How to Solve Futoshiki Puzzles
Futoshiki is a Japanese logic puzzle invented in 2001 by Tamaki Seto. The name literally means “inequality,” and this puzzle is sometimes also called “More or Less”.
This game is always played on a square grid. The goal is to place numbers in the cells so that each row and column contains the numbers 1 through however many cells there are on a side. For example, in a 5×5 puzzle, such as the one here, each row and column will contain the numbers 1-5. If it were a 7×7 grid, the numbers would range 1-7, and so on.
The clues will be a few pre-filled numbers, as well as some inequality symbols. The rules are as follows:
- Each row and column must contain all of the numbers from 1 up to the size of the puzzle
- There may not be duplicate numbers in any row or column
- Inequality signs between two adjacent cells must be followed. For example, in the puzzle pictured above, the top right cell must contain a larger number than the cell immediately to its left
- Adjacent cells without an inequality symbol do not have any requirements other than those in the first two rules
When solving Futoshiki, one of the most useful places to start is with a “chain” of inequalities. Since we have a rather large chain partially sharing the same row as the single number clue we were given, we can first consider these four cells.
In this sequence, the red cell must be larger than the blue, which is larger than yellow, which is larger than green. Because we know that this puzzle uses only the numbers 1-5, we know the sequence from green to red must be one of only five possible sets. For clarity, I’ll show them with the color of cell they would go in.
1
2
3
4
1
2
3
5
1
2
4
5
1
3
4
5
2
3
4
5
We can eliminate all but the first two options. The last three require a 4 to be placed in the blue cell, which is not allowed because of the 4 already in the row. Therefore, we know that green, yellow, blue must contain 1-2-3, and the red cell is still unknown, but must contain either 4 or 5.
Now we can look at the second sequence of 4 cells, but for now we’re just going to think about the first two places. Because it is a chain of 4 numbers, the possible sets are the same as above.
In this case, we can eliminate the last two options, because in both cases, a 3 would be placed into the yellow cell, which is illegal because of the 3 in the column. So for green, yellow, we know they must be 1-2.
Now let’s finish off this column. The only remaining numbers to use are 4 and 5, in both the green and blue cells.
In the green cell, it cannot be the 5, because the yellow cell must contain a larger number. So the green cell must be a 4, and the yellow cell can only be a 5.
This also leaves only a 5 possible in the blue cell.
Next, we’re going to finish the second row. Because of what is already filled in, the only numbers remaining are 1 and 5.
The yellow cell cannot contain the 5, because that number already exists in the column.
Therefore, yellow is 1, and green must be 5.
We have enough information to complete the fifth column. The green cell must be greater than 2, which means it is a 3, 4, or 5.
However, there is already a 5 in the column, which leave the 3 and 4. Because green must be less than the yellow cell, that means green is 3, and yellow is 4.
This leaves only the 2 available for the blue cell.
This is a small step, but demonstrates an important solving technique.
Notice that the green cell already has 1, 2, 4, and 5 in the combination of the row and column it is within.
Because of the rule that there be no duplications, the only remaining number possible for this cell is a 3.
Now, we can do several steps quickly with the process of elimination.
This brings us to our final four cells. In the top row, we need 1 and 2. The green column contains a 2, so green must be 1 and yellow must be 2.
In the bottom row, you’re just filling out the missing numbers from the first two columns. Since green must be 1, that only leaves a 5 for blue, which means the red cell can only contain a 1.
The completed puzzle.
