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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

In some cases it may come in handy to take a peek at the Hokuro Cheat Sheet.


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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  • 1
    $\begingroup$ 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 ↓. $\endgroup$ Dec 16 '20 at 14:13
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    $\begingroup$ I imported this into Google Sheets for anyone who prefers to solve it this way: docs.google.com/spreadsheets/d/… You will need to copy the whole thing and paste it into your own sheet, then change a few things like horizontal and vertical align the cells in the center, the row and column widths and heights etc. $\endgroup$
    – hb20007
    Dec 13 '21 at 19:35
  • $\begingroup$ @hb20007 And thanks again :) $\endgroup$ Dec 14 '21 at 8:17
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Graylocke has worked through an older version of the puzzle, which did not have a unique solution. If (s)he would had seen the updated puzzle, I’m sure that (s)he would have been able to solve it quickly using the same type of deductions which (s)he had been using.

However, since it’s been more than a year without any updates, I am posting my solution.


Complete Solution

Complete Solution

Walkthrough

Step 0

Step 0

I started by importing the puzzle into a spreadsheet, which is my preferred method of solving this type of puzzles.

Step 1

Step 1

The top-left, top-right and bottom-right corners have intersecting arrows which are completely opposite to each other. The only way to satisfy them is for the cell at the intersection to have ● in it. Then, there is only 1 choice when it comes to filling the rest of the cells.

Step 2

Step 2

Continuing in the bottom-right corner, looking at ↘(4), the only way to fulfil it given what we have is (●↗↙↘). ↘ cannot be on top, because then we would not be able to make horizontal ↖(4). We would need 2 ↖s to make that, which is illegal. So, ↙ must be on top.

Step 3

Step 3

Now we know that horizontal ↖(4) must be (●↙↗↖). Looking at vertical →(2), there would be no way to make it which involves a ↖, so that must be (↗↓).

Step 4

Step 4

To complete ↙(4), we need ↖ and ←, and there is only 1 legal way to place them.

Step 5

Step 5

The only combination of ↘(5) that fits is (●→↘↓↖). Neither ↓ nor ↘ can be at the top, because they contain a downward component. So, → must go at the top, which means that ↗(2) is (→↑).

Step 6

Step 6

To finish off ↘(5), we need ↘ and ↓. ↘ cannot be at the bottom because of horizontal ↖(3). This is because the only way to fulfil that using a ↘ would be (↘↖↖), which is illegal.

Step 7

Step 7

We need a → and ↘ to complete ↑(5) and there is only 1 way they can go.

Step 8

Step 8

Horizontal ↖(3) needs ↑ and ↖. We cannot have ↖ in the ●(2) column, because then ●(2) would have to be (↖↘), which causes a contradiction.

Step 9

Step 9

We need ↑, ↖ and ← to finish off ●(7). Only ↑ can be placed in the leftmost cell, since the other 2 options contain a leftward component and would clash with vertical →(2). This allows us to fill in the bottom-left corner.

Step 10

Step 10

Vertical ↓(5) has to be (↖←→↓↘). It intersects with ↑(2), which cannot have a downward component. So, ↑(2) must be (→↖).

Step 11

Step 11

To finish off ↓(5), we need ↘ and ↓. To finish off ●(4), we need ↘ and →. Since ↘ is common between both, this leaves us with 2 options. Here, we realize that horizontal ↓(3) cannot contain ↘ and → together, which eliminates one of the options.

Step 12

Step 12

Now we can complete the whole bottom half of the puzzle.

Step 13

Step 13

The only combination for ↘(5) that contains a ↖ is (●→↘↓↖). Out of all these options, the only one that can be used alongside the horizontal ↖(2) is ●. So ↖(2) must be (●↖). Then, the only remaining option that can be used alongside the horizontal ↗(4) is →, because the other 2 options have downward components.

Step 14

Step 14

Now we know that ↗(4) has to be (→↘↖↑), so we need to fill in a ↖ and ↑. ↖ does not fit with the vertical →(2), so the ↑ has to go there. We now know that the first 2 cells of ↓(4) are ↑ and ↘. Since we need ↘ and ↓ to complete the vertical ↘(5), we can now fill this area with the only possible permutation.

Step 15

Step 15

To finish off ↙(5), we need either ↙ and ●, or ↙ and →. The only combination of ↑(7) that fits horizontally is (↑↗→↘↙←↖). Combining these 2 facts, we can deduce that the intersection of these two has has to be →.

Step 16

Step 16

Horizontal →(3) has to be (↙↗→). The middle cell cannot be ↗ because we know that ↑(7) cannot have a ●. With this, we can fill in →(3) and ↑(7).

Step 17

Step 17

Horizontal ↖(4) has to be (↑↘←↖) since we already have a ↘. Looking at the intersection with ↗(5), we can see that there is a ↖ in that column, so the cell at the intersection has to be either ↑ or ←. If we assume it is ←, we would need 2 ↖s to fulfil ↗(5) then, which is illegal. So, we know that the cell at the intersection is ↑.

Step 18

Step 18

↗(5) is now missing a ↓ and ↘. ↘ cannot coincide with horizontal ↙(2), so we know that’s (↓←).

Step 19

Step 19

We can now fill ↖(4) in with the only permutation of (↑↘←↖) which remains legal. This leads to only 1 permutation for ↖(2), which leads to only 1 arrow which satisfies the final cell.

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    $\begingroup$ Well done, finally it got solved! $\endgroup$ Jan 13 at 7:54
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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. $\endgroup$ 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. $\endgroup$ 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. $\endgroup$ 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). $\endgroup$ Feb 1 '21 at 11:33

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