So as you might or might not know, I can be quite evil sometimes, especially with my Hidato puzzles that contain no numbers.

But, today, I decided to give you an ! This means that you will have to determine what type of puzzle this is and solve it completely. However, you do have this to help you:

$\color{white}.$ $\color{white}.$ $\color{white}.$ $\color{white}.$ $\color{white}.$ $\color{white}.$
$\color{white}.$ $\color{white}.$ $\color{white}.$ $\color{white}.$ $\color{white}.$ $\color{white}.$
$\color{white}.$ $\color{white}.$ $\color{white}.$ $\color{white}.$ $\color{white}.$ $\color{white}.$
$\color{white}.$ $\color{white}.$ $\color{white}.$ $\color{white}.$ $\color{white}.$ $\color{white}.$
$\color{white}.$ $\color{white}.$ $\color{white}.$ $\color{white}.$ $\color{white}.$ $\color{white}.$
$\color{white}.$ $\color{white}.$ $\color{white}.$ $\color{white}.$ $\color{white}.$ $\color{white}.$
R1: roobbo
R2: robobb
R3: brybro
R4: yooyrb
R5: bbobbb
R6: boborg

Hint 1:

The lowercase letters (except b) represent colors, but what do the colors represent?

Hint 2:

Look at the tags.


1 Answer 1


The completed grid:

enter image description here

Where do we start?

Adding the colours to the grid gives us the following:
enter image description here

First I guessed that this would probably be another coloured Hidato puzzle, similar to some of your recent puzzles.

Hopefully this puzzle works in the same way (place the numbers 1-36 in the cells of the grid, so that consecutive numbers are adjacent (orthogonally or diagonally)). So we need to work out what the different colours represent, and then solve the puzzle.

  • There are six red cells, which matches the number of square numbers in range (1, 4, 9, 16, 25, 36).
  • There are eleven orange cells, which matches the number of primes in range (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31).
  • There are three yellow cells and (based on previous puzzles) these could be the remaining Fibonacci numbers not already covered by earlier sets (8, 21, 34).
  • And finally there is a single green cell. This could be several things, but (again based on previous puzzles) this could be the single cube in range that is not already covered (27).

  • Let's start solving:

  • We need a connected sequence of red-orange-orange-red-orange for 1-2-3-4-5. There are only two ways to place 1-2-3-4 (with another orange cell adjacent for 5), and one of them completely isolates the top-left cell. So we can place 1-4.
  • In the opposite corner, 27 goes in the single green cell.
  • 29 must in an orange cell two cells away and there is only one possibility.
  • 28 must be in a white cell between those two.
  • and then 26 must be in the remaining white cell adjacent to 27.
    enter image description here

  • The red cell below the 1 cannot be 5 (orange), so is a dead-end. That means it must be the other end of our number chain, and 36 works.
  • 35 is then forced and 34 must be in the yellow cell below that.

  • In the bottom right, we have another red cell has only one free neighbour. We cannot have another dead-end, so it must join to the existing chain, so must be 25.
  • And then 24 is forced.

  • 8 must be in a yellow cell, and have a red neighbour for 9. So we can place 8 and 21.
    enter image description here

  • There is only one way to place 22, 23 (white,orange) to join up 21 and 24.
  • That forces 20 (white)
  • And 5 and 7 (both orange).
  • And then 6 (white) must join those.
  • Now the path from 29 to 34 in the bottom left is forced.
    enter image description here

  • Orange cells (primes) cannot contain consecutive numbers (except for 2 and 3 which are already placed). The orange cell above the 20 only has one free non-orange neighbour, so must join to 20. So that is 19.
  • That forces 18, 17, 16 (white, orange, red).
  • And that leaves no choices for 9, 10, 11.
    enter image description here

  • Finally, 13 must be in the single remaining orange cell.
  • And from here, there are unfortunately two ways to complete the chain from 11 to 16:
    enter image description here

  • $\endgroup$

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