The New Sudoku Players' Forum

[SpAce]

First of all, many thanks for detail guidance with patience. I will try to digest it slowly.

No problem. I also try to evolve and develop my patience

What I understand from yzfwsf's post is that, there are two different scenarios as per elimination point of view. I assumed BUG+n for elimination(s) from guardian candidates of BUG+cells. Whereas, "counter" Bivalue Universal Grave + n for elimination(s) from other cells that are based on BUG+cells/BUG+candidates (like Cenoman's BUG+3 move. But, it needs confirmation from yzfwsf.)

Ok. Let us know what you find out. We don't normally make such distinctions.

Btw, note that if you apply RW's BUG+3 logic and remove 9r1c7, the grid no longer has any BUG+ pattern:

Code: Select all

.----------.----------.--------------.
| 89  2  6 | 5  4   7 | 38   1   389 |
| 89  1  4 | 3  29  6 | 289  7   5   |
| 5   3  7 | 1  29  8 | 4    6   29  |
:----------+----------+--------------:
| 4   7  2 | 8  3   9 | 6    5   1   |
| 3   6  5 | 2  7   1 | 89   89  4   |
| 1   9  8 | 4  6   5 | 23   23  7   |
:----------+----------+--------------:
| 2   5  1 | 6  8   4 | 7    39  39  |
| 7   8  3 | 9  5   2 | 1    4   6   |
| 6   4  9 | 7  1   3 | 5    28  28  |
'----------'----------'--------------'


That's not a BUG+2. It's now impossible to choose guardians from the remaining non-bivalue cells r1c9 and r2c7 so that after removing them there would be two of every digit in every house.

For example, 8r1c9 should be a guardian based on row 1, because there are three 8s. However, there are only two 8s in column 9, so 8r1c9 can't be a guardian based on that -- it should be 9r1c9 as before (because there are three 9s in column 9). Yet 9r1c9 can't be a guardian because there are only two 9s in row 1... and so on.

[Cenoman]

Deleted. Irrelevant.

[Cenoman]

A short tutorial for rjamil.

By definition, the pattern BUG+n encompasses all not solved cells in a puzzle. Otherwise one must display the subset of the concerned pattern and call it BUG-lite+n
n is the number of candidates in excess to the BUG (some people rather count the number of cells having more than two candidates)
How to identify those excess candidates, that we call guardians ? In the BUG+1, it is easy: select the trivalue cell and spot the candidate in this cell that is present three times in the cell row, the cell column, the cell box, simultaneously.
For a BUG+n, just think that the pattern is a BUG+(n-1)+1 and search a cell that is alone in a sector with an excess candidate. Apply the process of the BUG+1 and then iterate with remaining BUG+(n-1) pattern.
So in the object puzzle, start with r2c7 (alone with 3 candidates in row 2) Which digit is present 3 times in r2 ? The 9, just check that it is 3 times in column 7, and 3 times in box 3. Keep aside 9r2c7. Remaining BUG+2: start with any of r1c7 or r1c9: they are alone in their column. Let's choose r1c7. Which digit is present 3 times in column 7 (9r2c7 excluded...) The 8, also present 3 times in r1 and b3. Keep aside 8r1c7. Remaining BUG+1: there must be in the last un-processed cell (r1c9) a digit present 3 times in r1, c9, b3. The 9, bingo!
No need of a software, this is easy to do with p&p.
The set of guardians is a SIS: at least one is true (but which one is don't known). You have to search for chains targetting a potentential elimination.

Such processing sometimes fails: either a cell alone in a sector has no candidate present more than 2 times. In such a case, the puzzle has no BUG configuration. Or there remains cells having more than two candidates, none is alone in a sector. Generally, such cells contain a potential UR and the puzzle has several BUG configurations. But let's keep complex things for later.

Reminder: search first cells having more than two candidates, alone in at least one of their sectors (row, column, box)

[SpAce]

Hi again, Steve,

Ignore basically everything I've said before in this thread, since I obviously didn't originally understand how your solution was supposed to work. I assumed it was another normal BUG+3 solution, so I didn't look at the guardians or the logic carefully enough. Now that I think I get it better, it gets even more interesting. In fact, I'm not quite sure what the best way to notate it is, which is... well, refreshing!

Yes, I like it. It was recognition of the fact that eliminating 8r2c7 changed things from an almost BUG with 3 guardians to an almost BUG with 2 guardians that led me to write BUG+2 initially, but I didn't have a way to say it. Your way serves nicely.

Actually, I don't think it does. Like I said, it was based on false assumptions, just like anything else I wrote before.

I could also have simply written 8r2c7 = (8,9)r21c1 - 9r1c79 = BUG => +8 r2c7 . Would that have been better?

Yes, it would have been better, as it made me see the logic you really intended (at least I hope so!). However, I don't think it's fully correct because the last strong link doesn't work independently. In other words, it requires memories from the previous steps. My first improvement would be:

8r2c7 = (8,9)r21c1 - 8r2c7|9r1c79 = BUG => +8 r2c7

However, that's not correct either. If it were, then it could be written like this too:

8r2c7 = (8,9)r21c1 - 9r1c79 =BUG+3= 8r2c7 => +8 r2c7

That's practically no different from the original, except for the orientation and the placement conclusion. What makes both that and the version above incorrect is that there's no BUG+3 in the current grid (with 16 unsolved cells) that has 9r1c79 and 8r2c7 as guardians. Thus the strong link is still invalid, either way it's written.

However, those candidates are indeed guardians for an implied BUG in a grid with only 14 unsolved cells after (8,9)r2c21 is assumed. That makes it a bit harder to write accurate strong links with it, because the assumed placements must be included too. Here's one attempt:

8r2c7 = (8,9)r21c1 - 8r2c7|9r1c79 = ((8,9)r21c1 & BUG) => +8 r2c7

What do you think of that? It's of course a bit complicated, but so far I don't see any simpler way to write it.

I think the important part (besides correct links) is that it makes it relatively clear that the BUG in question is not based on the current set of unsolved cells. It just got demonstrated that if the notation for an implied BUG is too similar to a normal BUG+N, it causes confusion and misunderstandings.

Added. A couple of less explicit but probably quite (perhaps more) understandable variants:

(8,9)r12c1 = ((9,8)r12c1 & BUG) => +(8,9)r12c1; stte

(9,8)r12c1 -> BUG => +(8,9)r12c1; stte

[SpAce]

Deleted. (Problem fixed.)

[rjamil]

Hi SpAce and Cenoman,

Ok. Let us know what you find out. We don't normally make such distinctions.

Well, there are two definitions for BUG+2 in sudopedia. And example puzzle taken from opening post where RW explain simple definition.

Cenoman, you mean, find BUG cells having more than 2 digits as candidates. Then, for each BUG cell, find trilocation guardian digit in all units (i.e., row, column and box).

Now, in my observation for your BUG+3 move, if guardian 8 present only in one BUG cell then apply BUG+1 for guardian 8. Similarly, if guardian 9 is present in more than one BUG cells then remove 9 from common peers of all BUG cells having guardian 9. In this case, 9 may be removed from r3c9 as well.

R. Jamil

[Cenoman]

That would be so nice... but unfortunately I don't see how it's supposed to work as a BUG-Lite. As far as I see, it has several candidates (3r6c7, 3r7c9, 8r5c7, 8r9c9) that don't pair up with anything in their columns. Aren't they supposed to, just like in a full BUG?

Yes. Obviously, posted too fast.

[SpAce]

Yes. Obviously, posted too fast.

Ok, thanks for confirming! It happens, no harm done. It just got me confused for a second because I'd tried to look for a BUG-Lite. At first glance it does look like it could have one, especially because of the paired 29s in b23, which made at least me expect that the loop could be closed in b69. No such luck.

[SpAce]

Hi rjamil,

Well, there are two definitions for BUG+2 in sudopedia. And example puzzle taken from opening post where RW explain simple definition.

There's only one definition of BUG+2, or any BUG+n for that matter. The Sudopedia link just shows two slightly different ways to use a BUG+2 to eliminate something. It doesn't mean BUG+2 itself has two different definitions.

Furthermore, the elimination logic is exactly the same in both cases, as it is with any BUG+n. It may just look different for someone who doesn't see how it works and depends on specific tricks for each situation.

Cenoman, Now, in my observation for your BUG+3 move, if guardian 8 present only in one BUG cell then apply BUG+1 for guardian 8. Similarly, if guardian 9 is present in more than one BUG cells then remove 9 from common peers of all BUG cells having guardian 9. In this case, 9 may be removed from r3c9 as well.

Probably not even close, but once again, I'm not sure what you even mean. Btw, I told you last time that "BUG cell" is an ambiguous term and should be avoided. In my terminology every unsolved cell is a "BUG cell" in a BUG+n situation. If you mean cells that contain guardians, you should call them "guardian cells" to avoid confusion. Then the others can be called "non-guardian cells".

In any case, that seems very far from any recognizable BUG logic, despite my best efforts in our last discussion. This time I'll let Cenoman explain his logic to you. Perhaps he has better luck than I did

[yzfwsf]

Hi all:
I uploaded examples of various bugs + n implemented by my solver. Please confirm whether the logic is correct.
http://forum.enjoysudoku.com/post292218.html

[SpAce]

Hi all:
I uploaded examples of various bugs + n implemented by my solver. Please confirm whether the logic is correct.
http://forum.enjoysudoku.com/post292218.html

Nice examples. All look good to me at first glance.

PS. These are easy to check just by eyeballing, but in general it would be nice if you attached puzzle strings and text-based pencil marks to your examples, so they can be easily pasted into software tools.

[rjamil]

Hi SpAce,

There's only one definition of BUG+2, or any BUG+n for that matter. The Sudopedia link just shows two slightly different ways to use a BUG+2 to eliminate something. It doesn't mean BUG+2 itself has two different definitions.

Furthermore, the elimination logic is exactly the same in both cases, as it is with any BUG+n. It may just look different for someone who doesn't see how it works and depends on specific tricks for each situation.

If I understand correctly then my interpretation is that both the definitions are separate despite end results are same. (Fight for My Way.)

1) According to sudopedia: "... where the extra candidates are 7 in r7c2 and 8 in r8c3. This implies that either r7c2=7 or r8c3=8." (It further means that either both are true or false.)
According to wapati: "BUG+2 suggests that R7C2 is 7 or R8C3 is 8. In either case R7C2 is not 8 which forces R4C2 = 8."

2) According to sudopedia: "Actually, for this example, a much simpler deduction can be made. Because r7c2 and r8c3 see each other, we can immediately conclude that r7c2<>8 and r8c3<>7 (by a simple proof by cases)." (It means that both the guardians must be occupy one of its guardian cells. In this case, since each guardian has one occurrence therefore it must go to that guardian cell.)
According to RW: "Ah, you make it so complicated, no need to look at any other cells than the ones with extra candidates. As you said, either r7c2=7 or r8c3=8 has to be true. Therefore r7c2<>8 and r8c3<>7, puzzle solved. This is actually the simplest kind of BUG+2 where both cells with extra candidates can see each other and allow direct eliminations.". (If either case is true then it is impossible to eliminate both directly from each other.)

I read 1st definition as "Either r7c2=7 or r8c3=8 has to be true. 2nd definition as "Both r7c2=7 and r8c3=8 have to be true.

Now, what I conclude from above quoted paras for BUG+n is that, each guardian candidate must be placed in one of its guardian cells. I mean if there are two guardian candidates, out of which one guardian candidate exists in one guardian cell and other in two; then according to my logic, both the guardian candidates must go to their specific guardian cell(s), i.e., one to its guardian cell and other to one of its two guardian cells.)

Let see both the scenarios elimination wise:
1) r4c2 = 8 (or r7c2 <> 8) (with complicated logic);
2) r7c2 <> 8 and r8c3 <> 7 (simply).

Otherwise, my question is that, is there any scenario of BUG+n where number of guardian cells > number of guardian candidates; and not all guardian candidates occupy from one of its respective guardian cells?

Similarly, there is no difference by assuming that both 8 and 9 guardian candidates in r1c7 for opening post's puzzle.

Hope that I convey my point of view.

R. Jamil

[SteveG48]

I could also have simply written 8r2c7 = (8,9)r21c1 - 9r1c79 = BUG => +8 r2c7 . Would that have been better?

Yes, it would have been better, as it made me see the logic you really intended (at least I hope so!). However, I don't think it's fully correct because the last strong link doesn't work independently. In other words, it requires memories from the previous steps. My first improvement would be:

8r2c7 = (8,9)r21c1 - 8r2c7|9r1c79 = BUG => +8 r2c7

However, that's not correct either. If it were, then it could be written like this too:

8r2c7 = (8,9)r21c1 - 9r1c79 =BUG+3= 8r2c7 => +8 r2c7

That's practically no different from the original, except for the orientation and the placement conclusion. What makes both that and the version above incorrect is that there's no BUG+3 in the current grid (with 16 unsolved cells) that has 9r1c79 and 8r2c7 as guardians. Thus the strong link is still invalid, either way it's written.

However, those candidates are indeed guardians for an implied BUG in a grid with only 14 unsolved cells after (8,9)r2c21 is assumed. That makes it a bit harder to write accurate strong links with it, because the assumed placements must be included too. Here's one attempt:

8r2c7 = (8,9)r21c1 - 8r2c7|9r1c79 = ((8,9)r21c1 & BUG) => +8 r2c7

What do you think of that? It's of course a bit complicated, but so far I don't see any simpler way to write it.

I think that works. As you say, it's pretty complicated. Maybe this is a case where simple words would do best.

[SpAce]

Hi rjamil,

Just a very quick reply for now. Your fundamental mistake is (once again) here, and it ripples to everything else you wrote:

I read 1st definition as "Either r7c2=7 or r8c3=8 has to be true. 2nd definition as "Both r7c2=7 and r8c3=8 have to be true.

There is no 2nd definition. You've misinterpreted what you've read. It's always OR, never AND.

It's possible that multiple guardians are true, but it's never guaranteed and you can't know which ones are true or false. The ONLY thing a BUG+n pattern guarantees is that AT LEAST ONE of the guardians is true. That is distinctly different from assuming that ALL GUARDIANS are true, which seems to be what you try to apply every time. It's the difference between OR and AND.

For example, if you have BUG+5, it could have 1..5 true guardians. You just can't know how many and which one(s) is/are true, so your logic can never depend on such assumptions. The only thing you know is that the number is not zero. If the five guardians are {a b c d e} you know that the boolean expression (a OR b OR c OR d OR e) is true.

What follows from that is that you actually have to prove any elimination for ALL of the guardians separately, because you don't know which one(s) is/are true. Every single one of the guardians must be able to eliminate the same victim on their own, or otherwise it's not a guaranteed elimination. For some unknown reason you seem to repeatedly think that it's somehow enough that a single guardian could eliminate something for it to be a valid elimination. It's not. The only time that's true is in a BUG+1 situation because there's only one guardian to worry about. There are no other exceptions.

No matter what you may (think you) have read elsewhere, that is the sole truth of this matter. If you think someone has said something else, you're almost certainly misunderstood what they've said, or otherwise someone has said something that is not true (very unlikely, because this is really basic).

This is not a matter of opinion. It's a fundamental fact. All BUG+n logic depends on understanding that simple truth. It's both necessary and sufficient for it.

I hope you finally accept this so that we'll never have to come back to it again.

[rjamil]

Hi SpAce,

I think, yzfwsf gives enough visual material with one line explanations that will clear the picture.

Let me study all aspects of BUG+n from just 4-5 examples and try to convert in to generic exemplars form.

R. Jamil