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rogers132's solution

to Spiral Matrix in the Julia Track

Published at Feb 22 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
@enum Orientation begin
    right
    left
    up
    down
end

struct Direction
    orientaion
    delta_i
    delta_j
    next_orientaion
end

const directions = Dict(right => Direction(right, 0, 1, down),
                        down => Direction(down, 1, 0, left),
                        left => Direction(left, 0, -1, up),
                        up => Direction(up, -1, 0, right),)

function spiral_matrix(n)
    matrix = zeros(Int, (n, n))
    (i, j) = (1, 1)
    side_length = n
    d = directions[right]

    for k in 1:n * n
        matrix[i, j] = k
        final_index = d.delta_i + d.delta_j > 0 ? side_length : n + 1 - side_length
        current_index = d.delta_i != 0 ? i : j

        if current_index == final_index
            d = directions[d.next_orientaion]
            if d.orientaion == up
                side_length -= 1
            end
        end
        
        i += d.delta_i
        j += d.delta_j
    end
    
    return matrix
end

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