The New Sudoku Players' Forum

[speter]

G'Day Folks,

I am studying puzzles with Uniqueness Test 3, and am solving 12/14 of my puzzles.

In the following puzzle, I am finding 2 candidate pairs r4c1 and r5c1, and am finding points 3/4 at r4c5 & r5c5.

Code: Select all

1345  2     357  | 146   156   356  | 567   8     9
9     457   357  | 146   8     356  | 567   146   2
1458  458   6    | 9     2     7    | 3     14    15
-----------------+------------------+-----------------
67    9     4    | 5     167   2    | 8     136   1367
67    3     2    | 168   1679  689  | 4     5     167
58    58    1    | 3     67    4    | 9     2     67
-----------------+------------------+-----------------
345   456   9    | 7     456   1    | 2     36    8
2     567   578  | 68    3     568  | 1     9     4
34    1     38   | 2     469   689  | 56    7     356

I then create a virtual column 5 (combining r4c5 & r5c5):

Code: Select all

156  8  2  19  -  67  456  3  469

in which I find a hidden quad (c=1,4,5,9) and discard r1c5,r7c5,r9c5<>6.

Unfortunately, this leads to no solution (no candidates in r1c7 & c6c9), so clearly my logic is faulty!?

BTW, Hodoku (at the same point) finds a different UT3 (5/8 in r3c12,r612 => r3c9<>1).

Any help would be appreciated.

cheers
S.

[jco]

Hi, speter

Code: Select all

,------------------------------------------------------------,
| 1345   2     357   | 146   156   356   | 567   8     9     |
| 9      457   357   | 146   8     356   | 567   146   2     |
|[58]14 [58]4  6     | 9     2     7     | 3    (14)   5-1   |
|--------------------+-------------------+-------------------|
| 67     9     4     | 5     167   2     | 8     136   1367  |
| 67     3     2     | 168   1679  689   | 4     5     167   |
|[58]   [58]   1     | 3     67    4     | 9     2     67    |
|--------------------+-------------------+-------------------|
| 345    456   9     | 7     456   1     | 2     36    8     |
| 2      567   578   | 68    3     568   | 1     9     4     |
| 34     1     38    | 2     469   689   | 56    7     356   |
'------------------------------------------------------------'

Regarding HoDoKu's move, the UR(58)r36c12 is prevented by the internal or external guardians.

For Row 3, there is only one external guardian (5)r3c9. If this was false, then (58) would be locked
in Row 3 into the cells r3c12, making the UR unavoidable, so to prevent this we must have +5 r3c9.

Looking at it from the perspective of using the internal guardians, we must have (1)r3c1 OR (4)r3c12,
in the first case of the OR, -1 r3c9; in the second case, (4)r3c12 eliminates (4)r3c8, so (1)r3c8 would
be true, eliminating again (1)r3c9. Written as a chain:

UR(58)r36c12 using internals: (1)r3c1 == (4)r3c12 - (4=1)r3c8 => -1 r3c9

If you try to apply this logic into the other UR mentioned, you will see that it doesn't work.

For this puzzle, I like more the DP(67)r45c1, r46c5, r56c9 (cells marked with *)

Code: Select all

,-----------------------------------------------------------,
| 1345  2     357   | 146   156   356   | 567   8     9     |
| 9     457   357   | 146   8     356   | 567   146   2     |
| 1458  458   6     | 9     2     7     | 3     14    15    |
|-------------------+-------------------+-------------------|
|*67    9     4     | 5    *67(1) 2     | 8     36-1  367-1 |
|*67    3     2     | 68-1  679-1 689   | 4     5    *67(1) |
| 58    58    1     | 3    *67    4     | 9     2    *67    |
|-------------------+-------------------+-------------------|
| 345   456   9     | 7     456   1     | 2     36    8     |
| 2     567   578   | 68    3     568   | 1     9     4     |
| 34    1     38    | 2     469   689   | 56    7     356   |
'-----------------------------------------------------------'

with internal guardians (1)r4c5, (1)r5c9.
One of them must be true to prevent that DP (Extended UR), so -1 r4c89 and -1 r5c45.

I hope this helps.

[Maxito_Bahiense]

Hi, Speter!

In the following puzzle, I am finding 2 candidate pairs r4c1 and r5c1, and am finding points 3/4 at r4c5 & r5c5.

Code: Select all

1345  2     357  | 146   156   356  | 567   8     9
9     457   357  | 146   8     356  | 567   146   2
1458  458   6    | 9     2     7    | 3     14    15
-----------------+------------------+-----------------
67    9     4    | 5     167   2    | 8     136   1367
67    3     2    | 168   1679  689  | 4     5     167
58    58    1    | 3     67    4    | 9     2     67
-----------------+------------------+-----------------
345   456   9    | 7     456   1    | 2     36    8
2     567   578  | 68    3     568  | 1     9     4
34    1     38   | 2     469   689  | 56    7     356

I then create a virtual column 5 (combining r4c5 & r5c5):

Code: Select all

156  8  2  19  -  67  456  3  469

in which I find a hidden quad (c=1,4,5,9) and discard r1c5,r7c5,r9c5<>6.

So, you are referring to 15r1c5, 19r45c5, 45r7c5 49r9c5 as the hidden quad with virtual group r45c5 holding internal guardians 1,9.

If you check Hodoku explanation of UR Type 3, you'll see that it uses naked sets with virtual cells. When using hidden sets, you must have c digits distributed over c sets, such that digits are not found elsewhere, and each set holds exactly one digit. Here, when you build the hidden quad, the virtual cell built from guardians fails to satisfy that last condition, since guardians are only strongly linked, and the virtual cell can hold both 1 and 9 in its two cells. So, if that is the case (it is precisely so here), you have no hidden quad, as the three remaining cells r1c5, r7c5 and r9c5 must hold only 4 and 5, leaving the possibility of holding one extra candidate [here 6 r9c5].
Edit: typos.

[speter]

My thanks to you both, for identifying my faulty logic.

cheers
S.