- r2n1{c4 c1}- c2n1{r1 r4}- r4n4{c2 c5}- c5n8{r4 r3}- b1n8{r3c1 .} => -5r1c2

Single(s): 5r1c3, 9r1c9, 9r3c1, 9r4c6, 9r6c8, 9r8c4, 9r9c3, 5r7c4, 1r7c8, 4r9c7, 4r3c5, 4r2c8, 4r4c2, 1r4c7, 1r5c1, 1r2c4, 1r1c2, 4r5c4, 4r7c3, 1r9c5
Box/Line: 5r6b4 => -5r4c1
Box/Line: 2c3b4 => -2r4c1 -2r6c1
Box/Line: 6b1r2 => -6r2c6 -6r2c9
Box/Line: 7b8c6 => -7r1c6 -7r2c6
Box/Line: 7r2b1 => -7r3c2
Single(s): 8r3c2
Naked pairs: 38c9r27 => -3r4c9 -8r4c9 -3r5c9 -8r5c9 -3r9c9
Box/Line: 8b6c8 => -8r1c8 -8r8c8
Naked pairs: 26r5c39 => -2r5c6 -6r5c6 -2r5c8 -6r5c8
Single(s): 2r1c6, 6r9c6, 3r8c5, 7r7c6, 5r8c2, 5r6c1
Box/Line: 6c9b6 => -6r4c8 -6r6c7
Box/Line: 6r6b5 => -6r4c4 -6r4c5

whip[2]: c2n3{r9 r6}- b6n3{r6c7 .} => -3r9c8
STTE

Remark:
without the initial basics we can start by eliminating 5r1c2 with this g-whip:
g-whip[8]: c2n1{r1 r4}- c7n1{r4 r9}- c5n1{r9 r1}- r2n1{c4 c1}- b1n8{r2c1 r3c12}- c5n8{r3 r4}- c5n4{r4 r3}- c7n4{r3 .} => -5r1c2
and then continue as above.

[P.O.]

hi Cenoman, DEFISE thank you for your answers;
i included the basic fish in the basics because i have often seen them equated with subsets and also the SER rating groups them together:
https://github.com/SudokuMonster/SukakuExplainer/wiki/Difficulty-ratings-in-Sudoku-Explainer-v1.2.1
3.0, 3.2, 3.4: Naked Pair, X-Wing, Hidden Pair
3.6, 3.8, 4.0: Naked Triplet, Swordfish, Hidden Triplet
5.0, 5.2, 5.4: Naked Quad, Jellyfish, Hidden Quad
as can be seen with DEFISE solution they can do a lot of eliminations; so i do basics before the first resolution state and after the last chain, what i'm not doing now but might in the future also as in DEFISE’s solution is to do basics after each chain; of course, not getting those 20 eliminations like with Cenoman’s solution makes things a bit harder and the one Cenoman found is really good.
my solution:

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n8r134c5 => r1c3 <> 9
 r1c5=8 - r1n1{c5 c2} - r1n5{c2 c3}
 r3c5=8 - r3c2{n8 n7} - r2c3{n7 n6} - r3c1{n678 n9}
 r4c5=8 - r4n4{c5 c2} - c2n1{r4 r1} - r1n5{c2 c3}

singles: ( n9r9c3  n9r3c1  n9r1c9  n9r8c4  n9r6c8  n9r4c6  n5r7c4  n1r9c5  n1r7c8  n1r2c4  n4r9c7  n4r7c3  n4r5c4  n4r3c5 n4r2c8  n1r4c7  n1r5c1  n1r1c2 n4r4c2  n5r1c3 )
intersections: n7r79c6 => r1c6 r2c6 <> 7
               n6r2c13 => r2c6 r2c9 <> 6
               n5r4c89 => r4c1 <> 5
               n2r89c1 => r4c1 r6c1 <> 2
               n7r2c13 => r3c2 <> 7
single: ( n8r3c2 )

n3r5c689 => r4c9 r6c7 <> 3
 r5c6=3 - r2n3{c6 c9} - r7n3{c9 c1} - r4n3{c1 c8}
 r5c8=3 -
 r5c9=3 -
 
bte.


evidently following the example of Cenoman i had to look for a solution without using the basics eliminations:

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n8r134c5 => r1c2 <> 5
 r1c5=8 - c5n1{r1 r9} - c7n1{r9 r4} - r5n1{c8 c1} - c2n1{r4 r1}
 r3c5=8 - r3c2{n8 n7} - r2c3{n7 n6} - r3c1{n678 n9} - r1c3{n679 n5}
 r3c5=8 - c5n4{r4 r3} - c7n4{r3 r9} - c7n1{r9 r4} - c2n1{r4 r1}


then the same as above.

[DEFISE]

... what i'm not doing now but might in the future also as in DEFISE’s solution is to do basics after each chain;

Hi P.O.
Indeed I think it's quite natural, unless you count subsets as one step (in addition to chains) and you want to minimize the number of steps. That's what Denis does with his "Fewer Steps" algorithm.
For my part, I only try to minimize the number of chains (of maximum length given) and I don't care about the number of subsets.

[P.O.]

hi DEFISE,
i agree, my goal is also to minimize the number and complexity of chains and i think i will get rid of the basics between chains even if sometimes it is useless as it is the case for some intersections too.

[denis_berthier]

... what i'm not doing now but might in the future also as in DEFISE’s solution is to do basics after each chain;

Indeed I think it's quite natural, unless you count subsets as one step (in addition to chains) and you want to minimize the number of steps. That's what Denis does with his "Fewer Steps" algorithm.
For my part, I only try to minimize the number of chains (of maximum length given) and I don't care about the number of subsets.

Hi François,
This is not a correct description of my version of the fewer step algorithm.
It takes as a parameter a resolution theory RT0 whose rules one wants to consider as zero-step. RT0 can be W1, S2, S3, or S4 (S for Subsets and the digit for their size). This is described in detail in the Basic User Manual for CSP-Rules.

I usually choose RT0=W1, because I consider any other choice is cheating on the number of steps, especially as a single Subset usually makes a lot of eliminations. How would you justify not counting steps that make most of the eliminations in a resolution path, e.g. in some of mith's puzzles that have tens of Subsets?

[DEFISE]

Hi François,
This is not a correct description of my version of the fewer step algorithm.
It takes as a parameter a resolution theory RT0 whose rules one wants to consider as zero-step. RT0 can be W1, S2, S3, or S4 (S for Subsets and the digit for their size). This is described in detail in the Basic User Manual for CSP-Rules.

Ok.

I usually choose RT0=W1, because I consider any other choice is cheating on the number of steps, especially as a single Subset usually makes a lot of eliminations. How would you justify not counting steps that make most of the eliminations in a resolution path, e.g. in some of mith's puzzles that have tens of Subsets?

It's not very satisfying not to count subsets, it's true.
But it's also true that a pair weighs very little compared to a whip[6], in terms of complexity.

[yzfwsf]

Swordfish in rows: 1r257c148 => -1r1c4 -1r4c1 -1r4c8 -1r9c8
Swordfish in rows: 4r257c348 => -4r3c4 -4r3c8 -4r4c3 -4r4c4 -4r9c3 -4r9c8
Hidden pairs: 14b9p27 => -3r7c8 -5r7c8 -8r7c8 -2r9c7 -3r9c7 -6r9c7
Swordfish in rows: 9r368c148 => -9r1c8 -9r4c4 -9r4c8 -9r9c1

whip[6]: r1n1{c2 c5}- r2n1{c4 c1}- c2n1{r1 r4}- r4n4{c2 c5}- c5n8{r4 r3}- b1n8{r3c1 .} => -5r1c2

Single(s): 5r1c3, 9r1c9, 9r3c1, 9r4c6, 9r6c8, 9r8c4, 9r9c3, 5r7c4, 1r7c8, 4r9c7, 4r3c5, 4r2c8, 4r4c2, 1r4c7, 1r5c1, 1r2c4, 1r1c2, 4r5c4, 4r7c3, 1r9c5
Box/Line: 5r6b4 => -5r4c1
Box/Line: 2c3b4 => -2r4c1 -2r6c1
Box/Line: 6b1r2 => -6r2c6 -6r2c9
Box/Line: 7b8c6 => -7r1c6 -7r2c6
Box/Line: 7r2b1 => -7r3c2
Single(s): 8r3c2
Naked pairs: 38c9r27 => -3r4c9 -8r4c9 -3r5c9 -8r5c9 -3r9c9
Box/Line: 8b6c8 => -8r1c8 -8r8c8
Naked pairs: 26r5c39 => -2r5c6 -6r5c6 -2r5c8 -6r5c8
Single(s): 2r1c6, 6r9c6, 3r8c5, 7r7c6, 5r8c2, 5r6c1
Box/Line: 6c9b6 => -6r4c8 -6r6c7
Box/Line: 6r6b5 => -6r4c4 -6r4c5

whip[2]: c2n3{r9 r6}- b6n3{r6c7 .} => -3r9c8
STTE

Remark:
without the initial basics we can start by eliminating 5r1c2 with this g-whip:
g-whip[8]: c2n1{r1 r4}- c7n1{r4 r9}- c5n1{r9 r1}- r2n1{c4 c1}- b1n8{r2c1 r3c12}- c5n8{r3 r4}- c5n4{r4 r3}- c7n4{r3 .} => -5r1c2
and then continue as above.

Maybe there is a loop here?
My solver output:

Hidden Text: Show

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Memory Chain: Start From 5r1c2 causes 8 to disappear in Box 1 => r1c2<>5
5r1c2- 1r1(c2=c5) - 1r2(c4=c1) - 1b4(p4=p2) - 4r4(c2=c5) - 8c5(r4=r3) - 8b1(p78=.)

[DEFISE]

whip[6]: r1n1{c2 c5}- r2n1{c4 c1}- c2n1{r1 r4}- r4n4{c2 c5}- c5n8{r4 r3}- b1n8{r3c1 .} => -5r1c2

Maybe there is a loop here?

Yes there is a loop because 1r1c2 is used two times as a left candidate: r1n1{c2 and c2n1{r1
Denis Berthier eliminated whips with these kind of loop for his statistics, but they meet the definition of a whip.
So, the corresponding elimination is justified.
I did not exclude such loops to simplify my software and not to do exactly like Denis Berthier in SudokuRules.