#440101
Postby modellingman » September 6th, 2021, 4:42 pm
I agree with Gengulphus, trial and error is not necessary.
Without looking at any spoilers so far...
Columns 2 and 6 provide a starting point. In column 2, the only possible location of the 5 is in row 2. In column 6, the pair of possibilities {3,4} must occupy rows 3 and 5, so these possibilities can be eliminated from rows 1 and 6, leaving solutions of 5 in row 1 and 6 in row 6.
The two remaining 5's can now be added to the mid-left and bottom-right blocks in cells (4,3) and (6,4) using a (row, column) convention.
In row 6, the 4 cannot be in column 5, so must be in either column 1 or column 3. However, in the bottom-left block, cell (5,2) has only possibilities of {1,4} so the 4 can be eliminated leaving a solution of 1 in cell (5,2), with consequential solutions of 1 in cell (6, 5) and 4 in cell (3,2).
The missing 3 and 4 can now be added to column 6 in cells (3,6) and (5,6) respectively, whilst the missing 1 and 2 can be added to the mid-left block in cells (3,3) and (4,1), respectively, allowing row 3 to be completed with a 2 in cell (3,4).
Completion of rows 4 and 5 follow easily with a 4 in cell(4,4), a 6 in cell (4,5), a 3 in (5,4) and a 2 in (5,5), allowing column 5 to be completed with a 5 in cell (2,5) and then column 4 to be completed with 1 in cell (1,4) and a 6 in (2,4).
The 1 in the top-left block can only be placed in cell (2,1), leaving row 2 to be completed with a 4 in cell (4,3) and the block is completed with a 3 in cell(1,1) and a 2 in (1,3). This leaves two remaining unsolved cells of (6,1) and (6,3) which are trivially solved as 4 and 3, respectively.