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katrina1232'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)
spiral = Matrix{Int}(undef, n, n)
seen = BitArray(undef, n, n)
dirlist = [[0, 1], [1, 0], [0, -1], [-1, 0]]
dir = 1
x = 1
y = 1
for i in 1:(n*n)
spiral[x, y] = i
seen[x, y] = true
x += dirlist[dir][1]
y += dirlist[dir][2]
if ((x > n || x < 1)||(y > n || y < 1)) || seen[x, y]
x -= dirlist[dir][1]
y -= dirlist[dir][2]
dir += 1
if dir == 5
dir = 1
end
x += dirlist[dir][1]
y += dirlist[dir][2]
end
end
return spiral
end``````