3 ms·

I tried to script that together after the same idea occurred to me. What's interesting is the selection plane affects the outcome quite heavily. As the plane of

I tried to script that together after the same idea occurred to me. What's interesting is the selection plane affects the outcome quite heavily. As the plane of random line selection increases in size, the probability of a random chord having length>sqrt3 seems to approach about 0.13 My math might be wrong; doesn't seem to be a floating point precision issue. from math import sqrt from random import random def randomPoints(width): return [ [random()*width-width/2, random()*width-width/2], [random()*width-width/2, random()*width-width/2] ] def chordCalc(points): def quadraticEquation(a,b,c): descriminant=(b**2 - 4*a*c) if descriminant <= 0: return None x1 = (-b+sqrt(descriminant))/(2*a) x2 = (-b-sqrt(descriminant))/(2*a) y1 = sqrt(1-x1**2) y2 = sqrt(1-x2**2) return [[x1, y1],[x2,y2]] m = (points[0][1] - points[1][1]) / (points[0][0] - points[1][0]) b = points[0][1] - m*points[0][0] qa = 1+m**2 qb = 2*b*m qc = b**2-1 return quadraticEquation(qa, qb, qc) def distance(points): return sqrt((points[0][1] - points[1][1])**2 + (points[0][0] - points[1][0])**2) chords=0 largeChords=0 sqrt3=sqrt(3) while True: if random() > 0.9999: print float(largeChords)/chords points = randomPoints(500) #adjust me down to 0.5 and see probability go to ~0.33 intersections=chordCalc(points) if intersections is None: continue chords += 1 if distance(intersections) > sqrt3: largeChords += 1

sampo 11y ago
You might have a bug in the code. >>> horizontal = chordCalc([[-1,0],[1,0]]) >>> distance(horizontal) 2.0 All good so far, but >>> diagonal = chordCalc([[-1,-1],[1,1]]) >>> distance(diagonal) 1.4142135623730951 shouldn't this give 2.0 as well?
dstyrb 11y ago
I was wrong. My initial approach appears to be the ninja approach like sampo said above. I got 50% -_- import numpy as np import random as rnd import math as m chords = [] samp = 10000 for c in range(samp): #random point on infinite plane is described by a single coordinate dist = rnd.uniform(1., 1000.) #DO NOT KNOW HOW TO DO INTERNAL SLICES #ignore all possible lines that do not intesect circle xsec_half = m.atan(1./dist) angle = rnd.uniform(-xsec_half, xsec_half) #make 2 points on the randomly chosen line x1, y1 = -dist, 0 x2, y2 = dist, 2*dist*m.tan(angle) #copy and paste some shit from wolfram alpha dx = x2-x1 dy = y2-y1 dr = (dx**2 + dy**2)**0.5 D = x1*y2 - x2*y1 sqr = (dr**2 - D**2)**0.5 xa = (D*dy + np.sign(dy)*dx*sqr) / (dr**2) xb = (D*dy - np.sign(dy)*dx*sqr) / (dr**2) ya = (-D*dx + np.abs(dy)*sqr) / (dr**2) yb = (-D*dx - np.abs(dy)*sqr) / (dr**2) chords.append(((xa-xb)**2 + (ya-yb)**2)**0.5) print len([a for a in chords if a > 3**0.5])/float(samp) Basically make the unit circle at 0,0. Pick any radius up to infinity (infinite plane). By symmetry, we don't need 2 coordinates, just a distance from the circle. Choose any angle from that point which will draw a vector that intersects the circle (to not waste CPU). Calculate the length of the intersecting chord. I was really surprised by the distribution of chord lengths, it peaks at chord lengths of 2.0, which is the maximum chord length, and falls exponentially with smaller chord lengths. I don't know what I expected, but not that.

sampo 11y ago

You might have a bug in the code. >>> horizontal = chordCalc([[-1,0],[1,0]]) >>> distance(horizontal) 2.0 All good so far, but >>> diagonal = chordCalc([[-1,-1],[1,1]]) >>> distance(diagonal) 1.4142135623730951 shouldn't this give 2.0 as well?

dstyrb 11y ago

I was wrong. My initial approach appears to be the ninja approach like sampo said above. I got 50% -_- import numpy as np import random as rnd import math as m chords = [] samp = 10000 for c in range(samp): #random point on infinite plane is described by a single coordinate dist = rnd.uniform(1., 1000.) #DO NOT KNOW HOW TO DO INTERNAL SLICES #ignore all possible lines that do not intesect circle xsec_half = m.atan(1./dist) angle = rnd.uniform(-xsec_half, xsec_half) #make 2 points on the randomly chosen line x1, y1 = -dist, 0 x2, y2 = dist, 2*dist*m.tan(angle) #copy and paste some shit from wolfram alpha dx = x2-x1 dy = y2-y1 dr = (dx**2 + dy**2)**0.5 D = x1*y2 - x2*y1 sqr = (dr**2 - D**2)**0.5 xa = (D*dy + np.sign(dy)*dx*sqr) / (dr**2) xb = (D*dy - np.sign(dy)*dx*sqr) / (dr**2) ya = (-D*dx + np.abs(dy)*sqr) / (dr**2) yb = (-D*dx - np.abs(dy)*sqr) / (dr**2) chords.append(((xa-xb)**2 + (ya-yb)**2)**0.5) print len([a for a in chords if a > 3**0.5])/float(samp) Basically make the unit circle at 0,0. Pick any radius up to infinity (infinite plane). By symmetry, we don't need 2 coordinates, just a distance from the circle. Choose any angle from that point which will draw a vector that intersects the circle (to not waste CPU). Calculate the length of the intersecting chord. I was really surprised by the distribution of chord lengths, it peaks at chord lengths of 2.0, which is the maximum chord length, and falls exponentially with smaller chord lengths. I don't know what I expected, but not that.