The New Sudoku Players' Forum

[keith]

If it looks like a BUG ...

Code: Select all

+-------+-------+-------+
| 5 . . | . . 8 | . 4 3 |
| . . . | 5 9 4 | . . . |
| 7 1 . | 3 . . | . . . |
+-------+-------+-------+
| . 7 5 | . 4 . | . . 8 |
| . . 6 | . . . | 1 . . |
| 9 . . | . 8 . | 5 2 . |
+-------+-------+-------+
| . . . | . . 3 | . 8 5 |
| . . . | 4 1 7 | . . . |
| 3 6 . | 8 . . | . . 2 |
+-------+-------+-------+

Keith

[Leren]

... it probably is one.

Code: Select all

*-----------------------------------*
| 5     29   29 | 1  7  8  | 6 4 3  |
| 6     38   38 | 5  9  4  | 2 7 1  |
| 7     1    4  | 3  26 26 | 8 5 9  |
|---------------+----------+--------|
| 12    7    5  | 9  4  12 | 3 6 8  |
|*28-4 *48-2 6  | 27 3  5  | 1 9 47 |
| 9     34   13 | 67 8  16 | 5 2 47 |
|---------------+----------+--------|
| 14    49   19 | 26 26 3  | 7 8 5  |
| 28    5    28 | 4  1  7  | 9 3 6  |
| 3     6    7  | 8  5  9  | 4 1 2  |
*-----------------------------------*

2 Digit BUG + 2: Bug Digits 2 in r5c1 and 4 in r5c2: Eliminate 4 in r5c1 and 2 in r5c2.

Leren

[keith]

Well, in a BUG every unsolved clue is a bivalue, and every digit appears twice in each house. Right?

After basics:

Code: Select all

+-------------+-------------+-------------+
| 5   29  29  | 1   7   8   | 6   4   3   |
| 6   38  38  | 5   9   4   | 2   7   1   |
| 7   1   4   | 3   26  26  | 8   5   9   |
+-------------+-------------+-------------+
| 12  7   5   | 9   4   12  | 3   6   8   |
| 248 248 6   | 27  3   5   | 1   9   47  |
| 9   34  13  | 67  8   16  | 5   2   47  |
+-------------+-------------+-------------+
| 14  49  19  | 26  26  3   | 7   8   5   |
| 28  5   28  | 4   1   7   | 9   3   6   |
| 3   6   7   | 8   5   9   | 4   1   2   |
+-------------+-------------+-------------+

In C1 the extra digit is 2 in 248 R5C1, in C2 it is 4 in 248 R5C2, in R5 it is 24 in R5C12, in B4 it is 24 in R5C12. In other words, the BUG is

Code: Select all

+----------+----------+----------+
| 5  29 29 | 1  7  8  | 6  4  3  |
| 6  38 38 | 5  9  4  | 2  7  1  |
| 7  1  4  | 3  26 26 | 8  5  9  |
+----------+----------+----------+
| 12 7  5  | 9  4  12 | 3  6  8  |
| 48 28 6  | 27 3  5  | 1  9  47 |
| 9  34 13 | 67 8  16 | 5  2  47 |
+----------+----------+----------+
| 14 49 19 | 26 26 3  | 7  8  5  |
| 28 5  28 | 4  1  7  | 9  3  6  |
| 3  6  7  | 8  5  9  | 4  1  2  |
+----------+----------+----------+

which my solver says is a valid puzzle, but after the XY-wing 34-8 in R5C2 the puzzle is invalid.

Clearly I am missing something. This fits the definition I know, and stated above. Further, there are supposed to be two solutions to a BUG, right? Pick any digit in one cell, and you will get a solution. Pick the other digit in that cell, you will get the other solution. To destroy the BUG in this puzzle, you have to pick R5C1=2 and R5C2=4.

Of course, that does not work. The solution is R5C1=2, R5C2=8.

[Leren]

keith wrote : Further, there are supposed to be two solutions to a BUG, right?

The BUG theorem should state that the solution validity of the first "solution" is the same as the state of the first. I common parlance, if a BUG situation is uncovered, the number of solutions of the puzzle is either 0 or 2, ie it is not one.

By almost universal agreement, it is considered "unfair" to post a puzzle with multiple solutions unless you warn the solver that this may be the case.

So, in a practical sense, the number of solutions of BUG situation is, in practice, "always" 0. Maybe this is what you are missing.

Leren
'

[keith]

No, what I am missing is the method to solve this BUG+2. R5C2=4 destroys the BUG but is incorrect.

Leren, I am afraid I do not understand the notation in your original diagram.

Keith

[rjamil]

Hi keith,

To destroy the BUG in this puzzle, you have to pick R5C1=2 and R5C2=4.

Wrong.

According to BUG+2 definition, at least one or both of the R5C1=2 and/or R5C2=4 is/are true.

R. Jamil

[keith]

Hi keith,

To destroy the BUG in this puzzle, you have to pick R5C1=2 and R5C2=4.

Wrong.

According to BUG+2 definition, at least one or both of the R5C1=2 and/or R5C2=4 is/are true.

R. Jamil

So, how do I use a BUG+2 to advance the solution of a puzzle?

[Leren]

Put in my full 2 Digit Bug+2 notation above. Now to answer Keith's question, eliminate a candidate that would violate Jamil's constraint "at least one or both of the R5C1=2 and/or R5C2=4 is/are true."

4 in r5c1 and 2 in r5c2 both do this so you can eliminate them. Leren

[rjamil]

Hi keith,

In addition to Leren's explanation, as per RW logic:

if r5c1=2 then r5c1<>4 AND r5c2<>2; and
if r5c2=4 then r5c1<>4 AND r5c2<>2.

In both cases, same elimination possible.

R. Jamil

[keith]

Thank you, Jamil and Leren.

This is the first time I have seen a BUG+n where the digits to break the BUG are "or", not "and".

Keith

[rjamil]

Hi keith,

Actually, I was also thinking about BUG+n wrongly.

Thanks to yzfwsf and SpAce who correct me in this thread.

R. Jamil

[Leren]

A simple way to look at it is to assume that r5c1<>2 and r5c2<>4. This is what you get.

Code: Select all

*-----------------------------------*
| 5     29   29 | 1  7  8  | 6 4 3  |
| 6     38   38 | 5  9  4  | 2 7 1  |
| 7     1    4  | 3  26 26 | 8 5 9  |
|---------------+----------+--------|
| 12    7    5  | 9  4  12 | 3 6 8  |
| 48    28   6  | 27 3  5  | 1 9 47 |
| 9     34   13 | 67 8  16 | 5 2 47 |
|---------------+----------+--------|
| 14    49   19 | 26 26 3  | 7 8 5  |
| 28    5    28 | 4  1  7  | 9 3 6  |
| 3     6    7  | 8  5  9  | 4 1 2  |
*-----------------------------------*

This is a fully exposed BUG, leading to 0 solutions, so at least one of the two assumptions must be False.

Leren

[Yogi]

Since it generally comes up near the end, I find BUG+2 fairly straight-forward with this two-step process:
1) Identify the essential candidate in the two BUG cells. That's the one that leads to multiple solutions if it's not there.
2) Look for a verity which is predicted by separately placing the BUG candidates one at a time in their own cell. If both cases lead to the same result in another cell somewhere else in the puzzle then that is known to be true and you can proceed from there without being concerned for the time being about which is the correct candidate in the BUG cells.
The reasoning is based on the assumption that at least one of the identified BUG candidates must be true in its cell (but not necessarily both.) So anything else that flows on from both of these start-points must be true anyway.
Of course it becomes much harder to do the same thing if you have a BUG+3, when you would need to find a verity for all three cases.
Much easier to find some other technique to simplify down to BUG+2 or 1.

[RSW]

In other words, the BUG is

Code: Select all

+----------+----------+----------+
| 5  29 29 | 1  7  8  | 6  4  3  |
| 6  38 38 | 5  9  4  | 2  7  1  |
| 7  1  4  | 3  26 26 | 8  5  9  |
+----------+----------+----------+
| 12 7  5  | 9  4  12 | 3  6  8  |
| 48 28 6  | 27 3  5  | 1  9  47 |
| 9  34 13 | 67 8  16 | 5  2  47 |
+----------+----------+----------+
| 14 49 19 | 26 26 3  | 7  8  5  |
| 28 5  28 | 4  1  7  | 9  3  6  |
| 3  6  7  | 8  5  9  | 4  1  2  |
+----------+----------+----------+

which my solver says is a valid puzzle.

I don't think anyone mentioned this specifically, but your solver is clearly wrong if it says this is a valid puzzle. This PM grid has no valid solution.