Sudoku has conquered the world, but I think futoshiki is more fun — more concentrated, more compact and quicker. When complete, a 5×5 futoshiki will be a Roman square in which every row and column contains the numbers 1 to 5, but no number is repeated in any row or column. You have to work out the missing numbers using logic and the “inequality” signs that show whether one square contains a number more than or less than a number in a neighboring square — futōshiki, 不等式, means “inequality” in Japanese (fu- is the negative prefix). Here’s an example of a futoshiki puzzle:
Futoshiki puzzle, with 4 in square (3,2) and 3 in square (1,1)
If you identify the squares by row and column, 4 is in (3,2) and 3 is in (1,1). And you can say, for example, that the empty square (3,4) dominates the empty square (3,3) or that (3,3) is dominated by (3,4). I’ll describe one route (not the best or most efficient) to completing the puzzle. Let’s start by considering the general rules that 1 cannot appear in any square that dominates another square and that 5 cannot appear in any square dominated by another square.
If you extend that logic, you’ll see that 4 cannot appear in any square that is at the end of what you might call a chain of dominations, where one square dominates a second square that in turn dominates a third square. Therefore, in the puzzle above, 4 cannot appear in squares (2,1) and (4,1) of column 1. And it can’t appear in square (3,1), because that would mean two 4s in the same row. This leaves one place for 4 to appear: square (5,1). And if 4 is there, 5 has to be in square (3,1):
Now look at row 3. Two of the remaining three empty squares are dominators: (3,4) dominates (3,2) and (3,5) dominates (2,5). 1 cannot appear in a dominating square, so 1 has to be in the dominated square (3,3):
The next step I’ll take is a bit more complicated. In row 2, the number 4 cannot be in (2,1) and (2,4). It can’t be in (2,1) because that would mean 4 was greater than itself. And it can’t be in (2,4), because (2,4) is dominated by (3,4) and (3,4) can’t contain 5, the only number that dominates 4. Therefore 4 must be in either (2,3) or (2,4). But so must 5. Therefore (2,3) contains either 4 or 5 and (2,4) contains either 4 or 5. That means that the numbers [1,2,3] must be in the other three squares of row 2. Now, 3 can’t be in square (2,1), because the chain of dominations is too long. And 3 can’t be in (2,5), because (2,5) is dominated by (3,5), which contains either 2 or 3. Therefore 3 must be in (2,2):
Now consider column 2. Square (4,2) cannot contain 1 or 5, because it’s both dominated and dominating. And if it can’t contain 1 or 5, there’s only one number it can contain: 2. And it immediately follows that (4,1) must contain 1, the only number less than 2. And if four squares of column 1 now contain the numbers [1,3,4,5], the remaining empty square (2,1) must contain 2:
Now consider row 2. Squares (2,3) and (2,4) contain either 4 or 5, therefore (2,5) must contain 1:
Now consider row 5. The number 1 is logically excluded from three squares: from (5,3), because there’s a 1 in (3,3); from (5,4), because (5,4) dominates (5,5); and from (5,5), because there’s a 1 in (2,5). Therefore 1 must be in (5,2). And if 1 is in (5,2), the number 5 must be in (1,2):
Now 1 pops up in row 1 because it can only be in (1,4):
And 5 pops up in row 4 because it can only be in (4,5):
Once 5 is in (4,5), the number 4 must be in (4,4) and the number 3 in (4,3):
The 4 of (4,4) immediately collapses the ambiguity of (2,4), which must contain 5. Therefore (2,3) contains 4:
Next, 5 pops up in (5,3):
And 3 must be in (5,4), dominating 2 in (5,5):
With 3 in (5,4) and 2 in (5,5), the ambiguity of (3,4) and (3,5) collapses:
And the square is completed like this:
⬇
Here’s an animated version of the steps to completion:
Futoshiki puzzle animated
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