I am trying to implement the closest pair problem in Python using divide and conquer, everything seems to work fine except that in some input cases, there is a wrong answer. My code is as follows:
def closestSplitPair(Px,Py,d):
X = Px[len(Px)-1][0]
Sy = [item for item in Py if item[0]>=X-d and item[0]<=X+d]
best,p3,q3 = d,None,None
for i in xrange(0,len(Sy)-2):
for j in xrange(1,min(7,len(Sy)-1-i)):
if dist(Sy[i],Sy[i+j]) < best:
best = (Sy[i],Sy[i+j])
p3,q3 = Sy[i],Sy[i+j]
return (p3,q3,best)
I am calling the above function through a recursive function which is as follows:
def closestPair(Px,Py): """Px and Py are input arrays sorted according to
their x and y coordinates respectively"""
if len(Px) <= 3:
return min_dist(Px)
else:
mid = len(Px)/2
Qx = Px[:mid] ### x-sorted left side of P
Qy = Py[:mid] ### y-sorted left side of P
Rx = Px[mid:] ### x-sorted right side of P
Ry = Py[mid:] ### y-sorted right side of P
(p1,q1,d1) = closestPair(Qx,Qy)
(p2,q2,d2) = closestPair(Rx,Ry)
d = min(d1,d2)
(p3,q3,d3) = closestSplitPair(Px,Py,d)
return min((p1,q1,d1),(p2,q2,d2),(p3,q3,d3),key=lambda tup: tup[2])
where min_dist(P) is the brute force implementation of the closest pair algorithm for a list P having 3 or less elements and returns a tuple containing the pair of closest points and their distance.
If my input is P = [(0,0),(7,6),(2,20),(12,5),(16,16),(5,8),(19,7),(14,22),(8,19),(7,29),(10,11),(1,13)], then my output is ((5,8),(7,6),2.8284271) which is the correct output. But when my input is P = [(94, 5), (96, -79), (20, 73), (8, -50), (78, 2), (100, 63), (-14, -69), (99, -8), (-11, -7), (-78, -46)] the output I get is ((78, 2), (94, 5), 16.278820596099706) whereas the correct output should be ((94, 5), (99, -8), 13.92838827718412)
You have two problems, you are forgetting to call dist to update the best distance. But the main problem is there is more than one recursive call happening so you can end up overwriting when you find a closer split pair with the default, best,p3,q3 = d,None,None. I passed the best pair from closest_pair as an argument to closest_split_pair so I would not potentially overwrite the value.
def closest_split_pair(p_x, p_y, delta, best_pair): # <- a parameter
ln_x = len(p_x)
mx_x = p_x[ln_x // 2][0]
s_y = [x for x in p_y if mx_x - delta <= x[0] <= mx_x + delta]
best = delta
for i in range(len(s_y) - 1):
for j in range(1, min(i + 7, (len(s_y) - i))):
p, q = s_y[i], s_y[i + j]
dst = dist(p, q)
if dst < best:
best_pair = p, q
best = dst
return best_pair
The end of closest_pair looks like the following:
p_1, q_1 = closest_pair(srt_q_x, srt_q_y)
p_2, q_2 = closest_pair(srt_r_x, srt_r_y)
closest = min(dist(p_1, q_1), dist(p_2, q_2))
# get min of both and then pass that as an arg to closest_split_pair
mn = min((p_1, q_1), (p_2, q_2), key=lambda x: dist(x[0], x[1]))
p_3, q_3 = closest_split_pair(p_x, p_y, closest,mn)
# either return mn or we have a closer split pair
return min(mn, (p_3, q_3), key=lambda x: dist(x[0], x[1]))
You also have some other logic issues, your slicing logic is not correct, I made some changes to your code where brute is just a simple bruteforce double loop:
def closestPair(Px, Py):
if len(Px) <= 3:
return brute(Px)
mid = len(Px) / 2
# get left and right half of Px
q, r = Px[:mid], Px[mid:]
# sorted versions of q and r by their x and y coordinates
Qx, Qy = [x for x in q if Py and x[0] <= Px[-1][0]], [x for x in q if x[1] <= Py[-1][1]]
Rx, Ry = [x for x in r if Py and x[0] <= Px[-1][0]], [x for x in r if x[1] <= Py[-1][1]]
(p1, q1) = closestPair(Qx, Qy)
(p2, q2) = closestPair(Rx, Ry)
d = min(dist(p1, p2), dist(p2, q2))
mn = min((p1, q1), (p2, q2), key=lambda x: dist(x[0], x[1]))
(p3, q3) = closest_split_pair(Px, Py, d, mn)
return min(mn, (p3, q3), key=lambda x: dist(x[0], x[1]))
I just did the algorithm today so there are no doubt some improvements to be made but this will get you the correct answer.
Here is a recursive divide-and-conquer python implementation of the closest point problem based on the heap data structure. It also accounts for the negative integers. It can return the k-closest point by popping k nodes in the heap using heappop().
from __future__ import division
from collections import namedtuple
from random import randint
import math as m
import heapq as hq
def get_key(item):
return(item[0])
def closest_point_problem(points):
point = []
heap = []
pt = namedtuple('pt', 'x y')
for i in range(len(points)):
point.append(pt(points[i][0], points[i][1]))
point = sorted(point, key=get_key)
visited_index = []
find_min(0, len(point) - 1, point, heap, visited_index)
print(hq.heappop(heap))
def find_min(start, end, point, heap, visited_index):
if len(point[start:end + 1]) & 1:
mid = start + ((len(point[start:end + 1]) + 1) >> 1)
else:
mid = start + (len(point[start:end + 1]) >> 1)
if start in visited_index:
start = start + 1
if end in visited_index:
end = end - 1
if len(point[start:end + 1]) > 3:
if start < mid - 1:
distance1 = m.sqrt((point[start].x - point[start + 1].x) ** 2 + (point[start].y - point[start + 1].y) ** 2)
distance2 = m.sqrt((point[mid].x - point[mid - 1].x) ** 2 + (point[mid].y - point[mid - 1].y) ** 2)
if distance1 < distance2:
hq.heappush(heap, (distance1, ((point[start].x, point[start].y), (point[start + 1].x, point[start + 1].y))))
else:
hq.heappush(heap, (distance2, ((point[mid].x, point[mid].y), (point[mid - 1].x, point[mid - 1].y))))
visited_index.append(start)
visited_index.append(start + 1)
visited_index.append(mid)
visited_index.append(mid - 1)
find_min(start, mid, point, heap, visited_index)
if mid + 1 < end:
distance1 = m.sqrt((point[mid].x - point[mid + 1].x) ** 2 + (point[mid].y - point[mid + 1].y) ** 2)
distance2 = m.sqrt((point[end].x - point[end - 1].x) ** 2 + (point[end].y - point[end - 1].y) ** 2)
if distance1 < distance2:
hq.heappush(heap, (distance1, ((point[mid].x, point[mid].y), (point[mid + 1].x, point[mid + 1].y))))
else:
hq.heappush(heap, (distance2, ((point[end].x, point[end].y), (point[end - 1].x, point[end - 1].y))))
visited_index.append(end)
visited_index.append(end - 1)
visited_index.append(mid)
visited_index.append(mid + 1)
find_min(mid, end, point, heap, visited_index)
x = []
num_points = 10
for i in range(num_points):
x.append((randint(- num_points << 2, num_points << 2), randint(- num_points << 2, num_points << 2)))
closest_point_problem(x)
:)
Brute force can work faster with stdlib functions. Therefore, it can be effectively applied to more than 3 points.
from itertools import combinations
def closest(points_list):
return min((dist(p1, p2), p1, p2)
for p1, p2 in combinations(points_list, r=2))
The most effective way to divide the points is to divide them on tiles. If you don't have outliers, you can just split your space on equal parts and compare points only in the same or in the neighbour tiles.
Number of tiles must be as large as it possible. But, to avoid isolated tiles, when each point doesn't have points in neighbour tiles, you must limit number of tiles by the number of points.
Full listing:
from math import sqrt
from itertools import combinations, product
from collections import defaultdict
import sys
max_float = sys.float_info.max
def dist((x1, y1), (x2, y2)):
return sqrt((x1 - x2) ** 2 + (y1 - y2) **2)
def closest(points_list):
if len(points_list) < 2:
return (max_float, None, None) # default value compatible with min function
return min((dist(p1, p2), p1, p2)
for p1, p2 in combinations(points_list, r=2))
def closest_between(pnt_lst1, pnt_lst2):
if not pnt_lst1 or not pnt_lst2:
return (max_float, None, None) # default value compatible with min function
return min((dist(p1, p2), p1, p2)
for p1, p2 in product(pnt_lst1, pnt_lst2))
def divide_on_tiles(points_list):
side = int(sqrt(len(points_list))) # number of tiles on one side of square
tiles = defaultdict(list)
min_x = min(x for x, y in points_list)
max_x = max(x for x, y in points_list)
min_y = min(x for x, y in points_list)
max_y = max(x for x, y in points_list)
tile_x_size = float(max_x - min_x) / side
tile_y_size = float(max_y - min_y) / side
for x, y in points_list:
x_tile = int((x - min_x) / tile_x_size)
y_tile = int((y - min_y) / tile_y_size)
tiles[(x_tile, y_tile)].append((x, y))
return tiles
def closest_for_tile(tiles, (x_tile, y_tile)):
points = tiles[(x_tile, y_tile)]
return min(closest(points),
# use dict.get to avoid creating empty tiles
# we compare current tile only with half of neighbours (right/top),
# because another half (left/bottom) make it in another iteration by themselves
closest_between(points, tiles.get((x_tile+1, y_tile))),
closest_between(points, tiles.get((x_tile, y_tile+1))),
closest_between(points, tiles.get((x_tile+1, y_tile+1))),
closest_between(points, tiles.get((x_tile-1, y_tile+1))))
def find_closest_in_tiles(tiles):
return min(closest_for_tile(tiles, coord) for coord in tiles.keys())
P1 = [(0,0),(7,6),(2,20),(12,5),(16,16),(5,8),(19,7),(14,22),(8,19),(7,29),(10,11),(1,13)]
P2 = [(94, 5), (96, -79), (20, 73), (8, -50), (78, 2), (100, 63), (-14, -69), (99, -8), (-11, -7), (-78, -46)]
print find_closest_in_tiles(divide_on_tiles(P1)) # (2.8284271247461903, (7, 6), (5, 8))
print find_closest_in_tiles(divide_on_tiles(P2)) # (13.92838827718412, (94, 5), (99, -8))
print find_closest_in_tiles(divide_on_tiles(P1 + P2)) # (2.8284271247461903, (7, 6), (5, 8))
You just need to change the seventh line in your closestSplitPair function def from best=(Sy[i],Sy[i+j]) to best=dist(Sy[i],Sy[i+j]) and you will get the correct answer: ((94, 5), (99, -8), 13.92838827718412). You were missing the calling to the dist function.
This was pointed out by Padraic Cunningham's answer as the first problem.
Best Regards.
Related
I am try to find the angle of line
i know the coordinate points
Start Point : 404, 119
Mid Point : 279, 214
End Point : 154, 310
import numpy as np
def findangle(x1,y1,x2,y2,x3,y3):
ria = np.arctan2(y2 - y1, x2 - x1) - np.arctan2(y3 - y1, x3 - x1)
webangle = int(np.abs(ria * 180 / np.pi))
return webangle
result
Its return 270. But actual angle is 85-90.
Now, I want formula to calculate the angle (Either i Will rotate the image clockwise or Anticlockwise that time also return actual angle) in python code
I think you need to adjust the returned angle based on the range you wish the result to be in.
The following assumes you desire results between -180 (inclusive) and 180 (exclusive).
def adjust(a, degrees=True):
v = 180 if degrees else np.pi
return (a + v) % (2 * v) - v
def angle(p, degrees=True):
assert p.shape == (3, 2), 'p should have 3 points, in 2D'
yx = p[:, ::-1]
a0 = np.arctan2(*(yx[1] - yx[0]))
a1 = np.arctan2(*(yx[2] - yx[1]))
a = np.rad2deg(a1 - a0) if degrees else a1 - a0
return adjust(a, degrees)
And here is some code to test validity:
def rot(a, degrees=True):
a = np.deg2rad(a) if degrees else a
sa, ca = np.sin(a), np.cos(a)
return np.array([[ca, sa], [-sa, ca]])
def genp(angle, initial=0, lengths=(1, 1), degrees=True):
p0 = np.array((0, 0))
p1 = np.array((lengths[0], 0))
p2 = p1 + np.array((lengths[1], 0)) # rot(angle, degrees=degrees)
p = np.vstack((p0, p1, p2)) # rot(initial, degrees=degrees)
return p
def plot_problem(p, ax=None):
ax = plt.gca() if ax is None else ax
ax.plot(*p.T, '-o')
for s, xy in zip(list('abc'), p):
ax.annotate(s, xy, xytext=(0, 5), textcoords='offset points')
ax.set_aspect('equal')
Example:
p = genp(60, initial=30, lengths=(2, 1))
>>> angle(p)
60.0
plot_problem(p)
Now, a stress test:
n = 1000
np.random.seed(0)
for _ in range(n):
a = np.random.uniform(-180, 180)
initial = np.random.uniform(-180, 180)
lengths = np.random.uniform(1, 10, size=2)
p = genp(a, initial, lengths)
ahat = angle(p)
assert np.allclose(a, ahat)
Which passes.
It's a simple math. The numpy.arctan2() returns a value in the range [-π, π]. So a difference of two values is in the range [-2π, 2π], and the absolute value of that is in the range [0, 2π]. You want an angle x in the range [0, π], and x is equivalent to (2π - x) in your context. So you can take x if x < π, and take (2π - x) otherwise.
So, just do like this using min() for example.
def findangle(x1,y1,x2,y2,x3,y3):
...
return min(webangle, 360 - webangle)
As a side note, you don't need to use Numpy functions for a scalar input. Just use the math.atan2() and math.fabs().
I have a function that is intended to rotate polygons by 5 degrees left or right and return their new points. This function is as follows, along with the function player_center that it requires.
# finds center of player shape
# finds slope and midpoint of each vertice-midpoint line on the longer sides,
# then the intercept of them all
def player_center(player):
left_mid = line_midpoint(player[0], player[1])
right_mid = line_midpoint(player[0], player[2])
left_mid_slope = line_slope(left_mid, player[2])
right_mid_slope = line_slope(right_mid, player[1])
left_mid_line = find_equation(player[2], left_mid_slope, True)
right_mid_line = find_equation(player[1], right_mid_slope, True)
standard_left_mid_line = slope_intercept_to_standard(left_mid_line[0], left_mid_line[1], left_mid_line[2])
standard_right_mid_line = slope_intercept_to_standard(right_mid_line[0], right_mid_line[1], right_mid_line[2])
lines = sym.Matrix([standard_left_mid_line, standard_right_mid_line])
return (float(lines.rref()[0].row(0).col(2)[0]), float(lines.rref()[0].row(1).col(2)[0]))
# rotates the player using SOHCAHTOA
# divides x coordinate by radius to find angle, then adds or subtracts increment of 5 to it depending on direction
# calculates the position of point at incremented angle, then appends to new set of points
# finally, new set is returned
# direction; 1 is left, 0 is right
def rotate_player(player, direction):
increment = math.pi/36 # radian equivalent of 5 degrees
full_circle = 2 * math.pi # radian equivalent of 360 degrees
center = player_center(player)
new_player = []
for point in player:
radius = line_distance(point, center)
point_sin = (center[1] - point[1])/radius
while (point_sin > 1):
point_sin -= 1
point_angle = math.asin(point_sin)
if (direction == 1):
if ((point_angle+increment) > math.pi * 2):
new_angle = (point_angle+increment) - math.pi * 2
else:
new_angle = point_angle + increment
else:
if ((point_angle-increment) < 0):
new_angle = 2 * math.pi + (point_angle-increment)
else:
new_angle = point_angle-increment
print("The angle was {}".format(math.degrees(point_angle)))
print("The angle is now {}".format(math.degrees(new_angle))) # print lines are for debug purposes
new_point = ((radius * math.cos(new_angle)) + center[0], -(radius * math.sin(new_angle)) + center[1])
new_player.append(new_point)
print(new_player)
return new_player
The geometric functions that it relies on are all defined in this file here:
import math
import sympy as sym
# shape is in form of list of tuples e.g [(1,1), (2,1), (1,0), (2,0)]
# angle is in degrees
# def rotate_shape(shape, angle):
def line_distance(first_point, second_point):
return math.sqrt( (second_point[0] - first_point[0]) ** 2 + (second_point[1] - first_point[1]) ** 2)
# undefined is represented by None in this program
def line_slope(first_point, second_point):
if (second_point[0] - first_point[0] == 0):
return None
elif (second_point[1] - first_point[1] == 0):
return 0
else:
return (second_point[1] - first_point[1])/(second_point[0] - first_point[0])
def line_midpoint(first_point, second_point):
return ( (first_point[0] + second_point[0])/2, (first_point[1] + second_point[1])/2 )
def find_equation(coord_pair, slope, array_form):
# Array form useful for conversion into standard form
if (array_form == True):
if (slope == 0):
intercept = coord_pair[1]
return [0, 1, intercept]
elif (slope == None):
intercept = coord_pair[0]
return [1, 0, intercept]
else:
intercept = coord_pair[1] - (coord_pair[0] * slope)
return [slope, 1, intercept]
else:
if (slope == 0):
intercept = coord_pair[1]
print("y = {0}".format(intercept))
return
elif (slope == None):
intercept = coord_pair[0]
print("x = {0}".format(intercept))
return
else:
intercept = coord_pair[1] - (coord_pair[0] * slope)
if (intercept >= 0):
print("y = {0}x + {1}".format(slope, intercept))
return
else:
print("y = {0}x - {1}".format(slope, intercept))
def find_perpendicular(slope):
if (slope == 0):
return None
elif (slope == None):
return 0
else:
return -(1/slope)
def find_perp_bisector(first_point, second_point):
# This function finds the perpendicular bisector between two points, using funcs made previously
midpoint = line_midpoint(first_point, second_point)
slope = line_slope(first_point, second_point)
return find_equation(midpoint, -(1/slope))
def find_perp_equation(x, y, m, array_form):
# returns the perpendicular equation of a given line
if (array_form == True):
return [find_perpendicular(x), y, m]
else:
if (m >= 0):
print("{0}y = {1}x + {2}".format(y, find_perpendicular(x), m))
else:
print("{0}y = {1}x - {2}".format(y, find_perpendicular(x), m))
def find_hyp(a, b):
return math.sqrt((a**2) + (b**2))
def find_tri_area(a, b, c):
# finds area of triangle using Heron's formula
semi = (a+b+c)/2
return math.sqrt(semi * (semi - a) * (semi - b) * (semi - c) )
def r_tri_check(a, b, c):
if (a**2) + (b**2) != (c**2):
print("This thing fake, bro.")
def find_point_section(first_point, second_point, ratio):
# I coded this half a year ago and can't remember what for, but kept it here anyway.
# separtions aren't necessary, but good for code readability
first_numerator = (ratio[0] * second_point[0]) + (ratio[1] * first_point[0])
second_numerator = (ratio[0] * second_point[1]) + (ratio[1] * first_point[1])
return ( first_numerator/(ratio[0]+ratio[1]), second_numerator/(ratio[0] + ratio[1]))
def slope_intercept_to_standard(x, y, b):
# x and y are the coeffients of the said variables, for example 5y = 3x + 8 would be inputted as (3, 5, 8)
if (x == 0):
return [0, 1, b]
elif (y == 0):
return [x, 0, b]
else:
return [x, -y, -b]
It mathematically seems sound, but when I try to apply it, all hell breaks loose.
For example when trying to apply it with the set polygon_points equal to [(400, 300), (385, 340), (415, 340)], All hell breaks loose.
An example of the output among repeated calls to the function upon polygon_points(outputs manually spaced for clarity):
The angle was 90.0
The angle is now 95.0
The angle was -41.633539336570394
The angle is now -36.633539336570394
The angle was -41.63353933657017
The angle is now -36.63353933657017
The angle was 64.4439547804165
The angle is now 69.4439547804165
The angle was -64.44395478041695
The angle is now -59.44395478041695
The angle was -64.44395478041623
The angle is now -59.44395478041624
The angle was 80.94458887142648
The angle is now 85.94458887142648
The angle was -80.9445888714264
The angle is now -75.9445888714264
The angle was -80.94458887142665
The angle is now -75.94458887142665
Can anyone explain this?
Too much irrelevant code, a lot of magic like this while (point_sin > 1): point_sin -= 1 - so hard to reproduce.
To rotate point around some center, you need just this (where cos(fi), sin(fi) are precalculated value in your case):
new_x = center_x + (old_x - center_x) * cos(fi) - (old_y - center_y) * sin(fi)
new_y = center_y + (old_x - center_x) * sin(fi) + (old_y - center_y) * cos(fi)
This is a built-in capability of RegularPolygon in SymPy:
>>> from sympy import RegularPolygon, rad
>>> p = RegularPolygon((0,0), 1, 5)
>>> p.vertices[0]
Point2D(1, 0)
>>> p.rotate(rad(30)) # or rad(-30)
>>> p.vertices[0]
Point2D(sqrt(3)/2, 1/2)
I have a simple 2D ray-casting routine that gets terribly slow as soon as the number of obstacles increases.
This routine is made up of:
2 for loops (outer loop iterates over each ray/direction, then inner loop iterates over each line obstacle)
multiple if statements (check if a value is > or < than another value or if an array is empty)
Question: How can I condense all these operations into 1 single block of vectorized instructions using Numpy ?
More specifically, I am facing 2 issues:
I have managed to vectorize the inner loop (intersection between a ray and each obstacle) but I am unable to run this operation for all rays at once.
The only workaround I found to deal with the if statements is to use masked arrays. Something tells me it is not the proper way to handle these statements in this case (it seems clumsy, cumbersome and unpythonic)
Original code:
from math import radians, cos, sin
import matplotlib.pyplot as plt
import numpy as np
N = 10 # dimensions of canvas (NxN)
sides = np.array([[0, N, 0, 0], [0, N, N, N], [0, 0, 0, N], [N, N, 0, N]])
edges = np.random.rand(5, 4) * N # coordinates of 5 random segments (x1, x2, y1, y2)
edges = np.concatenate((edges, sides))
center = np.array([N/2, N/2]) # coordinates of center point
directions = np.array([(cos(radians(a)), sin(radians(a))) for a in range(0, 360, 10)]) # vectors pointing in all directions
intersections = []
# for each direction
for d in directions:
min_dist = float('inf')
# for each edge
for e in edges:
p1x, p1y = e[0], e[2]
p2x, p2y = e[1], e[3]
p3x, p3y = center
p4x, p4y = center + d
# find intersection point
den = (p1x - p2x) * (p3y - p4y) - (p1y - p2y) * (p3x - p4x)
if den:
t = ((p1x - p3x) * (p3y - p4y) - (p1y - p3y) * (p3x - p4x)) / den
u = -((p1x - p2x) * (p1y - p3y) - (p1y - p2y) * (p1x - p3x)) / den
# if any:
if t > 0 and t < 1 and u > 0:
sx = p1x + t * (p2x - p1x)
sy = p1y + t * (p2y - p1y)
isec = np.array([sx, sy])
dist = np.linalg.norm(isec-center)
# make sure to select the nearest one (from center)
if dist < min_dist:
min_dist = dist
nearest = isec
# store nearest interesection point for each ray
intersections.append(nearest)
# Render
plt.axis('off')
for x, y in zip(edges[:,:2], edges[:,2:]):
plt.plot(x, y)
for isec in np.array(intersections):
plt.plot((center[0], isec[0]), (center[1], isec[1]), '--', color="#aaaaaa", linewidth=.8)
Vectorized version (attempt):
from math import radians, cos, sin
import matplotlib.pyplot as plt
from scipy import spatial
import numpy as np
N = 10 # dimensions of canvas (NxN)
sides = np.array([[0, N, 0, 0], [0, N, N, N], [0, 0, 0, N], [N, N, 0, N]])
edges = np.random.rand(5, 4) * N # coordinates of 5 random segments (x1, x2, y1, y2)
edges = np.concatenate((edges, sides))
center = np.array([N/2, N/2]) # coordinates of center point
directions = np.array([(cos(radians(a)), sin(radians(a))) for a in range(0, 360, 10)]) # vectors pointing in all directions
intersections = []
# Render edges
plt.axis('off')
for x, y in zip(edges[:,:2], edges[:,2:]):
plt.plot(x, y)
# for each direction
for d in directions:
p1x, p1y = edges[:,0], edges[:,2]
p2x, p2y = edges[:,1], edges[:,3]
p3x, p3y = center
p4x, p4y = center + d
# denominator
den = (p1x - p2x) * (p3y - p4y) - (p1y - p2y) * (p3x - p4x)
# first 'if' statement -> if den > 0
mask = den > 0
den = den[mask]
p1x = p1x[mask]
p1y = p1y[mask]
p2x = p2x[mask]
p2y = p2y[mask]
t = ((p1x - p3x) * (p3y - p4y) - (p1y - p3y) * (p3x - p4x)) / den
u = -((p1x - p2x) * (p1y - p3y) - (p1y - p2y) * (p1x - p3x)) / den
# second 'if' statement -> if (t>0) & (t<1) & (u>0)
mask2 = (t > 0) & (t < 1) & (u > 0)
t = t[mask2]
p1x = p1x[mask2]
p1y = p1y[mask2]
p2x = p2x[mask2]
p2y = p2y[mask2]
# x, y coordinates of all intersection points in the current direction
sx = p1x + t * (p2x - p1x)
sy = p1y + t * (p2y - p1y)
pts = np.c_[sx, sy]
# if any:
if pts.size > 0:
# find nearest intersection point
tree = spatial.KDTree(pts)
nearest = pts[tree.query(center)[1]]
# Render
plt.plot((center[0], nearest[0]), (center[1], nearest[1]), '--', color="#aaaaaa", linewidth=.8)
Reformulation of the problem – Finding the intersection between a line segment and a line ray
Let q and q2 be the endpoints of a segment (obstacle). For convenience let's define a class to represent points and vectors in the plane. In addition to the usual operations, a vector multiplication is defined by u × v = u.x * v.y - u.y * v.x.
Caution: here Coord(2, 1) * 3 returns Coord(6, 3) while Coord(2, 1) * Coord(-1, 4) outputs 9. To avoid this confusion it might have been possible to restrict * to the scalar multiplication and use ^ via __xor__ for the vector multiplication.
class Coord:
def __init__(self, x, y):
self.x = x
self.y = y
#property
def radius(self):
return np.sqrt(self.x ** 2 + self.y ** 2)
def _cross_product(self, other):
assert isinstance(other, Coord)
return self.x * other.y - self.y * other.x
def __mul__(self, other):
if isinstance(other, Coord):
# 2D "cross"-product
return self._cross_product(other)
elif isinstance(other, int) or isinstance(other, float):
# scalar multiplication
return Coord(self.x * other, self.y * other)
def __rmul__(self, other):
return self * other
def __sub__(self, other):
return Coord(self.x - other.x, self.y - other.y)
def __add__(self, other):
return Coord(self.x + other.x, self.y + other.y)
def __repr__(self):
return f"Coord({self.x}, {self.y})"
Now, I find it easier to handle a ray in polar coordinates: For a given angle theta (direction) the goal is to determine if it intersects the segment, and if so determine the corresponding radius. Here is a function to find that. See here for an explanation of why and how. I tried to use the same variable names as in the previous link.
def find_intersect_btw_ray_and_sgmt(q, q2, theta):
"""
Args:
q (Coord): first endpoint of the segment
q2 (Coord): second endpoint of the segment
theta (float): angle of the ray
Returns:
(float): np.inf if the ray does not intersect the segment,
the distance from the origin of the intersection otherwise
"""
assert isinstance(q, Coord) and isinstance(q2, Coord)
s = q2 - q
r = Coord(np.cos(theta), np.sin(theta))
cross = r * s # 2d cross-product
t_num = q * s
u_num = q * r
## the intersection point is roughly at a distance t_num / cross
## from the origin. But some cases must be checked beforehand.
## (1) the segment [PQ2] is aligned with the ray
if np.isclose(cross, 0) and np.isclose(u_num, 0):
return min(q.radius, q2.radius)
## (2) the segment [PQ2] is parallel with the ray
elif np.isclose(cross, 0):
return np.inf
t, u = t_num / cross, u_num / cross
## There is actually an intersection point
if t >= 0 and 0 <= u <= 1:
return t
## (3) No intersection point
return np.inf
For instance find_intersect_btw_ray_and_sgmt(Coord(1, 2), Coord(-1, 2), np.pi / 2) should returns 2.
Note that here for simplicity, I only considered the case where the origin of the rays is at Coord(0, 0). This can be easily extended to the general case by setting t_num = (q - origin) * s and u_num = (q - origin) * r.
Let's vectorize it!
What is very interesting here is that the operations defined in the Coord class also apply to cases where x and y are numpy arrays! Hence applying any defined operation on Coord(np.array([1, 2, 0]), np.array([2, -1, 3])) amounts applying it elementwise to the points (1, 2), (2, -1) and (0, 3). The operations of Coord are therefore already vectorized. The constructor can be modified into:
def __init__(self, x, y):
x, y = np.array(x), np.array(y)
assert x.shape == y.shape
self.x, self.y = x, y
self.shape = x.shape
Now, we would like the function find_intersect_btw_ray_and_sgmt to be able to handle the case where the parameters q and q2contains sequences of endpoints. Before the sanity checks, all the operations are working properly since, as we have mentioned, they are already vectorized. As you mentionned the conditional statements can be "vectorized" using masks. Here is what I propose:
def find_intersect_btw_ray_and_sgmts(q, q2, theta):
assert isinstance(q, Coord) and isinstance(q2, Coord)
assert q.shape == q2.shape
EPS = 1e-14
s = q2 - q
r = Coord(np.cos(theta), np.sin(theta))
cross = r * s
cross_sign = np.sign(cross)
cross = cross * cross_sign
t_num = (q * s) * cross_sign
u_num = (q * r) * cross_sign
radii = np.zeros_like(t_num)
mask = ~np.isclose(cross, 0) & (t_num >= -EPS) & (-EPS <= u_num) & (u_num <= cross + EPS)
radii[~mask] = np.inf # no intersection
radii[mask] = t_num[mask] / cross[mask] # intersection
return radii
Note that cross, t_num and u_num are multiplied by the sign of cross to ensure that the division by cross keeps the sign of the dividends. Hence conditions of the form ((t_num >= 0) & (cross >= 0)) | ((t_num <= 0) & (cross <= 0)) can be replaced by (t_num >= 0).
For simplicity, we omitted the case (1) where the radius and the segment were aligned ((cross == 0) & (u_num == 0)). This could be incorporated by carefully adding a second mask.
For a given value of theta, we are able to determine if the corresponing ray intersects with several segments at once.
## Some useful functions
def polar_to_cartesian(r, theta):
return Coord(r * np.cos(theta), r * np.sin(theta))
def plot_segments(p, q, *args, **kwargs):
plt.plot([p.x, q.x], [p.y, q.y], *args, **kwargs)
def plot_rays(radii, thetas, *args, **kwargs):
endpoints = polar_to_cartesian(radii, thetas)
n = endpoints.shape
origin = Coord(np.zeros(n), np.zeros(n))
plot_segments(origin, endpoints, *args, **kwargs)
## Data generation
M = 5 # size of the canvas
N = 10 # number of segments
K = 16 # number of rays
q = Coord(*np.random.uniform(-M/2, M/2, size=(2, N)))
p = q + Coord(*np.random.uniform(-M/2, M/2, size=(2, N)))
thetas = np.linspace(0, 2 * np.pi, K, endpoint=False)
## For each ray, find the minimal distance of intersection
## with all segments
plt.figure(figsize=(5, 5))
plot_segments(p, q, "royalblue", marker=".")
for theta in thetas:
radii = find_intersect_btw_ray_and_sgmts(p, q, theta)
radius = np.min(radii)
if not np.isinf(radius):
plot_rays(radius, theta, color="orange")
else:
plot_rays(2*M, theta, ':', c='orange')
plt.plot(0, 0, 'kx')
plt.xlim(-M, M)
plt.ylim(-M, M)
And that's not all! Thanks to the broadcasting of python, it is possible to avoid iteration on theta values. For example, recall that np.array([1, 2, 3]) * np.array([[1], [2], [3], [4]]) produces a matrix of size 4 × 3 of the pairwise products. In the same way Coord([[5],[7]], [[5],[1]]) * Coord([2, 4, 6], [-2, 4, 0]) outputs a 2 × 3 matrix containing all the pairwise cross product between vectors (5, 5), (7, 1) and (2, -2), (4, 4), (6, 0).
Finally, the intersections can be determined in the following way:
radii_all = find_intersect_btw_ray_and_sgmts(p, q, np.vstack(thetas))
# p and q have a shape of (N,) and np.vstack(thetas) of (K, 1)
# this radii_all have a shape of (K, N)
# radii_all[k, n] contains the distance from the origin of the intersection
# between k-th ray and n-th segment (or np.inf if there is no intersection point)
radii = np.min(radii_all, axis=1)
# radii[k] contains the distance from the origin of the closest intersection
# between k-th ray and all segments
do_intersect = ~np.isinf(radii)
plot_rays(radii[do_intersect], thetas[do_intersect], color="orange")
plot_rays(2*M, thetas[~do_intersect], ":", color="orange")
how to calculate the distance between multiple points in a list using a loop.
def create_list(x_range,y_range, locations):
generated_list = []
for x in range(locations):
x_range = random.randint(-300,300)
y_range = random.randint(-300,300)
generated_list.append([x_range, y_range])
return generated_list
above creates a random list and I need to calculate the total distance for all points returning back to the beginning using this code:
def calculate_distance(starting_x, starting_y, destination_x, destination_y):
distance = math.hypot(destination_x - starting_x, destination_y - starting_y) # calculates Euclidean distance (straight-line) distance between two points
return distance
here I need to calculate the distance between all points using a loop with the function above, how would I use a loop to calculate the distance between all points
Scipy can generate the matrix of distances without using a loop.
https://stackoverflow.com/a/19475995
Since the matrix is symmetrical around 0 we can divide the sum by two to get the one-way distances between points.
from scipy.spatial.distance import cdist
values = create_list(300, 300, 3)
distances = cdist(values, values)
print distances.sum() / 2
From a list of pairs (x, y) you may
use itertools.permutations to get the permutations inside this list. Here is an example with 3 points, you got 6 permutations by specifying a length of 2
values = [(2, 1), (0, 3), (3, 2)]
perm = [((2, 1), (0, 3)), ((2, 1), (3, 2)), ((0, 3), (2, 1)),
((0, 3), (3, 2)), ((3, 2), (2, 1)), ((3, 2), (0, 3))]
call your calculate_distance with the points
p1 = (1,2)
p2 = (2,3)
dist = calculate_distance(*p1, *p2) #or calculate_distance(p1[0], p1[1], p2[0], p2[1])
Use all of this
values = create_list(300, 300, 3)
for p1, p2 in permutations(values, r=2):
dist = calculate_distance(*p1, *p2)
print(p1, p2, dist)
To sum them all add a value to do total+=dist in the loop, or use sum that accepts an iterable and shorten in
values = create_list(300, 300, 3)
total = sum(calculate_distance(*p1, *p2) for p1, p2 in permutations(values, r=2))
Using loops
Sum consecutive points only
# Way 1 : d(a,b) + d(b,c) + d(c,d) + d(d,e) + d(e,a)
total = calculate_distance(*values[-1], *values[0]) # between last and first
for i in range(len(values) - 1):
total += calculate_distance(*values[i], *values[i + 1])
print(total)
Sum all combinations of points
# Way 2: d(a,b) + d(a,c) + d(a,d) + d(a,e) + d(b,c) + d(b,d) + d(b,e) + d(c,d) + d(c,e) + d(d,e)
total = 0
for p1 in values:
for p2 in values:
if p1 != p2:
total += calculate_distance(*p1, *p2)
print(total)
import math
def calculate_distance(x1, y1, x2, y2):
distance = math.hypot(x2 - x1, y2 - y1)
return distance
def total_dist(ml): # ml is a list of pairs of x,y points
n = len(ml)
total = 0.0
for i in range (n):
x1 = ml[i][0]
y1 = ml[i][1]
if i < n-1:
x2 = ml[i+1][0]
y2 = ml[i+1][1]
else:
x2 = ml[0][0]
y2 = ml[0][1]
dist = calculate_distance(x1,y1,x2,y2)
total += dist
return total
# main program
copy_list = [[110, -260], [284, -152], [-197, -167], [-253, 295], [-89, 26], [-165, 43], [-84, 148], [-188, 38], [201, 191], [232, 251]]
print total_dist(copy_list)
You are confronted with an enemy within a rectangular shaped room and you've got only a laser beam weapon, the room has no obstructions in it and the walls can completely reflect the laser beam. However the laser can only travels a certain distance before it become useless and if it hit a corner it would reflect back in the same direction it came from.
That's how the puzzle goes and you are given the coordinates of your location and the target's location, the room dimensions and the maximum distance the beam can travel. for example If the room is 3 by 2 and your location is (1, 1) and the target is (2, 1) then the possible solutions are:
I tried the following approach, start from the source (1, 1) and create a vector at angle 0 radians, trace the vector path and reflections until either it hits the target or the total length of the vectors exceeds the max allowed length, repeat with 0.001 radians interval until it completes a full cycle. This the code I have so far:
from math import *
UPRIGHT = 0
DOWNRIGHT = 1
DOWNLEFT = 2
UPLEFT = 3
UP = 4
RIGHT = 5
LEFT = 6
DOWN = 7
def roundDistance (a):
b = round (a * 100000)
return b / 100000.0
# only used for presenting and doesn't affect percision
def double (a):
b = round (a * 100)
if b / 100.0 == b: return int (b)
return b / 100.0
def roundAngle (a):
b = round (a * 1000)
return b / 1000.0
def isValid (point):
x,y = point
if x < 0 or x > width or y < 0 or y > height: return False
return True
def isCorner (point):
if point in corners: return True
return False
# Find the angle direction in relation to the origin (observer) point
def getDirection (a):
angle = roundAngle (a)
if angle == 0: return RIGHT
if angle > 0 and angle < pi / 2: return UPRIGHT
if angle == pi / 2: return UP
if angle > pi / 2 and angle < pi: return UPLEFT
if angle == pi: return LEFT
if angle > pi and angle < 3 * pi / 2: return DOWNLEFT
if angle == 3 * pi / 2: return DOWN
return DOWNRIGHT
# Measure reflected vector angle
def getReflectionAngle (tail, head):
v1 = (head[0] - tail[0], head[1] - tail[1])
vx,vy = v1
n = (0, 0)
# Determin the normal vector from the tail's position on the borders
if head[0] == 0: n = (1, 0)
if head[0] == width: n = (-1, 0)
if head[1] == 0: n = (0, 1)
if head[1] == height: n = (0, -1)
nx,ny = n
# Calculate the reflection vector using the formula:
# r = v - 2(v.n)n
r = (vx * (1 - 2 * nx * nx), vy * (1 - 2 * ny * ny))
# calculating the angle of the reflection vector using it's a and b values
# if b (adjacent) is zero that means the angle is either pi/2 or -pi/2
if r[0] == 0:
return pi / 2 if r[1] >= 0 else 3 * pi / 2
return (atan2 (r[1], r[0]) + (2 * pi)) % (2 * pi)
# Find the intersection point between the vector and borders
def getIntersection (tail, angle):
if angle < 0:
print "Negative angle: %f" % angle
direction = getDirection (angle)
if direction in [UP, RIGHT, LEFT, DOWN]: return None
borderX, borderY = corners[direction]
x0,y0 = tail
opp = borderY - tail[1]
adj = borderX - tail[0]
p1 = (x0 + opp / tan (angle), borderY)
p2 = (borderX, y0 + adj * tan (angle))
if isValid (p1) and isValid (p2):
print "Both intersections are valid: ", p1, p2
if isValid (p1) and p1 != tail: return p1
if isValid (p2) and p2 != tail: return p2
return None
# Check if the vector pass through the target point
def isHit (tail, head):
d = calcDistance (tail, head)
d1 = calcDistance (target, head)
d2 = calcDistance (target, tail)
return roundDistance (d) == roundDistance (d1 + d2)
# Measure distance between two points
def calcDistance (p1, p2):
x1,y1 = p1
x2,y2 = p2
return ((y2 - y1)**2 + (x2 - x1)**2)**0.5
# Trace the vector path and reflections and check if it can hit the target
def rayTrace (point, angle):
path = []
length = 0
tail = point
path.append ([tail, round (degrees (angle))])
while length < maxLength:
head = getIntersection (tail, angle)
if head is None:
#print "Direct reflection at angle (%d)" % angle
return None
length += calcDistance (tail, head)
if isHit (tail, head) and length <= maxLength:
path.append ([target])
return [path, double (length)]
if isCorner (head):
#print "Corner reflection at (%d, %d)" % (head[0], head[1])
return None
p = (double (head[0]), double (head[1]))
path.append ([p, double (degrees (angle))])
angle = getReflectionAngle (tail, head)
tail = head
return None
def solve (w, h, po, pt, m):
# Initialize global variables
global width, height, origin, target, maxLength, corners, borders
width = w
height = h
origin = po
target = pt
maxLength = m
corners = [(w, h), (w, 0), (0, 0), (0, h)]
angle = 0
solutions = []
# Loop in anti-clockwise direction for one cycle
while angle < 2 * pi:
angle += 0.001
path = rayTrace (origin, angle)
if path is not None:
# extract only the points coordinates
route = [x[0] for x in path[0]]
if route not in solutions:
solutions.append (route)
print path
# Anser is 7
solve (3, 2, (1, 1), (2, 1), 4)
# Answer is 9
#solve (300, 275, (150, 150), (185, 100), 500)
The code works somehow but it doesn't find all the possible solutions, I have a big precision problem in it, I dont' know how many decimals should I consider when comparing distances or angles. I'm not sure it's the right way to do it but that's the best I was able to do.
How can I fix my code to extract all solutions? I need it to be efficient because the room can get quite large (500 x 500). Is there a better way or maybe some sort of algorithm to do this?
what if you started by mirroring the target at all the walls; then mirror the mirror images at all the walls and so on until the distance gets too big for the laser to reach the target? any laser shot in any direction of a target mirrored that way will hit said target. (this is my comment from above; repeated here to make answer more self-contained...)
this is the mirroring part of the answer: get_mirrored will return the four mirror images of point with the mirror-box limited by BOTTOM_LEFT and TOP_RIGHT.
BOTTOM_LEFT = (0, 0)
TOP_RIGHT = (3, 2)
SOURCE = (1, 1)
TARGET = (2, 1)
def get_mirrored(point):
ret = []
# mirror at top wall
ret.append((point[0], point[1] - 2*(point[1] - TOP_RIGHT[1])))
# mirror at bottom wall
ret.append((point[0], point[1] - 2*(point[1] - BOTTOM_LEFT[1])))
# mirror at left wall
ret.append((point[0] - 2*(point[0] - BOTTOM_LEFT[0]), point[1]))
# mirror at right wall
ret.append((point[0] - 2*(point[0] - TOP_RIGHT[0]), point[1]))
return ret
print(get_mirrored(TARGET))
this will return the 4 mirror images of the given point:
[(2, 3), (2, -1), (-2, 1), (4, 1)]
which is the target mirrored one time.
then you could iterate that until all the mirrored targets are out of range. all the mirror images within range will give you a direction in which to point your laser.
this is a way how you could iteratively get to the mirrored targets within a given DISTANCE
def get_targets(start_point, distance):
all_targets = set((start_point, )) # will also be the return value
last_targets = all_targets # need to memorize the new points
while True:
new_level_targets = set() # if this is empty: break the loop
for tgt in last_targets: # loop over what the last iteration found
new_targets = get_mirrored(tgt)
# only keep the ones within range
new_targets = set(
t for t in new_targets
if math.hypot(SOURCE[0]-t[0], SOURCE[1]-t[1]) <= DISTANCE)
# subtract the ones we already have
new_targets -= all_targets
new_level_targets |= new_targets
if not new_level_targets:
break
# add the new targets
all_targets |= new_level_targets
last_targets = new_level_targets # need these for the next iteration
return all_targets
DISTANCE = 5
all_targets = get_targets(start_point=TARGET, distance=DISTANCE)
print(all_targets)
all_targets is now the set that contains all the reachable points.
(none of that has been thouroughly tested...)
small update for your counter example:
def ray_length(point_list):
d = sum((math.hypot(start[0]-end[0], start[1]-end[1])
for start, end in zip(point_list, point_list[1:])))
return d
d = ray_length(point_list=((1,1),(2.5,2),(3,1.67),(2,1)))
print(d) # -> 3.605560890844135
d = ray_length(point_list=((1,1),(4,3)))
print(d) # -> 3.605551275463989