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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:

```
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)
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
```

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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## Community comments

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.

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.