Posted on Oct 23, 2019

Daily Challenge #98 - Make a Spiral

#challenge

Your task is to create an NxN spiral with a given size.
For example, a spiral with size 5 should look like this:

00000 ....0 000.0 0...0 00000

and with the size 10:

0000000000 .........0 00000000.0 0......0.0 0.0000.0.0 0.0..0.0.0 0.0....0.0 0.000000.0 0........0 0000000000

Return value should contain array of arrays, of 0 and 1, for example for given size 5 result should be:

[[1,1,1,1,1],[0,0,0,0,1],[1,1,1,0,1],[1,0,0,0,1],[1,1,1,1,1]]

Because of the edge-cases for tiny spirals, the size will be at least 5.

General rule-of-a-thumb is, that the snake made with '1' cannot touch to itself.

This challenge comes from Bugari on CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge idea for a future post? Email yo+challenge@dev.to with your suggestions!

Top comments (6)

erezwanderman • Oct 23 '19 • Edited

Javascript:

const spiral = n => { return Array(n) .fill(Array(n).fill()) .map((x, r) => x.map((y, c) => { return +( // top (r < n / 2 && r % 2 === 0 && c >= r - 2 && c <= n - r - 1) || // right ((n - c) % 2 === 1 && r > n - c - 1 && r <= c) || // bottom ((n - r) % 2 === 1 && c > n - r - 1 && c < r) || // left (c % 2 === 0 && r > c + 1 && r < n - c) ); }) ); };

See it in action: codesandbox.io/s/hopeful-nash-wvvc...

erezwanderman • Oct 23 '19

Now I'm working on generalizing it to non square matrix

erezwanderman • Oct 23 '19

const spiral = (rowNum, colNum) => { return Array(rowNum) .fill(Array(colNum).fill()) .map((x, r) => x.map((_, c) => { return +( // top (r < rowNum / 2 && r % 2 === 0 && r - 2 <= c && c <= colNum - r - 1) || // right (c > (colNum - 2) / 2 && (colNum - c) % 2 === 1 && colNum - c - 2 < r && r <= rowNum - (colNum - c)) || // bottom (r > rowNum / 2 && (rowNum - r) % 2 === 1 && rowNum - r - 2 < c && c < colNum - (rowNum - r)) || // left (c < (colNum - 2) / 2 && c % 2 === 0 && r > c + 1 && r < rowNum - c) ); }) ); };

Ed Reeseg • Oct 23 '19 • Edited

Definitely not as efficient as iterating through each index once and deciding on a 1 or a 0 arithmetically, but I wanted to try solving this one in a different way to see how it worked out. I'd probably go back and refactor this a bit, given the choice, but it seems to work well enough.

If we had a way of initializing the array values at 0, this method could actually be more efficient, as it wouldn't necessitate visiting every single index, just those touched by the spiral.

C

#include <stdio.h> #include <stdlib.h> #include <math.h>  int main(int argc, char **argv) { int size = atoi(argv[1]); if (argc != 2 || size < 5) { printf("Usage: ./make-a-spiral.c int\n"); return 1; } int result[size][size]; int x = 0, y = 0, dir = 1, horizontal = 1, boundary_x = 0, boundary_y = 0, turns = 0; // Iterate through our result array and set the default to 0, // as otherwise we will end up with garbage values polluting the result. for (int i = 0; i < size; i++) { for (int j = 0; j < size; j++) { result[i][j] = 0; } } // Change our x and y location as is appropriate, and check when we  // need to turn. while (turns < size) { result[y][x] = 1; if (y == 0 + boundary_y && !horizontal && dir == -1) { horizontal = 1; dir *= -1; turns++; } else if (y == size - boundary_y - 1 && !horizontal && dir == 1) { horizontal = 1; boundary_y += 2; dir *= -1; turns++; } else if (x == size - boundary_x - 1 && horizontal && dir == 1) { horizontal = 0; turns++; } else if (y != 0 && x == 0 + boundary_x && horizontal && dir == -1) { horizontal = 0; boundary_x += 2; turns++; } // Handle movement if (horizontal) { x += dir; } else { y += dir; } } // Iterate through once more to print the resulting spiral. for (int i = 0; i < size; i++) { for (int j = 0; j < size; j++) { printf("%c", result[i][j] ? '0' : '.'); } printf("\n"); } return 0; }

Michael Tharrington • Oct 23 '19

Aschwin Wesselius • Oct 24 '19

Sprial or spiral?

The title says sprial, the article says spiral.