The New Sudoku Players' Forum

[Jasper32]

I have been working on the puzzle below for longer that I care to admit and I got stuck. At that point, I entered the puzzle in a “solver” and I arrived where I am now. At this point the “solver” gave the following move:

Triple Forcing Chain 2/5/8 in r1c3 implies r9c2 must be 7

Triple Forcing Chain 2/5/8 in r1c3 implies r9c7 must be 3

I have never heard of a TRIPLE forcing chain. Hopefully someone will be able to explain it to me and perhaps refer me to a link where I can read more about it.

Thanks

Code: Select all

 *-----------*
 |1..|...|..6|
 |.6.|4..|.3.|
 |..7|.2.|8..|
 |---+---+---|
 |..9|583|4..|
 |...|2.6|...|
 |..4|917|6..|
 |---+---+---|
 |..1|.3.|2..|
 |.8.|..4|.9.|
 |5..|...|..1|
 *-----------*


 *-----------*
 |14.|3..|..6|
 |.6.|4..|.3.|
 |..7|62.|8.4|
 |---+---+---|
 |6.9|583|4..|
 |...|246|...|
 |..4|917|6..|
 |---+---+---|
 |491|735|268|
 |.8.|164|.9.|
 |5.6|892|.41|
 *-----------*

 
 *-----------------------------------------------------------*
 | 1     4     258   | 3     57    89    | 579   27    6     |
 | 29    6     258   | 4     57    189   | 1579  3     2579  |
 | 39    35    7     | 6     2     19    | 8     15    4     |
 |-------------------+-------------------+-------------------|
 | 6     127   9     | 5     8     3     | 4     127   27    |
 | 378   1357  35    | 2     4     6     | 159   1578  359   |
 | 38    235   4     | 9     1     7     | 6     258   35    |
 |-------------------+-------------------+-------------------|
 | 4     9     1     | 7     3     5     | 2     6     8     |
 | 237   8     23    | 1     6     4     | 357   9     57    |
 | 5     37    6     | 8     9     2     | 37    4     1     |
 *-----------------------------------------------------------*


[eleven]

Dont see the triple forcing chain now, but if r9c2=3, then r8c3=2 and you have a pair 58 in r12c3, which leaves 3 in r3c2.

[Jasper32]

Thanks Eleven. What Iam primarily interested in is finding out what a Triple Forcing Chain is and what is the logic behind the step "Solvers" answer at this point.

[daj95376]

Thanks Eleven. What Iam primarily interested in is finding out what a Triple Forcing Chain is and what is the logic behind the step "Solvers" answer at this point.

Since a forcing chain is based on all of the candidates in a cell, and [r1c3] has three candidates, I suspect that the triple qualifier came from this property.

See Sudopedia for the logic behind forcing chains. Have booze handy!

[ronk]

Triple Forcing Chain 2/5/8 in r1c3 implies r9c2 must be 7
...
I have never heard of a TRIPLE forcing chain.
...

Code: Select all

 *-----------------------------------------------------------*
 | 1     4     258   | 3     57    89    | 579   27    6     |
 | 29    6     258   | 4     57    189   | 1579  3     2579  |
 | 39    35    7     | 6     2     19    | 8     15    4     |
 |-------------------+-------------------+-------------------|
 | 6     127   9     | 5     8     3     | 4     127   27    |
 | 378   1357  35    | 2     4     6     | 159   1578  359   |
 | 38    235   4     | 9     1     7     | 6     258   35    |
 |-------------------+-------------------+-------------------|
 | 4     9     1     | 7     3     5     | 2     6     8     |
 | 237   8     23    | 1     6     4     | 357   9     57    |
 | 5     37    6     | 8     9     2     | 37    4     1     |
 *-----------------------------------------------------------*


In this case, "triple forcing chains" just means that there is an implication chain from each of the three possibilities of cell r1c3 that leads to r9c2=7.

Code: Select all

Either r1c3=2 or r1c3=5 or r1c3=8

r1c3=2 -> r8c3=3 -> r9c2=7
r1c3=5 -> r3c2=3 -> r9c2=7
r1c3=8 -> r1c6=9 -> r3c6=1 -> r3c8=5 -> r3c2=3 -> r9c2=7

Since r9c2=7 for each of the three possibilities, r9c2 must ultimately be a 7

[edit: r3c2's were typos r2c2; thanks to Cec]

[eleven]

To search for triple forcing chains is long and boring. You have 13 cells with 3 candidates here. And - if counted right - 32 houses, where a number is 3 times. How do you decide, where to start ?

I like to start with unique rectangle. Here 17 in r45c29:
r1c8=7 -> r4c9=7 -> r8c9=5 -> r6c9=3 -> r6c1=8
r3c8=1 -> r3c6=9 -> r3c1=3 -> r6c1=8
r9c2=7 -> r5c1=7 -> r6c1=8

[Jasper32]

Thanks to all for your input. DAJ, A friend odf mine, Jack....Jack Daniels will be researching Forcing chains again. Especial thanks to Ronk for his detailed explanation to the "triple forcing chain". Without that I don't think I would have ever found it. And Eleven, I don't think I would ever know where to start looking to find such a thing. Baffling!...you bet.

[Sudtyro]

Code: Select all

 *-----------------------------------------------------------*
 | 1     4    B258   | 3     57    89    | 579   27    6     |
 |B29    6    B258   | 4     57    189   | 1579  3     2579  |
 |C39   B35    7     | 6     2     19    | 8     15    4     |
 |-------------------+-------------------+-------------------|
 | 6     127   9     | 5     8     3     | 4     127   27    |
 |C378   1357  35    | 2     4     6     | 159   1578  359   |
 |C38    235   4     | 9     1     7     | 6     258   35    |
 |-------------------+-------------------+-------------------|
 | 4     9     1     | 7     3     5     | 2     6     8     |
 | 23-7  8     23    | 1     6     4     | 357   9     57    |
 | 5    A37    6     | 8     9     2     | 37    4     1     |
 *-----------------------------------------------------------*

Forcing chains are certainly useful, but as eleven mentioned, how do you decide where to start when solving manually? My own preference is to look for Alternating Inference Chains (AIC). The placement of digit 7 in r9c2 can be derived from the following AIC that comprises mostly simple bivalue cells:
(7=3)r9c2 - (3)r3c2 = (3-9)r3c1 = (9-2)r2c1 = (2)r8c1 => r8c1 <> 7,
and therefore r9c2 = 7.
An AIC starts and ends with strong-inference links. Then, any candidate that can "see" (weakly link to) both ends of the chain can be eliminated. This AIC is a nice example of the case where the chain's starting and ending digits are different but belong to the same house (b7). Digit (7)r8c1 can "see" both (7)r9c2 and (2)r8c1, which have a derived strong inference (meaning at least one must be true). So, (7)r8c1 must be false.

A different AIC having only three nodes can be developed from Almost Locked Sets (ALS), denoted by letters A,B,C in the grid above. In Eureka notation the AIC is written [with help from Mike Barker] as:
(7=3)r9c2 - (3=2589)r1c3|r2c13|r3c2 - (9=387)r356c1 => r8c1 <> 7.
The first and last digits in each ALS are the ones used to form the weak links. The linking digits in each ALS occur only once, so these ALS are particularly easy to use in an AIC. This AIC is also known as an ALS-XY-Wing rule move.

Other interesting methods no doubt exist for the placement of (7)r9c2 or (3)r9c7. Perhaps other readers could contribute to the list.

[hobiwan]

Other interesting methods no doubt exist for the placement of (7)r9c2 or (3)r9c7. Perhaps other readers could contribute to the list.

I can't for r9c2 or r9c7, but there is a finned X-Wing in r36/c28, fin r6c9 => r5c8<>5

[daj95376]

Other interesting methods no doubt exist for the placement of (7)r9c2 or (3)r9c7. Perhaps other readers could contribute to the list.

Code: Select all

+--------------------------------------------------------------------+
|  1      4     c258   |  3      57     89    |  579    27     6     |
|  29     6     c258   |  4      57     189   |  1579   3      2579  |
|  39    b35     7     |  6      2      19    |  8      15     4     |
|----------------------+----------------------+----------------------|
|  6      127    9     |  5      8      3     |  4      127    27    |
|  378    1357   35    |  2      4      6     |  159    1578   359   |
|  38     235    4     |  9      1      7     |  6      258    35    |
|----------------------+----------------------+----------------------|
|  4      9      1     |  7      3      5     |  2      6      8     |
|  237    8     d23    |  1      6      4     |  357    9      57    |
|  5     a37     6     |  8      9      2     |  37     4      1     |
+--------------------------------------------------------------------+

  a        b        c          d        a
7-[r9c2]-3-[r3c2]-5-[r12c3]-28-[r8c3]-3-[r9c2] => [r9c2]=7


[eleven]

I already mentioned this above. For me all these notations and interpretations for the same logic are more confusing than helpful, but each notation might have its benefit. So please correct this try to write it as AIC. We have
r3c2=3 or r3c2=5 -> r12c3=28 -> r8c3=3
or
r8c3=3 or r8c3=2 -> r12c3=58 -> r3c2=3

(3=5)r3c2-(5=2&8)r12c3-(2=3)r8c3
So one of r3c2 (left side) and r8c3 (right side) must be 3.

And as ALS:
A={35}, B={2358},x=5,z=3

[ronk]

For me all these notations and interpretations for the same logic are more confusing than helpful, but each notation might have its benefit. So please correct this try to write it as AIC.

I think nice-loop (NL) notation, not Eureka notation, is the most common notation on this forum. After all, NL notation originated here. Although I would write it a little different, as ...

r9c2 -3- r3c2 -5- ALS:r12c3 -2- r8c3 -3- r9c2 => r9c2<>3

... IMO daj95376 was the only one to get the notation "right." Also in NL notation, your ALS xz-rule can be written ...

r9c2 -3- r3c2 -5- ALS:(r12c3 =5|3= r8c3) -3- r9c2 => r9c2<>3

[edit: replace r9c2=7 placement with r9c2<>3 elimination]

[eleven]

Ok, i am a player, no theory. So i solve this puzzle my way (how does program after r9c2=7 ?)

Code: Select all

 *-----------------------------------------------------------*
 | 1     4     258   | 3     57    89    | 579   27    6     |
 | 29    6     258   | 4     57    189   | 1579  3     2579  |
 | 39   #35    7     | 6     2     19    | 8     15    4     |
 |-------------------+-------------------+-------------------|
 | 6     127   9     | 5     8     3     | 4     127   27    |
 | 378   1357 #35    | 2     4     6     | 159   1578 -359   |
 |-38    2-35  4     | 9     1     7     | 6     258  #35    |
 |-------------------+-------------------+-------------------|
 | 4     9     1     | 7     3     5     | 2     6     8     |
 | 237   8     23    | 1     6     4     | 357   9     57    |
 | 5     37    6     | 8     9     2     | 37    4     1     |
 *-----------------------------------------------------------*

I look at the pairs:

35: r6c9=3 or r6c9=5 -> r3c8=5 -> r3c2=3: r6c2!=3
one more: r3c2=3 -> r12c3=5 -> r5c3=3: r5c9!=3, r6c12!=3

Now x-wing 5, xy-wing 257, y(w)-wing 37 (link 3), y(w)-wing 57 (link 5)

Code: Select all

 +---------------------------------------------+
 | 1   4   258  | 3 #57  89   |#579  27   6    |
 | 29  6   258  | 4 #57  189  | 159  3   #257  |
 | 39  35  7    | 6  2   19   | 8    15   4    |
 |--------------+-------------+----------------|
 | 6   12  9    | 5  8   3    | 4    127  27   |
 | 7   13  35   | 2  4   6    | 15   8    9    |
 | 8   25  4    | 9  1   7    | 6    25   3    |
 |--------------+-------------+----------------|
 | 4   9   1    | 7  3   5    | 2    6    8    |
 | 23  8   23   | 1  6   4    |#57   9   #57   |
 | 5   7   6    | 8  9   2    | 3    4    1    |
 +---------------------------------------------+


Here uniqueness pattern 57:
r2c7=5 or r3c8=5 or r1c8=7.

r1c8=7 -> r1c5=5: r1c7!=5
r3c8=5 -> r6c8=2 -> r1c8=7 -> r1c5=5: r2c5!=5
r2c7=5 -> r8c7=7: r1c7!=7

[daj95376]

I already mentioned this above.

Sorry eleven!!! I missed the fact that I'd followed your chain in reverse.

[Sudtyro]

So please correct this try to write it as AIC.
(3=5)r3c2-(5=2&8)r12c3-(2=3)r8c3

Your AIC appears correct. Per Eureka notation convention, I would suggest changing only the candidate order in the intermediate 2-cell ALS to make the linking digit clearer:
(3=5)r3c2 - (5=82)r12c3 - (2=3)r8c3 => r9c2 <> 3.
Your ampersand (preferred by Myth Jellies here) is optional.

To avoid the intermediate ALS, another (grouped) AIC for the same elimination is:
(3=5)r3c2 - (5)r12c3 = (5-3)r5c3 = (3)r8c3 => r9c2 <> 3.

In Eureka notation, the AIC for your ALS-XZ rule move would be:
(3=5)r3c2 - (5=283)r128c3 => r9c2 <> 3.
Regarding your notation issue, I generally prefer this AIC's notation to ronk's equivalent Nice Loop shown above. Myth Jellies provides an informative comparison of the two conventions here.

And to avoid chains altogether, one can also apply Subset Counting to subset (2358)r3c2|r12c3|r8c3, labeled b,c,d in daj95376's grid. Candidates 2,5 and 8 each have a multiplicity of one. Candidate 3 has a multiplicity of two, so the 4-cell subset has a total multiplicity of five. Placement of (3)r9c2 would remove the 3's in the subset and thereby reduce the subset multiplicity to a value of three, which is one less than the cell count. Hence, r9c2 <> 3. It's an interesting method, but, admittedly, I can usually only spot the subsets in hindsight.