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# SimonTreil's solution

## to Spiral Matrix in the Julia Track

Published at Feb 23 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 spiral_matrix(n)
ans = zeros(Int,n,n)
if n == 1
ans[1,1] = 1
elseif n>=2
i = 1
j = 1
it = 0
jt = 0
tour = 4*(n-1)
N=n
for k in 1:n^2
if k == tour+1
jt += 1
N = N-2
tour += 4*(N-1)
end
if k == tour
it += 1
end
println("K:",k,"   i:",i,"   j:",j,"   it:",it,"   jt:",jt)
if (j<n-jt) & (j>jt) & (i==it+1)
println(k)
ans[i,j] = k
j += 1
elseif (j == n-jt) & (i < n-it)
ans[i,j] = k
i += 1
elseif (i == n-it) & (j==n-jt)
ans[i,j]  = k
j -= 1
elseif (i==n-it) & (j<n-jt) & (j>jt+1)
ans[i,j] = k
j -= 1
elseif (i==n-it) & (j==jt+1)
ans[i,j] = k
i -= 1
elseif (i>it+1) & (i<n-it) & (j==jt+1)
ans[i,j] = k
i -= 1
end
end
end
return ans
end``````