选择距离最小的随机GPS点[关闭]

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 16 random points, each of which is at least 1km from each other point ( find the points that are in 1KM range )

it is a system that selects pharmacies with a minimum distance of 1 km between each pharmacy. I have 400 pharmacies in the database and every week I must select 16 pharmacies. I can not select two pharmacies very close.

example :

if the program return 3 pharmacies A B and C .

the ditance between pharmacies must be :

A and B = 1 KM

A and C = 1 KM

B and C = 1 KM

Let's try that the hard way, since you have only 400 records, It might only take a few hours ... Havn'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;
}

An off the hat answer here:

I would first create a view with the list of objects that are as close to the point you are interested in using the Cartesian distance formula and then apply 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 closest 16 points from your 'point of interest'. To check if the points are within 1 KM of your point of reference you could write a small PHP snippet after you get the results. This should help you with the snippet:

http://www.zipcodeworld.com/samples/distance.php.html

Here, I am using a Cartesian distance formula in the query which only serves the purpose of reducing the number of records you get to apply the spherical distance formula in PHP.x