## The BUG (Bivalue Universal Grave) principle

Advanced methods and approaches for solving Sudoku puzzles
15??

Pat

Posts: 3907
Joined: 18 July 2005

now this is "page 9" - but there's a non-existent "page 10" ??
just getting worse !!

udosuk wrote:If it's the database problem then only the IT department of the phpbb company can fix it, and I don't think the administrator of this forum can do much about it... However, I think we (the users) can fix it ourselves by posting 5-6 more posts on that thread to make the "p.9" really valid, and then delete the posts one by one and see if the "p.9" would disappear properly...

i did try deleting - doesn't help.

Pat

Posts: 3907
Joined: 18 July 2005

[quote="Pat"]now this is "page 9" - but there's a non-existent "page 10" ??
just getting worse !!

Maybe there is a really existing BUG type 10 in this thread. ;-)

Claudia
claudiarabia

Posts: 288
Joined: 14 May 2006

Pat,
Okay, my suggestion didn't work... I guess we'll have to wait for the IT dept of the phpbb forum services to fix it!
udosuk

Posts: 2698
Joined: 17 July 2005

Claudiarabia, I don't use Sudoku Explorer so I can't say what it will or will not find. Here's one example from the zoo. The extra digits in the BUG are 1,3,9. Combined with r2c6 and r4c6, these form a naked triple, therefore r8c6<>9. You should be able to work the other examples out by either forming a naked set as in this example or a hidden set using all of the candidates in the non-bivalued sets (same logic as a Type 3 UR). As always there are probably many techniques which can be used to solve this and the other puzzles, this is just one option.

Code: Select all
.35......8..6......1....6.3....1.827..3.....19..4.......9.742.....8...1.2.....79.
+----------+--------------+------------+
| 6  3  5  | 17* 24* 128* | 14* 78* 9  |
| 8  9  7  |  6  34*  13* | 14*  5  2  |
| 4  1  2  | 79* 59* 589* |  6  78* 3  |
+----------+--------------+------------+
| 5  4  6  | 39*  1   39* |  8   2  7  |
| 7  8  3  |  2  56*  56* |  9   4  1  |
| 9  2  1  |  4   8    7  |  3   6  5  |
+----------+--------------+------------+
| 1  6  9  |  5   7    4  |  2   3  8  |
| 3  7  4  |  8  29* 2-9* |  5   1  6  |
| 2  5  8  | 13* 36* 136* |  7   9  4  |
+----------+--------------+------------+
Mike Barker

Posts: 458
Joined: 22 January 2006

### Bug type 3

Hi Mike,

thank you for the example. I got it now. And I found it in the Zoo too. Is seems to be the most interesting bug of the bugs.

claudia
claudiarabia

Posts: 288
Joined: 14 May 2006

### A bug question

I have been attempting t o understand BUGs.

With this puzzle:
Code: Select all
*--------------------------------------------------*
| 6    5    18   | 78   4    9    | 17   2    3    |
| 49   2    18   | 78   3    5    | 147  6    79   |
| 3    7    49   | 6    1    2    | 45   8    59   |
|----------------+----------------+----------------|
| 5    8    6    | 9    2    1    | 3    7    4    |
| 24   1    47   | 3    5    78   | 28   9    6    |
| 29   3    79   | 4    6    78   | 258  1    58   |
|----------------+----------------+----------------|
| 1    6    3    | 2    9    4    | 78   5    78   |
| 8    9    5    | 1    7    3    | 6    4    2    |
| 7    4    2    | 5    8    6    | 9    3    1    |
*--------------------------------------------------*

There are two cells with three candidates (r2c7 and r6c7).

Which may be reduced and why?

This example is a BUG 3. How is it different from a BUG 1?

Appreciate any help.
dan

ArkieTech

Posts: 3355
Joined: 29 May 2006
Location: NW Arkansas USA

### Re: A bug question

ArkieTech wrote:I have been attempting t o understand BUGs.

With this puzzle:
Code: Select all
*--------------------------------------------------*
| 6    5    18   | 78   4    9    | 17   2    3    |
| 49   2    18   | 78   3    5    | 147  6    79   |
| 3    7    49   | 6    1    2    | 45   8    59   |
|----------------+----------------+----------------|
| 5    8    6    | 9    2    1    | 3    7    4    |
| 24   1    47   | 3    5    78   | 28   9    6    |
| 29   3    79   | 4    6    78   | 258  1    58   |
|----------------+----------------+----------------|
| 1    6    3    | 2    9    4    | 78   5    78   |
| 8    9    5    | 1    7    3    | 6    4    2    |
| 7    4    2    | 5    8    6    | 9    3    1    |
*--------------------------------------------------*

There are two cells with three candidates (r2c7 and r6c7).

Which may be reduced and why?

This example is a BUG 3. How is it different from a BUG 1?

Appreciate any help.

I'm not sure why it would be called a BUG 3, I would call it a BUG+2. Here is the argument I would make though:

If r2c7<>7 and r6c7<>8, then we would have a BUG. Therefore, we conclude that r2c7=7 or r6c7 = 8. However, if r1c7 = 7, then r7c7 = 8 and we will get the BUG grid. Hence, r1c7<>7 and the puzzle is solved.

Incidentally, there is also a BUG-lite pattern in r12c347 that implies that r2c7=4. This will also solve the puzzle.
re'born

Posts: 551
Joined: 31 May 2007

Code: Select all
With this puzzle:

*--------------------------------------------------*
| 6    5    18   | 78   4    9    | 17   2    3    |
| 49   2    18   | 78   3    5    | 147  6    79   |
| 3    7    49   | 6    1    2    | 45   8    59   |
|----------------+----------------+----------------|
| 5    8    6    | 9    2    1    | 3    7    4    |
| 24   1    47   | 3    5    78   | 28   9    6    |
| 29   3    79   | 4    6    78   | 258  1    58   |
|----------------+----------------+----------------|
| 1    6    3    | 2    9    4    | 78   5    78   |
| 8    9    5    | 1    7    3    | 6    4    2    |
| 7    4    2    | 5    8    6    | 9    3    1    |
*--------------------------------------------------*

rep'nA wrote:

I'm not sure why it would be called a BUG 3, I would call it a BUG+2. Here is the argument I would make though:

If r2c7<>7 and r6c7<>8, then we would have a BUG. Therefore, we conclude that r2c7=7 or r6c7 = 8. However, if r1c7 = 7, then r7c7 = 8 and we will get the BUG grid. Hence, r1c7<>7 and the puzzle is solved.

Incidentally, there is also a BUG-lite pattern in r12c347 that implies that r2c7=4. This will also solve the puzzle.

Thanks rep'nA

I guess the number after BUG represents the number of cells with candidates > 2.

And a BUG or BUG grid is where all unsolved cells have 2 candidates. Making an invalid puzzle.

I am going to tackle BUG lites next.

dan
dan

ArkieTech

Posts: 3355
Joined: 29 May 2006
Location: NW Arkansas USA

### Re: A bug question

rep'nA wrote:
ArkieTech wrote:I have been attempting t o understand BUGs.

With this puzzle:
Code: Select all
*--------------------------------------------------*
| 6    5    18   | 78   4    9    | 17   2    3    |
| 49   2    18   | 78   3    5    | 147  6    79   |
| 3    7    49   | 6    1    2    | 45   8    59   |
|----------------+----------------+----------------|
| 5    8    6    | 9    2    1    | 3    7    4    |
| 24   1    47   | 3    5    78   | 28   9    6    |
| 29   3    79   | 4    6    78   | 258  1    58   |
|----------------+----------------+----------------|
| 1    6    3    | 2    9    4    | 78   5    78   |
| 8    9    5    | 1    7    3    | 6    4    2    |
| 7    4    2    | 5    8    6    | 9    3    1    |
*--------------------------------------------------*

There are two cells with three candidates (r2c7 and r6c7).

Which may be reduced and why?

This example is a BUG 3. How is it different from a BUG 1?

Appreciate any help.

I'm not sure why it would be called a BUG 3, I would call it a BUG+2. Here is the argument I would make though:

If r2c7<>7 and r6c7<>8, then we would have a BUG. Therefore, we conclude that r2c7=7 or r6c7 = 8. However, if r1c7 = 7, then r7c7 = 8 and we will get the BUG grid. Hence, r1c7<>7 and the puzzle is solved.

Incidentally, there is also a BUG-lite pattern in r12c347 that implies that r2c7=4. This will also solve the puzzle.

I agree with the above, although I tend to look at them a little bit differently... As stated, r2c7=7 and/or r6c7 = 8. If I set r2c7 = 7, r7c7 = 8, r5c7 = 2, r6c7 = 5. If r2c7 <> 7, then r6c7 = 8 (to avoid Bug). This means that r6c7 <> 2, which will solve the puzzle.

Generally, I go for the Bug-lite pattern first when it is like the one stated in the previous response. I only stated the above is that it isn't that apparent to me (especially since I mostly work by hand) that setting r2c7=7 results in a Bug. Rather I set r2c7 = 7 to see what I can eliminate from the other tri-valued cell.

Erik
emalvick

Posts: 13
Joined: 01 August 2005

### Re: A bug question

emalvick wrote:
rep'nA wrote:If r2c7<>7 and r6c7<>8, then we would have a BUG. Therefore, we conclude that r2c7=7 or r6c7 = 8. However, if r1c7 = 7, then r7c7 = 8 and we will get the BUG grid. Hence, r1c7<>7 and the puzzle is solved.

I agree with the above, although I tend to look at them a little bit differently... As stated, r2c7=7 and/or r6c7 = 8. If I set r2c7 = 7, r7c7 = 8, r5c7 = 2, r6c7 = 5. If r2c7 <> 7, then r6c7 = 8 (to avoid Bug). This means that r6c7 <> 2, which will solve the puzzle.

Alternatively, one can use the "quantum cell" approach commonly seen for the Type 3 UR, whereby cells r26c7 are considered to be a single bivalued cell. Then the naked pair in c7 yields r1c7<>7 and r5c7<>8.
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Some of the confusion may be with nomenclature. This is a Type 3 BUG (analogous to a Type 3 UR). Because there are two nonBug cells this is a BUG+2X. (Capital "X" is shorthand for type 3, small "x" short hand for Type 2). You could have a Type 3 BUG with 3 nonBUG cells (BUG+3X), etc.
Mike Barker

Posts: 458
Joined: 22 January 2006

Mike Barker wrote:Some of the confusion may be with nomenclature. This is a Type 3 BUG (analogous to a Type 3 UR). Because there are two nonBug cells this is a BUG+2X. (Capital "X" is shorthand for type 3, small "x" short hand for Type 2). You could have a Type 3 BUG with 3 nonBUG cells (BUG+3X), etc.

I thought I was doing good with the BUGs, but I never realized there were specific Types. I've probably been using them all along, but is there anywhere where these types are specifically stated. I've read through this thread and haven't seen the types specifically defined. I haven't been able to find much searching the forum up to this point either.

Erik
emalvick

Posts: 13
Joined: 01 August 2005

Most anything you can do with a Unique Rectangle you can do with a BUG (and more since BUGs are bigger). Links for several threads discussing BUGs are found in the Collection of Solving Techniques. Check out the UR links as well (especially for the basic four types). Examples can be found in the Local Zoo.
Mike Barker

Posts: 458
Joined: 22 January 2006

### BUG Type 3 - a definition

ArkieTech wrote:
Code: Select all
*--------------------------------------------------*
| 6    5    18   | 78   4    9    | 17   2    3    |
| 49   2    18   | 78   3    5    | 14*7 6    79   |
| 3    7    49   | 6    1    2    | 45   8    59   |
|----------------+----------------+----------------|
| 5    8    6    | 9    2    1    | 3    7    4    |
| 24   1    47   | 3    5    78   | 28   9    6    |
| 29   3    79   | 4    6    78   | 25*8 1    58   |
|----------------+----------------+----------------|
| 1    6    3    | 2    9    4    | *78  5    78   |
| 8    9    5    | 1    7    3    | 6    4    2    |
| 7    4    2    | 5    8    6    | 9    3    1    |
*--------------------------------------------------*

I guess the number after BUG represents the number of cells with candidates > 2.
dan

Hi Dan,

I also was confused about the BUG Type 3. This BUG appears rather seldom. It's solution lays always in one line. (If somebody shows me an example with a box I take this back). In this line (column or row), in our example column 7, you have at least two cells with three candidates, disturbing the BUG-symmetry. BUG-Symmetrie means, that in every line and in every box every candidate has to appear exactly twice and every cell has 2 candidates.

You single out now the two Candidates disturbing the BUG-symmetry. In your case these are the 7 in r2c7 and the 8 in r6c7. These two candidates form together with another cell of the respective column, here with r7c7, a naked triplet. By this method r1c7 and r5c7 stay with one candidate only. The rest is placing singles.

You have this BUG also with 3 cells with 3candidates or 4. They form a naked quad or even quint then, leaving other cells in-line with a single candidate. This is the BUG3-principle.

With best regards

Claudia
claudiarabia

Posts: 288
Joined: 14 May 2006

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