# hallpaz-1596's solution

## to Spiral Matrix in the Julia Track

Published at Feb 25 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``````
``````function fillline!(m, line, rng, v, row=true)
for i = rng
if row
m[line, i] = v
else
m[i, line] = v
end
v = v + 1
end
v
end

fillrow!(m, row, rng, v) = fillline!(m, row, rng, v, true)

fillcol!(m, col, rng, v) = fillline!(m, col, rng, v, false)

function spiral_matrix(n)
if n == 0
return Matrix{Int}(undef,0,0)
elseif n == 1
return reshape([1],(1,1))
end

m = zeros(Integer, n, n)
v = 1
start = 1
final = n
while v < length(m)
v = fillrow!(m, start, start:final, v)
v = fillcol!(m, final, start+1:final, v)
v = fillrow!(m, final, final-1:-1:start, v)
v = fillcol!(m, start, final-1:-1:start+1, v)
start = start + 1
final = final - 1
end
if n % 2 == 1
mid = n Ã· 2 + 1
m[mid, mid] = v
end
m
end``````