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# rogers132's solution

## to Spiral Matrix in the Julia Track

Published at Feb 22 2021 · 0 comments
Instructions
Test suite
Solution

Given the size, return a square matrix of numbers in spiral order.

The matrix should be filled with natural numbers, starting from 1 in the top-left corner, increasing in an inward, clockwise spiral order, like these examples:

###### Spiral matrix of size 3
``````1 2 3
8 9 4
7 6 5
``````
###### Spiral matrix of size 4
`````` 1  2  3 4
12 13 14 5
11 16 15 6
10  9  8 7
``````

## Source

Reddit r/dailyprogrammer challenge #320 [Easy] Spiral Ascension. https://www.reddit.com/r/dailyprogrammer/comments/6i60lr/20170619_challenge_320_easy_spiral_ascension/

## Version compatibility

This exercise has been tested on Julia versions >=1.0.

## Submitting Incomplete Solutions

It's possible to submit an incomplete solution so you can see how others have completed the exercise.

### runtests.jl

``````using Test

include("spiral-matrix.jl")

@testset "Different valid values" begin
@testset "Empty spiral" begin
@test spiral_matrix(0) == Matrix{Int}(undef,0,0)
end
@testset "Trivial spiral" begin
@test spiral_matrix(1) == reshape([1],(1,1))
end
@testset "Spiral of size 2" begin
@test spiral_matrix(2) == [1 2; 4 3]
end
@testset "Spiral of size 3" begin
@test spiral_matrix(3) == [1 2 3; 8 9 4; 7 6 5]
end
@testset "Spiral of size 4" begin
@test spiral_matrix(4) == [1 2 3 4; 12 13 14 5; 11 16 15 6; 10 9 8 7]
end
@testset "Spiral of size 5" begin
@test spiral_matrix(5) == [1 2 3 4 5; 16 17 18 19 6; 15 24 25 20 7; 14 23 22 21 8; 13 12 11 10 9]
end
end``````
``````@enum Orientation begin
right
left
up
down
end

struct Direction
orientaion
delta_i
delta_j
next_orientaion
end

const directions = Dict(right => Direction(right, 0, 1, down),
down => Direction(down, 1, 0, left),
left => Direction(left, 0, -1, up),
up => Direction(up, -1, 0, right),)

function spiral_matrix(n)
matrix = zeros(Int, (n, n))
(i, j) = (1, 1)
side_length = n
d = directions[right]

for k in 1:n * n
matrix[i, j] = k
final_index = d.delta_i + d.delta_j > 0 ? side_length : n + 1 - side_length
current_index = d.delta_i != 0 ? i : j

if current_index == final_index
d = directions[d.next_orientaion]
if d.orientaion == up
side_length -= 1
end
end

i += d.delta_i
j += d.delta_j
end

return matrix
end``````