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Published at Mar 22 2021
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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:

```
1 2 3
8 9 4
7 6 5
```

```
1 2 3 4
12 13 14 5
11 16 15 6
10 9 8 7
```

Reddit r/dailyprogrammer challenge #320 [Easy] Spiral Ascension. https://www.reddit.com/r/dailyprogrammer/comments/6i60lr/20170619_challenge_320_easy_spiral_ascension/

This exercise has been tested on Julia versions >=1.0.

It's possible to submit an incomplete solution so you can see how others have completed the exercise.

```
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...] â‰ 0)
retreat!(R) |> turn_right! |> advance!
end
current_val += 1
end
return A
end
```

A huge amount can be learned from reading other peopleâ€™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?

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