A game concerning Sudoku 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? 

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} 
