The New Sudoku Players' Forum

[MikeF]

In this puzzle:

*-----------*
|...|34.|..6|
|.6.|78.|1..|
|...|.65|.72|
|---+---+---|
|.48|52.|361|
|615|.3.|72.|
|23.|61.|.8.|
|---+---+---|
|12.|47.|6..|
|..6|.9.|...|
|8..|.56|...|
*-----------*


*-----------*
|...|34.|..6|
|.6.|78.|1..|
|...|.65|.72|
|---+---+---|
|.48|52.|361|
|615|.3.|72.|
|23.|61.|.8.|
|---+---+---|
|12.|47.|6..|
|..6|.9.|...|
|8..|.56|...|
*-----------*


{579} {5789} {12} {3} {4} {12} {589} {59} {6}
{459} {6} {249} {7} {8} {29} {1} {3459} {3459}
{34} {89} {1349} {19} {6} {5} {489} {7} {2}
{79} {4} {8} {5} {2} {79} {3} {6} {1}
{6} {1} {5} {89} {3} {489} {7} {2} {49}
{2} {3} {79} {6} {1} {479} {459} {8} {459}
{1} {2} {39} {4} {7} {38} {6} {359} {3589}
{3457} {57} {6} {128} {9} {138} {24} {134} {3478}
{8} {79} {3479} {12} {5} {6} {249} {1349} {3479}

The candidate 4 in (9,8) can be eliminated. Assume it is 4. Then (9,4) is 1, and (9,7) is 2, and (8,7) is 4 -- which is impossible because there is already a 4 in that box.

Rubylips' solver program revealed this to me. Until then I was stuck.

But this should be easy to find, no? How does one spot such a pattern?

[ronk]

The candidate 4 in (9,8) can be eliminated. Assume it is 4. Then (9,4) is 1, and (9,7) is 2, and (8,7) is 4 -- which is impossible because there is already a 4 in that box.

Rubylips' solver program revealed this to me. Until then I was stuck.

Something must have been lost during transcription. If r9c8=4, then r8c7=2.

[edit: Oh, it must be ...Assume it is 4. Then (9,4) is 1, and (9,7) is 2 -- which is impossible because there is already a 2 in that box at (8,7).

But to answer the question, I think it's trial and error. Focusing on grid areas with conjugate pairs and bivalued cells helps speed it up a little. Good luck speeds it up even more. ]

[Jeff]

But to answer the question, I think it's trial and error. Focusing on grid areas with conjugate pairs and bivalued cells helps speed it up a little. Good luck speeds it up even more. ]

It can be T&E only if you want it to be. Bilocation/bivalue plot can help you to identify these types of double implication chains in a non-T&E and human executable manner.

[bennys]

Code: Select all

+----------------+----------------+----------------+
| 579  5789 12   | 3    4    12   | 589  59   6    |
| 459  6    249  | 7    8    29   | 1    3459 3459 |
| 34   89   1349 | 19   6    5    | 489  7    2    |
+----------------+----------------+----------------+
| 79   4    8    | 5    2    79   | 3    6    1    |
| 6    1    5    | 89   3    489  | 7    2    49   |
| 2    3    79   | 6    1    479  | 459  8    459  |
+----------------+----------------+----------------+
| 1    2    39   | 4    7    38   | 6    359  3589 |
| 3457 57   6    | 128  9    138  |*24   134  3478 |
| 8   *79  *3479 | 12   5    6    | 249  1349*3479 |
+----------------+----------------+----------------+

I dont know if you prefer to see it that way but here it is using the almost locked sets.
A={R9C2,R9C3,R9C9}
B={R8C7}
LETS LOOK NOW ON R9C8 AND R9C7
They cant be 4,9 because of A and the cant be 4,2 because of B.


[rubylips]

bennys,
I have nearly finished work on an artificial brain that will replicate you perfectly! Here's the output from the latest version of my solver:

Code: Select all

Consider the cell r9c7.
When it contains the value 9, the values 3, 4 and 7 in Row 9 must occupy the cells r9c2, r9c3 and r9c9 in some order.
When it contains the value 2, the value 4 in Box 9 must occupy the cell r8c7.
Whichever value it contains, the cell r9c8 cannot contain the value 4.


Unfortunately, even after this elimination the puzzle is still hard to to solve.
Just to clarify, my solver first attempts to build chains from strong links only (i.e. bilocation links - though it will connect such links at cells that have more than two values). When this fails, it will incorporate weak links. The first chain found is reported, which is generally the shortest, though this isn't necessarily the chain that leads to the most rapid solution. I quite like this method, however, as I think it replicates human techniques to a reasonable degree.

[ronk]

But to answer the question, I think it's trial and error. Focusing on grid areas with conjugate pairs and bivalued cells helps speed it up a little. Good luck speeds it up even more. ]

It can be T&E only if you want it to be. Bilocation/bivalue plot can help you to identify these types of double implication chains in a non-T&E and human executable manner.

I didn't mean "trial and error" in the sense that you're taking it. I meant that one locates a double implication chain (or whatever) at one grid location ... and if that chain doesn't result in a pin or an elimination ... then one tries a different value at that location or tries another location.

[Carcul]

MikeF,

You can also prove that r9c8 cannot be 4 through the following chain:

Chain 1: [r9c8]=1=[r9c4]=2=[r8c4]-2-[r8c7]-4-[r9c8].

It is also possible to perform several eliminations with the following continuous nice loop:

Chain 2: [r9c9]=7=[r8c9]=8=[r7c9]-8-[r7c6]-3-[r7c3]-9-[r9c2]-7-[r9c9],

which implies that

r8c9<>3,4;
r7c9<>3;
r7c8<>3;
r9c3<>9.

Regards, Carcul

[ronk]

Code: Select all

Consider the cell r9c7.
When it contains the value 9, the values 3, 4 and 7 in Row 9 must occupy the cells r9c2, r9c3 and r9c9 in some order.
When it contains the value 2, the value 4 in Box 9 must occupy the cell r8c7.
Whichever value it contains, the cell r9c8 cannot contain the value 4.


At the risk of sounding pedantic and suggesting verbose output for your solver ... since r9c7 is a tri-valued, for completeness sake shouldn't you include the trivial ...When it contains the value 4, ... in the report?

[ronk]

You can also prove that r9c8 cannot be 4 through the following chain:

Chain 1: [r9c8]=1=[r9c4]=2=[r8c4]-2-[r8c7]-4-[r9c8]

Would someone please explain that notation ... or point me to a topic that does?

I understand equals (=), not equals (<>) and implies (=>), so translate to ...
r9c8=1 => r9c4=2 => r8c4<>2 => r8c7=2
... but then I'm lost.

TIA, Ron

[Jeff]

Would someone please explain that notation ... or point me to a topic that does?

Try here.

[Carcul]

MikeF, ronk, Jeff, bennys, and rubylips:

Have someone of you found the solution to this puzzle in a straightforward way?
I found it, but, as usual, my method was too lengthy.
Could someone give me a hint?


Thanks, Carcul

[MikeF]

If there is a more straighforward solution, I don't know it. I'm still pondering the very interesting replies above. Thanks for all the input.

Mike

[Jeff]

Could someone give me a hint?

All logical solutions are good solutions. This is not a contest. Please list your chains.

[rubylips]

At the risk of sounding pedantic and suggesting verbose output for your solver ... since r9c7 is a tri-valued, for completeness sake shouldn't you include the trivial ...When it contains the value 4, ... in the report?

That's right! ... and I'd fully intended to make that change, except the opportunity to use the new bit of code cropped up before I'd fully completed development.

[Carcul]

Thanks Jeff. When I asked to someone give me a hint my purpose was to learn with your solutions. Nevertheless, here's how I reached the solution.
My first two chains are the ones posted above:

Chain 1: [r9c8]=1=[r9c4]=2=[r8c4]-2-[r8c7]-4-[r9c8] => r9c8<>4

Chain 2: [r9c9]=7=[r8c9]=8=[r7c9]-8-[r7c6]-3-[r7c3]-9-[r9c2]-7-[r9c9],
=> r8c9<>3,4;
r7c9<>3;
r7c8<>3;
r9c3<>9.

Now we have two Turbot Fishes:

Turbot Fish 1 (Chain3): [r1c2]-9-[r9c2]=9=[r7c3]-9-[r7c8]=9=[r1c8]-9-
[r1c2] => r1c2<>9.

Turbot Fish 2 (Chain 4): [r2c3]-4-[r9c3]=4=[r8c1]-4-[r8c8]=4=[r2c8]-4-
[r2c3] => r2c3<>4.

To complete the puzzle, I have used the following three nice loops:

Chain 5: [r9c3]=4=[r3c3]=1=[r3c4]=9=[r5c4]=8=[r5c6]-8-[r7c6]-3-
[r7c3]-9-[r9c2]-7-[r9c3] => r9c3<>7.

Chain 6: [r1c2]=8=[r3c2]=9=[r9c2]=7=[r9c9]=3=[r2c9]=5=[r2c1]-5-
[r1c2] => r1c2<>5.

Chain 7: [r7c3]=9=[r9c2]=7=[r9c9]=3=[r2c9]=5=[r2c1]-5-[r1c1]-7-
[r1c2]-8-[r3c2]-9-[r3c4]=9=[r5c4]=8=[r5c6]-8-[r7c6]-3-[r7c3]
=> r7c3<>3.

With this last conclusion, the remaining puzzle is easily solved. Although not necessary, we could also make the following deductions:

Chain 8: [r7c6]=3=[r8c6]=1=[r1c6]=2=[r2c6]=9=[r3c4]-9-[r3c2]=9=
[r9c2]-9-[r7c3]-3-[r7c6] => r8c6<>8.

Chain 9: [r3c3]=1=[r3c4]=9=[r5c4]=8=[r5c6]-8-[r7c6]-3-[r7c3]-9-[r3c3]
=> r3c3<>9.

Now, can you, Jeff and others, please post your solutions?

Regards, Carcul