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After introducing this new puzzle type, here is a slightly more challenging Hokuro puzzle.

Hokuro is a grid deduction puzzle inspired by Kakuro with the following 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 puzzle:

Hokuro really fixed


Please note that a previous version of this puzzle contained an error which made it unsolvable. I have now fixed it and made sure that there is one unique solution.

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    $\begingroup$ Can the sums represent multiple squares in that direction? For instance, would ↙↘ sum to ↓, or would that be completely disallowed? $\endgroup$ – Deusovi Dec 16 '20 at 14:04
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    $\begingroup$ @Deusovi As a pair, ↙↘ would be disallowed because it would add up to a double down arrow, whereas the clues are all one symbol. However, the triple ↙↘↑ is allowed, as it adds up to ↓. $\endgroup$ – sarsaparilla Dec 16 '20 at 14:13
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So I have an almost complete solution in which I am reasonably confident about my solve steps, but I end in an impossible state...

I am hoping someone can correct this because I know I am E/W or L/R dyslexic but I have gone over it too many times now! :)

Step 1:

enter image description here
purple corner squares have no choice, because they lie on exact opposite pairs.
orange squares follow directly on from purple as given.
blue needs 1W + 2N total, with one pair in two options.
green needs 2S + 0E/W total, also with one pair in two options.

Step 2:

enter image description here
if we choose the final yellow square as SE, we can't go NE twice in the central squares; therefore it is SW.
the two middle yellows are therefore NW and NE, and the red column tells us the order.
this gives us the blue boxes for free.
and the left over green squares must be W and NW, with the order is also fixed by yellow.

Step 3:

enter image description here
purple squares are a total of 2S and 2E in 3 squares, therefore S, E and SE
but the red row cannot contain S, so the top is E (and a free orange box!)
also the blue row can't contain SE, so the bottom is S (giving us some pencil marks for the rest)
lastly, the yellow column needs 2E and 1S, which can only be added in one order.

Step 4:

enter image description here
orange column can be resolved because we can't have a 2nd SE in the second bottom row.
this gives us the blue column for free.
the remaining boxes in the red row need to mirror the placed ones, so: N, NW, and W but we know the green column can't contain W, giving us the column as well.
yellow follows for free, and disambiguates the rest of the red row.

Step 5:

enter image description here
purple column must contain 2S and 2E (ie S, E and SE) but red row can't contain a S.
yellow column must contain E and SE as well. given we know we will have a SE in the blue row, we can't have 2 Es to complete it so we can choose both yellow and purple orders.
this solves blue, which solves green which solves orange... bottom half complete!

Step 6:

enter image description here
the purple column requires 2S and 2E, and only 1 combo in 4 boxes: S, E, SE, O also, the red row can't contain S or E which fixes the O and solves the red row.
the blue row can't have any more S, so we now know where the E is in the purple column.
blue has a N and a NW left, and the yellow column chooses the order giving us the yellow and green squares.

Step 7:

enter image description here
Slightly tricky, but there are only a few combos to get 1S in the purple column.
Using the red row, I note that I need to use all 3 N options to get my target, and what is left over needs to be E-W balanced (so can only be E and W!)
This means that the only thing that can go in the purple/red intersection is E, solving purple.
The blue row can only be NE and E... but NE can't be in the yellow column because I know the red row has no O in it.
This now solves yellow, green, and red can be selected from the only two N options available.

[EDIT] Started solve again from this point, as puzzle up until now remains the same!

Step 8: Some limitations:

enter image description here
There are a few limitations we can add now. Only one square in the orange column can contain an E, meaning the red row has only two options: [W | S] or [SW | x]
Yellow can't support anymore E, so the orange/yellow intersection is E-W neutral.
But purple still needs 2W in total, and so does the yellow row. So we have definite W-type arrows in the two remaining yellow squares.
And finally, this means the top green is E-W neutral as well.

Step 9: Uh-oh... But now, what..?

Unless I have gone crazy (again!) I don't have any further reason to eliminate options... This would mean that there are (at least) 3 solutions which fit the clues...

enter image description here

Sorry - I think I may have broken it again..?

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  • $\begingroup$ I can confirm that I reached the exact same position and faced the same contradiction. $\endgroup$ – Alaiko Dec 17 '20 at 11:45
  • $\begingroup$ In test-solving this, I also noticed that the final 8 squares are the hardest to fill. If you enumerate the left options, you should be able to eliminate all the ones that lead to contradictions. $\endgroup$ – sarsaparilla Dec 17 '20 at 12:13
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    $\begingroup$ @sarsaparilla Like Graylocke and Alaiko I hit the exact same place, and I'm pretty sure it is a contradiction. The three remaining columns need one excess down (D,D,U), while the three remaining rows need two excess downs (D,D,-). They cannot be simultaneously filled. $\endgroup$ – Jeremy Dover Dec 17 '20 at 14:14
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    $\begingroup$ I'm sorry guys for leading you down an impossible path. I will fix the puzzle when I get the chance. $\endgroup$ – sarsaparilla Dec 17 '20 at 15:10
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    $\begingroup$ @Graylocke At this point I don't trust my brain anymore in making the deductions necessary for creating a uniquely solvable puzzle, so I had the computer check the uniqueness of the solution (it is, now, really and finally, unique!). So here it is, my final edit. I hope you are still willing to solve the Hokuro. For me, it has been a learning curve to create a new type of puzzle, and everything that needs to be taken into account in order to make it uniquely solvable (and how easy it is to make a logical error, messing up the whole puzzle). $\endgroup$ – sarsaparilla Feb 1 at 11:33

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