A game concerning Sudoku

Question

We (me and you) start with a blank $9\times 9$ blank square (as an empty Sudoku) and I fill the first three rows legally according to Sudoku rules. Is it always possible that you complete the Sudoku with this given three rows?

Answer 1

Suppose you filled the first three rows according to Sudoku rules, let a1, a2, a3 be the 3x1 vectors in the top-left block; b1, b2, b3 be the next three 3x1 vectors in the top-middle block; and c1, c2, c3 be the last three 3x1 vectors in the top-right block. Then consider:

\begin{array}{|ccc|ccc|ccc|} \hline | & | & | & | & | & | & | & | & |\\ {\vec{a}_{1}} & {\vec{a}_{2}} & {\vec{a}_{3}} & {\vec{b}_{1}} & {\vec{b}_{2}} & {\vec{b}_{3}} & {\vec{c}_{1}} & {\vec{c}_{2}} & {\vec{c}_{3}}\\ | & | & | & | & | & | & | & | & |\\ \hline & & & & & & & & \\ & ? & & & ? & & & ? & \\ & & & & & & & & \\ \hline & & & & & & & & \\ & ? & & & ? & & & ? & \\ & & & & & & & & \\ \hline \end{array} and fill it with\begin{array}{|ccc|ccc|ccc|} \hline | & | & | & | & | & | & | & | & |\\ {\vec{a}_{1}} & {\vec{a}_{2}} & {\vec{a}_{3}} & {\vec{b}_{1}} & {\vec{b}_{2}} & {\vec{b}_{3}} & {\vec{c}_{1}} & {\vec{c}_{2}} & {\vec{c}_{3}}\\ | & | & | & | & | & | & | & | & |\\ \hline | & | & | & | & | & | & | & | & |\\ {\vec{a}_{2}} & {\vec{a}_{3}} & {\vec{a}_{1}} & {\vec{b}_{2}} & {\vec{b}_{3}} & {\vec{b}_{1}} & {\vec{c}_{2}} & {\vec{c}_{3}} & {\vec{c}_{1}}\\ | & | & | & | & | & | & | & | & |\\ \hline | & | & | & | & | & | & | & | & |\\ {\vec{a}_{3}} & {\vec{a}_{1}} & {\vec{a}_{2}} & {\vec{b}_{3}} & {\vec{b}_{1}} & {\vec{b}_{2}} & {\vec{c}_{3}} & {\vec{c}_{1}} & {\vec{c}_{2}}\\ | & | & | & | & | & | & | & | & | \\\hline \end{array}

Note the rest of the blocks are just permutations of the columns of the blocks above. It is done in a way so that in each block, we have all 1-9 numbers, since the first three rows are Sudoku-satisfied. And each column will also have 1-9, because we used the top three blocks to make the columns. Lastly, the rows are all satisfied with 1-9, because they are just the same rows as the first three rows except permuted again.

By way of an example, consider this:

\begin{array}{|ccc|ccc|ccc|} \hline 5 & 3 & 4 & 6 & 7 & 8 & 9 & 1 & 2\\ 6 & 7 & 2 & 1 & 9 & 5 & 3 & 4 & 8\\ 1 & 9 & 8 & 3 & 4 & 2 & 5 & 6 & 7\\ \hline & & & & & & & & \\ & ? & & & ? & & & ? & \\ & & & & & & & & \\ \hline & & & & & & & & \\ & ? & & & ? & & & ? & \\ & & & & & & & & \\ \hline \end{array}

We will fill the remaining blocks by permuting the columns of the 3x3 blocks directly above it cyclically:

\begin{array}{|ccc|ccc|ccc|} \hline 5 & 3 & 4 & 6 & 7 & 8 & 9 & 1 & 2\\ 6 & 7 & 2 & 1 & 9 & 5 & 3 & 4 & 8\\ 1 & 9 & 8 & 3 & 4 & 2 & 5 & 6 & 7\\ \hline 3 & 4 & 5 & 7 & 8 & 6 & 1 & 2 & 9\\ 7 & 2 & 6 & 9 & 5 & 1 & 4 & 8 & 3\\ 9 & 8 & 1 & 4 & 2 & 3 & 6 & 7 & 5\\ \hline 4 & 5 & 3 & 8 & 6 & 7 & 2 & 9 & 1\\ 2 & 6 & 7 & 5 & 1 & 9 & 8 & 3 & 4\\ 8 & 1 & 9 & 2 & 3 & 4 & 7 & 5 & 6 \\\hline \end{array}