The New Sudoku Players' Forum

[sudophil]

I stopped my Sudokus because I had decided finding chains is very little more than trial and error and that wasn't much fun as a puzzle.

But now I'm back and decided differently. The reason is cell region chains.

Although the sudokuexplainer classifies this as a hard technique, I think it is the only way to ID chain patterns and I think is easier to spot than most fishy patterns.

Find a bivalue cell and trace chains from both values and you'd be surprised how easy it becomes to find eliminations.

Look for cells claiming more than one conjugate pair. Or look for the almost locked sets and see what bivalue will lock the set.


BTW here's my method of play. I don't write in the candidates, my pencil marks are just dots in the obvious part of the cell to show what number it is.
First thing to do is eliminate with intersections and sets. The intersections should be possible without making pencil marks. Then make the pencil marks one number at a time. This way if you're patient you can easily spot the conjugate pairs and more intersections.

Once you done this use a felt tip marker to mark the bivalue cells and use the pencil to draw a square or box around the cojugate pair numbers.

Good luck and I hope those frustrated with chains give my method a go. At least you will be more pattern-identifying then just picking random numbers. Regional chains using bivalues and looking for ALS's in them is definitely easier for me than any of the fishy patterns. At least for my method of solving.

[sudophil]

for example here's the jellyfish example from sudopedia

Code: Select all

.------------------.------------------.------------------.
| 3     1     89   | 89    6     4    | 2     5     7    |
| 4     2     68   | 5     3     7    | 168   18    9    |
| 569   589   7    | 1     89    2    | 3     4     68   |
:------------------+------------------+------------------:
| 69    89    5    | 7     289   1    | 4     268   3    |
| 7     3     4689 | 89    24589 58   | 68    28    1    |
| 1     48    2    | 3     48    6    | 9     7     5    |
:------------------+------------------+------------------:
| 2     7     3    | 4     18    9    | 5     168   68   |
| 8     45    14   | 6     15    3    | 7     9     2    |
| 59    6     19   | 2     7     58   | 18    3     4    |
'------------------'------------------'------------------'


here's the link
http://www.sudopedia.org/wiki/Jellyfish

My first bivalue I looked at did the trick
8(r3c9) eliminates several 8's
6(r3c9) pins 6 to block 6 in r5c7 making an easy to see locked pair in block 6 that eliminates 8 in r2c8
everything else falls into place

I was looking at eliminating the 8's in the jellyfish but even eliminating the 6 was as just as simple
6(r3c9) -> (r2c7)<>6
8(r3c9) 1(r2c8)(r9c7) 9(r9c3) 6(r2c3) ->(r2c7)<>6

I find this to be easier to spot than the jellyfish. Fish just don't pop up you have to look at all the numbers to see if they have a fish. With my chain, most of the time the first bivalue cell you find can make an elimination and often more than one. Every link in the cell pins some value.
Of the two chains note the cells affected by the shortest then follow the longest until you see it cross paths with the other chain.

[Cec]

Hi sudophil,

Whilst finding your comments interesting I was unable to understand your following deductions - this of course is not to infer they are wrong but more than likely my own shortcomings

"...My first bivalue I looked at did the trick .."

Was your first bivalue the [68] naked pair in c9 or the [68] in r3c9 (box 3)?

"...8(r3c9) eliminates several 8's"

Which 8's are excluded and what is the logical explanation for these exclusions ?

"..6(r3c9) pins 6 to block 6 in r5c7 making an easy to see locked pair in block 6 that eliminates 8 in r2c8..."

This conclusion appears to result from placing candidate 6 in r3c9. How was this placement (6) arrived at?

Thanks in advance.

Cec

[RW]

I stopped my Sudokus because I had decided finding chains is very little more than trial and error and that wasn't much fun as a puzzle.

But now I'm back and decided differently. The reason is cell region chains.

Although the sudokuexplainer classifies this as a hard technique, I think it is the only way to ID chain patterns and I think is easier to spot than most fishy patterns.

Find a bivalue cell and trace chains from both values and you'd be surprised how easy it becomes to find eliminations.

Look for cells claiming more than one conjugate pair. Or look for the almost locked sets and see what bivalue will lock the set.

Hi sudophil,

many people do seem to have some ethical problems with chains, but I'm glad you found their usefulness. IMO chains are no worse than patterns, because patterns are merely common chains defined as patterns.

However, I hope you aren't limiting yourself too much to cell and region chains, they are also really no different from contradiction forcing chains. Most cell and region forcing chains can be explained as equally long contradiction forcing chains. In this case, perhaps if you first had looked at the bivalue cell in r2c8, you would have spotted the contradiction first (r2c8=8 => r3c9=6 => r7c9=8 => r9c7=1 => no candidates in r2c7). In case this would have happened, would you have accepted the elimination or went on to find a cell forcing chain?

RW

[daj95376]

Withdrawn: My question should have been in a separate post.

[sudophil]

Thanks RW for pointing out the usefull, contradicting chain pattern.

I'll try to clarify my post. Most programs like sudoexplainer and also definitions of chains lead you to believe the procedure of using chains is to pick a number and then find a chain that disproves it and to me this was more like trial and error, going through number by number or randomly picking one.

Cell forcing chains or contradicting chains like RW says start with a pattern or chain and then resolve a cell or part of a cell. This is no different then the simple patterns of intersections or sets. They both start with the pattern and use it to prove a point.

To clarify for Cec, region forcing chain states that if trying all possiblities in a region leads to a common result then that result is true. For a row or column or box it would be a common number, for a cell it would be all candidates in the cell.

In my example I'm trying all the candidates in r3c9, if r3c9=8 all other 8's are eliminated in the block and row and col, in particular the ones in the block for my example. Now if r3c9= 6 one can easily spot this simple chain
r2c7<>6 -> r5c7=6 -> c8 block 6 claims 8 and r2c8<>8
therefore r2c8 can't be 8 because all candidates of r3c9 prove this.

This is easier then starting with 8 in r2c8 and trying to find a chain that eliminates it simply because there's no real logic in picking it for trial.

Good luck I hope I help others in looking at chains once a puzzle gets to that point.

[Pat]

Code: Select all

.------------------.------------------.------------------.
| 3     1     89   | 89    6     4    | 2     5     7    |
| 4     2     68   | 5     3     7    | 168   18    9    |
| 569   589   7    | 1     89    2    | 3     4     68   |
:------------------+------------------+------------------:
| 69    89    5    | 7     289   1    | 4     268   3    |
| 7     3     4689 | 89    24589 58   | 68    28    1    |
| 1     48    2    | 3     48    6    | 9     7     5    |
:------------------+------------------+------------------:
| 2     7     3    | 4     18    9    | 5     168   68   |
| 8     45    14   | 6     15    3    | 7     9     2    |
| 59    6     19   | 2     7     58   | 18    3     4    |
'------------------'------------------'------------------'



perhaps if you first had looked at the bivalue cell in r2c8,
you would have spotted the contradiction first --

r2c8=8 => r3c9=6 => r7c9=8 => r9c7=1 => no candidates in r2c7

or slightly shorter --
r2c8=8 => r5c7=8 => r2c7=6 => no candidates in r2c3 and in r3c9

[Cec]

"In my example I'm trying all the candidates in r3c9, if r3c9=8 all other 8's are eliminated in the block and row and col, in particular the ones in the block for my example. Now if r3c9= 6 one can easily spot this simple chain
r2c7<>6 -> r5c7=6 -> c8 block 6 claims 8 and r2c8<>8
therefore r2c8 can't be 8 because all candidates of r3c9 prove this.

Thanks for your comments sudophil. From your initial post above I assumed you didn't use pencilmarks (candidates) and this intrigued me as to how you could solve these puzzles. I've bolded the above - which I confess makes me feel "relieved" - as this suggests you do use candidates which is what I would have expected when trying to solve "difficult" puzzles like this.

I can follow both yours and Pat's explanations as to why r2c8 can't be 8 which still leaves a naked pair [68] in box3 which still doesn't solve the puzzle - not to me anyway.

I notice you initially chose your first bivalue cell being the [68] in r3c9. Maybe it's my lack of "forward thinking" in not being able to readily identify the "best" choice but I can't help wonder why any of the following bivalue cells could not have been chosen ?
[89] in row1
[89] in box 2
[89] in column 4 and
[58] in column 6

Cec

[eleven]

Good question.

The first i see in the grid is a y(w)-wing 89 in r3c5/r4c2 with link for 8 in row 6. So r3c2 cannot be 9. That solves the puzzle and i would never look for chains.
Added: There is another one for 18.

[sudophil]

eleven's comment says a lot

The first i see in the grid

There is a lot of ways to solve this puzzle at this state with so many resolved cells and candidates. I'm not to solve this puzzle or even suggest how to best go about it.

I would have gave up on r1c3 just because I didn't resolve anything with obvious chains and I had lots of other bivalue cells to try. I went to r2c3 and found this one
6(r2c3)=6(r3c9)=8(r7)= 1(c5) = 5(r8) = 4(c2) =8(r6c2)-> r3c2<>8
This doesn't gain a lot but again I have no magic potion.

I'm trying to say for example, to randomly select 8 in r3c2 and find a chain to disprove it is starting out just that -- random.

Instead of forcing chains when it reaches that state, I've been looking at cell region chains.