by **Sue De Coq** » Wed Oct 19, 2005 8:36 am

Chris,

The puzzle has a unique solution, though it's necessary to invoke multiple forced chains in order just to make the first move. (However, once the first move has been made - a 3 in r2c2 - the rest of the puzzle follows straightforwardly).

Here's my solver log for the first move. (It might well be possible to find a simpler solution):

The value 6 in Box [1,3] must lie in Row 3.

- The moves (1,8):=6 and (1,9):=6 have been eliminated.

The value 9 in Box [2,2] must lie in Row 6.

- The move (5,6):=9 has been eliminated.

Consider the chains (4,3)~4~(4,4)~2~(8,4) and (4,3)~4~(7,3)~2~(7,6).

Whichever of the 2 candidates in Box [3,2] contains the value 2, the cell (4,3) does not contain the value 4.

- The move (4,3):=4 has been eliminated.

Consider the chains (1,3)~9~(1,9)~3~(2,9) and (2,3)~9~(2,6)~3~(2,9).

Whichever of the 2 candidates in Column 3 contains the value 9, the cell (2,9) does not contain the value 3.

- The move (2,9):=3 has been eliminated.

Consider the chains (2,9)~9~(1,9)~3~(7,9) and (2,9)~9~(2,6)~3~(7,6).

Whichever of the 2 candidates in Row 7 contains the value 3, the cell (2,9) does not contain the value 9.

- The move (2,9):=9 has been eliminated.

Consider the chains (2,8)~9~(1,9)~3~(7,9) and (2,8)~9~(2,6)~3~(7,6).

Whichever of the 2 candidates in Row 7 contains the value 3, the cell (2,8) does not contain the value 9.

- The move (2,8):=9 has been eliminated.

Consider the chains (2,3)-9-(2,6)-3-(3,4) and (2,3)~2~(3,1)~3~(3,4).

When the cell (2,3) contains the value 2, one chain states that the cell (3,4) contains the value 3 while the other says it doesn't - a contradiction.

Therefore, the cell (2,3) cannot contain the value 2.

- The move (2,3):=2 has been eliminated.

Consider the chains (1,3)~9~(1,9)~3~(2,8) and (2,3)-9-(2,6)~3~(2,8).

Whichever of the 2 candidates in Column 3 contains the value 9, the cell (2,8) does not contain the value 3.

- The move (2,8):=3 has been eliminated.

Consider the chains (2,2)~2~(3,1)~3~(3,4)-3-(2,6) and (2,2)-5-(2,3)-9-(2,6).

When the cell (2,2) contains the value 2, one chain states that the cell (2,6) contains the value 3 while the other says it doesn't - a contradiction.

Therefore, the cell (2,2) cannot contain the value 2.

- The move (2,2):=2 has been eliminated.

The value 2 in Column 1 must lie in Box [1,1].

- The move (6,1):=2 has been eliminated.

The values 1, 2 and 8 occupy the cells (2,1), (2,8) and (2,9) in some order.

- The move (2,1):=3 has been eliminated.

Consider the chains (2,2)~5~(6,2)-5-(6,6) and (2,2)-3-(2,6)-9-(6,6).

When the cell (2,2) contains the value 5, one chain states that the cell (6,6) contains the value 5 while the other says it doesn't - a contradiction.

Therefore, the cell (2,2) cannot contain the value 5.

- The move (2,2):=5 has been eliminated.

The value 3 is the only candidate for the cell (2,2).

PS I haven't bothered to changed the cell labels from the format e.g. (1,2) [standard mathematical matrix notation] to the forum standard of r1c2.