The New Sudoku Players' Forum

[SpAce]

It will be difficult to fill the shoes of a mathematician from Finland with skills like RW even if you were Arto Inkala

Are you saying the real RW is from Finland??? Then it's pretty easy to prove (Even in this day of relatively decent translators a non-native speaker should be easy to catch.)

[RW]

However it would be easy to prove RW is RW by confirming how he became a supporter of this site in 2010.

Look up PayPal receipt 4227-5753-8248-**** from June 2nd 2010.

eleven, I'm sorry if I confused you by not considering the variation of the puzzle in my first post. But if you know me at all, you should know that everything uniqueness related was always my main point of interest, and if you know extended uniqueness patterns you should know that in the first post I described how this 14 cell, 7 digit pattern is defined in a regular sudoku, and as you made me aware of my mistake I went on to describe how this technique needs to be altered to be applied in a pencilmark sudoku. The first one closely related to a technique I came up with in 2006, the second.. well, I don't remember if this topic was ever discussed, but since I have spent years studying these patterns, it is obvious to me that this is how it works. I would hope that would be proof enough for you that I am in fact me.

It will be difficult to fill the shoes of a mathematician from Finland with skills like RW even if you were Arto Inkala

Oh, that guy..

So thanks for the somewhat surprising welcome, I guess. I'm happy to see so many of you are still here. I haven't really done anything sudoku related in the past seven years, until I recently decided to write a little solver/puzzle generator of my own, which had me come back here to look up a few things. This place is still a goldmine of information!

RW

[eleven]

Puh, i really had a bad day
Excuse me, RW, you know what you are talking about.

In a normal sudoku with the 1,3's solved in the 4 cells, for any solution you could just switch the other cells of the 2 columns to get a second solution.

You might be interested in that thread, where the reverse bug was a bit extended.

[tarek]

So thanks for the somewhat surprising welcome, I guess.

What do you expect after disappearing for 7 years

BTW to atone for you absence maybe you should start by checking the posts from your famous thread ... Solving without pencilmarks

tarek

[RW]

BTW to atone for you absence maybe you should start by checking the posts from your famous thread ... Solving without pencilmarks

That old thing is still pinned here?

Thanks for the link, eleven! I'll check it out.

But now back to the topic. Did my corrected answer help OP?

The general rule is that all unavoidable sets that apply in a regular sudoku apply in a pencilmark sudoku as well, but some additional checks might be needed. An unavoidable set cannot be solved uniquely because cells outside the set cannot possibly provide the information to solve it. However, in a pencilmark sudoku it might already be partially solved from the inside by the puzzle generator, which has the power to remove whatever candidates it wants. Therefore we cannot rely on uniqueness techniques that only check if all candidates belong to some collection of candidates, we must also check that enough candidates are still present to make it an unavoidable set.

In this case the requirement is that when you focus only on the cells of the unavoidable set, every candidate that appears must appear at least twice in the row, column and box (within the set). If there is a candidate that appears in only one cell of a row within the set, it could eventually become a hidden single when you solve the rest of the puzzle. From this follows that the two columns here need to have pairwise identical sets of candidates. If a candidate appears in one cell, it has to appear in the other cell of the same row as well.

I believe this same check is all we need in pencilmark sudokus to validate just about all cases of simple unavoidable sets commonly used for solving. Have been trying to think of some counterexample, but can't come up with any now.

[creint]

Proof (using a solver) if it has multiple solutions if digit is removed from the blue cells. If so then you can use any logic that excludes all blue digits you find for exclusions.

[yzfwsf]

In this case the requirement is that when you focus only on the cells of the unavoidable set, every candidate that appears must appear at least twice in the row, column and box (within the set). If there is a candidate that appears in only one cell of a row within the set, it could eventually become a hidden single when you solve the rest of the puzzle. From this follows that the two columns here need to have pairwise identical sets of candidates. If a candidate appears in one cell, it has to appear in the other cell of the same row as well.

I believe this same check is all we need in pencilmark sudokus to validate just about all cases of simple unavoidable sets commonly used for solving. Have been trying to think of some counterexample, but can't come up with any now.

In fact, this is the code method I wrote, but there are still bugs for sukaku, so I think other conditions should be added. What I'm thinking about now is to enumerate the arrangement of cells involved in the dead pattern to see if it can form two permutations that can be exchanged, but this will affect the operation efficiency and see if experts have other simple methods. Thanks.

[RW]

In fact, this is the code method I wrote, but there are still bugs for sukaku, so I think other conditions should be added.

Do you have an example of some situation where this verification method fails? I'm not familiar with your solver and don't know how complex uniqueness patterns you have implemented.

[yzfwsf]

The sukaku in first post is a well example. I have implemented 2 rows*(3~7)cols or 2 cols *(3~7)rows uniqueness patterns.

[RW]

That example does not use this method. If you implement the check correctly, what you should see is this:

Code: Select all

.---------------------------------.---------------------------------.---------------------------------.
|  1          23678      23456789 |  2345678    234589     23567    |  3789       23458      4589     |
| #259+4      2357      #259+3678 |  12456      123458     369      |  1245678    23457      13689    |
| #357+246    2345678   #357+89   |  14567      134589     12345789 |  134579     34567      147      |
:---------------------------------+---------------------------------+---------------------------------:
| #24+8       9         #24+56    |  23567      12348      1278     |  12358      2345678    3456     |
|  2345678    2357       1        |  9          34         23456    |  23678      2356       234      |
| #26+5       234568    #26+478   |  1358       23478      124678   |  3457       9          12356    |
:---------------------------------+---------------------------------+---------------------------------:
| #24789      34        #24789+35 |  78         6          248      |  49         1          457      |
| #4568+27    1         #4568+9   |  23         34589      35       |  234679     234568     2789     |
| #245678+9   4567      #245678   |  134578     234789     13459    |  234579     245        345678   |
'---------------------------------'---------------------------------'---------------------------------'

Everything after the + signs are extra candidates that invalidate the unavoidable set. Those candidates only appear once in their rows within the set.

Actually, to make sure you find all potential unavoidable sets, this verification step should also take into account candidates that have been eliminated in previous steps. For example, if the starting grid did contain candidate 4 in r3c3, but this has since been eliminated, then the 4 in r3c1 can be moved to the left side of the + sign. So for every candidate in the set, you must check if it appeared in both columns (same row) in the starting grid.

[yzfwsf]

Thank you RW, I know what's wrong. I just check whether there are two or more common candidates in a pair of cells in each row (column), and then use these unions to find additional candidates. In fact, I should deal with the additional candidates of each cell pair separately.