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# gitblight1's solution

## to Spiral Matrix in the Julia Track

Published at Mar 18 2021 · 2 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)
m = Matrix{Int}(undef,n,n)
n == 0 && return m
m[1,:] = collect(1:n)
m[2:n,n] = collect(n+1:2*n-1)
# rest of matrix is an n-1 square, rotated and starting at 2n
m[2:n, 1:n-1] = rot180(spiral_matrix(n-1)) .+ (2*n-1)
return m
end``````

I could not resist to comment on this solution, even if I am not asking a question: smart! I have never really developed an eye for recursive functions and here comes a perfect opportunity to apply them. Exploiting rot180 is also neat.

Solution Author
commented 93 days ago

Thanks! This does have the issue of running out of stack size if you try to compute a huge spiral matrix (My box can handle `spiral_matrix(2000)`, but not 3000), but it was fun to play with once I spotted the pattern.

### What can you learn from this solution?

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?