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pedrobtz's solution

to Spiral Matrix in the Julia Track

Published at Apr 07 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)
    local x = CartesianIndex{2}[]
    local left = top = 1
    local right = bottom = n
    local y = zeros(Int,n,n)
    
    while left <= right && top <= bottom
        for i in left:right
            push!(x, CartesianIndex(top,i) )
        end
        top+=1
        for i in top:bottom
            push!(x, CartesianIndex(i,right) )
        end
        right-=1
        for i in right:-1:left
            push!(x, CartesianIndex(bottom,i) )
        end
        bottom-=1
        for i in bottom:-1:top
            push!(x, CartesianIndex(i,left) )
        end
        left+=1
    end
    y[x] = 1:(n*n)
    return y
end

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