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Published at Mar 18 2021
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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:

```
1 2 3
8 9 4
7 6 5
```

```
1 2 3 4
12 13 14 5
11 16 15 6
10 9 8 7
```

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

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

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

```
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::Int,
spiral= nothing,
cur_n= nothing,
count= 1
)
# we use recursion: create the matrix
spiral==nothing && return spiral_matrix(n, Matrix{Int64}(zeros(n,n)),n,count)
n==0 && return spiral
# calculate the run for a single side
# n: the size of the spiral
# cur_n: the size of the current turn of the spiral
offset= (n-cur_n)Ã·2
from= 1+offset
to= n-offset-1
# for uneven n we end with a turn of a single cell
if cur_n == 1
spiral[from,from]= count;
return spiral
end
# the first side of the turn: left to right
for xx= from:to
spiral[from,xx]= count; count+=1
end
# second side: top bottom on the right
for yy= from:to
spiral[yy,to+1]= count; count+= 1
end
# third side: right to left on the bottom
for xx= range(to+1,from+1,step= -1)
spiral[to+1,xx]= count; count+= 1
end
# fourth side: bottom to top on the left
for yy= range(to+1,from+1,step= -1)
spiral[yy,from]= count; count+= 1
end
# after a complete turn rerun the function for a turn that is 2 smaller than cur_n
cur_n > 0 && spiral_matrix(n,spiral, cur_n-2, count)
# return the spiral
spiral
end
```

A huge amount can be learned from reading other peopleâ€™s code. This is why we wanted to give exercism users the option of making their solutions public.

Here are some questions to help you reflect on this solution and learn the most from it.

- What compromises have been made?
- Are there new concepts here that you could read more about to improve your understanding?

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