How to Solve Comparison Sudoku Puzzles

Comparison Sudoku, which is sometimes named Greater Than Sudoku, Inequality Sudoku, or shortened to Compdoku, is basically a blend of the classic Sudoku puzzle with Futoshiki. The main difference between it and a regular 9×9 Futoshiki puzzle is that there is also the 3×3 region restriction.

As with a normal Sudoku puzzle, you will be using the numbers 1-9 to fill the grid in such a way that no number repeats in any row, column, or 3×3 region. The challenge is that you begin with no clue numbers.

Fortunately, the cell borders have inequality symbols, indicating whether the digit in the adjacent cell is greater or less than the number in that cell. Solve the puzzle while obeying all inequality restrictions.

Because you don’t start with any clue numbers, you will be using the inequalities in each individual region to try and deduce some starting numbers, or at least narrow down the possible numbers in each cell so you can use them to eliminate numbers in each row and column with normal Sudoku scanning techniques.

There are a few ways to do this, so I’m going to demonstrate each of 3 methods.

Method 1 : Pure Pencil Marks

Pencil marks are the tried and true method of putting all the potential numbers in each cell, then whittling away options. Typically, this is done one region at a time, so we’re starting with the region in the upper left of the grid.

This cell is shown to be greater than all three of the cells around it. This allows us to immediately eliminate 1, 2, and 3 as possibilities, because the numbers must all be different.

As a result, even if those three cells were as low as they could be, the highlighted space would have to be at least a 4.

We also know none of those three cells can contain a 9. We’ll look at other inequalities to find out more.

Now, we’ll look at the center cell of this region. We can see it is Less than the three adjacent cells at the top and right and below it, so it can’t be a 9, 8, or 7, and none of those cells can be a 1.

We can also see that the cell to the left must be less than the center square. Because the highest possible number in the center is now a 6, that means the cell to the left must be no more than a 5.

Next, we look at this cell, which is greater than the three adjacent cells, so once again, it can’t be less than 4, and the adjacent cells can’t contain a 9.

As you can imagine, this slow whittling of possible numbers, while effective, is very time consuming. Let’s look at a better way to do pencil marks.

Method 2 : Branching Pencil Marks

One thing you can use to your advantage is inequality chains – that is, a cell that is greater than another cell, which is greater than a third cell, and so on. Let’s start back at the top cell of the same region.

The yellow cell is the end point for several chains, but what is most significant is that it must be a greater number than all of them, because we’re working within a 3×3 region in a Sudoku puzzle – all the numbers between 1 and 9 must exist in this group of cells.

That means that if, for example, the green chain uses 1 and 2, those numbers now can’t be used by any of the other cells. Using this type of thinking, we can see that the yellow cell is greater than seven different cells in this region, so it cannot be lower than an 8. We’ll give it the pencil marks 8 and 9.

Next, we go back one cell in the blue chain, and now, it’s greater than five other cells, so it can’t be lower than a 6.

It’s also less than one other cell, so it can’t be higher than 8.

This cell is greater than four other cells, so we know it must be at least a 5. And it’s less than two other cells, so it can’t be more than 7.

The cell below it is the furthest one back in the blue arrow chain, so it will be less than three other cells, meaning that it can’t be higher than a 6. It’s also lower than the cell in the bottom center, to that makes four other cells it’s less than, changing that maximum number to a 5.

The center cell is greater than two cells (following the green chain back), and less than four cells (following the blue, green, and purple inequalities forward).

This means it has a potential range of 3 to 5.

By now, you should have a good idea of how this method works, so we’ll skip ahead a bit and use the branches to complete this region and see how it looks.

This looks pretty good – we’ve eliminated a lot of numbers. If we did this for the other 8 regions, it would take a while, but we’d then be able to start looking for patterns in the full grid with normal Sudoku techniques:

Scanning along rows and columns to see if a pair of numbers only exists in two cells – this eliminates all other options for those two cells.
Two or three numbers that exist only within the same row or column in the same region – this eliminates those number options everywhere else in their region.
Repeating these eliminations will eventually give you some definite numbers to start filling in the Sudoku grid.

Now, let’s look at one more method.

Method 3 : Extreme Hunting

In this method, you start out looking only for the extreme numbers, 1s or 9s, and work your way toward the middle. This is still a type of pencil marks, except you’re only looking at one number at a time.

Here, I’ve highlighted all the cells that don’t have the less-than side of the inequality symbol pointed at them, because they are all the possible placements of a 9.

Notice that by doing this, we quickly found a cell which is the only possible placement of a 9 in its column.

Now, remembering that this is still a Sudoku puzzle, we can eliminate possible 9s in the same row, column, or region. This reveals two more cells that are the only possible 9 remaining in their row or column (red dots).

The blue dot is the only placce for a 9 in its region, so we’ll be able to remove other 9s in that row and column, as well.

Here, the blue dots show that these two cells are the only possible placements for a 9 within their region.

This eliminates any other options for a 9 in that column, so the pink cell can’t contain a 9.

As a result, the cell with a red dot is definitely a 9.

After another couple of eliminations, we have two more cells where a 9 is certain. This will leave us with a couple of options in two rows and columns at the bottom of the grid, unable to be eliminated by our known 9s.

So let’s try the other extreme and look for 1s in the next step.

This time, using blue for cells that are are only on the less-than side of the inequality symbol, we found two places which are the only option for a 1 within their rows.

At the top center, we see only one cell left in its region, so it has to be a 1. Below it, the two cells marked with blue dots are the only options in their region, so they eliminate all other options on their row.

After removing the pink cell as a possibility, we’re left with the cell marked with a red dot in that region, so it also has to be a 1.

After placing those 1s and eliminating possible duplicates, we have these cells as the only option for a 1 left in their row, column, or region.

After we place them, that will leave only one more cell in the top left region to hold a 1, so we’ll go ahead and fill it, as well. And just like that, we’ve found all the 1s in the grid!

So now, we start the process of looking for the lowest possible cells in each region, which will be our possible placements for 2s. This time, a greater-than symbol is allowed, but only if that wall – and only that wall – is adjacent to a 1.

With slightly lower constraints, we ended up with a few more options, but there is still one cell that is the only option for a 2 in its column.

One important thing to note is that while this method of hunting for extremes is useful, it is not a perfect solving method for more difficult puzzles – they may have cells with no inequalities at all, and sometimes you’ll be forced to resort to pencil marks to start finding what numbers are certain. Looking for extremes is a great starting point, though, and then if you get stuck, you can combine it with the branching pencil marks to narrow down your possibilities.

Now, the cells with the blue dots are the only possibilities for their column, and they are in the same region, so that eliminates 2 as a possibility from the pink cell.

That leaves the cell with the red dot as the only option in its row, so it must hold a 2.

Once more, we have a cell that is the only possibility in its column, so we can place another 2.

And we have one more placement of a 2, but it doesn’t look like we can easily determine any other cells after that.

Before we resort to pencil marks, let’s try looking at the high extreme again. We didn’t find all the 9s, but we did find most of them, so maybe we’ll discover a few 8s in the rows and columns other than the uncertain ones.

The red dots are the highest possible cells remaining in their regions, so they have to contain 8s.

The blue dot, while the only cell marked in its column, isn’t as certain, because of the region below it. We didn’t determine where the 9 must be down there, so we don’t yet know if there is an 8 in this column.

However, now that it is the only possible placement left in its region, we can confirm this cell to be an 8.

Here is an interesting situation. Because of the two 8s we just found, we now know it has to be in the third column of the bottom region, where I’ve highlighted the highest possible cell green.

And even though we haven’t found a 9 yet, we know that it has to be in one of the two yellow cells, because of the two 9s we discovered earlier.

Therefore, this green cell must be an 8, and it’s going to set off a nice chain reaction.

The red dot is the last possible cell in its region that can be an 8, and when we set that, it will:

eliminate the red cell in the same column,
which leaves the blue dot as the last cell to be an 8,
which removes the blue cell in the same column,
which leaves the pink dot as an 8,
which removes the pink cell in the same row,
which means the green dot has to be an 8.

But wait, there’s more!

Because we found all the other 8s, we can use normal Sudoku techniques to eliminate the rows and columns in this region where an 8 can’t be, proving that the red dot must be an 8.

When we do that, we now know that the cell with a blue dot is the only remaining possibility for a 9 in this region. That eliminates the blue cell in the same row, meaning the pink dot is the final 9, and now we can move on to finding the 7s in this grid.

Now we’re looking for the highest remaining cells. They must be greater than any adjacent cell except 8 or 9.

The cells with the red dots are the only option in their row or region, so they must each contain a 7.

As the remaining cells in the grid get fewer, we find it easier to see multiple placements at once.

These cells are the only option remaining in their various rows, columns, or regions.

Looks like that leaves only one more cell to hold a 7, so in the next step, we’ll start looking for possible 6s.

We’ll repeat the same thing we’ve been doing – look for cells that are larger than any adjacent cell, except for 7, 8, or 9.

After marking the remaining high cells, here are the positions that are the only possibility in their row, column, or region.

The red dots are the only options left in their region, and they will eliminate the red cells, so the blue dots will be the final 6s.

After highlighting the remaining high cells, we find a couple of cells to place a 5.

But after doing so, there doesn’t appear to be any more cells singled out, so we’ll switch back to the low numbers and try to find the rest of the 2s.

Well, after accounting for the 2s we already found, it looks like we are only able to definitively place one more 2, and we’re still going to have some ambiguity after that.

Trying to skip forward to the 3s is unlikely to help, so it looks like the best next step is to change to conventional pencil marks and see what patterns we can discover.

For this step, I filled the empty cells with pencil marks using only the normal Sudoku method – eliminating only the numbers already used in the same row, column, or region.

This helps us see things like the two blue cells, which both contain the same two numbers, proving those two numbers can’t exist elsewhere in the same region. When we remove them from the yellow cell, we’re only left with a 5.

However, we cal also use inequalities to our advantage. The pink cell and the two orange cells have the same range – 3, 4, and 5. But the pink cell must be less than both of the orange cells, which will eliminate 4 and 5 as options, so we know it has to be a 3.

Let’s apply that thinking to the pencil marks in the rest of the grid.

When we remove duplicate numbers from the last two placements, and check where inequalities can remove digits from a range when two cells are adjacent, we discover three cells which only have one possibility remaining.

At this point, you’re simply repeating the same couple of steps – eliminate duplicates from the last placement, then look for cells where there’s only one possible number left.