The New Sudoku Players' Forum

[flip]

Does that mean "implication chains, tabling, nishio, T/E", etc. are totally disabled? Or do these advanced techniques make reductions when req'd ... then giving simpler techniques additional opportunity to advance the puzzles?

See the edited post. Yes, discontinuous nice loops with assertion will be tried when everything else fails. After any technique finds a reduction, everything is restarted again from simplest through more difficult techniques again.
I always try type 1/2/3 reductions before doing type 4. The reason is of course that a type 4 reduction eliminates the possibility of doing a type 3 reduction afterwards. That may account for the differences between our results. This is my immediate answer, but I will have to look into this more closely. Different sequences of applying solving techniques may give results that are quite different on the same puzzle. It would still be interesting to see what results are obtained by different solvers for the same set of puzzles.

[flip]

In what order do you check the uniqeness in? The order listed? (1, 2, 2b etc)
Because if it is done in that order I would be very interested to know how many of the 3's that could have been done with a 4, and also:
By swapping the order of the two methods would you be able to find a puzzle where you indeed find a 4, but finding that will lead you to a dead end, as opposed to first finding the 3 and then solving it. (to prove your point in a previous post)

As far as I can see, type 1 cannot be confused with anything else, and type 2 is a special case of type 3. Then I always do type 1/2/3 first before type 4, because doing type 4 eliminates the possibility of finding any of the other afterwards. Look at it this way: type 3 eliminates candidates outside the roof cells, while type 4 changes roof cells only, it is a win-win situation.
As to swapping the order of the eliminations, I will have to recode the solver to do this, and I cannot promise anything.

[tarek]

Although type 2 is a special case of type 3, I would always leave type 2A as a special type, however type 2B should really join Type 3

Tarek

[aeb]

In other words you can't have the two cells with "other" candidates across from each other!

Because otherwise it is of type 5?

Let me show an example. Starting with

Code: Select all

..794...8
....7.2..
9.......3
2.5......
...31...4
..3..2...
4....65..
....3..41
8.6......


one arrives, after having used an interesting collection of techniques, at

Code: Select all

  56    .   .     .    .   .   .   56    .
   .   18   .    56    .  18   .  569  569
   .   56  18    26  256  18   .    .    .
   .   48   .   467   89  79   .    .   69
  67   69  89     .    .   .  78    .    .
   .  478   .    46  689   .  78  569   59
   .    .  19   127   29   .   .    .   27
  57  579   .     .    .  79   .    .    .
   .   17   .  1257   25   .   .    .   27


Now the UR of type 5 eliminates the 6 at r1c8, finishing the puzzle.

[Havard]

aeb wrote:

Now the UR of type 5 eliminates the 6 at r1c8, finishing the puzzle.

I am not familiar with a type 5? Can you explain it, or give me a link to somewhere where it is?

havard

[vidarino]

aeb wrote:

Now the UR of type 5 eliminates the 6 at r1c8, finishing the puzzle.

I am not familiar with a type 5? Can you explain it, or give me a link to somewhere where it is?

If you look at the rectangle (R26C89), you'll see that it needs a 6 in either of the three cells to avoid a deadly pattern of 59s. All of these 6s can "see" R1C8, which makes it impossible for that one to contain a 6.

Or put the other way around; if R1C8 = 6, you'd immediately end up with a deadly pattern.

Vidar

[ronk]

Although type 2 is a special case of type 3, I would always leave type 2A as a special type, however type 2B should really join Type 3

For types 2A and 2B, the "extra candidate" is effectively a naked single, unconditionally causing eliminations. For types 3A and 3B, the "two extra candidates" must combine with other cell(s) as a naked pair or naked triple to cause eliminations.

In both types 2A and 3A, eliminations occur in one column and one box. In both type 2B and 3B, eliminations occur only in one row.

Therefore, I think of type 2A as a special case of type 3A ... and 2B as a special case of 3B.

[Havard]

Vidar:
Cool, I can see that! Tusen takk! But is this recognized as a type 5?

[ronk]

Thanks for your link!

As you pointed out, it was not a valid UR type 3 puzzle, so here are two replacements. The first is #6523 from the gfroyle32930 puzzle pack.

Code: Select all

 ...|2.1|...
 .4.|...|.9.
 ...|...|...
 ---+---+---
 ...|75.|.6.
 2.1|...|...
 8..|6..|...
 ---+---+---
 .9.|.3.|...
 ...|...|1.3
 ...|.4.|8..


 #5679  8     35679 | 2    #67    1     | 34567 45    567
  1     4     67    | 3     8     5     | 67    9     2
 #567   2     3567  | 9    #67    4     | 3567  1     8
 -------------------+-------------------+----------------
 %49    3     49    | 7     5     8     | 2     6     1
  2     6     1     | 4     9     3     | 57    8     57
  8     57    57    | 6     1     2     | 9     3     4
 -------------------+-------------------+----------------
 %45    9     8     | 1     3     7     | 456   2     56
 *4567  57    4567  | 8     2     9     | 1     45    3
  3     1     2     | 5     4     6     | 8     7     9

UR3A(67) with triplet 459 causing eliminations r8c1<>4 and r8c1<>5

The puzzle also contains a UR type 1.

Next is puzzle #16937 from the same source.

Code: Select all

 .5.|3.2|...
 ...|...|4.1
 .2.|...|...
 ---+---+---
 ...|5..|.9.
 1..|7..|...
 4.9|...|...
 ---+---+---
 .3.|...|.8.
 ...|.1.|7..
 6..|...|...


 78     5      1      | 3      4      2      | 89     67     689
 3      9      6      | 8     #57    #57     | 4      2      1
 78     2      4      | 1      6      9      | 38     57     358
----------------------+----------------------+---------------------
 2      7      3      | 5      8      1      | 6      9      4
 1      6      5      | 7      9      4      | 28     3      28
 4      8      9      | 26     23     36     | 5      1      7
----------------------+----------------------+---------------------
 59     3      27     | 4     #257   #567    | 1      8      2569
 59     4      28     |%269    1     *3568   | 7      56     23569
 6      1      278    |%29    *2357   3578   | 239    4      2359

UR3A(57) with triplet 269 causing eliminations r8c6<>6 and r9c5<>2

Both puzzles are "pure", i.e., the UR type 3 is not accompanied by a type 4.

[NeilB]

The fiendish puzzle on Feb. 6, 2006 at Sudoku of the Day has two occasions where using the Uniqueness test can greatly simplify finding the solution.

You can use Uniqueness for 4 & 9's in Blocks 1 & 3; and Uniqueness for 3 & 5's in Blocks 2 & 5.

[tarek]

Hi ronk,

I think we are talking about the same thing from your description of uniqueness types earlier.

The puzzles you posted are quite interesting especially the 1st one, after the 1st type 3 uniqueness followed by the type 1, the is a finned swordfish & then a possible hidden triple through the quantum cell in a another type 3

Code: Select all

*-----------------------------------------------------------------*
|*5679   8      35679 | 2     *67     1     | 34567  45     567   |
| 1      4      67    | 3      8      5     | 67     9      2     |
|*567    2      3567  | 9     *67     4     | 3567   1      8     |
|---------------------+---------------------+---------------------|
|#49     3      49    | 7      5      8     | 2      6      1     |
| 2      6      1     | 4      9      3     | 57     8      57    |
| 8      57     57    | 6      1      2     | 9      3      4     |
|---------------------+---------------------+---------------------|
|#45     9      8     | 1      3      7     | 456    2      56    |
|-4567   57     4567  | 8      2      9     | 1      45     3     |
| 3      1      2     | 5      4      6     | 8      7      9     |
*-----------------------------------------------------------------*
Candidates 59 form a Quantum cell as Candidates 67 in r1c5,r3c5,r1c1 & r3c1 form a unique quadrangle which leads to:
r8c1 Must only have 67 as valid Candidates (459 is a Naked Triple in Column 1)
*-----------------------------------------------------------------*
| 5679   8      35679 | 2      67     1     | 34567  45     567   |
| 1      4      67    | 3      8      5     | 67     9      2     |
| 567    2      3567  | 9      67     4     | 3567   1      8     |
|---------------------+---------------------+---------------------|
| 49     3      49    | 7      5      8     | 2      6      1     |
| 2      6      1     | 4      9      3     | 57     8      57    |
| 8     *57    *57    | 6      1      2     | 9      3      4     |
|---------------------+---------------------+---------------------|
| 45     9      8     | 1      3      7     | 456    2      56    |
| 67    *57   #*4567  | 8      2      9     | 1      45     3     |
| 3      1      2     | 5      4      6     | 8      7      9     |
*-----------------------------------------------------------------*
r8c3 must only contain 46 (Candidates 57 in r6c2,r6c3,r8c2 & r8c3 form a type-1 unique quadrangle)
*-----------------------------------------------------------------*
|-5679   8     *35679 | 2      67     1     | 34567 *45     567   |
| 1      4      67    | 3      8      5     | 67     9      2     |
| 567    2     #3567  | 9      67     4     | 3567   1      8     |
|---------------------+---------------------+---------------------|
| 49     3      49    | 7      5      8     | 2      6      1     |
| 2      6      1     | 4      9      3     | 57     8      57    |
| 8     *57    *57    | 6      1      2     | 9      3      4     |
|---------------------+---------------------+---------------------|
| 45     9      8     | 1      3      7     | 456    2      56    |
| 67    *57     46    | 8      2      9     | 1     *45     3     |
| 3      1      2     | 5      4      6     | 8      7      9     |
*-----------------------------------------------------------------*
Eliminating 5 From r1c1 (Finned Swordfish in Columns 238)
*-----------------------------------------------------------------*
|*679    8     -35679 | 2     *67     1     | 34567  45     567   |
| 1      4      67    | 3      8      5     | 67     9      2     |
|*567    2     -3567  | 9     *67     4     | 3567   1      8     |
|---------------------+---------------------+---------------------|
| 49     3      49    | 7      5      8     | 2      6      1     |
| 2      6      1     | 4      9      3     | 57     8      57    |
| 8      57     57    | 6      1      2     | 9      3      4     |
|---------------------+---------------------+---------------------|
| 45     9      8     | 1      3      7     | 456    2      56    |
| 67     57     46    | 8      2      9     | 1      45     3     |
| 3      1      2     | 5      4      6     | 8      7      9     |
*-----------------------------------------------------------------*
Candidates 59 form a Quantum cell as Candidates 67 in r1c5,r3c5,r1c1 & r3c1 form a unique quadrangle which leads to:
r1c3 Must only have 359 as valid Candidates (359 is a Hidden Triple in Box 1)
r3c3 Must only have 35 as valid Candidates (359 is a Hidden Triple in Box 1)

[ronk]

Code: Select all

*-----------------------------------------------------------------*
|*679    8     -35679 | 2     *67     1     | 34567  45     567   |
| 1      4      67    | 3      8      5     | 67     9      2     |
|*567    2     -3567  | 9     *67     4     | 3567   1      8     |
|---------------------+---------------------+---------------------|
| 49     3      49    | 7      5      8     | 2      6      1     |
| 2      6      1     | 4      9      3     | 57     8      57    |
| 8      57     57    | 6      1      2     | 9      3      4     |
|---------------------+---------------------+---------------------|
| 45     9      8     | 1      3      7     | 456    2      56    |
| 67     57     46    | 8      2      9     | 1      45     3     |
| 3      1      2     | 5      4      6     | 8      7      9     |
*-----------------------------------------------------------------*
Candidates 59 form a Quantum cell as Candidates 67 in r1c5,r3c5,r1c1 & r3c1 form a unique quadrangle which leads to:
r1c3 Must only have 359 as valid Candidates (359 is a Hidden Triple in Box 1)
r3c3 Must only have 35 as valid Candidates (359 is a Hidden Triple in Box 1)

Hi tarek, that's interesting as I hadn't gotten to unique rectangles (quadrangles) and hidden sets yet. (Isn't a quadrangle some sort of building? ) But I've always had trouble spotting hidden sets, so I'm wondering whether another approach might not also be valid.

Since the extra UR candidates 59 can be considered to be of "one quantum cell", it seems the UR candidates 67 should be of "the other quantum cell". From that perspective, there's a 67 naked pair in box 1 ... eliminating those candidates from r13c3 just like your hidden triple does.

Ron

[tarek]

(Isn't a quadrangle some sort of building? )

Well, sort of........

if the FLOOR cells are not sharing a line then you will need some FOUNDATION cells to SUPPORT them, the quadrangle then would be a more appropriate term as it harbours any closed shape with foru points


Tarek

[ronk]

Here are my solver results for UR with the top95.
The UR reductions are only attempted after all simpler reductions,
and also after all xwing/swordfish/jellyfish, conjugate coloring, bug and xy-wing. No implication chains, tabling, nishio, T/E before the UR..

............

I hope you will find this interesting, and would appreciate feedback.
I would like to know that the players on the forum agree with this.
If required I can post the candidates matrix for any of the solutions. If there is sufficient interest, we can do this also for top 870, top 520, ...

I think any T&E will make comparisons difficult ... because of the almost random eliminations made by different T&E algorithms. However, only those UR eliminations made before the first T&E step should be reasonably similar ... depending upon the definition of simpler reductions, of course.

On that basis, my solver finds in the top95 only the following:

pzl 11, uniqueness4A: pass 6, excludes 1 from r8c1 = 159
pzl 11, uniqueness4A: pass 6, excludes 1 from r9c1 = 1459
pzl 14, uniqueness3A2: pass 9, excludes 39 from r5c5 = 139
pzl 32, uniqueness1: pass 18, excludes 79 from r5c5 = 4679
pzl 39, uniqueness4A: pass 19, excludes 8 from r7c9 = 2468
pzl 39, uniqueness4A: pass 19, excludes 8 from r9c9 = 24689
pzl 56, uniqueness4A: pass 30, excludes 4 from r4c1 = 3469
pzl 56, uniqueness4A: pass 30, excludes 4 from r4c2 = 4569
pzl 56, uniqueness4A: pass 31, excludes 6 from r2c4 = 1689
pzl 56, uniqueness4A: pass 31, excludes 6 from r2c6 = 1679
pzl 61, uniqueness3A2: pass 37, excludes 49 from r3c6 = 59
pzl 82, uniqueness2C: pass 48, excludes 6 from r2c8 = 36
pzl 91, uniqueness1: pass 57, excludes 79 from r3c5 = 379
pzl 92, uniqueness4A: pass 54, excludes 7 from r1c8 = 4567
pzl 92, uniqueness4A: pass 54, excludes 7 from r3c8 = 2467

For comparison, can you easily capture eliminations on a similar basis?

TIA, Ron

P.S. The 'C' part of type 2C is meaningless. The '2' part of 3A2 means a naked pair was found. A naked triple would have been shown as 3A3.

[Myth Jellies]

Code: Select all

*-----------------------------------------------------------------*
|*679    8     -35679 | 2     *67     1     | 34567  45     567   |
| 1      4      67    | 3      8      5     | 67     9      2     |
|*567    2     -3567  | 9     *67     4     | 3567   1      8     |
|---------------------+---------------------+---------------------|
| 49     3      49    | 7      5      8     | 2      6      1     |
| 2      6      1     | 4      9      3     | 57     8      57    |
| 8      57     57    | 6      1      2     | 9      3      4     |
|---------------------+---------------------+---------------------|
| 45     9      8     | 1      3      7     | 456    2      56    |
| 67     57     46    | 8      2      9     | 1      45     3     |
| 3      1      2     | 5      4      6     | 8      7      9     |
*-----------------------------------------------------------------*
Candidates 59 form a Quantum cell as Candidates 67 in r1c5,r3c5,r1c1 & r3c1 form a unique quadrangle which leads to:
r1c3 Must only have 359 as valid Candidates (359 is a Hidden Triple in Box 1)
r3c3 Must only have 35 as valid Candidates (359 is a Hidden Triple in Box 1)

In general, this is an invalid assumption. The reason being that uniqueness allows r1c1 = 9, or r1c3 = 5, or BOTH. Thus, because of that BOTH case, you can't assume that one of the two cells, r13c1, will get a 6 or a 7, and you have three digits, 3-5-9, spread across 4 cells. The only reason it works here is because both 6 and 7 show up in only one other cell in column one, thus one of those two cells, r13c1, must also be a 6 or 7. That is the additional constraint needed for you to be able to work your quantum hidden triple, or for ronk to apply his quantum naked pair.