Code: Select all
357 1 8 | 57 9 67 | 4 2 3567
257 9 6 | 4 3 12 | 157 17 8
2357 57 4 | 18 567 128 | 13567 167 9
------------+--------------+-----------------
9 568 2 | 3 157 4 | 15678 1678 567
1 56 3 | 578 2 78 | 567 9 4
45 458 7 | 6 15 9 | 1258 3 25
------------+--------------+-----------------
4678 3 19 | 2 467 167 | 6789 5 67
4678 2 5 | 79 467 3 | 6789 4678 1
467 47 19 | 19 8 5 | 23 467 23
Some 3D coloring of the 3’s cluster, plus the locked 7’s and 4’s in c2, leads to AIC1:
(3)r1c1 = (3)r1c9 – (3=2)r9c9 – (2=5)r6c9 – (5=4)r6c1 – (4)r6c2 = (4-7)r9c2 = (7)r3c2 => r1c1 <> 7.
AIC1 yields a second elimination by replacing the first node with (3)r3c7 to form AIC2:
(3)r3c7 = (3)r1c9 – (3=2)r9c9 – (2=5)r6c9 – (5=4)r6c1 – (4)r6c2 = (4-7)r9c2 = (7)r3c2 => r3c7 <> 7.
And the new (35)r1c1 bi-value then allows for AIC3:
(5)r3c5 = (5)r1c4 – (5=3)r1c1 – (3)r1c9 = (3)r3c7 => r3c7 <> 5.
QUESTION 1: Is it possible to combine the first two AIC's using notation something like:
(3)r1c1|r3c7 = (3)r1c9 – (3=2)r9c9 – (2=5)r6c9 – (5=4)r6c1 – (4)r6c2 = (4-7)r9c2 = (7)r3c2 => r1c1|r3c7 <> 7?
QUESTION 2: Would the notation, (3)r1c9|r3c1, be OK to designate alternate paths available at the fourth node in AIC3?
QUESTION 3: Any suggestions for alternative techniques to get these same three eliminations (e.g., ALS’s, Wings, etc.)?
The resulting grid now appears as:
Code: Select all
35 1 8 | 57 9 67 | 4 2 3567
257 9 6 | 4 3 12 | 157 17 8
2357 57 4 | 18 567 128 | 136 167 9
------------+--------------+-----------------
9 568 2 | 3 157 4 | 15678 1678 567
1 56 3 | 578 2 78 | 567 9 4
45 458 7 | 6 15 9 | 1258 3 25
------------+--------------+-----------------
4678 3 19 | 2 467 167 | 6789 5 67
4678 2 5 | 79 467 3 | 6789 4678 1
467 47 19 | 19 8 5 | 23 467 23
And I hit a brick wall at this point.
QUESTION 4: Can anyone provide a hint or two to manually advance from here?