Can a standard Sudoku puzzle contain a miniature Sudoku puzzle?

Question

The following is an example of "Miniature 4x4 Sudoku puzzle" in which the goal is to fill each empty square with a number (1, 2, 3 or 4) so that there are no duplicated numbers in any row, column or 2x2 block.

The solution (as a 4x4 matrix) to the above puzzle is:

[ 2 1 3 4 ]
[ 4 3 1 2 ]
[ 3 4 2 1 ]
[ 1 2 4 3 ]

Does there exist a standard 9x9 Sudoku puzzle (with standard rules) such that its solution has the property that some 4x4 submatrix of the solution (taken from four consecutive rows and four consecutive columns) is equal to the 4x4 solution matrix of some "Miniature 4x4 Sudoku puzzle"?

Answer 1

The answer is

no.

Let's try it out:

Obviously we'll need to align the boxes of the mini-sudoku with the maxi-sudoku - if we didn't, we'd end up with at least six numbers from the 4x4 in the same box, and with only four distinct digits, that won't work. WLOG, let's fill in the solution to the 4x4 that you've provided in the top-left boxes:



Now, we need to fit the numbers 5 through 9 in the remainder of those boxes. Since there are none of them in the grid, it doesn't matter where we put which number, so WLOG, we'll start in the top left box and put 5, 6, and 7 along the top and 5, 8, and 9 along the left side:



And we've immediately run into a problem. Look at the top center box: based on what's already filled in, the 5, 6, and 7 cannot be in the top row of that box, so are constrained to the two cells in the bottom right. (The same is of course true of the center left box and the 5, 8, and 9.) Therefore, we cannot build a 9x9 sudoku with a 4x4 sub-sudoku.

Answer 2

To propose an alternate perspective on things using the pigeonhole principle:

• There are two ways to align the mini-sudoku with the big sudoku. I say this because all the alternate alignments directly overlap a 3x3 corner of the mini-sudoku with a nonet. First, let's tackle this way:

In this case our board is aligned with four nonets, like this. The remaining numbers in each nonet (or, the cells right next to the mini-sudoku) have to be 5, 6, 7, 8 or 9 - that is 5 numbers. But the red row shown above has six cells. Using the pigeonhole principle, there has to be a repeating number, which contradicts the sudoku.

• Similarly let us tackle the other problem:

Once again, in the red 2x3 grid, we have only $4$ numbers to put in the grid, and 6 grid cells. Therefore there have to be repeats, which contradicts the sudoku again.