Divide a rectangle into a number of unequal triangles: Difference between revisions

* Explain, or refer to the algorithm employed.

* Give a sample of sets of triangles produced from running the algorithm, on this page.

 

=={{header|Python}}==

It is assumed that the rectangle has its bottom left coordinate at (0,0) and is not rotated in the coordinate space, meaning that the location of the top-right corner of the rectangle is then enough to define it.

 

import random

from typing import List, Union, Tuple

D C

 

 

If we instead inserted **two** points between A-B, p0 and p1, we can insert **one** point q0, along D-C

 

We think of the L-to-R ordered top points as A p0 p1 then B; and the ordered l-to-R bottom points as D q0 then C.

* Ensure the distances between successive top points, B-C, and successive bottom points are all different to get different triangles.

* If you insert `n`top points p, then you need `n-1` bottom points q.

print("\nrect_into_top_tri #2")

pp(rect_into_top_tri((2, 1), 4, 10))

 

{{out}}

((0, 0), (1.2, 1), (2, 1)),

((0, 0), (2, 1), (2, 0))]</pre>

 

=={{header|Wren}}==

Assuming the rectangle is to be split into 'n' triangles where n >= 3, we first bisect the rectangle into two triangles, add the top one to the result list and then divide the bottom one into (n-1) triangles. We do this by choosing (n - 2) points at random on the bottom side of the rectangle which together with the two bottom vertices are such that, after sorting, the difference between successive pairs is never the same. We then link each pair of points with the upper right vertex of the rectangle to form the requisite number of triangles.

 

import "./seq" for Lst

 

var rand = Random.new()

var pointsOfRect = Fn.new { |w, h| [[0, 0], [h, 0], [h, w], [0, w]] }

 

var divideRectIntoTris = Fn.new { |w, h, n|

System.print("A rectangle with a lower left vertex at (0, 0), width %(w) and height %(h)")

System.print("can be split into the following %(n) triangles:")

System.print()

 

{{out}}