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.