The New Sudoku Players' Forum

[JC Van Hay]

However, to find such a solution, it looks easier to me to follow this chain and observe, that it kills all 4's in box 9.

Code: Select all

4r3c9 -> 2r3c6->8r8c6 -> 4r8c4 -> 4r9c3 -> 4r5c7
r7c9                     r8c8     r9c78    r7c7 <> 4  => contradiction, r3c9<>4


Writing the inferences as a Forbidding Matrix, one can see more :

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4r3c9
4r3c6=2r3c6
      2r8c6=8r8c6
            8r8c4=4r8c4 ---------------> [4r3c6==4r8c4]-4r2c4
                  4r8c2=4r9c3 ---------> [4r3c6==4r9c3]-4r9c6
                        4r5c3==4r5c7
4r7c9=============4r8c8=4r9c78=4r7c7 --> [4r3c6==4r7c9]-4r3c9

As the 3 last rows decribe an Almost Swordfish, one can write :

4r7c9=*Swordfish(4R5B79*)-(4=8)r8c4-(8=24)r83c6 :=> -4r3c9,r2c4,r9c6(?)

[DonM]

Solution using an AAIC (Kraken) cell:

1. (7)r3c3=r1c3-r1c7=(7-9)r9c7=r7c7-(9=5)r7c5-(5=2)r3c5 => r3c3<>2=7, r1c3=2
2. AAIC cell:
(4)r7c7
II
(8)r7c7-(4=8)r7c9-(4=5)r3c9-r5c5=r1c6-(5=3)r6c6-r5c6=(3)r5c7
II
(9)r7c7-(9=5)r7c5-r5c5=r1c6-(5=3)r6c6-r5c6=(3)r5c7
=> r5c7<>4 -> 7 singles
3. (5=9)r7c5-als(9=48)r7c79-(4=3)r9c8-r9c5=(3-2)r8c5=(2)r3c5) => r3c5<>5=2

stte

[Luke]

I don't think I've seen too many SE 7.3 puzzles that required nets, kraken, AAIC etc to solve! I am/was working on an AALS idea but it's just not coming together so far.

To be honest, I would probably end up chipping away at this with 7 or 8 or 27 moves, so kudos to those who made it in less. Surprising depth, indeed.