Post the puzzle or solving technique that's causing you trouble and someone will help

I understand that Bug+1 is when all cells in a block have 2 possible candidates except 1 cell which have 3.
If I use tips when solving the help text will be that the candidate which figures 3 times in the row, column and cell is the solution.
This is not always true. How can I know when it is true?
EsbenHarder

Posts: 1
Joined: 06 December 2019

Consider the following pic:

https://imgur.com/I70Pv0M

The cell at R6C3 has exactly 3 candidates (1, 7, and 8). In the house making up column 3 there is exactly two 7's. In the house making up column 3 there is exactly two 8's. In the house making up row 6 there is exactly two 7's. And in the house making up row 6 there is exactly two 8's. Both the 7's and the 8's in the two house form an "Either-Or" type link between them. This means if one is false the other must be true. But both can't both be false. And both can't both be true.

So here is how the logic works. Assume the possible 1 candidate is not present in the cell R6C3 as shown:

https://imgur.com/NtIbuwT

Then you could have the following situation. imagine we have a 7 in R4C3 and R6C8, And we have an 8 in R3C3 and R6C1. This would result in the cell at R6C3 being blank! Since no cell can be blank this situation is not possible. Hence we can conclude the following result:

https://imgur.com/L2IMh6e

We were able to get 5 values along with a bunch of Naked Singles opening up with this puzzle.

The thing is even though I understand the logic I'm not sure I believe it! It could be the case that a 7 or 8 will shows up in cell R6C3. Maybe someone else with more brain power than I have can go further with why than I can.
dxSudoku

Posts: 27
Joined: 06 April 2020

Here's a better example because it's a naked triple in both shared houses:

https://imgur.com/ooYiDO0

Now assume R3C8 is does not have a value of 2. Then if we choose 5 for the value of R3C8 we get a number of contradictions:

https://imgur.com/shiMCTf

And if we choose a value of 9 for R3C8 we also get a number of contradictions:

https://imgur.com/lfi1GsX

So I think I figured it out. Check out the "not" example here:

http://hodoku.sourceforge.net/en/tech_ur.php

Any candidate in the two shared houses must not occur more than twice. Except the one in the pivot cell which the candidate to be remove must occur exactly 3 times in each direction. I almost see it!

No I don't see it. In the example above it was like brute force was needed to prove it was a BUG+1.

Maybe it means every possible candidate can only occur twice in all 27 houses except the one to be removed as non-possible.
dxSudoku

Posts: 27
Joined: 06 April 2020

Leren

Posts: 3915
Joined: 03 June 2012

Here is another explanation which is pretty good:
http://www.sudoku9981.com/sudoku-solvin ... -grave.php
dxSudoku

Posts: 27
Joined: 06 April 2020

Sorry to belabor this discussion. But I think I finally have a handle on it.

Take a look at this "not" example show here:

http://hodoku.sourceforge.net/en/tech_ur.php

There is possible 9 candidate occurring in the puzzle that somehow means the logic doesn't work. And then I read this in one of the links I posted above:

" A Bivalue Universal Grave or BUG is a state of the grid in which every unsolved cell has 2 candidates, without the presence of Hidden Singles."

[ http://hodoku.sourceforge.net/en/tech_ur.php ] Wiki Link 1

And then I read this text for "proof" of how this works:

" In a BUG, each unsolved cell has two candidates, and in each row, column and box constraint, each candidate value occurs exactly twice. If we can find a valid solution using half the number of remaining candidates, all the eliminated candidates would also satisfy each constraint exactly once. This other half will also give us a solution to the puzzle. There is also the possibility that the puzzle does not have a solution at all. In that case, the puzzle is also invalid. The existence of the BUG pattern allows us to conclude that the puzzle is invalid. It has either no solution or 2 solutions. "

[ http://sudopedia.enjoysudoku.com/Bivalu ... Grave.html ] Wiki Link 2

Again, I hate to belabor point but consider my first first example:

https://imgur.com/I70Pv0M

Cell R6C3 has three possibilities (1, 7, 8). This is what got me hung up. It seemed to me 7 or 8 could solve the puzzle. I did not see why either the 7 or 8 could not solve the puzzle. So I tried the value of 8:

https://imgur.com/Rt961P9

The value of 8 resulted in a contradiction in the puzzle as expected. BUT, the value 7 in R6C3 solved the puzzle:

https://imgur.com/7ipONK4

And the value of 1 in cell R6C3 solved the puzzle:

https://imgur.com/ZXEZf1n

So my initial problem in understanding this is I did not see why having a 7 or 8 in the cell could not be a possible solution to the puzzle. So I think the position taken with this technique is even though the 7 worked, just assume both will result in failure and pick the +1 candidate.

I think the must have condition for the Bug+1 to work is no other candidate in the entire puzzle can have more than 2 possible candidates in any house except the Bug+1 candidate which has 3 in the intersecting shared houses with the pivot cell.

Even though the 7 resulted in a positive result I think I now am on-board with the Bug+1. It's not too hard to find once you accept the conditions as I see them although I may be wrong. But I think I am right on this.

Here's an example where choosing either one of the non Bug+1 candidates resulted in a contradiction:

https://imgur.com/ooYiDO0
dxSudoku

Posts: 27
Joined: 06 April 2020

Something just occurred to me that simplifies the argument why choosing the Bug+1 candidate is the only valid choice. If you remove the Bug+1 candidate as a possible choice in the Bug+1 pivot cell, then every remaining cell in the puzzle has only two possible candidates. At this point, you just have to guess the value of a cell to complete the puzzle. Guessing in any Sudoku puzzle automatically disqualifies it as a valid Sudoku puzzle.
dxSudoku

Posts: 27
Joined: 06 April 2020