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DzungTmd's solution

to Spiral Matrix in the Julia Track

Published at Mar 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
"""
Spiral matrix: The solution is motivated by the Robot-simulation exercise
and the turtle programming on Reddit.
Note: One can easily modify it to make a counter clockwise spiral matrix
"""

@enum Direction N=0 E=1 S=2 W=3
mutable struct Robot
    Pos
    Head::Direction
    function Robot(Pos,Head::Direction)
        new(Pos,E) #head East by default
    end
end
function turn_right!(r::Robot) 
    r.Head = Direction(mod(Int(r.Head)+1,4))
    return r
end
function turn_left!(r::Robot) 
    r.Head = Direction(mod(Int(r.Head)-1,4))
    return r
end
function advance!(r::Robot)
    (r.Head == N) && (r.Pos[1] = r.Pos[1]-1);
    (r.Head == E) && (r.Pos[2] = r.Pos[2]+1);
    (r.Head == S) && (r.Pos[1] = r.Pos[1]+1);
    (r.Head == W) && (r.Pos[2] = r.Pos[2]-1);
    return r
end
function retreat!(r::Robot)
    (r.Head == N) && (r.Pos[1] = r.Pos[1]+1);
    (r.Head == E) && (r.Pos[2] = r.Pos[2]-1);
    (r.Head == S) && (r.Pos[1] = r.Pos[1]-1);
    (r.Head == W) && (r.Pos[2] = r.Pos[2]+1);
    return r
end

function spiral_matrix(n)
    A = zeros(Int,n,n);
    R = Robot([1,1],E)
    current_val = 1;
    while (current_val <= n^2)
        A[R.Pos...] = current_val;
        advance!(R)
        if (max(R.Pos...)>n) || (min(R.Pos...)<1) || (A[R.Pos...] ≠ 0)
            retreat!(R) |> turn_right! |> advance!
        end
        current_val += 1
    end
    return A
end

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