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## to Spiral Matrix in the Julia Track

Published at Mar 04 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))
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 spiral_matrix(n)
if n > 0
# initialize 2D array
a = zeros(Int, n, n)

# countdown array values from highest to lowest
counter = n^2 + 1

# initial spiral diameter
spiral = n % 2

# for odd n, initialize with unique inner spiral of 1 x 1
if spiral  == 1
a[div(n+1, 2), div(n+1, 2)] = counter - 1
counter -= 1
end

# step through outer spirals from innermost to outermost
spiral += 2
while spiral <= n
start = floor(Int, (n - spiral)/2) + 1
length = spiral - 1
for i = 1 : length
# update left side of this spiral
a[start + i, start] = counter - i
# update bottom side of this spiral
a[start + length, start + i] = counter - length - i
# update right side of this spiral
a[start + length - i, start + length] = counter  - 2 * length- i
# update top side of this spiral
a[start, start + length - i] = counter - 3 * length - i
end
counter -= 4 * length
spiral += 2
end
a
else
Matrix{Int}(undef,0,0)
end
end``````

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