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Hokuro is a grid deduction puzzle inspired by Kakuro.

Basic Hokuro rules

  • Each cell contains one of the following symbols:

    all hokuro symbols

  • Each arrow indicates a step in that direction, while the dot indicates no movement.

  • The clues in the black cells show the sum of the movements indicated by the symbols in the corresponding row or column.

  • Symbols in consecutive white cells must be unique (a 'sum' cannot contain the same symbol more than once).

For example, one of the ways to get the symbol ↑ in 4 steps:

example hokuro sum


The cheat sheet


The answer

  • These three Hokuro puzzles are linked: same colored cells contain the same symbol.
  • Use the numbered cells and the Polybius square to find the final answer.
  • The answer to this puzzle is a 9 letter word.

The puzzles

first linked hokuro

second linked hokuro

third linked hokuro

polybius square

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1 Answer 1

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The first break comes at the bottom of puzzle 3, where there is only one way to fill 3 and 17 without immediately running out of options.

The row with 22 must now have the NE arrow, and it can only go on 43. Combined with the 46-column this forces a few more cells; some pencil marks arise from considering possible combinations (one set of which is transported from basic deductions on board 1).

We now move to board 2:

It is fairly easy to show that the four starred cells below must collectively contain C, W, N and NW. Then the cell above 9 can only be NE or E and 9 itself N or C – but if 9 is N there is a contradiction since the other two cells in its row could only be S and SW, whereas 34 is also locked to C/W/N/NW. Hence 9 is C and we get a lot of deductions from there:

Now on board 1

the arrows from other grids together with possible combinations allow filling in the top row and more pencil marks: From here we deduce that the cell below 18 can only be N:

This fills in a cell on board 3 which leaves only two possible combinations for the column containing it. Both have NW and neither have SW, which eventually fills in yet another coloured cell among a lot of others: Now the cell right of 39 can only be N or W (column combination), which implies that SE must be in its row. SE cannot be in the cell on its left, so it must be in the cell on its right – another coloured cell:

Back to board 1, where the information from elsewhere allows us to finish all but the upper-right pocket:

At last we return to

board 2, where its fresh new value allows a certain deduction: cell 2 can only be SW or S, where choosing S (or SW and then NW for 47) makes the row with 47 unable to be completed. A chain of simple deductions then allows its complete solution:

This provides a value for board 3 allowing us to solve it completely too, leading to board 1 as well:


Interpret the symbols in numbered squares in order as movement commands on the Polybius square, with dots meaning "print letter"; there is only one starting point that keeps the whole path in the square:

This spells out KNUTH DOUBLE ARROW = TETRATION, the final answer.

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  • $\begingroup$ Well done. You make it sound like an easy solve :) $\endgroup$ Jan 24 at 19:46

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