Consider the following code which is a solution to the "circle stabbing problem" of computational geometry, namely finding the line that intersects a maximal number of non-intersecting circles.
The main question here is why the Python code is 42x slower than the C++ code. Is it because some improper use of Python? Or is it due to Python being inherently this slower than C++ at arithmetic and math operations? Is there anything that can be done to make it a bit faster?
First, the Python code:
from __future__ import division
from math import sqrt, atan2, pi
from sys import stdin
__author__ = "Sahand Saba"
EPS = 1e-6
class Point(object):
__slots__ = ['x', 'y']
def __init__(self, xx=0.0, yy=0.0):
self.x = float(xx)
self.y = float(yy)
def angle(self):
"""
Returns the angle the point makes with the x-axis,
counter-clockwise. Result is in the [0, 2*pi) range.
"""
a = atan2(self.y, self.x)
if a < 0:
a = 2 * pi + a
return a
class Circle(object):
__slots__ = ['center', 'radius']
def __init__(self, center, radius):
self.center = center
self.radius = float(radius)
def common_tangents(self, other):
"""
Returns [[p1,p2],[q1,q2]] with p1, p2, q1, q2 all Point objects
representing points on C1 such that the tangent lines to C1 at p1, p2,
q1, q2 are tangent to C2 as well. Further more, p1 and p2 represent
external tangent lines, while q1 and q2 represent internal ones. It is
also guaranteed that p1 and q1 are both on the left-side of the line
connecting C1.center to C2.center, and p2 and q2 are on the right-side
of it.
"""
C1, C2 = self, other
mu = C1.center.x - C2.center.x
eta = C1.center.y - C2.center.y
r1 = C1.radius
r2 = C2.radius
r1r1 = r1 * r1
r1r2 = r1 * r2
delta1 = r1r1 - r1r2
delta2 = r1r1 + r1r2
mumu = mu * mu
etaeta = eta * eta
D = etaeta + mumu
result = [[], []]
if abs(D) < EPS:
return result
if abs(eta) < EPS:
# In this case there is symmetry along the x-axis and we can
# not divide by eta. Use x^2 + y^2 = r^2 to find y.
dmu = -1 if mu < 0 else 1
x = (-delta1 * mu) / D
y = -dmu * sqrt(r1r1 - x * x)
result[0].append(Point(x, y))
result[0].append(Point(x, -y))
x = (-delta2 * mu) / D
y = -dmu * sqrt(r1r1 - x * x)
result[1].append(Point(x, y))
result[1].append(Point(x, -y))
else:
# Here, the symmetry is along the line connecting the two centers.
# Use the equation eta*y + mu *x + r1^2 - r1 * r2 = 0 to derive y
# since we can divide by eta.
dd1 = delta1 * delta1
dd2 = delta2 * delta2
Delta1 = sqrt(dd1 * mumu - D*(dd1 - etaeta * r1r1))
Delta2 = sqrt(dd2 * mumu - D*(dd2 - etaeta * r1r1))
deta = -1 if eta < 0 else 1
x = (-delta1 * mu + deta * Delta1) / D
result[0].append(Point(x, -(mu*x + delta1)/eta))
x = (-delta1 * mu - deta * Delta1) / D
result[0].append(Point(x, -(mu*x + delta1)/eta))
x = (-delta2 * mu + deta * Delta2) / D
result[1].append(Point(x, -(mu*x + delta2)/eta))
x = (-delta2 * mu - deta * Delta2) / D
result[1].append(Point(x, -(mu*x + delta2)/eta))
return result
def add_events(A, p, q):
start = p.angle()
end = q.angle()
A.append((start, 1, p))
A.append((end, -1, q))
return 1 if start > end else 0
def max_intersecting_line(C):
"""
Given a list of circles, returns (m, c, p) where m is the maximal number of
circles in C any line can intersect, and p is a point on a circle c in C
such that the tangent line to c at p intersects m circles in C.
"""
global_max = 1
solution_circle = C[0]
solution_point = Point(C[0].radius, 0.0)
for c1 in C:
local_max = 1
A = []
for c2 in (c for c in C if c is not c1):
Q = c1.common_tangents(c2)
t1 = add_events(A, Q[1][0], Q[0][0])
t2 = add_events(A, Q[0][1], Q[1][1])
local_max += max(t1, t2)
if local_max > global_max:
global_max = local_max
solution_point = Point(c1.radius, 0.0)
solution_circle = c1
A.sort(key=lambda a: a[0])
for a in A:
local_max += a[1]
if local_max > global_max:
global_max = local_max
solution_point = Point(c1.center.x + a[2].x,
c1.center.y + a[2].y)
solution_circle = c1
return global_max, solution_circle, solution_point
if __name__ == '__main__':
T = int(stdin.readline())
for __ in xrange(T):
n = int(stdin.readline())
C = []
for i in xrange(n):
x, y, r = tuple(stdin.readline().split(' '))
C.append(Circle(Point(x, y), r))
print max_intersecting_line(C)[0]
And the almost line-for-line equivalent C++ code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
double EPS = 1e-6;
class Point {
public:
double x, y;
Point(double x=0.0, double y=0.0) : x(x), y(y) {}
double angle() const {
double a = atan2(y, x);
if (a < 0) {
a = atan(1) * 8.0 + a;
}
return a;
}
};
class Event {
public:
double angle;
double count;
Event(double angle = 0, int count = 1) : angle(angle), count(count) {}
bool operator<(const Event &o) const {
return angle < o.angle;
}
};
struct CircleCircleTangents {
public:
Point external[2];
Point internal[2];
};
class Circle {
public:
Point center;
double radius;
Circle(double x=0.0, double y=0.0, double r=1.0) : radius(r), center(x,y) {}
// external[0] and internal[0] are guaranteed to be on the left-side of
// the directed line contennting C1.center to C2.center
CircleCircleTangents commonTangents(const Circle& C2) const {
const Circle& C1 = *this;
double mu = C1.center.x - C2.center.x;
double eta = C1.center.y - C2.center.y;
double r1 = C1.radius;
double r2 = C2.radius;
double r1r1 = r1 * r1;
double r1r2 = r1 * r2;
double delta1 = r1r1 - r1r2;
double delta2 = r1r1 + r1r2;
double D = eta*eta + mu*mu;
CircleCircleTangents result;
if (abs(eta) < EPS){
// Do not divide by eta! Use x^2 + y^2 = r^2 to find y.
double dmu = mu < 0? -1 : 1;
double x = (-delta1 * mu) / D;
double y = -dmu * sqrt(r1r1 - x * x);
result.external[0].x = x;
result.external[0].y = y;
result.external[1].x = x;
result.external[1].y = -y;
x = (-delta2 * mu) / D;
y = -dmu * sqrt(r1r1 - x * x);
result.internal[0].x = x;
result.internal[0].y = y;
result.internal[1].x = x;
result.internal[1].y = -y;
} else {
// Dividing by eta is ok. Use mu*x + eta*y + delta = 0 to find y.
double mumu = mu * mu;
double etaeta = eta * eta;
double dd1 = delta1 * delta1;
double dd2 = delta2 * delta2;
double deta = eta < 0? -1 : 1;
double Delta1 = deta * sqrt(dd1 * mumu - D*(dd1 - etaeta * r1r1));
double Delta2 = deta * sqrt(dd2 * mumu - D*(dd2 - etaeta * r1r1));
double x = (-delta1 * mu + Delta1) / D;
result.external[0].x = x;
result.external[0].y = -(mu*x + delta1)/eta;
x = (-delta1 * mu - Delta1) / D;
result.external[1].x = x;
result.external[1].y = -(mu*x + delta1)/eta;
x = (-delta2 * mu + Delta2) / D;
result.internal[0].x = x;
result.internal[0].y = -(mu*x + delta2)/eta;
x = (-delta2 * mu - Delta2) / D;
result.internal[1].x = x;
result.internal[1].y = -(mu*x + delta2)/eta;
}
return result;
}
};
bool add_events(vector<Event>& A, const Point& p, const Point& q) {
double start = p.angle();
double end = q.angle();
A.push_back(Event(start, 1));
A.push_back(Event(end, -1));
return start > end;
}
// Given a list of circles, returns (m, c, p) where m is the maximal number of
// circles in C any line can intersect, and p is a point on a circle c in C
// such that the tangent line to c at p intersects m circles in C.
int max_intersecting_line(const Circle* C, int n) {
int global_max = 1;
vector<Event> A;
for(int i = 0; i < n; i++) {
const Circle& c1 = C[i];
A.clear();
int local_max = 1;
for(int j = 0; j < n; j++) {
if(j == i) continue;
const Circle& c2 = C[j];
CircleCircleTangents Q = c1.commonTangents(c2);
bool t1 = add_events(A, Q.internal[0], Q.external[0]);
bool t2 = add_events(A, Q.external[1], Q.internal[1]);
if(t1 || t2) {
local_max++;
}
}
if (local_max > global_max) {
global_max = local_max;
}
sort(A.begin(), A.end());
for(int i = 0; i < A.size(); i++) {
local_max += A[i].count;
if(local_max > global_max) {
global_max = local_max;
}
}
}
return global_max;
}
int main() {
Circle C[2000];
int T;
cin >> T;
for (int t = 0; t < T; t++) {
int n;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> C[i].center.x >> C[i].center.y >> C[i].radius;
}
cout << max_intersecting_line(C, n) << endl;
}
return 0;
}
And their performance difference:
$ time ./janeway < io/Janeway.in > /dev/null
real 0m8.436s
user 0m8.430s
sys 0m0.003s
$ time python janeway.py < io/Janeway.in > /dev/null
real 5m57.899s
user 5m57.217s
sys 0m0.165s
As you can see, the C++ code is about 42 times faster.
(Test input is taken from ACM ICPC regional's problem from 2013. See http://www.acmicpc-pacnw.org/results.htm problem "Janeway" of 2013.)
EDIT: Here is the cProfile output for the Python code:
799780565 function calls in 507.293 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 0.000 0.000 __future__.py:48(<module>)
1 0.000 0.000 0.000 0.000 __future__.py:74(_Feature)
7 0.000 0.000 0.000 0.000 __future__.py:75(__init__)
1 0.047 0.047 507.293 507.293 janeway.py:1(<module>)
25 63.703 2.548 507.207 20.288 janeway.py:103(max_intersecting_line)
1 0.000 0.000 0.000 0.000 janeway.py:11(Point)
24250014 5.671 0.000 5.671 0.000 janeway.py:116(<genexpr>)
96926032 9.343 0.000 9.343 0.000 janeway.py:127(<lambda>)
96955733 57.902 0.000 57.902 0.000 janeway.py:14(__init__)
96926032 46.840 0.000 63.156 0.000 janeway.py:18(angle)
1 0.000 0.000 0.000 0.000 janeway.py:29(Circle)
18506 0.009 0.000 0.009 0.000 janeway.py:32(__init__)
24231508 167.128 0.000 245.945 0.000 janeway.py:36(common_tangents)
48463016 59.402 0.000 129.139 0.000 janeway.py:95(add_events)
48463016 4.106 0.000 4.106 0.000 {abs}
96926032 16.315 0.000 16.315 0.000 {math.atan2}
48463016 4.908 0.000 4.908 0.000 {math.sqrt}
24231508 9.483 0.000 9.483 0.000 {max}
193870570 18.503 0.000 18.503 0.000 {method 'append' of 'list' objects}
1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}
18532 0.009 0.000 0.009 0.000 {method 'readline' of 'file' objects}
18506 43.918 0.002 53.261 0.003 {method 'sort' of 'list' objects}
18506 0.007 0.000 0.007 0.000 {method 'split' of 'str' objects}
