[quote="Jeff]
Since r5c8=2 will promote a bivalue universal death situation, r5c8 must be 4. And this completes the grid.[/quote]
I don't understand this -- could you explain?
tso wrote:In this case, r7c2 is still the only tri-value cell in its row, column and box BUT none of the other cells in its three houses share a house with another tri-value or greater cell. Would we be able to correctly apply the tactic to r7c2 in this case?
Nick70 wrote:.....we haven't proven that a grid where all cells are bivalues cannot have a single solution.
Case 1
. . . | . . . | . . .
. xy . | . . . | . xy .
. . . | . . . | . . .
-----------------+-------------------+-----------------
. . . | . . . | . . .
. xy . | . . . | . xy .
. . . | . . . | . . .
-----------------+-------------------+-----------------
. . . | . . . | . . .
. . . | . . . | . . .
. . . | . . . | . . .
This case can't happen because r2c2 is the only unsolved cell in box 1, and likewise for r5c2, r2c8 & r5c8.
Case 2
. . . | . . . | . . .
. xy . | . . . | . xy .
. . xy | . . . | . xy .
-----------------+-------------------+-----------------
. . . | . . . | . . .
. . . | . . . | . . .
. . . | . . . | . . .
-----------------+-------------------+-----------------
. . . | . . . | . . .
. . . | . . . | . . .
. . . | . . . | . . .
This case can't happen because r2c2 is the only unsolved cell in col 2, and likewise for r3c3.
Case 3
. . . | . . . | . . .
. xy . | . . . | . xy .
. xy . | . . . | . xy .
-----------------+-------------------+-----------------
. . . | . . . | . . .
. . . | . . . | . . .
. . . | . . . | . . .
-----------------+-------------------+-----------------
. . . | . . . | . . .
. . . | . . . | . . .
. . . | . . . | . . .
This case implies that there is no unique solution.
. 1 3 | . . 9 | . . .
8 . . | 1 . 5 | . . .
. . 9 | . 8 . | 5 . .
------+-------+------
1 . 5 | . . 8 | . 7 .
. 9 . | . . . | . 4 .
. 7 . | 9 . . | 1 . 6
------+-------+------
. . 6 | . 1 . | 3 . .
. . . | 2 . 7 | . . 8
. . . | 8 . . | 9 1 .
5 1 3 | . . 9 | . 8 .
8 6 4 | 1 . 5 | 7 9 .
7 2 9 | 4 8 . | 5 . 1
------+-------+------
1 . 5 | . . 8 | 2 7 9
6 9 2 | . . 1 | 8 4 .
. 7 8 | 9 . 2 | 1 . 6
------+-------+------
9 8 6 | 5 1 4 | 3 2 7
. . 1 | 2 9 7 | . . 8
2 . 7 | 8 . . | 9 1 .
5 1 3 | 67 267 9 | 46 8 24
8 6 4 | 1 23 5 | 7 9 23
7 2 9 | 4 8 36 | 5 36 1
---------------+----------------+-------------
1 34 5 | 36 46 8 | 2 7 9
6 9 2 | 37 57 1 | 8 4 35
34 7 8 | 9 45 2 | 1 35 6
---------------+----------------+-------------
9 8 6 | 5 1 4 | 3 2 7
34 345 1 | 2 9 7 | 46 56 8
2 45 7 | 8 36 36 | 9 1 45
tso wrote:All of these implications turn out to be true and reduce the puzzle to singles.
Is this just a coincidence? Am I just as likely to find a puzzles with two tri-value cells left in which the implications of this type will be false? Or is there a reason why it *does* work in this case -- the position and/or the relationship of the two tri-value cells. Maybe the fact that the two tri-value cells have no candidates in common?
Original grid with link-cells highlighted:
5 1 3 | 67 267 9 |<46> 8 24
8 6 4 | 1 23 5 | 7 9 23
7 2 9 | 4 8 36 | 5 36 1
---------------+----------------+-------------
1 <34> 5 | 36 <46> 8 | 2 7 9
6 9 2 | 37 57 1 | 8 4 35
34 7 8 | 9 45 2 | 1 35 6
---------------+----------------+-------------
9 8 6 | 5 1 4 | 3 2 7
34 345 1 | 2 9 7 |<46> 56 8
2 45 7 | 8 36 36 | 9 1 45
Grid reduced to a BUG:
5 1 3 | 67 [27] 9 |<46> 8 24
8 6 4 | 1 23 5 | 7 9 23
7 2 9 | 4 8 36 | 5 36 1
---------------+----------------+-------------
1 <34> 5 | 36 <46> 8 | 2 7 9
6 9 2 | 37 57 1 | 8 4 35
34 7 8 | 9 45 2 | 1 35 6
---------------+----------------+-------------
9 8 6 | 5 1 4 | 3 2 7
34 [35] 1 | 2 9 7 |<46> 56 8
2 45 7 | 8 36 36 | 9 1 45
In order to avoid the BUG, r1c5=6 and r8c2=4.
BUG reduction through reduction of a bivalue link-cell:
1 7 4 | 8 3 2 | 5 9 6
5 9 3 | 4 6 1 | 2 7 8
6 8 2 | 9 5 7 | 34 34 1
----------------+----------------+----------------
28 6 7 | 5 12 48 | 9 123 34
248 1 9 | 7 48 3 | 6 <24> 5
24 3 5 | 12 9 6 | 8 124 7
----------------+----------------+----------------
3 24 1 | 6 28 48 | 7 5 9
9 24 8 | 12 7 5 | 13 6 34
7 5 6 | 3 14 9 | 14 8 2
Since r5c8=2 will promote a BUG. In order to avoid the BUG, r5c8=4.
Jeff wrote:I believe the principle works in this 2-trivalue case because these link-cells remain unchanged when the trivalue cells are reduced to a BUG. In other words, this 2-trivalue case is acting as if it was 2 individual trivalue cases.
Nick70 wrote:If r4c2 = 4, then r8c2 reduces to 35, and r1c5 reduces (through r4c5) to 27, leaving us in a BUD situation. Therefore r4c2 must be 3.
Jeff wrote:Could you post the grid with such BUD situation. I suspect that, by putting r4c2=4, it won't be much of a BUD situation left.