Since ½C1 C2 C3 form the first sub-grid, which contains all nine numbers,
it is clear that each of the first three columns, as well as each leftmost subgrid,
will also contain exactly those nine numbers. Similarly, the fourth to
sixth columns will be completed, and so will be the seventh–ninth columns.
Note that with these choices, the fourth row and seventh row are just
permutations of the first row. Similarly, fifth row and eighth row are
permutations of the second row, and sixth and ninth rows are permutations
of the third row. Therefore, each row will contain exactly those nine
numbers. Hence, we know that the master grid generated here is really a
solution that satisfies the requirements (1)–(3). Also note that at this step, we
have (2)3¼8 choices.