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to Spiral Matrix in the Julia Track

Published at Mar 28 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)
    if n == 0
        return Matrix{Int}(undef, 0, 0)
    elseif n == 1
        return reshape([1], (1, 1))
    end
    spiral_matrix = zeros(Int64, (n, n))
    for i in 1:n - 1
        spiral_matrix[1,i] = i
    end
    for i in 1:n - 1
        spiral_matrix[i,n] = i + (n - 1)
    end
    for i in 1:n - 1
        spiral_matrix[n,n + 1 - i] = i + 2 * (n - 1)
    end
    for i in 1:n - 1
        spiral_matrix[n + 1 - i,1] = i + 3 * (n - 1)
    end
    spiral_matrix[2:n - 1, 2:n - 1] = 4 * (n - 1) .+ recursive_spiral_matrix(n - 2)
    return spiral_matrix
end

What can you learn from this solution?

A huge amount can be learned from reading other people’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?
  • Are there new concepts here that you could read more about to improve your understanding?