Hokuro Puzzle: Introduction

Question

^{This is an entry for Fortnightly Topic Challenge #44: Introduce a new grid deduction genre to the community.}

This puzzle type is inspired by Kakuro, and it has similar rules.

Kakuro is short for 'kasan kurosu' (加算クロス) meaning 'addition cross'. Using Google Translate, I get 'hōkō kurosu' (方向クロス) when translating 'direction cross' into Japanese.

Unless someone with a better grasp of the Japanese language has serious objections, I would like to call this puzzle 'Hokuro'.

Hokuro is a grid deduction puzzle with the following rules:

• Each cell contains one of the following 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.

• The sum of the movements will always be an exact multiple of 45 degrees.

• 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:

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

The puzzle:

Answer 1

Step 1

Look at arrows in black cells, which are in the intercardinal direction and are the sum of two white cells and such black cells intersect horizontally and vertically in one of the two white cells.

The key realisation here is that these cells cannot contain

a) any of the cardinal direction arrows, because even if one of the black cells can be satisfied, the other one cannot be. (Essentially, intercardinal arrows pointing in opposite directions do not share cardinal components)

b) Any other intercardinal direction other than the one listed in the black cells

Based on these two points, we can fill every such intersection with a dot and the remaining cell with the intercardinal direction of that row or column.

Step 2

Now, look at column 2. The final direction of the black cell of that column should be NW, but it already has a SE and SW direction. This means that the remaining 4 arrows must be NW, NE, N and W in some order.

Now, row 3 cannot have an arrow that is pointing in the northern direction (otherwise, you will never get an arrow in the South direction just using two cells). Therefore, the West arrow must be in R3C2 and we can make some knock-on deductions there as well.

Row 5's current sum is currently a dot. Since its eventual sum needs to point North, the remaining components must be the N arrow and a dot. The N arrow is common to both column 2 and row 5, so it must go in that intersection.

The grid at this point:

Step 3

Now, look at the cells in R12C45. The SW black cell that faces two white cells horizontally can only be satisfied in two ways. Either it is W-Arrow + S-Arrow or it is a SW-arrow + dot. The other thing to notice the current sum in column 5. That column needs a final NE-Arrow, but its current sum is 2 * N-Arrow.

Now, if R1C45 must be SW-Arrow + dot, we must place the dot in R1C4 and not R1C5, since there is already a dot in R5C5. However, we now run into a contradiction: Adding SW + 2 * N-Arrows gives result of NW and there is no way to achieve the final result of NE with the remaining arrows.

Therefore, those two cells must be W-Arrow + S-Arrow. Since placing a W-arrow in R1C4 means placing a dot in R2C4 and there is already a dot in R2C3, we must place S-Arrow in R1C4.

Step 4:

On the upper left side of the grid, we can make some observations. R12C2 must be the NE-Arrow and NW-Arrow in some order. However, since there is already a NW Arrow in row 2, then R2C2 must be NE-Arrow.

Step 5:

Now, column 5's current sum is 2 * N-arrow + W-Arrow, so it needs the SE-Arrow and E-Arrow. This means R3C5 must be either a SE-Arrow or E-Arrow. Based on that cell, R2C6 must be either a W-Arrow or SW-Arrow. However, since there is already a W-Arrow in Row 2, R2C6 must be SW-Arrow.

This finishes the grid: