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to Spiral Matrix in the Julia Track

Published at Mar 18 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::Int, 
        spiral= nothing, 
        cur_n= nothing,
        count= 1
    )
    # we use recursion: create the matrix
    spiral==nothing && return spiral_matrix(n, Matrix{Int64}(zeros(n,n)),n,count)
    n==0 && return spiral
    
    # calculate the run for a single side
    # n: the size of the spiral
    # cur_n: the size of the current turn of the spiral
    offset= (n-cur_n)÷2
    from= 1+offset
    to=  n-offset-1
    
    # for uneven n we end with a turn of a single cell
    if cur_n == 1
        spiral[from,from]= count;
        return spiral
    end
    
    # the first side of the turn: left to right
    for xx= from:to
        spiral[from,xx]= count; count+=1
    end
      
    # second side: top bottom on the right
    for yy= from:to
        spiral[yy,to+1]= count; count+= 1
    end

    # third side: right to left on the bottom
    for xx= range(to+1,from+1,step= -1)
        spiral[to+1,xx]= count; count+= 1
    end
    
    # fourth side: bottom to top on the left
    for yy= range(to+1,from+1,step= -1)
        spiral[yy,from]= count; count+= 1
    end    

    # after a complete turn rerun the function for a turn that is 2 smaller than cur_n
    cur_n > 0 && spiral_matrix(n,spiral, cur_n-2, count)      
    
    # return the spiral
    spiral
end

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