Note: this is not in conjunction with my Minesweeper puzzles

In all honesty, I don't know why I decided to look up this type of puzzle since I also have never heard about it, and there doesn't appear to be any puzzles on here mentioning it, so I don't know why I looked it up.

From Wikipedia:

Hidato (Hebrew חידאתו, originating from the Hebrew word Hida = Riddle), also known as "Hidoku", is a logic puzzle game invented by Dr. Gyora M. Benedek, an Israeli mathematician. The goal of Hidato is to fill the grid with consecutive numbers that connect horizontally, vertically, or diagonally.

From the section "About the Puzzle":

About the puzzle

In Hidato, a grid of cells is given. It is usually square-shaped, like Sudoku or Kakuro, but it can also include hexagons or any shape that forms a tessellation. It can have inner holes (like a disc), but it has to be made of only one piece.

What is the goal of Hidato?

To fill the grid with a series of consecutive numbers adjacent to each other orthogonally or diagonally. All tiles are required to be filled in.

"In every Hidato puzzle the smallest and the highest numbers are given on the grid. There are also other given numbers on the grid (with values between the smallest and the highest) to help direct the player how to start the solution and to ensure that Hidato has a single solution."

However, the above conditions of the smallest or highest numbers sometimes are relaxed. When they are, only the numbers are given, not their position on the grid.

Every well-formed Hidato puzzle has a unique solution. A Hidato puzzle intended for human solvers should have a solution that can be found through logic and deductions. However, it is possible for there to be difficult puzzles, even of small size.

You can find plenty of well-formed Hidato puzzles here

The puzzle (Created in Google Docs)

Difficulty: ★☆☆☆☆

enter image description here

Note: To get the green checkmark, you must solve this puzzle and show the process through which you solved it.

  • $\begingroup$ wait oh no there are two possible solutions (at least) whoops I'll update the image quick :\ Edit: looked at the solution, never mind $\endgroup$
    – CrSb0001
    Oct 31, 2023 at 17:29
  • 1
    $\begingroup$ Actually, a Hidato puzzle has been posted here at least once before... $\endgroup$
    – Stiv
    Oct 31, 2023 at 20:02
  • $\begingroup$ @Stiv Yeah, realized that a bit too late. :\ $\endgroup$
    – CrSb0001
    Oct 31, 2023 at 20:08

1 Answer 1


Completed Grid:

Note: the answer is not unique. 20/33 can be interchanged and still yield a solution to the puzzle:

Completed Grid

Let's start with some easy deductions:

13 and 15 are both given, and only one cell touches them both, which must be 14. With this placed, we can look at how to get from 4 to 8. The 5 cannot go directly left or over the 4, since there is no chain of cells to reach 8. If the 5 goes northeast of 4, then all of 5, 6 and 7 must be orthogonal to the 15, which forces the 17 to go southwest of 16. But this would leave no path for the 17 to get out of the lower right corner. So the 5 must go southeast of the 4. This forces the 6 and 7, as well as the 9 and 10, filling out the lower right corner. The grid thus far:

Progress 1

Moving on:

Now there is only one place for the 17. If the 18 went northwest of the 17, then the cell north of 17 could only be 19, 33, or 35, but all of these are impossible since they would not be adjacent to 20, 32 or 36, respectively. Thus the 18 must go north of 17, which forces the 19 as well. The grid thus far:

Progress 2

In the twenties and thirties:

There are two places we could put 20, but moving further into the sequence, we are given all of the numbers from 21 to 26, and only have one path to get to the given 29. There is then one possible cell for 30.

Now to get from 31 to 34, 32 has to go east of 31, but there remain two possible cells for 33 to get from 32 to 34, which are the same two cells which could get us from 19 to 31. Now, the upper right corner cell must be 35 and the cell to its west 36. The grid thus far:

Progress 3

Jump around!

Since the cell north of 4 is neither 2 nor 3, we may now place the 2 and 3 . Bouncing back to the 38, we have only one possible placement for 39 and 40. Now jump to the 57 (the largest number in the puzzle). It must be adjacent to the 56, leaving one possible cell for the 55. This forces the locations of the 42 and 43, and the rest falls apart.

  • $\begingroup$ Ah, so my puzzle wouldn't be considered a good Hidato puzzle (since it's required that it has a unique solution for it to be considered a good Hidato puzzle) but good job on solving the puzzle! $\endgroup$
    – CrSb0001
    Oct 31, 2023 at 17:59

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