The New Sudoku Players' Forum

[thomas129]

Hi,

I used this a while ago and wanted to know what tecnique it would be. . .

Its the sudocue daily nightmare for 22/06/06 (!)

Code: Select all

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

After some basic stuff, the puzzle gets to this point, i'm exclusivly focusing on the middle section of the puzzle so thats filled in in full :-

Code: Select all


.------------.------------.------------.
|    .    6      . |    .      .     . |  .    .     8 |
|    2    .      . |    8      .     3 |  .    7     6 |
|    .    1      . |    .      9     . |  5    .     . |
:------------+------------+------------:
|  157    2  14567 | 3567    357     9 |  8   14   145 |
|  579  579    567 |    4      8     1 | 67   23    23 |
| 1578 4578      3 | 2567    257  2567 | 67    9   145 |
:------------+------------+------------:
|    .    .      9 |    .      6     . |  .    5     . |
|    6    3      . |    9      .     4 |  .    8     7 |
|    4    .      . |    .      .     . |  .    6     9 |
'------------'------------'------------'



We can eliminate both [r6c12] <> 7 and [r6c9] <> 5 The reasoning go's something like this:-

T/M/B= top/middle/bottom of the middle set of 3 boxes.

6 and 7 together go in the middle and bottom of box 6. - the 67 locked set at [r56c7]

In the 5th box 6 and 7 must go one on the top and one on the bottom, if they both went in the top row, 3 could go nowhere in box 5 and if 6 and 7 both went on the bottom it would force an empty square at [r6c7] Therefore Reading across boxes 4,5 and 6:

Given that 6 cannot go on the bottom row of box 4 or the top row of box 6, 6 can be placed:-

TBM or MTB

and

7 can go TBM or MTB or BTM, (the only restriction being not in the midlle row of box 5 or the top row of box 6)

Combining the positions of 6 and 7:

If 6 goes TBM then 7 goes MTB
If 6 goes MTB then 7 goes TBM

Therefore 7 cannot go BTM as its "controlled" by 6 and . . .

Since one of 6 or 7 and 3 goes on the top in box 5, then the digit 5 go's on the bottom of box 5, thus the eliminations,

I was wondering if anyone had anymore similar examples as it would good to easily determine the exact conditions and number of restrictions that would need to be applied for the technique to work, is the logic of this anything like braids analysis?

Thanks, T.

[David P Bird]

Thomas129, Yes you are applying some braid analysis logic, but first you're missing a more fundamental preliminary check. I know it as a line/box elimination but it has other aliases which I forget. In row5 digit 5 must be in box4, it can't therefore be in row4 or row6 in that box - it's obvious when you think about it that the set of true candidates in r5c456789 must equal the set in r4c123 + r6c123.

The band now reduces to:

Code: Select all

*------------------*-------------------*---------------*
|   17    2   1467 | 3567    357     9 |  8   14   145 |
|  579  579    567 |    4      8     1 | 67   23    23 |
|  178  478      3 | 2567    257  2567 | 67    9   145 |
*------------------*-------------------*---------------*

Now to the braid analysis bit; in a band of boxes digits will repeat (or travel) in a diagonal direction as going from box 3 to 4 to 5. In your grid digit 2 repeats in the next row up (they're considered to wrap round) and 3 repeats in the next row down. We therefore know that digits are repeating in opposite directions. However the digits in every mini-row of 3 cells in the band must adhere to the same pattern of 1 down and 2 up or 2 down and 1 up as if they didn't there would be four digits competing for three cells somewhere.

In row5 box4 digits 6,7, & 9 can only repeat upwards, so we know the pattern is 1 down 2 up, and digit 5 must be the one that repeats down allowing it to be placed in r4c9.

We can only do this because digits 2 and 3 prove that digits are repeating in opposite directions (braid patterns) but without that knowledge we couldn't rule out all three digits in each set repeating in the same direction (rope patterns).

[RW]

There's a naked quad/hidden triplet that takes care of all the same eliminations. In situations when you can make all these eliminations, there is always a naked quad that does the same eliminations, so I don't think there's any need to formulate this as a new technique.

In a more generalized case with more unsolved cells we could have:

Code: Select all

| abcY    abcY     Y |
|    Y       Y     Y |
|  abY     abY   abY | ab

( Y <> abc )

In which case we can eliminate 'a' and 'b' from the remaining cells in the bottom row and Y from both cells 'abcY'.

RW