# What is the maximum number of empty 3x3 blocks a proper Sudoku can have?

It is possible to have a proper Sudoku with one empty block (consisting of 3x3 cells). The trivial (and very boring) example would be a puzzle where all other cells are given.

It's impossible to have a proper Sudoku with an empty chute (a band of 3 blocks), since units within the chute that are aligned with it, could be swapped around while the puzzle remained valid. So we need at least three non-empty blocks.

But what is the maximum number of empty blocks a proper Sudoku can have? I don't think I've seen Sudoku's with more than one block empty; is this even possible?

A standard Sudoku consists of 81 cells in 9 3x3 blocks. Those 9 blocks live in 6 chutes; 3 bands (horizontal) and 3 stacks (vertical).
A block is a unit; other units are rows and columns.
For chutes and other terminology, see this glossary.

• Just fyi: the problem cases are 4 or 5 empty blocks only. 6 Empty blocks becomes trivial to prove as having multiple solutions, as you'll have a fully empty chute, in which the orthogonals are interchangeable. – Weckar E. Mar 15 '17 at 6:44
• @WeckarE. not necessarily so. You can arrange three non-empty blocks so that there's one of them in each band and each stack. – SQB Mar 16 '17 at 6:54

Four is provably the least number of empty boxes. I will proceed to demonstrate this here.

Start by defining a 3x3 block on the Sudoku grid. This has no restrictions, as all the numbers are currently interchangeable without issue. From here out, I will refer to the purple boxes as "odd" boxes, and the green ones as "even." The images were generated using a Sudoku solver, so I trust the listings of possibilities. We want to create a sequence of squares which will fully define other squares. We know a couple important properties:

• That if three consecutive boxes are empty, then there is not a singular solution
• That if all the other boxes in a given row and column are not defined, then the target box is not defined.
• That it is trivially irrelevant what the permutation of this puzzle is, but rather that one exists.
• That the removal of clues does not help to solve the puzzle, so we can assume the four known boxes are fully solved.

I also assume the following vernacular:

• A rigorously defined box is one for which all values are known.
• A rigorously defined box is one for which all other boxes in the same row and column are rigorously defined.

We want to demonstrate that it is impossible to fully define the remaining five boxes with four others. To do this, we need to show another couple properties:

• Using only the numbers in two adjacent boxes, it is impossible to fully define any numbers in a given box.
• Using three boxes, it is impossible to define any more than three numbers in a given box, and that those three numbers must lie on a diagonal (or trivial permutation of a diagonal).

Because the order of the columns is trivially irrelevant, as is the second selected box, we fill in another box as such: We can see that no matter which three boxes we select, the numbers must fall along a diagonal. As a consequence of two boxes being collinear, the possibilities of the target box must be inherently limited to the diagonal. The possibilities in each column are such that, horizontally, they are unique, and vertically, they are distinct. This proves the first lemma.

Thus, the maximum solution to this situation is to place two numbers in a row to cancel two possibilities. The sudoku puzzle we currently have is: None of the remaining numbers in the third box are relevant to those in the second. Therefore, as this situation has trivially covered all permutations, the second lemma is (admittedly, intuitively) proven.

Now we can actually move on to demonstrating why five empty boxes is impossible.

To execute this proof, I will first prove that all four boxes must fall on odd squares. Once that is done, I will prove that under no configuration of boxes in odd squares can a functional Sudoku be achieved.

• If the four filled boxes are placed on the green squares above (the edge squares), then the center box is rigorously defined, and none of the corner boxes are.
• If three edge boxes are filled, and any corner is filled, then no boxes are rigorously defined.

As a result, we can conclude easily that at least two boxes must not be on the edge squares.

• If two boxes are placed on the corners, then there exists no combination of boxes that can define any other box. (The proof of this is intuitive, but has a lot of cases, so is left to the reader.)
• If one box is placed on the center, and one is placed in the corner, then it is at most possible to place the even boxes such that the diagonals of two other even boxes are known. However, at this point, it is not possible to continue, as no other boxes will be defined.

This is a point useful to demonstrate with an image: Note that even though we now have the diagonals on two boxes, we only have $1\frac{1}{3}$ of a definition for the surrounding boxes. Thus, we cannot continue.

As a result, we now know that at least three boxes must be on the odd squares. Now, let us consider the possibilities where there are three boxes on odd squares.

• If there are three boxes on odd squares, then if the three boxes are on corners, we have three cases:
• If the remaining green box falls on the edge between two defined corners, then the box partially defines the green box opposite it, but the solutions end here - no other boxes have the 3 required to be solvable.
• If the green box is not between two solved corners, then it partially defines the remaining unsolved corner box, as well as the edge opposite it. However, no other values can be derived.
• If there are two boxes on adjacent corners, and one box in the center:
• If the remaining box falls between the solved corners, it partially defines the opposite edge box, but that is all.
• If the remaining box is adjacent to a single solved corner, then it partially defines the corner next to it, as well as the edge across from it, but that is all.
• If the remaining box is placed on an edge opposite the solved corners, then the edge opposite it is rigorously defined, but no other boxes can be solved.
• If there are two boxes on opposite corners, and one in the center:
• Placing the green box anywhere on the grid can partially define two boxes: the opposite edge and the unsolved adjacent corner. Nothing else can be derived.

As a result of these three cases, it is clear that all four boxes must lie on the odd spaces - the corners and center.

This is trivial. If all four boxes lie on corners, then the puzzle cannot be solved, as no box can be defined. If three boxes lie on corners and one lies in the center, then two boxes are rigorously defined, but nothing* else can be derived.

*: In some cases, it may be possible to derive the diagonals of the opposite boxes by the Unique Rectangles or Hidden Unique Rectangles reduction methods, but it goes no farther than this (I swear!).

In summary, the four solved boxes must be placed on odd boxes (corners and center), but no combination exists there for the boxes to be solved. Thus, the puzzle requires more than four boxes to be solved.

To see why it is possible to have four boxes unsolved, this puzzle is trivially solvable: Proof: complete.

You can have four $3 \times 3$ blocks empty. I took the example from Wikipedia, blanked four blocks to get the below, and it solved easily. I would be surprised but not shocked at five. Wikipedia has a good discussion of this. However, given the sheer size of the number of possible puzzles, it's not yet known whether the below puzzles can do better.

The largest rectangular orthogonal "hole" (region with no clues) in a proper sudoku is believed to be a rectangle of 30 cells (a 5 x 6 rectangular area). One such example is the following with 22 clues: The largest total number of empty groups (rows, columns, and squares) in a sudoku is believed to be nine. One example is the following; a sudoku which has 3 empty squares, 3 empty rows, and 3 empty columns and gives 22 clues. • Would've accepted if it hadn't been 'believed to be' (which is not your fault). – SQB May 21 '14 at 13:37
• On better reading, the second part only claims there are probably no more than 9 empty units possible, of which in the example given 3 are empty blocks. That indeed doesn't mean the maximum number of empty blocks is 3. – SQB May 21 '14 at 13:45
• This answer does not contain a number for the maximum number of empty 3x3 blocks in the puzzle, which is what was asked in the question. – Aza May 21 '14 at 14:30