Selecting random GPS points with minimum distance
I want to write a php program to select 16 random gps points from 400 points in my database
(point table: id - title - latitude - longitude).
lat 37.9824
lon -87.5781547
The only requirement is 16 random points, each at least 1 km apart (find points that are in the 1KM range)
it is a system that selects pharmacies with a minimum distance of 1 km between each pharmacy. I have 400 pharmacies in my database and I have to select 16 pharmacies every week. I cannot pick two pharmacies very close.
example:
if the program returns 3 pharmacies AB and C.
the cost between pharmacies should be:
A and B = 1 KM
A and C = 1 KM
B and C = 1 KM
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Try this because you only have 400 entries. It may only take a few hours ... Haven't tried it but it might give you an idea
$min =1;
$n =16;
$pharmas = fillUp();
// main function
function fillUp(){
$points = array();
while(count($points)< $n){
$tmp = getRandomPoint();
if(checkAll($tmp, $points){
$points[] = $tmp;
}
}
return $points; // after a few hours ??
}
// get a random point
// after all we might get lucky
function getRandomPoint(){
//...
// return array with ['latitude'] & ['longitude']
}
// check that all points meet the requirements
function checkAll($pt, $points){
foreach($points as $point){
if(distance($point, $pt) < $min {
return false;
}
}
return true;
}
// calculate the distance between 2 points
function distance ($point1, $point2, $uom='km') {
// Use Haversine formula to calculate the great circle distance
// between two points identified by longitude and latitude
switch (strtolower($uom)) {
case 'km' :
$earthMeanRadius = 6371.009; // km
break;
case 'm' :
$earthMeanRadius = 6371.009 * 1000; // km
break;
case 'miles' :
$earthMeanRadius = 3958.761; // miles
break;
case 'yards' :
case 'yds' :
$earthMeanRadius = 3958.761 * 1760; // miles
break;
case 'feet' :
case 'ft' :
$earthMeanRadius = 3958.761 * 1760 * 3; // miles
break;
case 'nm' :
$earthMeanRadius = 3440.069; // miles
break;
}
$deltaLatitude = deg2rad($point2['latitude'] - $point1['latitude']);
$deltaLongitude = deg2rad($point2['longitude'] - $point1['longitude']);
$a = sin($deltaLatitude / 2) * sin($deltaLatitude / 2) +
cos(deg2rad($point1['latitude'])) * cos(deg2rad($point2['latitude'])) *
sin($deltaLongitude / 2) * sin($deltaLongitude / 2);
$c = 2 * atan2(sqrt($a), sqrt(1 - $a));
$distance = $earthMeanRadius * $c;
return $distance;
}
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From the hat, answer here:
First, I create a view with a list of objects that are as close as possible to the point where you are interested in using the Cartesian distance formula, and then I use PHP code to calculate the actual spherical distance.
@MY_LAT = 37.9824;
@MY_LONG = -87.5781547;
SELECT *, SQRT(
ABS((latitude - @MY_LAT) * (latitude - @MY_LAT) +
(longitude - @MY_LONG) * (longitude - @MY_LONG)))
AS DIST
FROM POINT_TABLE
ORDER BY DIST ASC
Select the top n rows from this view to get the next 16 points from your point of interest. To check if the points are within 1KM of your reference point, you can write a small PHP snippet after getting the results. This should help you with the snippet:
http://www.zipcodeworld.com/samples/distance.php.html
Here I am using the Cartesian distance formula in the query, which only serves to reduce the number of records you get to apply the spherical distance formula in PHP.x
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