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# nicholasvoltani's solution

## to Spiral Matrix in the Julia Track

Published at Mar 25 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)
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``````