The New Sudoku Players' Forum

[tarek]

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.

I do understand that this is a special case, but I found it works (for the hidden tuple) if the quantum cell is exactly the extra candidates in both roof cells (no repititions) & the tuple formed has no candidates belonging to the FLOOR cells. If these conditions are met then (& i tried these on other puzzles) we can construct a hidden tuple safely.

Tarek

[ronk]

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.

tarek may well be using that constraint and just not putting it in his solver log. However, that limitation escaped me. Thanks, Ron

[edit: Oops, I see tarek responded long ago.]

[NeilB]

For those of you whom build slover programs... why would you test for uniqueness (type 1) after all the other elimination methods? For 'human' slovers it is much easier to see a pattern of 3 pairs with the fourth block containing the candiate pair plus several other canidates. My point is that it is so easy to spot.

My recommendation would be to test for uniqueness (type 1) before you test for X-Wing, Swordfish, Forcing Chains, Coloring. I would never try these methods without first scanning the board to see if I can find a uniqueness case.

[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     |
*-----------------------------------------------------------------*

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.

After a re-read, that is no longer a convincing argument.

Consider col 1 instead of box 1. In the BOTH case, you would also have the three naked triple digits 459 spread across 4 cells ... and there seems to be no question that that naked triple elimination is valid. IOW, conclusions involving the quantum cell are not exactly intuitive.

I don't know how to prove it, but I think eliminations based on the quantum pair 59 (from the roof cells) are valid ... whether constructing a naked N-tuple or a hidden N-tuple. However, eliminations based on the quantum pair 67 (from the floor cells) ... as I hypothesized ... are not.

[tarek]

I don't know how to prove it, but I think eliminations based on the quantum pair 59 (from the roof cells) are valid ... whether constructing a naked N-tuple or a hidden N-tuple. However, eliminations based on the quantum pair 67 (from the floor cells) ... as I hypothesized ... are not.

if the roof cells share only the floor cells' candidates, then you can safely assume a hidden tuple from the quantum cell, the simple reason would be that they won't be HIDDEN if that condition is not met.

I have no idea if the ANTI-QUANTUM CELL theory holds

Tarek

[ronk]

if the roof cells share only the floor cells' candidates, then .................

What does that mean? The roof cells always contain the floor cells. And they can't contain ONLY the floor cells, else we'd already have non-uniqueness.

[tarek]

if the roof cells share only the floor cells' candidates, then .................

What does that mean? The roof cells always contain the floor cells. And they can't contain ONLY the floor cells, else we'd already have non-uniqueness.

]

No, what i said is that the roof cells SHARE only the candidates appearing in the floor cells, meaning that the extra candidates appear only once in both roof cells. all roof cells should share the Floor cell candidates but it is not a must to share extra candidates.

Tarek

[ronk]

What is needed for the Local BUG Principle that this forum seems to be in the process of rediscovering is a collection of positions not fixed by having one of the original clues, and a collection of candidate values at those positions, two candidate values at each, where each row, column and box contains 0 or 2 candidates with any given value. Those candidates may be freely invented, they need not have any relation with the current list of candidates that some solver maintains.

"candidates may be freely invented"?

In the BUG context ... preferably for a uniqueness rectangle ... please provide an example of how "free invention" may be used to advance a puzzle's solution.

TIA, Ron

[aeb]

What is needed for the Local BUG Principle that this forum seems to be in the process of rediscovering is a collection of positions not fixed by having one of the original clues, and a collection of candidate values at those positions, two candidate values at each, where each row, column and box contains 0 or 2 candidates with any given value. Those candidates may be freely invented, they need not have any relation with the current list of candidates that some solver maintains.

"candidates may be freely invented"?

In the BUG context ... preferably for a uniqueness rectangle ... please provide an example of how "free invention" may be used to advance a puzzle's solution.

TIA, Ron

Whenever you invent a 02-setup as described above, the number of solutions that 'fit' the setup, in the sense that for each cell with two affixed numbers the solution digit there is one of these two, is even. For puzzles with a unique solution that means that the number of solutions that fit the 02-setup is zero. This is true for any 02-setup you might think of. It is up to you to think of applications. The above quote was contradicting people with the mistaken belief that uniqueness rectangles can be destroyed. The validity of the argument does not depend on the small numbers that one may write in the puzzle during the solving process.
If you want a more concrete example, consider a rectangle with corners with possibilities 123,123,123,123. You can apply the UR argument for values 1, 2 and conclude that there is a corner that is 3. You can also apply it for values 2, 3 and conclude that there is a corner that is 1. In other words, all three values must occur. You knew that, but this is not an application of UR in a situation where the current possibilities have a pair at at least one corner.

[ronk]

The above quote was contradicting people with the mistaken belief that uniqueness rectangles can be destroyed. The validity of the argument does not depend on the small numbers that one may write in the puzzle during the solving process.

So the "free invention" is *not* really all that free ... since one can only "invent" (read mentally replace) candidates of a valid UR pattern that existed before it was destroyed ... assuming the pattern was even noticed before the destruction.

[ronk]

No, what i said is that the roof cells SHARE only the candidates appearing in the floor cells, meaning that the extra candidates appear only once in both roof cells.

Wow, was I asleep at the switch or what?

I didn't realize my UR type 3 example was also a SueDeCoq, which produces exactly the same eliminations we've discussed.

Code: Select all

 A5679  8    *35679 | 2     67    1     | 34567 45    567
  1     4    B67    | 3     8     5     | 67    9     2
 A567   2    *3567  | 9     67    4     | 3567  1     8
 -------------------+-------------------+----------------
 C49    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
 -------------------+-------------------+----------------
 C45    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

set A = {r1c1,r3c1} = {5679}
    B = {r2c3} = {67}
    C = {r4c1,r7c1} = {459}

yielding 459 eliminations in col 1 and 67 eliminations in box 1, specifically, r8c1<>45, r1c3<>67 and r3c3<>67

For the SueDeCoq, it matters not how candidates are shared within set A.

Ron

[tarek]

Apparantly I was wrong, here are 2 examples where the hiddin principle is wrong:

Code: Select all

*--------------------------------------------------------*
| 5     89    4    | 1     2     3    | 6     89    7    |
| 7     3     6    | 5     89    89   | 2     4     1    |
| 28    1     289  | 4     7     6    | 5     89    3    |
|------------------+------------------+------------------|
| 48    689   3    | 678   489   5    | 1     67    2    |
| 248   2568  7    | 68    3     1    | 9     56    48   |
| 1     5689  89   | 2     4689  789  | 3     567   48   |
|------------------+------------------+------------------|
| 6     278   28   | 9     1     78   | 4     3     5    |
| 3     4     5    | 678   68    2    | 78    1     9    |
| 9     78    1    | 3     5     4    | 78    2     6    |
*--------------------------------------------------------*
Candidates 467 form a Quantum cell as Candidates 89 in r2c5,r2c6,r6c5 & r6c6 form a unique quadrangle which leads to:
r6c2 Must only have 56 as valid Candidates (4567 is a Hidden Quad in Row 6)
r6c9 Must only have 4 as valid Candidates (4567 is a Hidden Quad in Row 6)
*-----------------------------------------------------------------*
| 79     3      48    | 2      6      45    | 1      57     89    |
| 1      6      5     | 8      79     79    | 3      2      4     |
| 79     28     24    | 345    34     1     | 89     57     6     |
|---------------------+---------------------+---------------------|
| 35     7      1     | 345    8      45    | 6      9      2     |
| 235    28     236   | 13579  13     679   | 48     134    17    |
| 4      9      368   | 137    2      67    | 5      13     178   |
|---------------------+---------------------+---------------------|
| 23     14     23    | 6      79     8     | 79     14     5     |
| 8      5      9     | 147    14     2     | 47     6      3     |
| 6      14     7     | 49     5      3     | 2      8      19    |
*-----------------------------------------------------------------*
Candidates 56 form a Quantum cell as Candidates 23 in r7c1,r7c3,r5c1 & r5c3 form a unique quadrangle which leads to:
r5c4 Must only have 59 as valid Candidates (569 is a Hidden Triple in Row 5)
r5c6 Must only have 69 as valid Candidates (569 is a Hidden Triple in Row 5)

These will lead to a contradiction

Tarek