How to Solve From 1 to X Puzzles
From 1 to X doesn’t have a clear history as to who invented it, although it’s been around since at least 2006, when it was shown to be by Joachim Vetter in the publication Logisch. It’s not certain if that was the first appearance, though.
This puzzle is a little different from the Sudoku-type puzzles that have you filling in numbers and avoiding repeats. Here repeats in the same row and column are allowed.
In this grid, you will be filling in numbers from 1 to X in which X is the size of the region, not the grid. So, a region that only encloses 3 cells will have the digits 1, 2, and 3, while one containing 5 cells will use 1-5.
- Numbers may not repeat within the same region.
- Duplicates are not allowed in orthogonally adjacent cells, but diagonally is fine.
- Numbers outside the grid indicate the sum of all the digits in that row or column.
As with any puzzle, start out with the easy spaces. If your puzzle contains any regions with only 1 cell, you can enter a 1 there immediately.
In our example, we have a few cells that are the only ones in their regions without a number, making it easy to see which digit is missing.
The cell in the first row must contain a 2, while the third row holds a 1. The fourth row cell is a 2, and at the bottom of the puzzle, the only missing number for the region is a 5.
This is another pretty simple placement. The region in the bottom left has to contain the numbers 1-4, and 3 is already given.
The highlighted space is the only possible position for a 2, because the other two cells are orthogonally adjacent to a 2 in a neighboring region.
Now, we’ll see how arithmetic is part of solving this puzzle. The number 15 at the top is the sum of all the numbers in this column. The brackets show the totals of the regions.
Notice that the green region has a total even though it has no numbers placed. This is because it is entirely contained within the column, and is composed of three cells, so we know it will contain the digits 1, 2, and 3, which add up to 6.
The two cells in the pink region will be equal to 15-6-5-1, or 15-12=3. The only possible combination of usable numbers that add up to 3 is 1+2. Because of the 1 in the top row, we know that we can’t place a 1 in the upper pink cell, so it has to be a 2, with a 1 below it.
It looks like we can do that again for the third row. The empty cells in green have to hold 4 and 5, because those are the only numbers left in that region, so we know that total is 1+3+4+5=13.
In the yellow, we have 3+1+2=6. That only leave one cell, so we can do some subtraction. 23-13-6, or 23-19=4. Therefore, the pink cell holds a 4.
This also means that in the green cells, the 4 may not be adjacent to the one we just placed in the pink space, so the 5 must be on the right, and 4 will be on the left, completing the region, and the row.
Let’s take a break from math for a moment to do some simple deduction. This region must contain the digits 1, 2, and 3, because it has three cells.
The 2 cannot be placed on either end, because of adjacency to a 2 in other regions, so it has to go in the middle.
This leaves a 1 and 3 for the ends, and we know the 3 can’t be placed on the left, because there is a 3 adjacent to that cell. That means that in this region, the sequence of digits is simply 1-2-3.
When you know, or can deduce the totals most cells within a row or column, look for the smallest and largest sums, as this will let you reduce the options for remaining cells, and adjacency will often reduce them further.
Here, we have a total of 16, and known cells in the row totaling 8 and 5. When we subtract them, we get 16-8-5, or 16-13=3.
Of course, that means the pink cells will have to be 1+2, and the 2 can’t be placed on the right, because if another 2 that is orthogonally adjacent. So we will have a 2 on the left, and a 1 on the right, and now we’ve solved another row.
Sometimes, you’ll need to combine deduction techniques. Here, we have a sum of 22, and known cells in the column totaling 7+5+5=17. We can subtract 22-17=5 for the remaining two spaces in the column. They aren’t part of the same region, and our two options to total 5 are either 1+4, or 2+3.
But notice the pink cell is in a region that only has two spaces. This means it must be either a 1 or a 2. If it is a 2, then the green cell would have to hold a 3 to reach the correct sum. However, it can’t, because there is already a 3 directly below it. This means that 2+3 is not a valid option.
That leave us 1+4, and the 4 can’t go in the pink cell, because it’s part of a 2-space region. So the pink cell must be a 1, the green cell must be a 4, and we can also place the 2 to complete the 2-cell region.
A technique that is often helpful in these types of puzzles is to use pencil marks – basically fill in cells with the remaining possible numbers. This can often reveal digits that can be known, but aren’t obvious.
I’ve filled in the grid with pencil marks with all possible numbers for each region, then removed options that couldn’t be placed either because of an orthogonally-adjacent duplicate, or because it was already used in that region.
In the yellow cell, we’re down to only a 3 as a possible number, so that was easy. In the green cells, the only options are 2 or 3, which means neither can be possible in the pink space, which is in the same region. That leaves only a 1 as a possibility there, so we can fill in that cell as well.
Pencil marks can also be useful when checking sums in a row or column. Because they show you all the numbers possible in their cells, it can save a lot of time when calculating combinations to reach the remaining amount to complete a sum.
In this column, we need a total of 22, and we already have known numbers totaling 10 and 1, with an additional pair of numbers that must total 5, even if we don’t yet know their exact position. 22-10-5-1, or 22-16=6, so the pink cells must add up to 6.
It doesn’t look like we’ll be able to immediately solve either of these cells, but we can eliminate some possibilities in our pencil marks. Since they share the same region, we know it can’t be 3+3, and 3 isn’t an option in both cells, regardless. That leaves us with either 2+4 or 1+5.
If it will be 2+4, the 2 only exists as an option in the top cell, so the 4 would have to be in the bottom cell in that situation. So we can eliminate 4 as a possibility in the upper cell, leaving only 2 or 5.
Similarly, a 1 is only an option in the bottom cell, so if we were using 1+5, the 5 would have to be in the top cell – it could not be in the bottom one. And we know the 3 can’t be used at all here, so we can eliminate that option as well. So the bottom space can now only be either a 1 or 4.
Now, by looking at the pencil marks for this region, we can see that there is only one placement option left for the number 3. It might not seem like much, but every number we can be certain of makes things easier.
Notice that this column has the largest sum on the grid, but so far, it only has small numbers that have been placed. Here is a chance for our pencil marks to shine by letting us test options quickly.
So far, our known numbers total 5+3+3=11. Subtract 11 from 24, and we need another 13 in the remaining three cells.
The only possible numbers in the pink cell are 1 or 4. In cyan are 2 or 5, and green can be 3 or 4.
If the pink cell were to be a 1, we would still need another 12, which cannot be reached with any combination in the cyan and green cells. That means the pink space must be 4.
This leaves us with 9 for the total that cyan and green must reach. Now we can eliminate 2 from the cyan space, because 7 is not an option in green. So cyan must be a 5.
This tells us that the green cell must be 4. Better still, by solving all three of these cells, each of their regions is left with only one space remaining, so we can solve those, too.
Finally, pencil marks allow you to quickly scan the grid and solve cells where options are eliminated due to newly-discovered adjacency.
In the yellow cells, there can no longer be a 4 on the left, so it has to be a 1, and the 4 will go on the right.
In the pink cells, we now see there is a 2 adjacent to the top space, so it must be the 3, and the 2 is placed below it.
In the row with the yellow cells, the known numbers add up to 16, and 21-16=5. The only combination possible in the yellow cells that adds up to 5 is 2+3, with the 2 on the left, and the 3 on the right.
In the row with the green cells, the known numbers sum 14, and 17-14=3. This must be reached with 1+2, and here, 2 is only an option on the right, so the 1 must be on the left.
After cleaning up the pencil marks, we quickly see a couple cells with only one possibility left, and as a bonus, the green cell is the only remaining placement option for a 4 in that region.
After removing a possibility due to adjacency, we’re left with only a 2 in the yellow cell. Once we add that to the other known numbers in this row, we reach a total of 19. Subtract that from 22, and we now know that the green cell must be 3.
Once we fill those in, there will only be one cell left in these two regions, so from here, we can simply solve the puzzle.
And here we have the completed puzzle.
