The New Sudoku Players' Forum

[yzfwsf]

If you have a confirmed BUG+n situation and exactly one guardian cell which holds multiple guardians (i.e. all of them), then yes, it is safe to eliminate all the other candidates (BUG candidates) from it because all of the guardians see them. It's like BUG+1 generalized (because in BUG+1 the sole guardian also eliminates all the other candidates from its cell). However, I'm not sure if that can ever happen in a real grid. I think all situations with just one guardian-cell must be BUG+1s, but I can't be sure. Do you have an example in mind? In any case, your logic is valid in theory, even if it can't happen in a real grid. It's basically minimal kraken logic.

I will provide you with a random puzzle example searched by my program as shown below.In fact, this is also a uniqueness rectangle.

Code: Select all

.3...4..97..6..1....9.5..2.8....5..3..2.63.1....8..5.....9...6.6...1...2..7...8..

Code: Select all

.---------------------------------.---------------------------------.---------------------------------.
|  15         3          15       |  2          7          4        |  6          8          9        |
|  7          2          8        |  6          3          9        |  1          45         45       |
|  4          6          9        |  1          5          8        |  3          2          7        |
:---------------------------------+---------------------------------+---------------------------------:
|  8          1          6        |  47         9          5        |  2          47         3        |
|  59         57         2        |  47         6          3        |  49         1          8        |
|  39         47         34       |  8          2          1        |  5          79         6        |
:---------------------------------+---------------------------------+---------------------------------:
|  13         45         13       |  9          8          2        |  7          6          45       |
|  6          8          45       |  35         1          7        |  49         3459       2        |
|  2          9          7        |  35         4          6        |  8          35         1        |
'---------------------------------'---------------------------------'---------------------------------'

bug+2_1.png (32.12 KiB) Viewed 578 times

[SpAce]

Hi yzfwsf,

If you have a confirmed BUG+n situation and exactly one guardian cell which holds multiple guardians (i.e. all of them), then yes, it is safe to eliminate all the other candidates (BUG candidates) from it because all of the guardians see them. It's like BUG+1 generalized (because in BUG+1 the sole guardian also eliminates all the other candidates from its cell). However, I'm not sure if that can ever happen in a real grid. I think all situations with just one guardian-cell must be BUG+1s, but I can't be sure. Do you have an example in mind? In any case, your logic is valid in theory, even if it can't happen in a real grid. It's basically minimal kraken logic.

I will provide you with a random puzzle example searched by my program as shown below.

Thanks for that example! I really appreciate it. This has been literally bugging me since I wrote that! I'd already started to think that I probably must have seen cases like that but simply couldn't remember for sure. Now it's confirmed, so my initial instincts were indeed wrong. Of course the logic was valid in any case, like I said.

For rjamil: the confirmation that these cases exist means that I was also wrong here:

If you implement those two rules I showed (two different guardian digits in cells that see each other, or two same guardian digits in cells anywhere), you should find all BUG+2 eliminations that don't require chains.

Obviously those two rules don't cover every simple case of BUG+2, after all, since we now have three of them. This is actually the simplest of them all, and it should probably be the first rule to implement in your catalog of BUG+2 tricks that don't need chains.

In fact, the logic works with any number of guardians (as I said originally), so if you happen to see BUG+3s or higher with all the guardians concentrated in just one cell, you can safely eliminate all other candidates. yzfwsf, do you possibly have any examples of such bigger BUGs in a similar situation?

In fact, this is also a uniqueness rectangle.

Indeed, though it gives different eliminations in r8c8 (-35 instead of -39). Either set gives stte. If you combine both the BUG+2 and UR Type 1 into a single move, you get a direct placement +4 r8c8 (since -359). In fact, eliminating the 9 is enough alone for stte, but neither 3 or 5 does the trick alone (they yield BUG+1s and an AR or a UR with a missing candidate, respectively).

Here are those possibilities (at least the ones I could immediately see):

uniqueness solving options for this puzzle: Show

Code: Select all

.------------.----------.----------------.
| 15  3   15 | 2   7  4 | 6    8      9  |
| 7   2   8  | 6   3  9 | 1    45     45 |
| 4   6   9  | 1   5  8 | 3    2      7  |
:------------+----------+----------------:
| 8   1   6  | 47  9  5 | 2    47     3  |
| 59  57  2  | 47  6  3 | 49   1      8  |
| 39  47  34 | 8   2  1 | 5    79     6  |
:------------+----------+----------------:
| 13  45  13 | 9   8  2 | 7    6      45 |
| 6   8   45 | 35  1  7 | 49  +45-39  2  |
| 2   9   7  | 35  4  6 | 8    35     1  |
'------------'----------'----------------'

BUG+2: (4==5)r8c8 => -39 r8c8; stte

Code: Select all

.------------.-----------.----------------.
| 15  3   15 | 2    7  4 | 6    8      9  |
| 7   2   8  | 6    3  9 | 1    45     45 |
| 4   6   9  | 1    5  8 | 3    2      7  |
:------------+-----------+----------------:
| 8   1   6  | 47   9  5 | 2    47     3  |
| 59  57  2  | 47   6  3 | 49   1      8  |
| 39  47  34 | 8    2  1 | 5    79     6  |
:------------+-----------+----------------:
| 13  45  13 | 9    8  2 | 7    6      45 |
| 6   8   45 | 35+  1  7 | 49  +49-35  2  |
| 2   9   7  | 35+  4  6 | 8    35+    1  |
'------------'-----------'----------------'

UR Type 1 (35): (4==9)r8c8 => -35 r8c8; stte

Or overkill: BUG+2 & UR Type 1 => -359 r8c8 => +4 r8c8; stte

If we only eliminate 3r8c8 and solve basics, we get this:

Code: Select all

.------------.----------.---------------.
| 15  3   15 | 2   7  4 | 6    8     9  |
| 7   2   8  | 6   3  9 | 1    45    45 |
| 4   6   9  | 1   5  8 | 3    2     7  |
:------------+----------+---------------:
| 8   1   6  | 47  9  5 | 2    47    3  |
| 59  57  2  | 47  6  3 | 49   1     8  |
| 39  47  34 | 8   2  1 | 5    79    6  |
:------------+----------+---------------:
| 13  45  13 | 9   8  2 | 7    6     45 |
| 6   8   45 | 3   1  7 | 49  +4-59  2  |
| 2   9   7  | 5   4  6 | 8    3     1  |
'------------'----------'---------------'

BUG+1: +4 r8c8; stte

It could also be solved with an Avoidable Rectangle (which Hodoku doesn't report for some reason):

Code: Select all

.------------.----------.---------------.
| 15  3   15 | 2   7  4 | 6   8      9  |
| 7   2   8  | 6   3  9 | 1   45     45 |
| 4   6   9  | 1   5  8 | 3   2      7  |
:------------+----------+---------------:
| 8   1   6  | 47  9  5 | 2   47     3  |
| 59  57  2  | 47  6  3 | 49  1      8  |
| 39  47  34 | 8   2  1 | 5   79     6  |
:------------+----------+---------------:
| 13  45  13 | 9   8  2 | 7   6      45 |
| 6   8   45 | 3*  1  7 | 49  49-5*  2  |
| 2   9   7  | 5*  4  6 | 8   3*     1  |
'------------'----------'---------------'

AR (35) => -5 r8c8; stte

Or if we only eliminate 5r8c8, we get:

Code: Select all

.------------.----------.---------------.
| 15  3   15 | 2   7  4 | 6    8     9  |
| 7   2   8  | 6   3  9 | 1    45    45 |
| 4   6   9  | 1   5  8 | 3    2     7  |
:------------+----------+---------------:
| 8   1   6  | 47  9  5 | 2    47    3  |
| 59  57  2  | 47  6  3 | 49   1     8  |
| 39  47  34 | 8   2  1 | 5    79    6  |
:------------+----------+---------------:
| 13  45  13 | 9   8  2 | 7    6     45 |
| 6   8   45 | 35  1  7 | 49  +4-39  2  |
| 2   9   7  | 35  4  6 | 8    35    1  |
'------------'----------'---------------'


BUG+1 => +4 r8c8; stte

Or:

Code: Select all

.------------.-----------.---------------.
| 15  3   15 | 2    7  4 | 6    8     9  |
| 7   2   8  | 6    3  9 | 1    45    45 |
| 4   6   9  | 1    5  8 | 3    2     7  |
:------------+-----------+---------------:
| 8   1   6  | 47   9  5 | 2    47    3  |
| 59  57  2  | 47   6  3 | 49   1     8  |
| 39  47  34 | 8    2  1 | 5    79    6  |
:------------+-----------+---------------:
| 13  45  13 | 9    8  2 | 7    6     45 |
| 6   8   45 | 35+  1  7 | 49  +49-3  2  |
| 2   9   7  | 35+  4  6 | 8    35+   1  |
'------------'-----------'---------------'

UR Type 1 (35) (missing candidate 5 in r8c8) => -3 r8c8; stte

[rjamil]

Hi SpAce,

Just look in to your OTP move for November 14, 2018 puzzle:
[Start correction as on 20200531]

Code: Select all

.----------.--------------.------------------------.
| 5  4   1 | 2     6  c38 |b78+3    37      9      |
| 9  38  2 | 1     7   5  | 38      4       6      |
| 7  38  6 | 38    9   4  | 5       1       2      |
:----------+--------------+------------------------:
| 4  2   8 | 9     1  d37 | 6      a35(+7)  57     |
| 6  9   5 |e3(7)  4   2  |a13(+7)  8       1-7    |
| 1  7   3 | 5     8   6  | 2       9       4      |
:----------+--------------+------------------------:
| 2  1   4 | 6     3   78 | 9       57     a58(+7) |
| 3  5   7 | 48    2   9  | 14      6       18     |
| 8  6   9 | 47    5   1  | 47      2       3      |
'----------'--------------'------------------------'


BUG+4

(7)r4c8,r5c7,r7c9 =BUG= (3)r1c7 - r1c6 = r4c6 - (3=7)r5c4 => -7 r5c9; stte

Code: Select all

+----------+------------+-------------------+
| 5  4   1 | 2   6   38 | 78+3   37    9    |
| 9  38  2 | 1   7   5  | 38     4     6    |
| 7  38  6 | 38  9   4  | 5      1     2    |
+----------+------------+-------------------+
| 4  2   8 | 9   1   37 | 6      35+7  5-7  |
| 6  9   5 | 37  4   2  | 13+7   8     1-7  |
| 1  7   3 | 5   8   6  | 2      9     4    |
+----------+------------+-------------------+
| 2  1   4 | 6   3   78 | 9      57    58+7 |
| 3  5   7 | 48  2   9  | 14     6     18   |
| 8  6   9 | 47  5   1  | 47     2     3    |
+----------+------------+-------------------+

3 @ r1c7
7 @ r4c8 r5c7 r7c9 => -7 @ r45c9
(If a guardian candidate exists more than one BUG+cells then that guardian candidate may be removed from all common peers, including from other BUG+cells' BUG+candidates.)[End correction as on 20200531]

Similarly, look in to three BUG+2 and one BUG+3 examples from this thread/topic:

1) BUG+2:

Code: Select all

 *--------------------------------------------------*
 | 9    3    7    | 1    2    8    | 5    4    6    |
 | 15   2    58   | 4    3    6    | 7    18   9    |
 | 16   68   4    | 7    5    9    | 18   3    2    |
 |----------------+----------------+----------------|
 | 8    5    1    | 9    4    7    | 26   26   3    |
 | 3    9    6    | 2    1    5    | 4    7    8    |
 | 4    7    2    | 6    8    3    | 9    5    1    |
 |----------------+----------------+----------------|
 | 26   1    9    | 38   7    4    |#36+8 #28+6 5   |
 | 25   4    3    | 58   6    1    | 28   9    7    |
 | 7    68   58   | 35   9    2    | 13   16   4    |
 *--------------------------------------------------*

Since both of the guardian candidate sees other BUG+cell's BUG candidate, therefore, both the guardian candidate may be remove from opposite BUG+cell's BUG candidate.

2) BUG+3:

Code: Select all

 *-----------------------------------------------------------*
 | 17    17    9     | 6     4     3     | 2     58    58    |
 | 6     4     8     | 2     9     5     | 3     7     1     |
 | 2     3     5     | 18    18    7     | 6     4     9     |
 |-------------------+-------------------+-------------------|
 | 58    58    4     | 9     3     1     | 7     26    26    |
 | 17    2     3     | 47    6     48    | 15    9     58    |
 | 9     17    6     | 5     78    2     | 4     18    3     |
 |-------------------+-------------------+-------------------|
 | 4     6     1     | 3     57    9     | 8     25    27    |
 | 58    9     7     | 18    2     6     | 15    3     4     |
 | 3     58    2     | 47   #15+78 48    | 9    #16+5  67    |
 *-----------------------------------------------------------*

Since BUG+cell r9c8 guardian candidate 5 sees BUG+cell r9c5 BUG candidate, therefore, 5 may be remove from r9c5.

3) BUG+2:

Code: Select all

 *--------------------------------------------------*
 | 6    4    7    | 38   1    28   | 9    23   5    |
 | 23   5    1    | 6    4    9    | 37   8    27   |
 | 8    23   9    | 37   27   5    | 4    1    6    |
 |----------------+----------------+----------------|
 | 7    6    4    | 2    9    1    | 8    5    3    |
 | 1    28   5    | 4    3    68   | 26   7    9    |
 | 23   9    38   | 5    68   7    | 26   4    1    |
 |----------------+----------------+----------------|
 | 4    37   2    | 1    67   36   | 5    9    8    |
 | 5    1    38   | 9    28   4    | 37   6    27   |
 | 9   78+3  6    | 78   5   23+8  | 1    23   4    |
 *--------------------------------------------------*

Same rule applied for both the guardian candidates vs BUG candidates.

4) BUG+2:

Code: Select all

 *--------------------------------------------------*
 | 8    5    79   | 17   4    19   | 6    3    2    |
 | 3    6    29   | 28   5    89   | 7    1    4    |
 | 1    27   4    | 3    27   6    | 5    89   89   |
 |----------------+----------------+----------------|
 | 79   28   3    | 6    1    5    | 4    27   89   |
 | 4    1    28   | 9    3    7    | 28   6    5    |
 | 6    79   5    | 4    8    2    | 3    79   1    |
 |----------------+----------------+----------------|
 | 79  89+7  6    | 5    27   4    | 1    28   3    |
 | 5    3   17+8  | 27   9    18   | 28   4    6    |
 | 2    4    18   | 18   6    3    | 9    5    7    |
 *--------------------------------------------------*

Since one of the guardian candidate sees opposite BUG candidate, therefore, it may be remove from respective cell.

[Added as on 20200531]Similarly, for opening post's puzzle:

Code: Select all

+---------+-----------------+---------+
| 4  8  1 | 9      7   2    | 3  6  5 |
| 7  9  3 | 56     68  58   | 2  4  1 |
| 6  2  5 | 4+3-1  13  34+1 | 7  8  9 |
+---------+-----------------+---------+
| 8  1  2 |  36    36  9    | 4  5  7 |
| 3  5  6 |  8     4   7    | 9  1  2 |
| 9  4  7 | 25+1   12  5-1  | 8  3  6 |
+---------+-----------------+---------+
| 5  3  8 |  7     9   6    | 1  2  4 |
| 1  6  9 |  24    28  48   | 5  7  3 |
| 2  7  4 |  13    5   13   | 6  9  8 |
+---------+-----------------+---------+

1 @ r3c6 r6c4 => -1 @ r3c3 r6c6
3 @ r3c3[End added as on 20200531]

Note: I am digging BUG related posts and studying slowly. However, my point of view is limited.

R. Jamil
Note: correction are based on below given SpAce post.

[SpAce]

Hi rjamil,

Just look in to your OTP move for November 14, 2018 puzzle:

Hidden Text: Show

Code: Select all

.----------.--------------.------------------------.
| 5  4   1 | 2     6  c38 |b78+3    37      9      |
| 9  38  2 | 1     7   5  | 38      4       6      |
| 7  38  6 | 38    9   4  | 5       1       2      |
:----------+--------------+------------------------:
| 4  2   8 | 9     1  d37 | 6      a35(+7)  57     |
| 6  9   5 |e3(7)  4   2  |a13(+7)  8       1-7    |
| 1  7   3 | 5     8   6  | 2       9       4      |
:----------+--------------+------------------------:
| 2  1   4 | 6     3   78 | 9       57     a58(+7) |
| 3  5   7 | 48    2   9  | 14      6       18     |
| 8  6   9 | 47    5   1  | 47      2       3      |
'----------'--------------'------------------------'


BUG+4

(7)r4c8,r5c7,r7c9 =BUG= (3)r1c7 - r1c6 = r4c6 - (3=7)r5c4 => -7 r5c9; stte

Lets build guardian candidates vs BUG+cells relationship:
3 @ r1c8; and
7 @ r4c7 r5c47 r7c9.

You have mistakes there. First, a couple of typos; should be: 3 @ r1c7 and 7 @ r4c8. A bigger problem is that you've listed 7r5c4 as a BUG+ cell. Look again. It only has two candidates so it certainly can't hold any guardians. The marked (7) in it is a result of the chain from the guardian 3r1c7 which doesn't see the elimination directly (so it must imply a remote candidate that does).

Since all BUG+cells, of guardian candidate 7, sees r5c9, therefore 7 may be removed from r5c9. (if all BUG+cells of guardian candidate 7 sees any other BUG-cell that contain 7 as BUG candidate, then 7 may also be removed from that BUG+cell.

There will be no more need to build a chain to prove anything. Simply each guardian candidate's BUG+cells are enough to prove the removals.

None of that makes any sense, sorry to say. Read again what I've written before. Don't you think I would have used a simpler BUG move if one existed?

Similarly, look in to three BUG+2 and one BUG+3 examples from this thread/topic:

They're all valid examples of RW's trick and its extensions I mentioned. No problem at all with them. However, the BUG+4 situation above can't be solved with that trick. My original one is probably the simplest possible BUG solution for it.

Note: I am digging BUG related posts and studying slowly.

That's good for me, because I like the examples and interesting discussions you find. However, I'm not sure how much it helps you to dig up old threads with possibly questionable and confusing explanations, until you understand what I've said, because it seems that you keep making the same systemic mistakes.

Please, try to read and internalize what I've written in this thread. It's probably as good as it gets, as far as basic BUG explanations go. Yet it seems that you've still completely missed the most important point of BUG solving, even though I've been constantly repeating it. Without it you can't understand how my BUG+4 solution above works (and why it's probably the simplest one available).

Before I continue with this thread, I want you to answer this question: what is the fundamental (and very simple) rule of BUG solving?

If you don't know it by now, please reread everything I've said, and find the answer from there. It's the key to everything. Until you understand that, there's no hope of real progress. You will not find anything simpler or better in the old threads, I guarantee you. (But you will find it mentioned there too, of course, because it's all there is to it.)

[rjamil]

Hi SpAce,

Thanks for the corrections that I have made in your OTP move for http://forum.enjoysudoku.com/post272325.html#p272325November%2014,%202018 puzzle. It is due to my first attempt to convert BUG+n (where n > 3) move.

I have just picked the puzzles from this post as an example; and not for the logic to find the BUG+n move.

However, I am digging old post containing BUG+2 and BUG+3 examples. Trying to resolve them with RW style without even reading explanation and understanding of the provided solution.

Since I have yet to code for finding more than two cells common peer/buddy cells, therefore, I am concentrating BUG+2 and BUG+3 moves only. My limit is to find a guardian candidate that lying no more than two BUG+cells only. I'll then easily find common peers of two BUG+cells as per XY-Wing move elimination patterns and/or Naked pair move.

R. Jamil

[yzfwsf]

Give an example of bug + 6.

Code: Select all

..7....8.1...5...6.5...61...9......12.....7....547.2....68.5....1..9...83...146..

Code: Select all

.---------------------------------.---------------------------------.---------------------------------.
|  6          24         7        |  1          24         39       |  5          8          39       |
|  1          24         38       |  79         5          28       |  34         79         6        |
|  9          5          38       |  37         48         6        |  1          2          47       |
:---------------------------------+---------------------------------+---------------------------------:
|  7          9          4        |  35         26         23       |  8          356        1        |
|  2          36         1        |  359        68         389      |  7          3456       45       |
|  8          36         5        |  4          7          1        |  2          369        39       |
:---------------------------------+---------------------------------+---------------------------------:
|  4          7          6        |  8          3          5        |  9          1          2        |
|  5          1          2        |  6          9          7        |  34         34         8        |
|  3          8          9        |  2          1          4        |  6          57         57       |
'---------------------------------'---------------------------------'---------------------------------'

bug+6.png (31.37 KiB) Viewed 528 times

[SpAce]

Give an example of bug + 6.

Very nice! Just like I said here:

That rule could be further generalized by allowing multiple guardians (different digits obviously) in the elimination cell, as long they don't include the external guardian digit (which is to be eliminated from that cell).

[rjamil]

Hi yzfwsf,

Give an example of bug + 6.

Just to add one more information regarding beautiful BUG+6 example.

Is your logic to detect the elimination from common peers of all guardian cells contain same guardian candidate, i.e., in this case, r5c789?

R. Jamil

[yzfwsf]

Just to add one more information regarding beautiful BUG+6 example.

Is your logic to detect the elimination from common peers of all guardian cells contain same guardian candidate, i.e., in this case, r5c789?

In the bug + n technique, at least one candidate as a guardian must be true. My solver calculates the intersection of the elimination sets produced by each guardian as true. This puzzle is 3r5c8. No matter which of 3r5c46, 3r46c8, 56r5c8 is true, 3r5c8 can be eliminated.

[rjamil]

In the bug + n technique, at least one candidate as a guardian must be true. My solver calculates the intersection of the elimination sets produced by each guardian as true. This puzzle is 3r5c8. No matter which of 3r5c46, 3r46c8, 56r5c8 is true, 3r5c8 can be eliminated.

Thanks for answering.

However, please explain, how 56r5c8 checked for guardian candidate 3?

Well, as per RW's derived theory (maybe I interpreted wrongly), each guardian candidate may be removed from cells that sees all guardian cells of that particular guardian candidate. In your example, 3 may be removed from r5c8 because it sees all guardian cells containing guardian candidate 3. In fact, essentially, 3 may be removed from r5c789 if exist. (Since a guardian candidate exists in more than one BUG+cells; therefore at least one of the BUG+cell MUST contain that guardian candidate. That particular guardian candidate may safely be removed from any cell that sees all BUG+cells of that particular guardian candidate.)

Similarly, since guardian candidates 5 and 6 are lying only in one BUG+cell, therefore no essential removal possible. (However, it may be treated as BUG+2 case, where guardian candidates are lying in one cell and all BUG+candidates may be removed from that BUG+cell too.)

R. Jamil

[yzfwsf]

However, please explain, how 56r5c8 checked for guardian candidate 3?

If all guardians are false, then the board will become all bi-valued cells, and each candidate appears in each row, column, and box exactly 2 times.

bug + n is that at least one of all the guardians (may have multiple types of candidates) is true, not to say that at least one guardian of each candidate number is true, specifically this puzzle is 3r5c46, 3r46c8 56r5c8 at least one is true.Mybe you mistake here.

[rjamil]

Hi SpAce and yzfwsf,

Note: This is another excellent example of BUG+2 move by RW.

I repeat my question again. Why not each guardian candidate is enough to remove from common peers? Why all guardian candidates are required to remove one guardian candidate from common peers?

Here is BUG+3 removal shown by RW as follows:

Code: Select all

+--------------+--------------+-------------+
| 2   5   1    | 7   4    8   | 9   6   3   |
| 3   9   4    | 6   5    2   | 7   18  18  |
| 6   7   8    | 1   3    9   | 4   5   2   |
+--------------+--------------+-------------+
| 5   26  67+2 | 9   8    3   | 12  4   17  |
| 1   8   3    | 4   2    7   | 5   9   6   |
| 9   4   27   | 5   1    6   | 28  3   78  |
+--------------+--------------+-------------+
| 8   3   9-6  | 2   79+6 4   | 16  17  5   |
| 4   26  5    | 8   67   1   | 3   27  9   |
| 7   1   29+6 | 3   69   5   | 68  28  4   |
+--------------+--------------+-------------+

[r7c3](-6-[r9c3])(-6-[r7c5])-6-[r7c7]-1-[r4c7]-2-[r4c3] => r7c3<>6

Whereas, according to my observation: 6 may also be eliminated from r7c12 and r9c456. Just assume two BUG+cells containing same guardian candidate 6, form two wings of XY-Wing move. (BTW, YZF_Sudoku 610 wont detect BUG+3 move at current pencil mark state.)

Similarly, if guardian candidate exists in only one BUG+cell then that guardian cell may be removed from buddy other BUG+cells' BUG+candidate. In that case, since BUG+candidate 2r9c3 sees guardian 2r4c3, therefore 2 may safely be removed from r9c3.

Note: Detailed discussion about BUG+n found in The Effortless Extremes thread.

R. Jamil

[SpAce]

Hi rjamil,

I repeat my question again. Why not each guardian candidate is enough to remove from common peers? Why all guardian candidates are required to remove one guardian candidate from common peers?

If you still have to ask that, after all the thorough explanations both I and yzfwsf have given and what you've read in the old threads, I don't think we can help you. The answer is there, very clearly stated and many times, but you either refuse to read it or to accept it. Either way, this is going nowhere.

Here is BUG+3 removal shown by RW as follows:

Hidden Text: Show

Code: Select all

+--------------+--------------+-------------+
| 2   5   1    | 7   4    8   | 9   6   3   |
| 3   9   4    | 6   5    2   | 7   18  18  |
| 6   7   8    | 1   3    9   | 4   5   2   |
+--------------+--------------+-------------+
| 5   26  67+2 | 9   8    3   | 12  4   17  |
| 1   8   3    | 4   2    7   | 5   9   6   |
| 9   4   27   | 5   1    6   | 28  3   78  |
+--------------+--------------+-------------+
| 8   3   9-6  | 2   79+6 4   | 16  17  5   |
| 4   26  5    | 8   67   1   | 3   27  9   |
| 7   1   29+6 | 3   69   5   | 68  28  4   |
+--------------+--------------+-------------+

[r7c3](-6-[r9c3])(-6-[r7c5])-6-[r7c7]-1-[r4c7]-2-[r4c3] => r7c3<>6

Here's the same with modern chain notation:

Code: Select all

+---------------+--------------+--------------+
| 2   5    1    | 7   4    8   |  9   6   3   |
| 3   9    4    | 6   5    2   |  7   18  18  |
| 6   7    8    | 1   3    9   |  4   5   2   |
+---------------+--------------+--------------+
| 5   26  b67+2 | 9   8    3   | c12  4   17  |
| 1   8    3    | 4   2    7   |  5   9   6   |
| 9   4    27   | 5   1    6   |  28  3   78  |
+---------------+--------------+--------------+
| 8   3    9-6  | 2  a79+6 4   | d16  17  5   |
| 4   26   5    | 8   67   1   |  3   27  9   |
| 7   1   a29+6 | 3   69   5   |  68  28  4   |
+---------------+--------------+--------------+

(6)r7c5|r9c3 == (2)r4c3 - (2=1)r4c7 - (1=6)r7c7 => -6 r7c3

That works, but it doesn't solve the puzzle. Here's a more effective, though (very) slightly more complicated, BUG elimination:

Code: Select all

.-----------------.-------------.--------------.
| 2    5     1    | 7   4     8 |   9   6   3  |
| 3    9     4    | 6   5     2 |   7   18  18 |
| 6    7     8    | 1   3     9 |   4   5   2  |
:-----------------+-------------+--------------:
| 5    6-2  a67+2 | 9   8     3 |  d12  4   17 |
| 1    8     3    | 4   2     7 |   5   9   6  |
| 9    4     27   | 5   1     6 |   28  3   78 |
:-----------------+-------------+--------------:
| 8    3     69   | 2  b79+6  4 | cd16  17  5  |
| 4  cd26    5    | 8   67    1 |   3   27  9  |
| 7    1    b29+6 | 3   69    5 |   68  28  4  |
'-----------------'-------------'--------------'

(2)r4c3
||
(6)r7c5 - (6=1)r7c7 - (1=2)r4c7
||
(6)r9c3 - (6=2)r8c2

=> -2 r4c2; stte

Or as a chain: (2)r4c3 == (6)r7c5|r9c3 - r7c7&r8c2 = (12)r74c7|(2)r8c2 => -2 r4c2; stte

Whereas, according to my observation: 6 may also be eliminated from r7c12 and r9c456. Just assume two BUG+cells containing same guardian candidate 6, form two wings of XY-Wing move. (BTW, YZF_Sudoku 610 wont detect BUG+3 move at current pencil mark state.)

Similarly, if guardian candidate exists in only one BUG+cell then that guardian cell may be removed from buddy other BUG+cells' BUG+candidate. In that case, since BUG+candidate 2r9c3 sees guardian 2r4c3, therefore 2 may safely be removed from r9c3.

I have no idea what you're talking about, and your cell coordinates seem to be wrong again. Anyhow, if you're interested in an XY-Wing style move with BUGs or URs, May 15 and May 16 are examples (added: and most recently June 2). It works if you have a BUG+2 or a UR+2 with two different guardian digits. Then it can be used like a virtual bivalue cell, e.g. as one of the three bivalue cells in an XY-Wing.

That said, none of that matters until you accept the fundamentals we've tried to teach you. Until then we're all just wasting our time.

[yzfwsf]

Hi SpAce and yzfwsf
I repeat my question again. Why not each guardian candidate is enough to remove from common peers? Why all guardian candidates are required to remove one guardian candidate from common peers?

You need to try various combinations of guardians, and then check and confirm. After excluding all the guardians, the unsolved cells are all bi-value cells and the unsettled candidates appear exactly twice in each house (row \ column \ box). After confirmation, only one of all guardians can be ensured to be true. Not every guardian must have one to be true.