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gitblight1's solution

to Spiral Matrix in the Julia Track

Published at Mar 18 2021 · 2 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)
    m = Matrix{Int}(undef,n,n)
    n == 0 && return m
    m[1,:] = collect(1:n)
    m[2:n,n] = collect(n+1:2*n-1)
    # rest of matrix is an n-1 square, rotated and starting at 2n
    m[2:n, 1:n-1] = rot180(spiral_matrix(n-1)) .+ (2*n-1)
    return m
end

Community comments

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Avatar of hurak

I could not resist to comment on this solution, even if I am not asking a question: smart! I have never really developed an eye for recursive functions and here comes a perfect opportunity to apply them. Exploiting rot180 is also neat.

Avatar of gitblight1

Thanks! This does have the issue of running out of stack size if you try to compute a huge spiral matrix (My box can handle spiral_matrix(2000), but not 3000), but it was fun to play with once I spotted the pattern.

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