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Published at Mar 25 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)
function anyzeros(mat)
return any(i->i==0, mat)
end
if n == 0
## ????
return Matrix{Int}(undef, 0, 0)
end
mat = zeros(Int,n,n)
## First row: adds n elements
mat[1,:] = 1:n
up = 1
down = n
left = 1
right = n
while anyzeros(mat)
## Always consider lastvalue to be the upper-rightmost, then do next loop to next upper-rightmost
lastvalue = mat[up, right]
## Going down the rows
mat[up+1:down, right] = lastvalue + 1: lastvalue + (down-up)
lastvalue += (down-up)
## Going left
mat[down, left:right-1] = lastvalue + (right-left):-1:lastvalue + 1
lastvalue += (right-left) ## == (down-up) in symmetric matrices
## Going up the rows
mat[up+1:down-1, left] = lastvalue + (down-up-1):-1:lastvalue + 1
lastvalue += (down-up) - 1
## Going right
mat[up+1, left+1:right-1] = lastvalue + 1: lastvalue + (right-left-2) + 1
lastvalue += (right-left) - 1
## Update for next loop, to make it spiral inward
up += 1
down -= 1
left += 1
right -= 1
end
return mat
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?
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