Nothing interesting to add, so just a couple of alternate expressions for fun.
Except for the last one, all yield the same conclusion: C = "-3 r2c156,b3p1389; -4 r2c13; -136 r1c5; btte"
1) MSLS (7 cells, 6 digits, irregularly shaped):
7X7 (Rank 0): {1N4 2N5678 3N46 \ 48r2 1356b2 3r2|3b3} => C
That's the same as Leren's solution (which, btw, is missing two valid cannibal eliminations).
- Code: Select all
3r2|
8r2 3b3 4r2 1b2 3b2 5b2 6b2
-------------------------------------------------
8r2c8 3r2c8 4r2c8 | 2N8
8r2c7 3r2c7 4r2c7 | 2N7
8r2c5 4r2c5 1b2p5 3b2p5 6b2p5 | 2N5
8r2c6 1b2p6 3b2p6 5b2p6 6b2p6 | 2N6
1b2p1 3b2p1 6b2p1 | 1N4
3b2p7 5b2p7 6b2p7 | 3N4
3b2p9 5b2p9 6b2p9 | 3N6
-------------------------------------------------
-3r2c1 -4r2c1 -1b2p2 -3b2p2 -6b2p2
-3r2c5 -4r2c3
-3r2c6
-3b3p1
-3b3p3
-3b3p8
-3b3p9
As an AIC:
(8=3'4)r2c78 - (4=1356'8)b2p51679 - loop => C
2) Mixed type Rank 0 (5 cells, 3 digits):
4x4 (Rank 0): {2N78 48B2 \ 48r2 1n5 3r2|3b3} => C
That's practically the same as Cenoman's, though with a bit different sets.
- Code: Select all
3r2|
8r2 3b3 4r2 1n5
----------------------------
8r2c8 3r2c8 4r2c8 | 2N8
8r2c7 3r2c7 4r2c7 | 2N7
4r2c5 4r1c5 | 4B2
8r2c56 8r1c5 | 8B2
----------------------------
-3r2c1 -4r2c1 -1r1c5
-3r2c5 -4r2c3 -3r1c5
-3r2c6 -6r1c5
-3b3p1
-3b3p3
-3b3p8
-3b3p9
As an AIC:
(8=3'4)r2c78 - r2c5 = (4-8)r1c5 = (8)r2c56 - loop => C
3) Lastly a joker:
[(4-8)r1c5] -> (4|8)r2c56 - (48=3)r2c78 => +3 r2c78; btte