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This topic concerns sudoku puzzles with arrangements of six or nine 3×3 boxes, such as those posted as You can find all of the puzzle configurations in the sudoku archive and some other variations, too.

Sudoku Variants

These puzzles are composed of 3×3 boxes, just like the classic sudoku. But it is not necessary to arrange the 3×3 boxes in a 3×3 grid. In fact, other arrangements make truly excellent puzzles, and each arrangement has its own flavor.

Consider the puzzle below, composed of six 3×3 boxes:

 4    
 1    
   8  
     2
      
 3    
   3  
   5  
 4   7
   3 7
      
   5 9
      
 7    
     6
     8
   9  
     4

The puzzle has 3×3 boxes with nine cells each. There are twelve strips with nine cells each, and each strip passes through three 3×3 boxes. These strips are just like the rows and columns of the classic sudoku, but a whole new world is opened up when rows can intersect rows and columns can intersect columns.

It's interesting to consider what sudoku variants can be made by joining 3×3 boxes. If you keep the strips (i.e., rows and columns) in groups of three, where each group of three strips passes through the same three 3×3 boxes, then there are a limited number of puzzle configurations that you can make from a fixed number of 3×3 boxes. Classic sodoku with nine 3×3 boxes is one, and the wedge configuration above with six 3×3 boxes is another. What is the smallest puzzle that you can make in this way? How many sudoku configurations can you make with no more than nine 3×3 boxes? (To be clear, I'm talking about puzzles where each cell is in exactly three groups: a 3×3 box, a strip passing through the box vertically, and another strip passing through the box horizontally. Each strip passes through three 3×3 boxes, and there are two other parallel strips that pass through the same three 3×3 boxes.)

Solution

Let's start by considering all puzzles composed of 3×3 boxes connected by groups of three row/columns. Each 3×3 box is intersected by two sets of row/columns, and each set of row/columns intersects three 3×3 boxes. If N is the number of 3×3 boxes, and M is the number of sets of row/columns, then 3M = 2N. Therefore, N must be divisble by three.

We can eliminate configurations where a set of row/columns passes through a 3×3 box horizontally and vertically. Why? Consider a single strip. That strip passes through the box as a row and as a column. That means that the strip must intersect itself. Therefore, the cell at the intersection appears twice in that strip, violating the fundamental rule of sudoku—that no value appears twice in the same row, column or box.

Let's label the 3×3 boxes A, B, C, etc. Each set of row/columns corresponds to a group of three 3×3 boxes. Every puzzle corresponds to a list of row/columns described by the three boxes that they intersect, with the following rules:

  • Each set of row/columns has three distinct boxes
  • Each box is used twice
  • The puzzle is fully connected. (You can't separate it into two separate puzzles.)
  • Two lists are not distinct if one can be transformed into the other by any permutation of the boxes.

There is no need to specify the order of the boxes or to distinguish the "vertical" row/columns from the "horizontal" row/columns. The same puzzle will have different graphical representations, but we are only interested in the logically distinct configurations.

Puzzles with Three 3×3 Boxes

There is one puzzle with three 3×3 boxes. The three boxes are labeled A, B and C. There is only one possible set of three distinct boxes: ABC. There must be two sets of row/columns, so the other set must also be ABC. This doesn't seem like a very interesting puzzle, but I might make one someday.

Puzzles with Six 3×3 Boxes

We can always label one of the sets ABC. From there we need to start considering the different ways that other sets can be formed. For six boxes, there cannot be another ABC, because that would form a diconnected group of three boxes. There could be an ABx, and we can always choose to label the x as D. In fact, we don't need to consider ACx or BCx, because those can be transformed to ABx by relabeling.

Let's suppose we have ABC and ABD. We need two more sets. We must use E and F twice each, so the other sets can only be CEF and DEF. Here is a puzzle with the arrangement ABC/ABD/CEF/DEF.

Let's suppose that we do not have ABC and ABD (or any instance where two sets intersect twice). We can start with ABC and ADE. We must use F twice, once with B and once with C. Similarly, we must use F once with D and once with E. At this point, B and C are indistinuishable—meaning that we could switch the labels B and C without changing the configuration. Similarly D and E are indistinguishable. Therefore, there is only one such configuration. Here is a puzzle with the arrangement ABC/ADE/BDF/CEF.

We have shown that there are exactly two configurations with six 3×3 boxes.

Puzzles with Nine 3×3 Boxes

We now need to be a little more careful. Let's name some topological features of potential puzzles:

  • ABC/ABD = loop. Two sets intersect each other twice.
  • ABC/ADE/BDF = triangle. Three sets all intersect each other.
  • ABC/ABD/CDE = loop/triangle. A third set includes both open ends of a loop.
  • ABC/ADE/BDF/CEG = double triangle. Two sets (ABC and ADE) are sides of two triangles with different bases (BDF or CEG)

Puzzle 1: Contains a loop/triangle. ABC/ABD/CDE. Box E must connect with something, so we label that set EFG. There are two more sets to make, and both must contain H and I. This leads to only one configuration. Here is a puzzle with the arrangement ABC/ABD/CDE/EFG/FHI/GHI.

Puzzle 2: This is the first of three puzzles that contain a loop, but not a loop/triangle. We label the loop ABC/ABD. We know that C and D are not in the same set, but we can consider two cases. The other set for C can be labeled CEF. The other set for D may include E or F, or not. First consider the case where the other set for C intersects the other set for D (and label the intersection E). Now we have ABC/ABD/CEF/DEG. The other two sets must both contain H and I, so there is only one configuration. Here is a puzzle with the arrangement ABC/ABD/CEF/DEG/FHI/GHI.

Puzzle 3: This is the second of three puzzles that contain a loop, but not a loop/triangle. We start with ABC/BCD, and we've found the only configuration where the other set for C intersects the other set for D. Now we consider ABC/ABD/CEF/DGH. The remaining two sets each contain I. Consider the set that contains E. One possibility is EFI, which means the other set is GHI. Here is a puzzle with the arrangement ABC/ABD/CEF/DGH/EFI/GHI. Note that this puzzle has three separate loops.

Puzzle 4: This is the third of three puzzles that contain a loop, but not a loop/triangle. The remaining possibility contains ABC/ABD/CEF/DGH and the other set containing E is EGI or EHI, which are equivalent choices. Here is a puzzle with the arrangement ABC/ABD/CEF/DGH/EGI/FHI.

Puzzle 5: This is the only puzzle that contains a triangle, but does not contain a loop. Start with a triangle: ABC/ADE/BDF. If we try to make a configuration with a double triangle (at least one set contains CE, CF or EF), we'll see that the configuration must contain a loop. All three choices are equivalent, so let the next set be CEG. We need two more sets, and both must contain H and I, which makes a loop. Therefore, a set with a triangle and no loop must include the configuration ABC/ADE/BDF/CGH. There are two remaining sets. Both sets contain I. One set contains E and the other contains F (because we can't have EF in the same set). One set contains G and the other contains H. Since E and F are indistinguishable and G and H are insistinguishable, this leads to only one configuration. Here is a puzzle with the arrangement ABC/ADE/BDF/CGH/EGI/FHI.

Puzzle 6: There is only one configuration of nine 3×3 boxes with no loops and no triangles. In order to avoid loops and triangles, we must have ABC/ADE/BFG/CHI, because the other sets for A, B and C cannot intersect each other. In order to avoid creating loops, we must create the other two sets by choosing one each from DE, FG and HI. It doesn't matter how we make the choice, because it will always result in classical sudoku ABC/ADE/BFG/CHI/DFH/EGI.

Puzzles with Twelve 3×3 Boxes

It is possible to make puzzles with more than nine 3×3 boxes. Some of these have compact layouts that work well on a two-dimensional page, and others are more difficult to render. There are twenty distinct configurations with twelve 3×3 boxes. I've made actual puzzles from six of them. You can find them from time to time as Sudoku of the Day or in the archive.

Puzzles with Aribitrary Size

There are many ways to make puzzles with an arbitrary number of 3×3 boxes, as long as that number is divisible by three. Look at this puzzle. The middle section can be lengthened by adding another group of three boxes. Now look at this puzzle. The same cascading pattern of boxes appears in the diagonal from upper left to lower right, but the row in the upper left intersects with a column from the lower right, instead of with a column frm the upper left. This puzzle can also be expanded to any length adding groups of three boxes in the diagonal. Alternatively, the ends of the diagonal can be finished by connecting the row in the upper left with the row from the lower right, and connecting the column from the upper left to the column from the lower right, as in this puzzle. Look at the other puzzles in the archive and see if you can extend some of those to arbitrary length by adding groups of three boxes.

Puzzles with 4×4 Boxes

A puzzle with 4×4 boxes must have an even number of boxes. The smallest such puzzle has four boxes in the ABCD/ABCD configuration. The next smallest configuration has six boxes, and there is only one possibility, which is cube sudoku. The configuration is equivalent to six 4×4 boxes on the six faces of a cube. The row/columns run around the cube, so that each strip crosses four faces.


Please send responses to my email address is mathrec at this domain

Thanks,
Steve