## Sudoku time to guess?

Advanced methods and approaches for solving Sudoku puzzles

### Sudoku time to guess?

Is there a pattern here that I can get without trial and error?

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`   |  1     2     3     4     5     6     7     8     9     ---+---------------------------------------------------A  |  4     18    7     58    6     15    3     2     9     B  |  6     28    9     278   2378  23    4     5     1     C  |  5     12    3     4     12    9     6     8     7     D  |  7     9     5     6     13    13    2     4     8     E  |  8     4     6     257   27    25    9     1     3     F  |  1     3     2     9     4     8     5     7     6     G  |  9     6     4     1     5     7     8     3     2     H  |  3     5     1     28    28    6     7     9     4     I  |  2     7     8     3     9     4     1     6     5 `
TopRank

Posts: 11
Joined: 22 December 2005

Welcome. Refer here for a solution making use of the BUG principle.
Jeff

Posts: 708
Joined: 01 August 2005

And look for xy wing too
bennys

Posts: 156
Joined: 28 September 2005

I don't know the term for this (I'm new to the Sudoku thing...) but:

D5 and D6 are a 13 pair. Setting D5 to 1 makes D6 3 and then B6 must be 2. It also makes C5 2. This can't be true, so D5 must be 3 and D6 must be 1.
natedogg635

Posts: 1
Joined: 27 December 2005

Nate:

Thanks. That looks like the xy wing, using the 12 as a pivot, either d5 or b6 has to be 3 so nothing that is connected to both can be a 3. So d6 has to be 1.

Did that pop out at you?
TopRank

Posts: 11
Joined: 22 December 2005

Top Rank
For an XYwing you look for 2 cells that share one candidate (XY, XZ)
Then look for a third cell with the two unshared candidates (YZ)
Then all cells that are buddies of the 2 cells with Z cannot be Z

In this case the XYwing = 15, 12, 25 = A6 C5 E6
You can remove all the 2s that are in the same group as C5 and E6 i.e. E5 and B6.

You might like to change your notation to that used by everyone in this forum i.e. A1 = r1c1.
sweetbix

Posts: 58
Joined: 10 December 2005

Nate
I think that your kind of one arm chain is considered guessing. For chains you need to test all candidates in a cell and get to the same result in another cell.

r1c2=8 => r2c2=2 => r2c6=3
r1c2=1 => r1c6=5 => r5c6=2 => r2c6=3
Therefore r2c6=3

r1c6=1 => r4c6=3 => r2c6=2 => r3c5=1 => r1c6=5
Therefore r1c6 does not = 1
sweetbix

Posts: 58
Joined: 10 December 2005

Jeff
Would you please show me how the BUG principle works here. Thanks.
sweetbix

Posts: 58
Joined: 10 December 2005

sweetbix wrote:For an XYwing you look for 2 cells that share one candidate (XY, XZ)
- Look where? Anywhere in the 9x9?, same box?, ???

Then look for a third cell with the two unshared candidates (YZ)
- Look where? Anywhere in the 9x9?, same box?, ???

- OK, now I have XY, XZ, AND YZ.

Then all cells that are buddies of the 2 cells with Z cannot be Z
- "buddies"? You mean in the same row or column or box? Or in the same row AND box?

In this case the XYwing = 15, 12, 25 = A6 C5 E6
------------------------------xy, xz, zy

You can remove all the 2s that are in the same group as C5 and E6 i.e. E5 and B6.
- Remove 2's because z=2? If there were 5's, they couldn't be removed because y's don't count?

I don't even get it a little bit. Sorry.

Mac
QBasicMac

Posts: 441
Joined: 13 July 2005

The squeeze that r3c5(12), r2c6(23) and r4c5(13) apply that prevents r4c6 from being 3 is certainly is not a guess.

Finding a pivot and squeezing is just another way of looking at the xy-wing. For example, using r1c6(15) as a pivot, either r3c5(12) is a two or r5c6(25) is a two and not both. Those two possible twos prevent both r5e5(27) and r2e6(23) from being twos since those two cells "see" both of the arms and one of the arms has to be two.

My take on xy-wings is that you need one cell that has two choices (xy), that can see two different cells (arms) that split the choices and share their other value. (xz and yz). The pivot means one of the arms is definitely z, so no cell that can see both arms can be z.

Both a6, c5, e6 and b6, c5, d6 are valid xy-wings.
TopRank

Posts: 11
Joined: 22 December 2005

sweetbix wrote:Would you please show me how the BUG principle works here.

Hi sweetbix, No knowing where to start, I have revised the post for the BUG principle with examples added for easier understanding. Could you gain some background understanding first and let us know you specific concerns, if there is still any.
Jeff

Posts: 708
Joined: 01 August 2005

TopRank, you’re right about the XYwings. Sorry, I misinterpreted which cells you were referring to.

Jeff, I re-read your example and can see how 8 works at r8c5. The hard part (for me) is seeing the implications without having them pointed out!
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`*-----------------------------------------------------------*  | 4     18    7     | 58    6     15    | 3     2     9     |  | 6     28    9     | 78+2  37+28 23    | 4     5     1     |  | 5     12    3     | 4     12    9     | 6     8     7     |  |-------------------+-------------------+-------------------|  | 7     9     5     | 6     13    13    | 2     4     8     |  | 8     4     6     | 57+2  27    25    | 9     1     3     |  | 1     3     2     | 9     4     8     | 5     7     6     |  |-------------------+-------------------+-------------------|  | 9     6     4     | 1     5     7     | 8     3     2     |  | 3     5     1     | 2+8   8+2   6     | 7     9     4     |  | 2     7     8     | 3     9     4     | 1     6     5     |  *-----------------------------------------------------------*`

Am I right that when you have many poly valued cells you have to set up the BUG grid and then make implications ( e.g. by chains) from all the non-BUG candidates? There isn't one simple application as tso described for a single poly value cell.

If it’s simple to explain can you tell me how in this example from the BUG thread you know that the 9 is a non-BUG in r1c4 and a BUG in r2c4. Is it related to the number of poly valued cells there are with that candidate?

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`+-----------------+------------------+----------------+ | 69   35   49    | 56+49 1     2    | 34   7    8    | | 26+9 38   28+49 | 69+4  7     49   | 34   5    1    | | 7    45   1     | 3     8     45   | 2    9    6    | +-----------------+------------------+----------------+ | 8    6    59    | 7     23    35   | 1    4    29   | | 4    2    7     | 8     9     1    | 6    3    5    | | 59   1    3     | 45+2  24    6    | 7    8    29   | +-----------------+------------------+----------------+ | 3    9    6     | 1     5     7    | 8    2    4    | | 25   7    45+2  | 24    6     8    | 9    1    3    | | 1    48   28+4  | 29+4  34+2  39+4 | 5    6    7    | +-----------------+------------------+----------------+`

Thanks.
sweetbix

Posts: 58
Joined: 10 December 2005

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` *-----------------------------------------------------------*  | 4     18    7     | 58    6     15    | 3     2     9     |  | 6     28    9     | 78+2  37+28 23    | 4     5     1     |  | 5     12    3     | 4     12    9     | 6     8     7     |  |-------------------+-------------------+-------------------|  | 7     9     5     | 6     13    13    | 2     4     8     |  | 8     4     6     | 57+2  27    25    | 9     1     3     |  | 1     3     2     | 9     4     8     | 5     7     6     |  |-------------------+-------------------+-------------------|  | 9     6     4     | 1     5     7     | 8     3     2     |  | 3     5     1     | 2+8   8+2   6     | 7     9     4     |  | 2     7     8     | 3     9     4     | 1     6     5     |  *-----------------------------------------------------------*`

sweetbix wrote:Am I right that when you have many poly valued cells you have to set up the BUG grid and then make implications ( e.g. by chains) from all the non-BUG candidates? There isn't one simple application as tso described for a single poly value cell.

Making implications from all non-BUG candidates is just one way to go. In this particular case, it's much easier to see that the grid reduces to a BUG+1 when a 2 is placed in r8c4 or an 8 is placed in r8c5; therefore these placements are valid. BTW, a grid with one single poly-valued cell is called a BUG+1.

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`+-----------------+------------------+----------------+ | 69   35   49    | 56+49 1     2    | 34   7    8    | | 26+9 38   28+49 | 69+4  7     49   | 34   5    1    | | 7    45   1     | 3     8     45   | 2    9    6    | +-----------------+------------------+----------------+ | 8    6    59    | 7     23    35   | 1    4    29   | | 4    2    7     | 8     9     1    | 6    3    5    | | 59   1    3     | 45+2  24    6    | 7    8    29   | +-----------------+------------------+----------------+ | 3    9    6     | 1     5     7    | 8    2    4    | | 25   7    45+2  | 24    6     8    | 9    1    3    | | 1    48   28+4  | 29+4  34+2  39+4 | 5    6    7    | +-----------------+------------------+----------------+`

sweetbix wrote:If it’s simple to explain can you tell me how in this example from the BUG thread you know that the 9 is a non-BUG in r1c4 and a BUG in r2c4. Is it related to the number of poly valued cells there are with that candidate?

To set a BUG grid, you should always start from a unit with a single poly-valued cell. For this particular grid, the starting cell is r9c6 and the sequence after that is:

R9C6=<39>, R9C5=<34>, R9C4=<29>, R9C3=<28>, R8C3=<45>, R6C4=<45>, R2C3=<28>, R2C1=<26>, R1C4=<56>, R2C4=<69>.

As it turned out, the 9 is a non-BUG candidate in r1c4 and a BUG candidate in r2c4.
Jeff

Posts: 708
Joined: 01 August 2005

Thanks for this reply Jeff. I'm going to study the sequence (and maybe get back to you! )
sweetbix

Posts: 58
Joined: 10 December 2005

sweetbix - did you see my comment to you? It was actually a request for clarification.

Mac
QBasicMac

Posts: 441
Joined: 13 July 2005

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