The four parachutists

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The four parachutists
Three parachutists land at random positions within a large circular cornfield. Albert fears that Barbara hurt herself on landing, so he dumps his gear and runs straight to her; but it turns out that she is fine, so he retraces his steps. Then both Albert and Barbara walk straight to Charlie, who happens to have landed closest to the field gate.
"The farmer will be annoyed", says Charlie, "Look at the triangle you've trodden out in his corn!"
Just then, Dennis, the first parachutist of the second drop, floats down into the field. If he lands in a random position, what are the chances that he will be inside the triangle?
"The farmer will be annoyed", says Charlie, "Look at the triangle you've trodden out in his corn!"
Just then, Dennis, the first parachutist of the second drop, floats down into the field. If he lands in a random position, what are the chances that he will be inside the triangle?

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Assume the field is a circle with radius 1, centered at (0,0). The area is then pi.
Albert lands at (r1, theta1), Barbara lands at (r2, theta2), and Charlie lands at (r3, theta3), where r1,r2,r3 are uniformly distributed variables over [0,1] and theta1, theta2, theta3 are UDVs over [0,2pi).
Let x1 = sqrt(r1)*cos(theta1) and y1 = sqrt(r1)*sin(theta1); similarly for x2,y2,x3,y3.
At this point the only way I see to finish is to solve a hell of a lot of ugly integrals. Any hints on an easier way?
Albert lands at (r1, theta1), Barbara lands at (r2, theta2), and Charlie lands at (r3, theta3), where r1,r2,r3 are uniformly distributed variables over [0,1] and theta1, theta2, theta3 are UDVs over [0,2pi).
Let x1 = sqrt(r1)*cos(theta1) and y1 = sqrt(r1)*sin(theta1); similarly for x2,y2,x3,y3.
At this point the only way I see to finish is to solve a hell of a lot of ugly integrals. Any hints on an easier way?

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The method you suggest won't give an even distribution of points  they will tend to cluster near the centre of the circle. If you are going to use polar coordinates, you'll need a nonuniform distribution of the radii to account for the fact that the area increases in proportion to the radius squared.Cecil wrote:Assume the field is a circle with radius 1, centered at (0,0). The area is then pi.
Albert lands at (r1, theta1), Barbara lands at (r2, theta2), and Charlie lands at (r3, theta3), where r1,r2,r3 are uniformly distributed variables over [0,1] and theta1, theta2, theta3 are UDVs over [0,2pi).
Let x1 = sqrt(r1)*cos(theta1) and y1 = sqrt(r1)*sin(theta1); similarly for x2,y2,x3,y3.
At this point the only way I see to finish is to solve a hell of a lot of ugly integrals. Any hints on an easier way?
An alternative method is to use cartesian coordinates, and then the distributions on x and y are uniform. Of course, you have to deal with the boundary problem, as the field is circular, and not square. :)
Or you could do it the easy way. ;)

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Hmm. My math, trig and geometry are very rusty so please excuse me if I say something stupid. So maybe we can ignore the points themselves and say that the area of the triangle can itself vary randomly from zero to the maximum size of a triangle inscribed in the circle. If it's a linear distribution, or even a smooth function, that seems to give an answer. I can't take it further than that yet. Am I on the right track at all?
Last edited by Sundog on Wed Jul 21, 2004 7:14 pm, edited 1 time in total.

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Aye, that's why I used the sqrt(r) term above. I believe the formulae I gave for x1,y1 give an even distribution on a circle.ceptimus wrote:The method you suggest won't give an even distribution of points  they will tend to cluster near the centre of the circle. If you are going to use polar coordinates, you'll need a nonuniform distribution of the radii to account for the fact that the area increases in proportion to the radius squared.Cecil wrote:Assume the field is a circle with radius 1, centered at (0,0). The area is then pi.
Albert lands at (r1, theta1), Barbara lands at (r2, theta2), and Charlie lands at (r3, theta3), where r1,r2,r3 are uniformly distributed variables over [0,1] and theta1, theta2, theta3 are UDVs over [0,2pi).
Let x1 = sqrt(r1)*cos(theta1) and y1 = sqrt(r1)*sin(theta1); similarly for x2,y2,x3,y3.
At this point the only way I see to finish is to solve a hell of a lot of ugly integrals. Any hints on an easier way?
*slaps forehead*Or you could do it the easy way. ;)
Of course, the easy way. Why didn't I think of that? :wink:

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You should know by now I always use computer programs whenever they are the easiest way:DanishDynamite wrote:So, ceptimus, when are you going to reveal the easy way to solve this one?
Code: Select all
/* numerical simulation of 'The four parachutists' puzzle posted on skepticalcommunity.com
* generate a random triangle inside a circle. What is the probability that another random
* point inside the circle is inside the triangle?
*/
#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
#include <math.h>
#include <time.h>
bool insideTriangle
(
double x, double y,
double x1, double y1,
double x2, double y2,
double x3, double y3
)
{
double b;
b = (x2  x1) * (y3  y1)  (x3  x1) * (y2  y1);
if ((((x2  x) * (y3  y)  (x3  x) * (y2  y)) / b) <= 0.0)
return false;
if ((((x3  x) * (y1  y)  (x1  x) * (y3  y)) / b) <= 0.0)
return false;
if ((((x1  x) * (y2  y)  (x2  x) * (y1  y)) / b) <= 0.0)
return false;
return true;
}
void randomPointInUnitCircle(double* x, double* y)
{
do
{
*x = 2.0 * rand() / RAND_MAX  1.0;
*y = 2.0 * rand() / RAND_MAX  1.0;
} while (*x * *x + *y * *y > 1.0);
}
int get_key()
{
int key = 0;
if (kbhit())
if ((key = getch()) == 0)
key = getch();
return key;
}
int main(int argc, char* argv[])
{
unsigned long trials = 0;
unsigned long hits = 0;
double x1, y1, x2, y2, x3, y3, x, y;
int key;
srand((unsigned)time(NULL));
while (true)
{
randomPointInUnitCircle(&x1, &y1);
randomPointInUnitCircle(&x2, &y2);
randomPointInUnitCircle(&x3, &y3);
randomPointInUnitCircle(&x, &y);
trials++;
if (insideTriangle(x, y, x1, y1, x2, y2, x3, y3))
hits++;
if (!(trials % 100000))
{
printf("Trials:%9lu Hits:%7.3f%%\n", trials, 100.0 * hits / trials);
if ((key = get_key()) == 0X1B) // Esc
exit(0);
if (key == 'p') // pause
while (!get_key())
;
}
}
return 0;
}
Code: Select all
Trials:794700000 Hits: 7.387%
Trials:794800000 Hits: 7.387%
Trials:794900000 Hits: 7.387%
Trials:795000000 Hits: 7.387%
Trials:795100000 Hits: 7.387%
Trials:795200000 Hits: 7.387%
Trials:795300000 Hits: 7.387%
Trials:795400000 Hits: 7.387%
Trials:795500000 Hits: 7.387%
Trials:795600000 Hits: 7.387%
Trials:795700000 Hits: 7.387%
Trials:795800000 Hits: 7.387%
Trials:795900000 Hits: 7.387%
Trials:796000000 Hits: 7.387%
Trials:796100000 Hits: 7.387%
Trials:796200000 Hits: 7.387%
Trials:796300000 Hits: 7.387%
Trials:796400000 Hits: 7.387%
Trials:796500000 Hits: 7.387%
Trials:796600000 Hits: 7.387%
Trials:796700000 Hits: 7.387%

 Posts: 2608
 Joined: Mon Jun 07, 2004 4:58 pm
 Location: Copenhagen
Thanks, ceptimus.ceptimus wrote:You should know by now I always use computer programs whenever they are the easiest way:DanishDynamite wrote:So, ceptimus, when are you going to reveal the easy way to solve this one?After a few minutes run time (it's been going while I prepared this post) the output looks like:Code: Select all
/* numerical simulation of 'The four parachutists' puzzle posted on skepticalcommunity.com * generate a random triangle inside a circle. What is the probability that another random * point inside the circle is inside the triangle? */ #include <stdio.h> #include <stdlib.h> #include <conio.h> #include <math.h> #include <time.h> bool insideTriangle ( double x, double y, double x1, double y1, double x2, double y2, double x3, double y3 ) { double b; b = (x2  x1) * (y3  y1)  (x3  x1) * (y2  y1); if ((((x2  x) * (y3  y)  (x3  x) * (y2  y)) / b) <= 0.0) return false; if ((((x3  x) * (y1  y)  (x1  x) * (y3  y)) / b) <= 0.0) return false; if ((((x1  x) * (y2  y)  (x2  x) * (y1  y)) / b) <= 0.0) return false; return true; } void randomPointInUnitCircle(double* x, double* y) { do { *x = 2.0 * rand() / RAND_MAX  1.0; *y = 2.0 * rand() / RAND_MAX  1.0; } while (*x * *x + *y * *y > 1.0); } int get_key() { int key = 0; if (kbhit()) if ((key = getch()) == 0) key = getch(); return key; } int main(int argc, char* argv[]) { unsigned long trials = 0; unsigned long hits = 0; double x1, y1, x2, y2, x3, y3, x, y; int key; srand((unsigned)time(NULL)); while (true) { randomPointInUnitCircle(&x1, &y1); randomPointInUnitCircle(&x2, &y2); randomPointInUnitCircle(&x3, &y3); randomPointInUnitCircle(&x, &y); trials++; if (insideTriangle(x, y, x1, y1, x2, y2, x3, y3)) hits++; if (!(trials % 100000)) { printf("Trials:%9lu Hits:%7.3f%%\n", trials, 100.0 * hits / trials); if ((key = get_key()) == 0X1B) // Esc exit(0); if (key == 'p') // pause while (!get_key()) ; } } return 0; }
I compared this result with the one from the mathworld link you provided  that link suggests the answer is 35 / (48 * PI^2)  and I found that it agrees pretty nearly.Code: Select all
Trials:794700000 Hits: 7.387% Trials:794800000 Hits: 7.387% Trials:794900000 Hits: 7.387% Trials:795000000 Hits: 7.387% Trials:795100000 Hits: 7.387% Trials:795200000 Hits: 7.387% Trials:795300000 Hits: 7.387% Trials:795400000 Hits: 7.387% Trials:795500000 Hits: 7.387% Trials:795600000 Hits: 7.387% Trials:795700000 Hits: 7.387% Trials:795800000 Hits: 7.387% Trials:795900000 Hits: 7.387% Trials:796000000 Hits: 7.387% Trials:796100000 Hits: 7.387% Trials:796200000 Hits: 7.387% Trials:796300000 Hits: 7.387% Trials:796400000 Hits: 7.387% Trials:796500000 Hits: 7.387% Trials:796600000 Hits: 7.387% Trials:796700000 Hits: 7.387%
(You could have just said "I did a computer simulation". ) :)

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 Title: mere ghost of his former self
Re: The four parachutists
See, right away, you're off to a bad start since parachutists don't choose random landing spots. They choose to land where it will minimize the distance they have to walk back to the packing area.ceptimus wrote:Three parachutists land at random positions within a large circular cornfield. . . .
:shock: :D :shock: :D

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Re: The four parachutists
Goodness me, perhaps I have been too hard on you, you're the first person to step back and see what the real problems with the question are. I seem to remember that Mel Frank told me that there was some way to derive the value of pi from some random kind of thing like this, although I thought it was dropping a pin. It has, I fear, been many years since I've had that conversation, would anyone remember what I should be able to remember here?xouper wrote:See, right away, you're off to a bad start since parachutists don't choose random landing spots. They choose to land where it will minimize the distance they have to walk back to the packing area.ceptimus wrote:Three parachutists land at random positions within a large circular cornfield. . . .
:shock: :D :shock: :D
I'm not that much of a mathematics person. Although I did well enough in it, I greatly preferred rhetoric and natural languages.

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Re: The four parachutists
A quick google reveals this method:Skeeve wrote:Goodness me, perhaps I have been too hard on you, you're the first person to step back and see what the real problems with the question are. I seem to remember that Mel Frank told me that there was some way to derive the value of pi from some random kind of thing like this, although I thought it was dropping a pin. It has, I fear, been many years since I've had that conversation, would anyone remember what I should be able to remember here?xouper wrote:See, right away, you're off to a bad start since parachutists don't choose random landing spots. They choose to land where it will minimize the distance they have to walk back to the packing area.ceptimus wrote:Three parachutists land at random positions within a large circular cornfield. . . .
:shock: :D :shock: :D
I'm not that much of a mathematics person. Although I did well enough in it, I greatly preferred rhetoric and natural languages.
Sample random points in a square, and count those that fall within a circle in the square. The ratio is proportional to the ratio of area of circle to a square.
From this ratio, you can derive Pi.