The New Sudoku Players' Forum

[Brendan]

Following on from the Sherlock line technique, suggested as Convergence in "Is this a new technique?" comes the Sherlock block technique. If the simple techniques are gone, and you don't want to spend time holding multiple contingencies in you head to use the advanced techniques, this technique may help. I will use the example provided by Sue de Coq for another technique.

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        4      3|8|9      3|5 |  2  3|9  6 |      1      7    5|8
  1|3|5|9  1|2|3|8|9        6 |  7  3|9  4 |  2|8|9  2|3|8  2|5|8
        7      2|3|9      2|3 |  8    5  1 |  2|4|9  2|3|4      6
------------------------------+------------+---------------------
        2          5        9 |  4    8  7 |      3      6      1
      3|6      3|6|7      3|7 |  9    1  5 |  2|4|8  2|4|8    2|8
        8        1|4      1|4 |  6    2  3 |      5      9      7
------------------------------+------------+---------------------
      1|6    1|2|4|6      1|4 |  5    7  9 |    2|8  1|2|8      3
      3|5      2|3|7  2|3|5|7 |  1    4  8 |      6    2|5      9
    1|5|9        1|9        8 |  3    6  2 |      7    1|5      4


Sherlocking row 1 on row 3 we get the following (where the numbers represent candidate options, the spaces represent the separation between blocks and the dot represents no unknowns in the block) Here, the capitol letters represent the possibilities in row 1 and the small letters represent possibilities in row 3.

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Row 1
35 9 8 A
83 9 5 B
95 3 8 C

Row 3
23 . 94 C    a
32 . 94 C    b
92 . 43 AB   c
93 . 42 (Crossed out)
93 . 24 (Crossed out)



This immediately eliminates 2 from R3 in block 3, and consequently 2 from R2 in block 1. Now the Sherlock block techique can be applied.

This technique begins by writing out all the options for two rows in a single block. In this example we will use row 1 and 3, and will use only those that have survived the Sherlock line comparison. More importantly, we can use the linkages that have resulted from the above comparison. For example, A only links to c, while C links to both a and b. In full

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Block
Row 1  Row 3 Linkage
35       92  Ac
83       92  Bc
95       23  Ca
95       32  Cb


Now we look for common factors. In this example, {2,3,9} are common to all possible combinations. Therefore they must be in either row 1 or row 3 of this block, and we can eliminate them from row 2. This leaves us with the top 3 rows looking like

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


The elimination of the 9 at r2c1 provides the break that we need, and the rest is resolved largely through finding the singles and hidden singles.

Does this explanation make sense? Is the Sherlock block technique simply a rehashing of another technique?

Brendan

[Nick67]

Does this explanation make sense? Is the Sherlock block technique simply a rehashing of another technique?

The explanation makes sense to me: seems likes a good technique!
But I can't say whether or not it's related to an existing technique
(just not enough knowledge on my part).

[Jeff]

Hi Brendan,

Your technique has similar features as Myth Jellies' POM, both require to map out the entire solution set(s).

Listed below are the solution sets for row 1 and row 2 of the toughest puzzle even known. I got stuck from here suspecting that mistakes have been made at some points. Could you demonstate how to move forward?

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Row 1
1 2 5 8 * 3 * * 6 A
1 2 5 8 * 6 * * 3 B
2 1 5 8 * 3 * * 6 C
2 1 5 8 * 6 * * 3 D
5 1 2 8 * 3 * * 6 E
5 1 2 8 * 6 * * 3 F
5 1 8 3 * 2 * * 6 G
5 1 8 3 * 6 * * 2 H
5 1 8 6 * 2 * * 3 I
5 1 8 6 * 3 * * 2 J
5 2 1 8 * 3 * * 6 K
5 2 1 8 * 6 * * 3 L
5 2 8 1 * 3 * * 6 M
5 2 8 1 * 6 * * 3 N
5 2 8 3 * 1 * * 6 O
5 2 8 3 * 6 * * 1 P
5 2 8 6 * 1 * * 3 Q
5 2 8 6 * 3 * * 1 R
6 1 5 8 * 2 * * 3 S
6 1 5 8 * 3 * * 2 T
6 2 5 8 * 1 * * 3 U
6 2 5 8 * 3 * * 1 V

Row 2
1 * 2 3 * 4 8 6 *
1 * 2 6 * 4 3 8 *
1 * 2 6 * 4 8 3 *
1 * 2 8 * 4 3 6 *
1 * 2 8 * 4 6 3 *
1 * 4 3 * 2 8 6 *
1 * 4 6 * 2 8 3 *
1 * 4 8 * 2 3 6 *
1 * 4 8 * 2 6 3 *
1 * 4 8 * 3 2 6 *
1 * 4 8 * 6 2 3 *
1 * 8 3 * 4 2 6 *
1 * 8 6 * 4 2 3 *
2 * 1 3 * 4 8 6 *
2 * 1 6 * 4 8 3 *
2 * 1 8 * 4 3 6 *
2 * 1 8 * 4 6 3 *
2 * 4 1 * 3 8 6 *
2 * 4 1 * 6 8 3 *
2 * 4 3 * 1 8 6 *
2 * 4 3 * 6 8 1 *
2 * 4 6 * 1 8 3 *
2 * 4 6 * 3 8 1 *
2 * 4 8 * 1 3 6 *
2 * 4 8 * 1 6 3 *
2 * 4 8 * 3 6 1 *
2 * 4 8 * 6 3 1 *
2 * 8 1 * 4 3 6 *
2 * 8 1 * 4 6 3 *
2 * 8 3 * 4 6 1 *
2 * 8 6 * 4 3 1 *
6 * 1 8 * 4 2 3 *
6 * 2 1 * 4 8 3 *
6 * 2 3 * 4 8 1 *
6 * 2 8 * 4 3 1 *
6 * 4 1 * 2 8 3 *
6 * 4 3 * 2 8 1 *
6 * 4 8 * 1 2 3 *
6 * 4 8 * 2 3 1 *
6 * 4 8 * 3 2 1 *
6 * 8 1 * 4 2 3 *
6 * 8 3 * 4 2 1 *

Thanks in anticipation

[DanO]

Sherlocking is equivalent to any of the unbounded solving techniques It most closely matches recursion except it's processed in parallel.

You can arrive at the same results by following implication chains which a skilled Sudoku solver can do without writing the whole thing down. For example, for Brendan's puzzle, this chain shows that R2C3 cannot be 2 or 3...

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(R2C2=3->R1C3=5,(R2C3=2->R3C2=9)->R1C2=8->R1C9=X) -> R2C3<>3
(R2C2=2->R1C3=5,R3C4=3->R3C2=9->R1C2=8->R1C9=X) -> R2C2<>2


These implication chains also go by the name Trial and Error

[Sue De Coq]

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Sherlocking is equivalent to any of the unbounded solving techniques.

I don't see why Sherlocking is equivalent to Guessing - to start with, Guessing will solve any puzzle whereas I'm sure there must be some puzzles (though I don't have an example) that Sherlocking won't. Although inelegant, it's based on static analysis of the current grid so - unlike Guessing - won't ever require a hypothesized move to be backtracked. As Brendan pointed out, the method is perhaps closest in spirit in Nishio in that it's an inelegant technique that should be invoked as a last-ditch resource before Guessing.

These implication chains also go by the name Trial and Error

I appreciate that the precise dividing line between Logical and Trial & Error methods is a matter of opinion but I believe my above comment applies even more strongly to Forced Chains.

BTW An elegant solution to the given puzzle is posted under the topic 'Two-Sector Disjoint Subsets'.

[Sue De Coq]

I've now implemented Sherlocking (I call the strategy 'Permutate Adjacent Sectors') in Rows and Columns (not yet Boxes). Unfortunately, it appears to eliminate very few candidates missed by Forced Chains. I tried the strategy on my collection of Unsolvable puzzles but, although it eliminated a few new candidates, it didn't help sufficiently in order to solve any of them. It didn't find anything in the Toughest-Known Puzzle.

Brendan introduced the technique as an alternative to Forced Chains and it's true that, when Forced Chains is swtiched off, Permutate Adjacent Sectors sometimes proves effective. Consider the following puzzle:

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 . . . | . . . | . . .
 . 9 . | 8 . 5 | . 4 .
 . . 6 | . 7 . | 8 . .
-------+-------+------
 . 5 . | . . . | . 3 .
 . . 1 | . 8 . | 6 . .
 . 4 . | . . . | . 2 .
-------+-------+------
 . . 2 | . 6 . | 7 . .
 . 6 . | 1 . 9 | . 5 .
 . . . | . . . | . . .


25 straightforward moves bring us here:

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      1    8  4|5 |      2    3|4|9    6 |  3|5|9  7      3|5|9
      7    9    3 |      8        1    5 |      2  4          6
    4|5    2    6 |  3|4|9        7  3|4 |      8  1      3|5|9
------------------+----------------------+---------------------
      2    5  8|9 |      6      4|9  1|7 |    1|4  3        7|8
      3    7    1 |    4|5        8    2 |      6  9        4|5
      6    4  8|9 |  3|5|9    3|5|9  1|7 |    1|5  2        7|8
------------------+----------------------+---------------------
  4|5|9  1|3    2 |  3|4|5        6  3|4 |      7  8    1|3|4|9
      8    6    7 |      1    2|3|4    9 |    3|4  5      2|3|4
  4|5|9  1|3  4|5 |      7  2|3|4|5    8 |  3|4|9  6  1|2|3|4|9


For the next move, Chains (without Tables) gives us:

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Consider the chain r3c9-5-r3c1~5~r7c1-5-r7c4~5~r5c4-5-r5c9.
The cell r5c9 must contain the value 5 if the cell r3c9 doesn't.
Therefore, these two cells are the only candidates for the value 5 in Column 9.
- The move r1c9:=5 has been eliminated.
Consider the chain r3c1-5-r3c9-5-r5c9-5-r5c4~5~r7c4-5-r7c1.
The cell r7c1 must contain the value 5 if the cell r3c1 doesn't.
Therefore, these two cells are the only candidates for the value 5 in Column 1.
- The move r9c1:=5 has been eliminated.
Consider the chain r5c4-5-r5c9-5-r3c9-5-r3c1-5-r7c1-5-r7c4.
The cell r7c4 must contain the value 5 if the cell r5c4 doesn't.
Therefore, these two cells are the only candidates for the value 5 in Column 4.
- The move r6c4:=5 has been eliminated.
Consider the chain r9c9-1-r7c9-9-r7c1-9-r9c1~4~r9c9.
When the cell r9c9 contains the value 4, it likewise contains the value 1 - a contradiction.
Therefore, the cell r9c9 cannot contain the value 4.
- The move r9c9:=4 has been eliminated.
Consider the chain r3c4~4~r3c1-5-r3c9-5-r5c9-4-r5c4.
When the cell r3c4 contains the value 4, so does the cell r5c4 - a contradiction.
Therefore, the cell r3c4 cannot contain the value 4.
- The move r3c4:=4 has been eliminated.
The values 4 and 5 occupy the cells r5c4 and r7c4 in some order.
- The move r7c4:=3 has been eliminated.
Consider the chain r9c7~4~r9c3-4-r1c3-5-r1c7-9-r9c7.
When the cell r9c7 contains the value 4, it likewise contains the value 9 - a contradiction.
Therefore, the cell r9c7 cannot contain the value 4.
- The move r9c7:=4 has been eliminated.
Consider the chain r7c9-1-r7c2-1-r9c2~3~r9c7~9~r7c9.
When the cell r7c9 contains the value 9, it likewise contains the value 1 - a contradiction.
Therefore, the cell r7c9 cannot contain the value 9.
- The move r7c9:=9 has been eliminated.
The cell r7c1 is the only candidate for the value 9 in Row 7.


Whereas Sherlocking gives us:

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The 6 permutations of Row 7 and 14 permutations of Row 9 combine legally in 11 different ways.
Row 7:
5-1-2-3-6-4-7-8-9
4-1-2-5-6-3-7-8-9
5-1-2-4-6-3-7-8-9
9-1-2-5-6-3-7-8-4
9-1-2-5-6-4-7-8-3
9-3-2-5-6-4-7-8-1
Row 9:
4-1-5-7-2-8-3-6-9
5-1-4-7-2-8-3-6-9
9-1-5-7-2-8-3-6-4
4-1-5-7-2-8-9-6-3
5-1-4-7-2-8-9-6-3
9-1-5-7-2-8-4-6-3
4-1-5-7-3-8-9-6-2
5-1-4-7-3-8-9-6-2
9-1-5-7-3-8-4-6-2
9-1-4-7-5-8-3-6-2
9-1-5-7-4-8-3-6-2
4-3-5-7-2-8-9-6-1
5-3-4-7-2-8-9-6-1
9-3-5-7-2-8-4-6-1
No combination has the value 3 as a candidate for the cell r7c4.
No combination has the value 4 as a candidate for the cell r7c4.
No combination has the value 4 as a candidate for the cell r9c5.
No combination has the value 4 as a candidate for the cell r9c9.
No combination has the value 5 as a candidate for the cell r7c1.
No combination has the value 5 as a candidate for the cell r9c5.
- The moves r7c4:=3, r7c4:=4, r9c5:=4, r9c9:=4, r7c1:=5 and r9c5:=5 have been eliminated.
The value 5 is the only candidate for the cell r7c4.


It's a matter of taste which you prefer. (Personally, I think they're both damn difficult)

[DanO]

Sherlocking will find any pattern that is contained in the rows and columns sherlocked. If extended to the entire board, sherlock will solve any valid puzzle. It's doing exactly the same thing as recursion only in a slightly different order. And since recursion is a structured form of guessing...

I still think that Sherlocking is best used as a tool to search for simple patterns that have so far been overlooked. You would need to extend your program so it can apply sherlock (or recursion) to an arbitrary set of cells. Then start with a large portion of an unsolvable (with simple rules) puzzle and see if sherlock makes progress. Then try to reduce the area involved until you find a minimum set of cells required to progress.


I tried again to sherlock the first 3 rows of the Toughest Known Puzzle. This time I double checked each step and still made errors. I wouldn't recommend sherlocking to anyone that isn't very methodical.

Anyhow, I reduced the first three rows down to

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A   518|6.3|..2         518|3.6|..2
B   528|6.3|..1         528|3.6|..1
L   518|3.2|..6
Q   512|8.6|..3         521|8.6|..3   125|8.6|..3   215|8.6|..3         
U   615|8.2|..3
V   625|8.1|..3

b   1.4|3.2|86.   : DV
i   2.4|1.3|86.   : CU          2.4|3.1|86.
s   6.2|1.4|83.   : AL
q   6.1|8.4|23.   : BM
u   6.8|3.4|21.   : Q     

    .49|82.|6.1   : AD   sA      .94|82.|6.1   .94|82.|1.6
    .94|12.|8.6   : ABQST   qBuQ               .49|12.|8.6
    .94|68.|2.1   : LM   sL      .49|68.|2.1
    .98|64.|2.1   : OUV   bViU


Which can be recombined into the (now reduced) permitted values:

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R1:   5612 12 1285 | 638  .  1236 | .   .  1236
R2:   126  .  1248 | 138  .  1234 | 28  136 .
R3:   .    49 489  | 168  248  .  | 268 .   16


And after all that, the puzzle still doesn't progress

[Brendan]

I see Sherlocking to be similar to similtaneous equations. If applied in the correct way to the whole of the nine rows or columns, then it may be able determine the whole puzzle, though I haven't tried that yet. It's not guessing, but it certainly can be cumbersome. Using it on just two rows is the equivalent of using just two equations in a multivariate set of equations. It may be immediately useful, as the following example shows, but the immediacy of its usefulness isn't guaranteed.

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a + b + c + d + e = 0  -> Eq 1
a + b + c + d = 1      -> Eq 2

Eq 1 - Eq 2 gives

e = -1  Immediately useful

a + b + c = 0  -> Eq 1
d + b + c = 1  -> Eq 2

Eq 1 - Eq 2 gives

a - d = -1  Not immediately useful



Is this trial and error? In the sense that the partial approach will not guarantee a useful result, yes. But that is true of all techniques (e.g. if I look for hidden triples, will I find any?) The very act of looking for something that is not guaranteed to exist is trial and error. In this case, looking requires some systematic work. However, it is far from "What if I place this number here?" as most people understand trial and error to be.

Since simultaneous equations is the algebraic parallel, I wonder if there is an eigenvalue parallel in Sudoku? Any mathematicians out there that can answer?

Now for the Toughest Known Sudoku. I didn't find anything immediately useful using the first two rows. However, using the first two columns yields

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Column 1
16.2.5..9 A
16.5.2..9 B
26.5.1..9 C
26.5.9..1 D
26.9.5..1 E
56.2.1..9 F
56.2.9..1 G
56.9.2..1 H
61.2.5..9 I
61.5.2..9 J
62.5.1..9 K
62.5.9..1 L
62.9.5..1 M

Column 2
1.2..934. a F
1.2..943. b F
1.4..239. c DELM
1.4..293. d DELM
1.4..923. e CFK
1.9..234. f CDEKLM
1.9..243. g CDEKLM
2.1..934. h F
2.1..943. i F
2.4..139. j GH
2.4..193. k GH
2.4..913. l ABFIJ
2.4..931. m ABFIJ
2.9..134. n ABGHIJ
2.9..143. o ABGHIJ

No eliminations found
 
Sherlock Block Box1 C1 on C2
56 12 F     ab
25 14 EDFC  cde
62 14 LMK   cde
26 19 CDE   fg
62 19 KLM   fg
56 21 F     hi
56 24 FGH   jklm
16 24 AB    lm
61 24 IJ    lm
16 29 AB    no
56 29 GH    no
61 29 IJ    no

56 14 F     e (Added on correction - thanks Sue)

Common Factors - {6,2}

6 is a trivial solution as it is limited to the two columns used for the technique (actually, it is only in C1)

There are no common factors
 


I was later able to eliminate 9 from R7C3 and R8C3 by Sherlocking (line) the C3 with the results of C1 on C2. However, as I am doing it manually, I worry that I am beginning to make mistakes.

{Correction, looks like I made a mistake already}

Brendan

[Sue De Coq]

Brendan,

I don't agree with your results. I agree with your permutations for C1 and C2 but I have many more legal pairings (53), one of which (#25) has no 2 in the first box. Personally, I can't make any progress with Toughest.

I've upgraded my implementation to consider all pairs of rows (or columns) - not just those that share a box. Things don't change much though, as the comparison is less stringent due to the lack of a box constraint.

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Column 1:
1-6-3-2-4-5-8-7-9
1-6-3-5-4-2-8-7-9
6-1-3-2-4-5-8-7-9
6-1-3-5-4-2-8-7-9
2-6-3-5-4-1-8-7-9
6-2-3-5-4-1-8-7-9
5-6-3-2-4-1-8-7-9
2-6-3-5-4-9-8-7-1
2-6-3-9-4-5-8-7-1
6-2-3-5-4-9-8-7-1
6-2-3-9-4-5-8-7-1
5-6-3-2-4-9-8-7-1
5-6-3-9-4-2-8-7-1

Column 2:
1-7-2-8-6-9-3-4-5
1-7-2-8-6-9-4-3-5
1-7-4-8-6-2-3-9-5
1-7-9-8-6-2-3-4-5
1-7-4-8-6-2-9-3-5
1-7-9-8-6-2-4-3-5
1-7-4-8-6-9-2-3-5
2-7-1-8-6-9-3-4-5
2-7-1-8-6-9-4-3-5
2-7-4-8-6-1-3-9-5
2-7-9-8-6-1-3-4-5
2-7-4-8-6-1-9-3-5
2-7-9-8-6-1-4-3-5
2-7-4-8-6-9-1-3-5
2-7-4-8-6-9-3-1-5

1. 1-6-3-2-4-5-8-7-9 X 2-7-9-8-6-1-3-4-5
2. 1-6-3-2-4-5-8-7-9 X 2-7-9-8-6-1-4-3-5
3. 1-6-3-2-4-5-8-7-9 X 2-7-4-8-6-9-1-3-5
4. 1-6-3-2-4-5-8-7-9 X 2-7-4-8-6-9-3-1-5
5. 1-6-3-5-4-2-8-7-9 X 2-7-9-8-6-1-3-4-5
6. 1-6-3-5-4-2-8-7-9 X 2-7-9-8-6-1-4-3-5
7. 1-6-3-5-4-2-8-7-9 X 2-7-4-8-6-9-1-3-5
8. 1-6-3-5-4-2-8-7-9 X 2-7-4-8-6-9-3-1-5
9. 6-1-3-2-4-5-8-7-9 X 2-7-9-8-6-1-3-4-5
10. 6-1-3-2-4-5-8-7-9 X 2-7-9-8-6-1-4-3-5
11. 6-1-3-2-4-5-8-7-9 X 2-7-4-8-6-9-1-3-5
12. 6-1-3-2-4-5-8-7-9 X 2-7-4-8-6-9-3-1-5
13. 6-1-3-5-4-2-8-7-9 X 2-7-9-8-6-1-3-4-5
14. 6-1-3-5-4-2-8-7-9 X 2-7-9-8-6-1-4-3-5
15. 6-1-3-5-4-2-8-7-9 X 2-7-4-8-6-9-1-3-5
16. 6-1-3-5-4-2-8-7-9 X 2-7-4-8-6-9-3-1-5
17. 2-6-3-5-4-1-8-7-9 X 1-7-9-8-6-2-3-4-5
18. 2-6-3-5-4-1-8-7-9 X 1-7-9-8-6-2-4-3-5
19. 2-6-3-5-4-1-8-7-9 X 1-7-4-8-6-9-2-3-5
20. 6-2-3-5-4-1-8-7-9 X 1-7-9-8-6-2-3-4-5
21. 6-2-3-5-4-1-8-7-9 X 1-7-9-8-6-2-4-3-5
22. 6-2-3-5-4-1-8-7-9 X 1-7-4-8-6-9-2-3-5
23. 5-6-3-2-4-1-8-7-9 X 1-7-2-8-6-9-3-4-5
24. 5-6-3-2-4-1-8-7-9 X 1-7-2-8-6-9-4-3-5
25. 5-6-3-2-4-1-8-7-9 X 1-7-4-8-6-9-2-3-5
26. 5-6-3-2-4-1-8-7-9 X 2-7-1-8-6-9-3-4-5
27. 5-6-3-2-4-1-8-7-9 X 2-7-1-8-6-9-4-3-5
28. 5-6-3-2-4-1-8-7-9 X 2-7-4-8-6-9-1-3-5
29. 5-6-3-2-4-1-8-7-9 X 2-7-4-8-6-9-3-1-5
30. 2-6-3-5-4-9-8-7-1 X 1-7-4-8-6-2-3-9-5
31. 2-6-3-5-4-9-8-7-1 X 1-7-9-8-6-2-3-4-5
32. 2-6-3-5-4-9-8-7-1 X 1-7-4-8-6-2-9-3-5
33. 2-6-3-5-4-9-8-7-1 X 1-7-9-8-6-2-4-3-5
34. 2-6-3-9-4-5-8-7-1 X 1-7-4-8-6-2-3-9-5
35. 2-6-3-9-4-5-8-7-1 X 1-7-9-8-6-2-3-4-5
36. 2-6-3-9-4-5-8-7-1 X 1-7-4-8-6-2-9-3-5
37. 2-6-3-9-4-5-8-7-1 X 1-7-9-8-6-2-4-3-5
38. 6-2-3-5-4-9-8-7-1 X 1-7-4-8-6-2-3-9-5
39. 6-2-3-5-4-9-8-7-1 X 1-7-9-8-6-2-3-4-5
40. 6-2-3-5-4-9-8-7-1 X 1-7-4-8-6-2-9-3-5
41. 6-2-3-5-4-9-8-7-1 X 1-7-9-8-6-2-4-3-5
42. 6-2-3-9-4-5-8-7-1 X 1-7-4-8-6-2-3-9-5
43. 6-2-3-9-4-5-8-7-1 X 1-7-9-8-6-2-3-4-5
44. 6-2-3-9-4-5-8-7-1 X 1-7-4-8-6-2-9-3-5
45. 6-2-3-9-4-5-8-7-1 X 1-7-9-8-6-2-4-3-5
46. 5-6-3-2-4-9-8-7-1 X 2-7-4-8-6-1-3-9-5
47. 5-6-3-2-4-9-8-7-1 X 2-7-9-8-6-1-3-4-5
48. 5-6-3-2-4-9-8-7-1 X 2-7-4-8-6-1-9-3-5
49. 5-6-3-2-4-9-8-7-1 X 2-7-9-8-6-1-4-3-5
50. 5-6-3-9-4-2-8-7-1 X 2-7-4-8-6-1-3-9-5
51. 5-6-3-9-4-2-8-7-1 X 2-7-9-8-6-1-3-4-5
52. 5-6-3-9-4-2-8-7-1 X 2-7-4-8-6-1-9-3-5
53. 5-6-3-9-4-2-8-7-1 X 2-7-9-8-6-1-4-3-5


[Brendan]

Thanks Sue

Looks like I did make a mistake. I had the 53 legal duets, but missed making the 56 14 coupling in the SBT. It was staring me in the face, and shows, along with a some previous postings by others, that because of the shear number of options, the method may be prone to making mistakes if worked out manually. (I suppose that forced chains are similar with these extreme Sudoku.) Brute force works, but it takes some methodical brute to do it manually.

Brendan

[Jeff]

......because of the shear number of options, the method may be prone to making mistakes if worked out manually. (I suppose that forced chains are similar with these extreme Sudoku.) Brute force works, but it takes some methodical brute to do it manually.

Two points regarding forcing chains:

1. Forcing chains developed so far do not garantee to solve every puzzle.
2. The Forcing chain technique is less painstaking in terms of manual operation, but 1 in every 5 chains may be missed out due to human errors.
   

[Myth Jellies]

I don't see why Sherlocking is equivalent to Guessing

If I understand Sherlock correctly:

Step 1: You come up with every possible legal combination in one row/column to create a few restrictions. This is hardly necessary since you can equivalently just see if each of your trials from step 2 allow you to legally fill in this row/column or result in a conflict.

Step 2: You apply as trials, or guesses as it were, every possible legal combination of digits that fill out a second row/column.

Step 3: You eliminate the erroneous trials that violate the restrictions from step 1.

Step 4: You examine the surviving trials for common traits--i.e. solving for or eliminating candidate digits from cells.

Obviously if steps 1 and 2 took the entire puzzle into consideration instead of just 22.222% of it, there would be no need for a step 4, and there would be no question about it being a guessing method.

The question is, what percentage of the puzzle would you let steps 1 & 2 apply to before you consider the method trial and error and/or guessing? If 22% is good, what about 33%, 50%, 66%, 98%?

It's up to each individual to decide, of course, but for some of us applying a trial and error method over any percentage of the puzzle would be unacceptable.

[Brendan]

The question is, what percentage of the puzzle would you let steps 1 & 2 apply to before you consider the method trial and error and/or guessing? If 22% is good, what about 33%, 50%, 66%, 98%?

It's up to each individual to decide, of course, but for some of us applying a trial and error method over any percentage of the puzzle would be unacceptable.

The same could be said about naked pairs. In both cases, Sherlock and naked pairs, they are making inferences about the range of candidates through knowledge of the restrictions inherent in the methodology. I don't think that it is a question of the percentage of the puzzle that determines whether a method is trial and error, but rather, it is the substitution for result and the abandonment of the search for inference and understanding, that is the dividing line. Although I argued above that Sherlock may technically have elements of trial and error, by the same arguments, so do all other techniques, and I don't believe that Sherlock violates the journey that is Sudoku. Of course, as Sue de Coq pointed out, elegence is another story.

Brendan

[PaulIQ164]

I disagree about the similarity with naked pairs. Yes, all techniques can be thought of in terms of trial and error, but I think there's an objective difference between the 'normal', strictly deductive methods, and Sherlocking (presuming of course that I understand what Sherlocking is exactly, which is far from certain).

[Brendan]

I think there's an objective difference between the 'normal', strictly deductive methods, and Sherlocking (presuming of course that I understand what Sherlocking is exactly, which is far from certain).

Could you elaborate further.

Thanks

Brendan