I develop code performing Schur decomposition.
I test it with eigen corresponding stuff.
I found out that my code gives result different form those of eigen.
Most of matrix elements of my and egein's output are the same to 2 or 4 decimal places.
However worst difference available is about 70%: for example my code gives certain matrix element equal to 0.3 and eigen 0.19.
I decided to look deeper into eigen sources and figured out that if I change following eigen code
void ::applyHouseholderOnTheLeft(....)
{
......
Map<typename internal::plain_row_type<PlainObject>::type> tmp(workspace,cols());
Block<Derived, EssentialPart::SizeAtCompileTime, Derived::ColsAtCompileTime> bottom(derived(), 1, 0, rows()-1, cols());
tmp.noalias() = essential.adjoint() * bottom;
tmp += this->row(0);
this->row(0) -= tau * tmp;
bottom.noalias() -= tau * essential * tmp;
}
to this one (same was done for applyHouseholderOnTheRight):
void ::applyHouseholderOnTheLeft(....)
{
......
Map<typename internal::plain_row_type<PlainObject>::type> tmp(workspace,cols());
Block<Derived, EssentialPart::SizeAtCompileTime, Derived::ColsAtCompileTime> bottom(derived(), 1, 0, rows()-1, cols());
tmp.noalias() = essential.adjoint() * bottom;
tmp += this->row(0);
tmp *= tau;
this->row(0) -= tmp;
bottom.noalias() -= essential * tmp;
}
i get eigen output equal to mine (within 7-6 decimal places) !!
Mathematically these two pieces of code are equivalent.
So the question is - why there is so big difference in outputs of equivalent code ?
And what result is actually true (0.3 or 0.19 :-) ) ?
Original test code:
Matrix<double, Dynamic, Dynamic, RowMajor> A(10,10);
A<<6.9 ,4.8 ,9.5 ,3.1 ,6.5 ,5.8 ,-0.9 ,-7.3 ,-8.1 ,3.0 ,0.1 ,9.9 ,-3.2 ,6.4 ,6.2 ,-7.0 ,5.5 ,-2.2 ,-4.0 ,3.7 ,-3.6 ,9.0 ,-1.4 ,-2.4 ,1.7 ,-6.1 ,-4.2 , -2.5 ,-5.6 ,-0.4 ,0.4 ,9.1 ,-2.1 ,-5.4 ,7.3 ,3.6 ,-1.7 ,-5.7 ,-8.0 ,8.8 ,-3.0 ,-0.5 ,1.1 ,10.0 ,8.0 ,0.8 ,1.0 ,7.5 ,3.5 ,-1.8 ,0.3 ,-0.6 ,-6.3 ,-4.5 , -1.1 ,1.8 ,0.6 ,9.6 ,9.2 ,9.7 ,-2.6 ,4.3 ,-3.4 ,0.0 ,-6.7 ,5.0 ,10.5 ,1.5 ,-7.8 ,-4.1 ,-5.3 ,-5.0 ,2.0 ,-4.4 ,-8.4 ,6.0 ,-9.4 ,-4.8 ,8.2 ,7.8 ,5.2 ,-9.5 , -3.9 ,0.2 ,6.8 ,5.7 ,-8.5 ,-1.9 ,-0.3 ,7.4 ,-8.7 ,7.2 ,1.3 ,6.3 ,-3.7 ,3.9 ,3.3 ,-6.0 ,-9.1 ,5.9;
RealSchur<Matrix<double, Dynamic, Dynamic, RowMajor>>schur(A);
Matrix<double, Dynamic, Dynamic, RowMajor> T = schur.matrixT();
// and ,for example, element T(row_0, col_2) has notable difference: 0.19 (first code), 0.3 (second code)
P.S. vectorization in eigen is disabled in my case (macros EIGEN_DONT_VECTORIZE is defined)
Let's say I plotted the position of a helicopter every day for the past year and came up with the following map:
Any human looking at this would be able to tell me that this helicopter is based out of Chicago.
How can I find the same result in code?
I'm looking for something like this:
$geoCodeArray = array([GET=http://pastebin.com/grVsbgL9]);
function findHome($geoCodeArray) {
// magic
return $geoCode;
}
Ultimately generating something like this:
UPDATE: Sample Dataset
Here's a map with a sample dataset: http://batchgeo.com/map/c3676fe29985f00e1605cd4f86920179
Here's a pastebin of 150 geocodes: http://pastebin.com/grVsbgL9
The above contains 150 geocodes. The first 50 are in a few clusters close to Chicago. The remaining are scattered throughout the country, including some small clusters in New York, Los Angeles, and San Francisco.
I have about a million (seriously) datasets like this that I'll need to iterate through and identify the most likely "home". Your help is greatly appreciated.
UPDATE 2: Airplane switched to Helicopter
The airplane concept was drawing too much attention toward physical airports. The coordinates can be anywhere in the world, not just airports. Let's assume it's a super helicopter not bound by physics, fuel, or anything else. It can land where it wants. ;)
The following solution works even if the points are scattered all over the Earth, by converting latitude and longitude to Cartesian coordinates. It does a kind of KDE (kernel density estimation), but in a first pass the sum of kernels is evaluated only at the data points. The kernel should be chosen to fit the problem. In the code below it is what I could jokingly/presumptuously call a Trossian, i.e., 2-d²/h² for d≤h and h²/d² for d>h (where d is the Euclidean distance and h is the "bandwidth" $global_kernel_radius), but it could also be a Gaussian (e-d²/2h²), an Epanechnikov kernel (1-d²/h² for d<h, 0 otherwise), or another kernel. An optional second pass refines the search locally, either by summing an independent kernel on a local grid, or by calculating the centroid, in both cases in a surrounding defined by $local_grid_radius.
In essence, each point sums all the points it has around (including itself), weighing them more if they are closer (by the bell curve), and also weighing them by the optional weight array $w_arr. The winner is the point with the maximum sum. Once the winner has been found, the "home" we are looking for can be found by repeating the same process locally around the winner (using another bell curve), or it can be estimated to be the "center of mass" of all points within a given radius from the winner, where the radius can be zero.
The algorithm must be adapted to the problem by choosing the appropriate kernels, by choosing how to refine the search locally, and by tuning the parameters. For the example dataset, the Trossian kernel for the first pass and the Epanechnikov kernel for the second pass, with all 3 radii set to 30 mi and a grid step of 1 mi could be a good starting point, but only if the two sub-clusters of Chicago should be seen as one big cluster. Otherwise smaller radii must be chosen.
function find_home($lat_arr, $lng_arr, $global_kernel_radius,
$local_kernel_radius,
$local_grid_radius, // 0 for no 2nd pass
$local_grid_step, // 0 for centroid
$units='mi',
$w_arr=null)
{
// for lat,lng <-> x,y,z see http://en.wikipedia.org/wiki/Geodetic_datum
// for K and h see http://en.wikipedia.org/wiki/Kernel_density_estimation
switch (strtolower($units)) {
/* */case 'nm' :
/*or*/case 'nmi': $m_divisor = 1852;
break;case 'mi': $m_divisor = 1609.344;
break;case 'km': $m_divisor = 1000;
break;case 'm': $m_divisor = 1;
break;default: return false;
}
$a = 6378137 / $m_divisor; // Earth semi-major axis (WGS84)
$e2 = 6.69437999014E-3; // First eccentricity squared (WGS84)
$lat_lng_count = count($lat_arr);
if ( !$w_arr) {
$w_arr = array_fill(0, $lat_lng_count, 1.0);
}
$x_arr = array();
$y_arr = array();
$z_arr = array();
$rad = M_PI / 180;
$one_e2 = 1 - $e2;
for ($i = 0; $i < $lat_lng_count; $i++) {
$lat = $lat_arr[$i];
$lng = $lng_arr[$i];
$sin_lat = sin($lat * $rad);
$sin_lng = sin($lng * $rad);
$cos_lat = cos($lat * $rad);
$cos_lng = cos($lng * $rad);
// height = 0 (!)
$N = $a / sqrt(1 - $e2 * $sin_lat * $sin_lat);
$x_arr[$i] = $N * $cos_lat * $cos_lng;
$y_arr[$i] = $N * $cos_lat * $sin_lng;
$z_arr[$i] = $N * $one_e2 * $sin_lat;
}
$h = $global_kernel_radius;
$h2 = $h * $h;
$max_K_sum = -1;
$max_K_sum_idx = -1;
for ($i = 0; $i < $lat_lng_count; $i++) {
$xi = $x_arr[$i];
$yi = $y_arr[$i];
$zi = $z_arr[$i];
$K_sum = 0;
for ($j = 0; $j < $lat_lng_count; $j++) {
$dx = $xi - $x_arr[$j];
$dy = $yi - $y_arr[$j];
$dz = $zi - $z_arr[$j];
$d2 = $dx * $dx + $dy * $dy + $dz * $dz;
$K_sum += $w_arr[$j] * ($d2 <= $h2 ? (2 - $d2 / $h2) : $h2 / $d2); // Trossian ;-)
// $K_sum += $w_arr[$j] * exp(-0.5 * $d2 / $h2); // Gaussian
}
if ($max_K_sum < $K_sum) {
$max_K_sum = $K_sum;
$max_K_sum_i = $i;
}
}
$winner_x = $x_arr [$max_K_sum_i];
$winner_y = $y_arr [$max_K_sum_i];
$winner_z = $z_arr [$max_K_sum_i];
$winner_lat = $lat_arr[$max_K_sum_i];
$winner_lng = $lng_arr[$max_K_sum_i];
$sin_winner_lat = sin($winner_lat * $rad);
$cos_winner_lat = cos($winner_lat * $rad);
$sin_winner_lng = sin($winner_lng * $rad);
$cos_winner_lng = cos($winner_lng * $rad);
$east_x = -$local_grid_step * $sin_winner_lng;
$east_y = $local_grid_step * $cos_winner_lng;
$east_z = 0;
$north_x = -$local_grid_step * $sin_winner_lat * $cos_winner_lng;
$north_y = -$local_grid_step * $sin_winner_lat * $sin_winner_lng;
$north_z = $local_grid_step * $cos_winner_lat;
if ($local_grid_radius > 0 && $local_grid_step > 0) {
$r = intval($local_grid_radius / $local_grid_step);
$r2 = $r * $r;
$h = $local_kernel_radius;
$h2 = $h * $h;
$max_L_sum = -1;
$max_L_sum_idx = -1;
for ($i = -$r; $i <= $r; $i++) {
$winner_east_x = $winner_x + $i * $east_x;
$winner_east_y = $winner_y + $i * $east_y;
$winner_east_z = $winner_z + $i * $east_z;
$j_max = intval(sqrt($r2 - $i * $i));
for ($j = -$j_max; $j <= $j_max; $j++) {
$x = $winner_east_x + $j * $north_x;
$y = $winner_east_y + $j * $north_y;
$z = $winner_east_z + $j * $north_z;
$L_sum = 0;
for ($k = 0; $k < $lat_lng_count; $k++) {
$dx = $x - $x_arr[$k];
$dy = $y - $y_arr[$k];
$dz = $z - $z_arr[$k];
$d2 = $dx * $dx + $dy * $dy + $dz * $dz;
if ($d2 < $h2) {
$L_sum += $w_arr[$k] * ($h2 - $d2); // Epanechnikov
}
}
if ($max_L_sum < $L_sum) {
$max_L_sum = $L_sum;
$max_L_sum_i = $i;
$max_L_sum_j = $j;
}
}
}
$x = $winner_x + $max_L_sum_i * $east_x + $max_L_sum_j * $north_x;
$y = $winner_y + $max_L_sum_i * $east_y + $max_L_sum_j * $north_y;
$z = $winner_z + $max_L_sum_i * $east_z + $max_L_sum_j * $north_z;
} else if ($local_grid_radius > 0) {
$r = $local_grid_radius;
$r2 = $r * $r;
$wx_sum = 0;
$wy_sum = 0;
$wz_sum = 0;
$w_sum = 0;
for ($k = 0; $k < $lat_lng_count; $k++) {
$xk = $x_arr[$k];
$yk = $y_arr[$k];
$zk = $z_arr[$k];
$dx = $winner_x - $xk;
$dy = $winner_y - $yk;
$dz = $winner_z - $zk;
$d2 = $dx * $dx + $dy * $dy + $dz * $dz;
if ($d2 <= $r2) {
$wk = $w_arr[$k];
$wx_sum += $wk * $xk;
$wy_sum += $wk * $yk;
$wz_sum += $wk * $zk;
$w_sum += $wk;
}
}
$x = $wx_sum / $w_sum;
$y = $wy_sum / $w_sum;
$z = $wz_sum / $w_sum;
$max_L_sum_i = false;
$max_L_sum_j = false;
} else {
return array($winner_lat, $winner_lng, $max_K_sum_i, false, false);
}
$deg = 180 / M_PI;
$a2 = $a * $a;
$e4 = $e2 * $e2;
$p = sqrt($x * $x + $y * $y);
$zeta = (1 - $e2) * $z * $z / $a2;
$rho = ($p * $p / $a2 + $zeta - $e4) / 6;
$rho3 = $rho * $rho * $rho;
$s = $e4 * $zeta * $p * $p / (4 * $a2);
$t = pow($s + $rho3 + sqrt($s * ($s + 2 * $rho3)), 1 / 3);
$u = $rho + $t + $rho * $rho / $t;
$v = sqrt($u * $u + $e4 * $zeta);
$w = $e2 * ($u + $v - $zeta) / (2 * $v);
$k = 1 + $e2 * (sqrt($u + $v + $w * $w) + $w) / ($u + $v);
$lat = atan($k * $z / $p) * $deg;
$lng = atan2($y, $x) * $deg;
return array($lat, $lng, $max_K_sum_i, $max_L_sum_i, $max_L_sum_j);
}
The fact that distances are Euclidean and not great-circle should have negligible effects for the task at hand. Calculating great-circle distances would be much more cumbersome, and would cause only the weight of very far points to be significantly lower - but these points already have a very low weight. In principle, the same effect could be achieved by a different kernel. Kernels that have a complete cut-off beyond some distance, like the Epanechnikov kernel, don't have this problem at all (in practice).
The conversion between lat,lng and x,y,z for the WGS84 datum is given exactly (although without guarantee of numerical stability) more as a reference than because of a true need. If the height is to be taken into account, or if a faster back-conversion is needed, please refer to the Wikipedia article.
The Epanechnikov kernel, besides being "more local" than the Gaussian and Trossian kernels, has the advantage of being the fastest for the second loop, which is O(ng), where g is the number of points of the local grid, and can also be employed in the first loop, which is O(n²), if n is big.
This can be solved by finding a jeopardy surface. See Rossmo's Formula.
This is the predator problem. Given a set of geographically-located carcasses, where is the lair of the predator? Rossmo's formula solves this problem.
Find the point with the largest density estimate.
Should be pretty much straightforward. Use a kernel radius that roughly covers a large airport in diameter. A 2D Gaussian or Epanechnikov kernel should be fine.
http://en.wikipedia.org/wiki/Multivariate_kernel_density_estimation
This is similar to computing a Heap Map: http://en.wikipedia.org/wiki/Heat_map
and then finding the brightest spot there. Except it computes the brightness right away.
For fun I read a 1% sample of the Geocoordinates of DBpedia (i.e. Wikipedia) into ELKI, projected it into 3D space and enabled the density estimation overlay (hidden in the visualizers scatterplot menu). You can see there is a hotspot on Europe, and to a lesser extend in the US. The hotspot in Europe is Poland I believe. Last I checked, someone apparently had created a Wikipedia article with Geocoordinates for pretty much any town in Poland. The ELKI visualizer, unfortunately, neither allows you to zoom in, rotate, or reduce the kernel bandwidth to visually find the most dense point. But it's straightforward to implement yourself; you probably also don't need to go into 3D space, but can just use latitudes and longitudes.
Kernel Density Estimation should be available in tons of applications. The one in R is probably much more powerful. I just recently discovered this heatmap in ELKI, so I knew how to quickly access it. See e.g. http://stat.ethz.ch/R-manual/R-devel/library/stats/html/density.html for a related R function.
On your data, in R, try for example:
library(kernSmooth)
smoothScatter(data, nbin=512, bandwidth=c(.25,.25))
this should show a strong preference for Chicago.
library(kernSmooth)
dens=bkde2D(data, gridsize=c(512, 512), bandwidth=c(.25,.25))
contour(dens$x1, dens$x2, dens$fhat)
maxpos = which(dens$fhat == max(dens$fhat), arr.ind=TRUE)
c(dens$x1[maxpos[1]], dens$x2[maxpos[2]])
yields [1] 42.14697 -88.09508, which is less than 10 miles from Chicago airport.
To get better coordinates try:
rerunning on a 20x20 miles area around the estimated coordinates
a non-binned KDE in that area
better bandwidth selection with dpik
higher grid resolution
in Astrophysics we use the so called "half mass radius". Given a distribution and its center, the half mass radius is the minimum radius of a circle that contains half of the points of your distribution.
This quantity is a characteristic length of a distribution of points.
If you want that the home of the helicopter is where the points are maximally concentrated so it is the point that has the minimum half mass radius!
My algorithm is as follows: for each point you compute this half mass radius centring the distribution in the current point. The "home" of the helicopter will be the point with the minimum half mass radius.
I've implemented it and the computed center is 42.149994 -88.133698 (which is in Chicago)
I've also used the 0.2 of the total mass instead of the 0.5(half) usually used in Astrophysics.
This is my (in python) alghorithm that finds the home of the helicopter:
import math
import numpy
def inside(points,center,radius):
ids=(((points[:,0]-center[0])**2.+(points[:,1]-center[1])**2.)<=radius**2.)
return points[ids]
points = numpy.loadtxt(open('points.txt'),comments='#')
npoints=len(points)
deltar=0.1
idcenter=None
halfrmin=None
for i in xrange(0,npoints):
center=points[i]
radius=0.
stayHere=True
while stayHere:
radius=radius+deltar
ninside=len(inside(points,center,radius))
#print 'point',i,'r',radius,'in',ninside,'center',center
if(ninside>=npoints*0.2):
if(halfrmin==None or radius<halfrmin):
halfrmin=radius
idcenter=i
print 'point',i,halfrmin,idcenter,points[idcenter]
stayHere=False
#print halfrmin,idcenter
print points[idcenter]
You can use DBSCAN for that task.
DBSCAN is a density based clustering with a notion of noise. You need two parameters:
First the number of points a cluster should have at minimum "minpoints".
And second a neighbourhood parameter called "epsilon" that sets a distance threshold to the surrounding points that should be included in your cluster.
The whole algorithm works like this:
Start with an arbitrary point in your set that hasn't been visited yet
Retrieve points from the epsilon neighbourhood mark all as visited
if you have found enough points in this neighbourhood (> minpoints parameter) you start a new cluster and assign those points. Now recurse into step 2 again for every point in this cluster.
if you don't have, declare this point as noise
go all over again until you've visited all points
It is really simple to implement and there are lots of frameworks that support this algorithm already. To find the mean of your cluster, you can simply take the mean of all the assigned points from its neighbourhood.
However, unlike the method that #TylerDurden proposes, this needs a parameterization- so you need to find some hand tuned parameters that fit your problem.
In your case, you can try to set the minpoints to 10% of your total points if the plane is likely to stay 10% of the time you track at an airport. The density parameter epsilon depends on the resolution of your geographic sensor and the distance metric you use- I would suggest the haversine distance for geographic data.
How about divide the map into many zones and then find the center of plane in zone with the most plane. Algorithm will be something like this
set Zones[40]
foreach Plane in Planes
Zones[GetZone(Plane.position)].Add(Plane)
set MaxZone = Zones[0]
foreach Zone in Zones
if MaxZone.Length() < Zone.Length()
MaxZone = Zone
set Center
foreach Plane in MaxZone
Center.X += Plane.X
Center.Y += Plane.Y
Center.X /= MaxZone.Length
Center.Y /= MaxZone.Length
All I have on this machine is an old compiler so I made an ASCII version of this. It "draws" (in ASCII) a map - dots are points, X is where the real source is, G is where the guessed source is. If the two overlap, only X is shown.
Examples (DIFFICULTY 1.5 and 3 respectively):
The points are generated by picking a random point as the source, then randomly distributing points, making them more likely to be closer to the source.
DIFFICULTY is a floating point constant that regulates the initial point generation - how much more likely the points are to be closer to the source - if it is 1 or less, the program should be able to guess the exact source, or very close. At 2.5, it should still be pretty decent. At 4+, it will start to guess worse, but I think it still guesses better than a human would.
It could be optimized by using binary search over X, then Y - this would make the guess worse, but would be much, much faster. Or by starting with larger blocks, then splitting the best block further (or the best block and the 8 surrounding it). For a higher resolution system, one of these would be necessary. This is quite a naive approach, though, but it seems to work well in an 80x24 system. :D
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>
#define Y 24
#define X 80
#define DIFFICULTY 1 // Try different values...
static int point[Y][X];
double dist(int x1, int y1, int x2, int y2)
{
return sqrt((y1 - y2)*(y1 - y2) + (x1 - x2)*(x1 - x2));
}
main()
{
srand(time(0));
int y = rand()%Y;
int x = rand()%X;
// Generate points
for (int i = 0; i < Y; i++)
{
for (int j = 0; j < X; j++)
{
double u = DIFFICULTY * pow(dist(x, y, j, i), 1.0 / DIFFICULTY);
if ((int)u == 0)
u = 1;
point[i][j] = !(rand()%(int)u);
}
}
// Find best source
int maxX = -1;
int maxY = -1;
double maxScore = -1;
for (int cy = 0; cy < Y; cy++)
{
for (int cx = 0; cx < X; cx++)
{
double score = 0;
for (int i = 0; i < Y; i++)
{
for (int j = 0; j < X; j++)
{
if (point[i][j] == 1)
{
double d = dist(cx, cy, j, i);
if (d == 0)
d = 0.5;
score += 1000 / d;
}
}
}
if (score > maxScore || maxScore == -1)
{
maxScore = score;
maxX = cx;
maxY = cy;
}
}
}
// Print out results
for (int i = 0; i < Y; i++)
{
for (int j = 0; j < X; j++)
{
if (i == y && j == x)
printf("X");
else if (i == maxY && j == maxX)
printf("G");
else if (point[i][j] == 0)
printf(" ");
else if (point[i][j] == 1)
printf(".");
}
}
printf("Distance from real source: %f", dist(maxX, maxY, x, y));
scanf("%d", 0);
}
Virtual earth has a very good explanation of how you can do it relatively quick. They also have provided code examples. Please have a look at http://soulsolutions.com.au/Articles/ClusteringVirtualEarthPart1.aspx
A simple mixture model seems to work pretty well for this problem.
In general, to get a point that minimizes the distance to all other points in a dataset, you can just take the mean. In this case, you want to find a point that minimizes the distance from a subset of concentrated points. If you postulate that a point can either come from the concentrated set of points of interest or from a diffuse set of background points, then this gives a mixture model.
I have included some python code below. The concentrated area is modeled by a high-precision normal distribution and the background point are modeled by either a low-precision normal distribution or a uniform distribution over a bounding box on the dataset (there is a line of code that can be commented out to switch between these options). Also, mixture models can be somewhat unstable, so running the EM algorithm a few times with random initial conditions and choosing the run with the highest log-likelihood gives better results.
If you are actually looking at airplanes, then adding some sort of time dependent dynamics will probably improve your ability to infer the home base immensely.
I would also be wary of Rossimo's formula because it includes some pretty-strong assumptions about crime distributions.
#the dataset
sdata='''41.892694,-87.670898
42.056048,-88.000488
41.941744,-88.000488
42.072361,-88.209229
42.091933,-87.982635
42.149994,-88.133698
42.171371,-88.286133
42.23241,-88.305359
42.196811,-88.099365
42.189689,-88.188629
42.17646,-88.173523
42.180531,-88.209229
42.18168,-88.187943
42.185496,-88.166656
42.170485,-88.150864
42.150634,-88.140564
42.156743,-88.123741
42.118555,-88.105545
42.121356,-88.112755
42.115499,-88.102112
42.119319,-88.112411
42.118046,-88.110695
42.117791,-88.109322
42.182189,-88.182449
42.194145,-88.183823
42.189057,-88.196182
42.186513,-88.200645
42.180917,-88.197899
42.178881,-88.192062
41.881656,-87.6297
41.875521,-87.6297
41.87872,-87.636566
41.872073,-87.62661
41.868239,-87.634506
41.86875,-87.624893
41.883065,-87.62352
41.881021,-87.619743
41.879998,-87.620087
41.8915,-87.633476
41.875163,-87.620773
41.879125,-87.62558
41.862763,-87.608757
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'''
import StringIO
import numpy as np
import re
import matplotlib.pyplot as plt
def lp(l):
return map(lambda m: float(m.group()),re.finditer('[^, \n]+',l))
data=np.array(map(lp,StringIO.StringIO(sdata)))
xmn=np.min(data[:,0])
xmx=np.max(data[:,0])
ymn=np.min(data[:,1])
ymx=np.max(data[:,1])
# area of the point set bounding box
area=(xmx-xmn)*(ymx-ymn)
M_ITER=100 #maximum number of iterations
THRESH=1e-10 # stopping threshold
def em(x):
print '\nSTART EM'
mlst=[]
mu0=np.mean( data , 0 ) # the sample mean of the data - use this as the mean of the low-precision gaussian
# the mean of the high-precision Gaussian - this is what we are looking for
mu=np.random.rand( 2 )*np.array([xmx-xmn,ymx-ymn])+np.array([xmn,ymn])
lam_lo=.001 # precision of the low-precision Gaussian
lam_hi=.1 # precision of the high-precision Gaussian
prz=np.random.rand( 1 ) # probability of choosing the high-precision Gaussian mixture component
for i in xrange(M_ITER):
mlst.append(mu[:])
l_hi=np.log(prz)+np.log(lam_hi)-.5*lam_hi*np.sum((x-mu)**2,1)
#low-precision normal background distribution
l_lo=np.log(1.0-prz)+np.log(lam_lo)-.5*lam_lo*np.sum((x-mu0)**2,1)
#uncomment for the uniform background distribution
#l_lo=np.log(1.0-prz)-np.log(area)
#expectation step
zs=1.0/(1.0+np.exp(l_lo-l_hi))
#compute bound on the likelihood
lh=np.sum(zs*l_hi+(1.0-zs)*l_lo)
print i,lh
#maximization step
mu=np.sum(zs[:,None]*x,0)/np.sum(zs) #mean
lam_hi=np.sum(zs)/np.sum(zs*.5*np.sum((x-mu)**2,1)) #precision
prz=1.0/(1.0+np.sum(1.0-zs)/np.sum(zs)) #mixure component probability
try:
if np.abs((lh-old_lh)/lh)<THRESH:
break
except:
pass
old_lh=lh
mlst.append(mu[:])
return lh,lam_hi,mlst
if __name__=='__main__':
#repeat the EM algorithm a number of times and get the run with the best log likelihood
mx_prm=em(data)
for i in xrange(4):
prm=em(data)
if prm[0]>mx_prm[0]:
mx_prm=prm
print prm[0]
print mx_prm[0]
lh,lam_hi,mlst=mx_prm
mu=mlst[-1]
print 'best loglikelihood:', lh
#print 'final precision value:', lam_hi
print 'point of interest:', mu
plt.plot(data[:,0],data[:,1],'.b')
for m in mlst:
plt.plot(m[0],m[1],'xr')
plt.show()
You can easily adapt the Rossmo's formula, quoted by Tyler Durden to your case with few simple notes:
The formula :
This formula give something close to a probability of presence of the base operation for a predator or a serial killer. In your case it could give the probability of a base to be in a certain point. I'll explain later how to use it. U can write it this way :
Proba(base on point A)= Sum{on all spots} ( Phi/(dist^f)+(1-Phi)(B*(g-f))/(2B-dist)^g )
Using Euclidian distance
You want an Euclidian distance and not the Manhattan's one because an airplane or helicopter is not bound to road/streets. So using Euclidian distance is the correct way, if you are tracking an airplane & not a serial killer. So "dist" in the formula is the euclidian distance between the spot ur testing and the spot considered
Taking reasonable variable B
Variable B was used to represent the rule "reasonably smart killer will not kill his neighbor". In your case the will also applied because no one use an airplane/roflcopter to get to the next street corner. we can suppose that the minimal journey is for example 10km or anything reasonable when applied to your case.
Exponential factor f
Factor f is used to add a weight to the distance. For example if all the spots are in a small area you could want a big factor f because the probability of the airport/base/HQ will decrease fast if all your datapoint are in the same sector. g works in a similar way, allow to choose the size of "base is unlikely to be just next to the spot" area
Factor Phi :
Again this factor has to be determined using your knowledge of the problem. It permits to choose the most accurate factor between "base is close to spots" and "i'll not use the plane to make 5 m" if for example u think that the second one is almost irrelevent you can set Phi to 0.95 (0<Phi<1) If both are interesting phi will be around 0.5
How to implement it as something usefull :
First you want to divide your map into little squares : meshing the map ( just like invisal did) (the smaller the squares ,the more accurate the result (in general)) then using the formula to find the more probable location. In fact the mesh is just an array with all possible locations. (if u want to be accurate you increase the number of possible spots but it will require more computational time and PhP is not well-known for it's amazing speed)
Algorithm :
//define all the factors you need(B , f , g , phi)
for(i=0..mesh_size) // computing the probability of presence for each square of the mesh
{
P(i)=0;
geocode squarePosition;//GeoCode of the square's center
for(j=0..geocodearray_size)//sum on all the known spots
{
dist=Distance(geocodearray[j],squarePosition);//small function returning distance between two geocodes
P(i)+=(Phi/pow(dist,f))+(1-Phi)*pow(B,g-f)/pow(2B-dist,g);
}
}
return geocode corresponding to max(P(i))
Hope that it will help you
First I would like to express my fondness of your method in illustrating and explaining the problem ..
If I were in your shoes, I would go for a density based algorithm like DBSCAN
and then after clustering the areas and removing the noise points a few areas (choices) will remain .. then I'll take the cluster with the highest density of points and calculate the average point and find the nearest real point to it . done, found the place! :).
Regards,
Why not something like this:
For each point, calculate it's distance from all other points and sum the total.
The point with the smallest sum is your center.
Maybe sum isn't the best metric to use. Possibly the point with the most "small distances"?
Sum over the distances. Take the point with the smallest summed distance.
function () {
for i in points P:
S[i] = 0
for j in points P:
S[i] += distance(P[i], P[j])
return min(S);
}
You can take a minimum spanning tree and remove the longest edges. The smaller trees give you the centeroid to lookup. The algorithm name is single-link k-clustering. There is a post here: https://stats.stackexchange.com/questions/1475/visualization-software-for-clustering.
I have to implement asin, acos and atan in environment where I have only following math tools:
sine
cosine
elementary fixed point arithmetic (floating point numbers are not available)
I also already have reasonably good square root function.
Can I use those to implement reasonably efficient inverse trigonometric functions?
I don't need too big precision (the floating point numbers have very limited precision anyways), basic approximation will do.
I'm already half decided to go with table lookup, but I would like to know if there is some neater option (that doesn't need several hundred lines of code just to implement basic math).
EDIT:
To clear things up: I need to run the function hundreds of times per frame at 35 frames per second.
In a fixed-point environment (S15.16) I successfully used the CORDIC algorithm (see Wikipedia for a general description) to compute atan2(y,x), then derived asin() and acos() from that using well-known functional identities that involve the square root:
asin(x) = atan2 (x, sqrt ((1.0 + x) * (1.0 - x)))
acos(x) = atan2 (sqrt ((1.0 + x) * (1.0 - x)), x)
It turns out that finding a useful description of the CORDIC iteration for atan2() on the double is harder than I thought. The following website appears to contain a sufficiently detailed description, and also discusses two alternative approaches, polynomial approximation and lookup tables:
http://ch.mathworks.com/examples/matlab-fixed-point-designer/615-calculate-fixed-point-arctangent
Do you need a large precision for arcsin(x) function? If no you may calculate arcsin in N nodes, and keep values in memory. I suggest using line aproximation. if x = A*x_(N) + (1-A)*x_(N+1) then x = A*arcsin(x_(N)) + (1-A)*arcsin(x_(N+1)) where arcsin(x_(N)) is known.
you might want to use approximation: use an infinite series until the solution is close enough for you.
for example:
arcsin(z) = Sigma((2n!)/((2^2n)*(n!)^2)*((z^(2n+1))/(2n+1))) where n in [0,infinity)
http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Expression_as_definite_integrals
You could do that integration numerically with your square root function, approximating with an infinite series:
Submitting here my answer from this other similar question.
nVidia has some great resources I've used for my own uses, few examples: acos asin atan2 etc etc...
These algorithms produce precise enough results. Here's a straight up Python example with their code copy pasted in:
import math
def nVidia_acos(x):
negate = float(x<0)
x=abs(x)
ret = -0.0187293
ret = ret * x
ret = ret + 0.0742610
ret = ret * x
ret = ret - 0.2121144
ret = ret * x
ret = ret + 1.5707288
ret = ret * math.sqrt(1.0-x)
ret = ret - 2 * negate * ret
return negate * 3.14159265358979 + ret
And here are the results for comparison:
nVidia_acos(0.5) result: 1.0471513828611643
math.acos(0.5) result: 1.0471975511965976
That's pretty close! Multiply by 57.29577951 to get results in degrees, which is also from their "degrees" formula.
It should be easy to addapt the following code to fixed point. It employs a rational approximation to calculate the arctangent normalized to the [0 1) interval (you can multiply it by Pi/2 to get the real arctangent). Then, you can use well known identities to get the arcsin/arccos from the arctangent.
normalized_atan(x) ~ (b x + x^2) / (1 + 2 b x + x^2)
where b = 0.596227
The maximum error is 0.1620º
#include <stdint.h>
#include <math.h>
// Approximates atan(x) normalized to the [-1,1] range
// with a maximum error of 0.1620 degrees.
float norm_atan( float x )
{
static const uint32_t sign_mask = 0x80000000;
static const float b = 0.596227f;
// Extract the sign bit
uint32_t ux_s = sign_mask & (uint32_t &)x;
// Calculate the arctangent in the first quadrant
float bx_a = ::fabs( b * x );
float num = bx_a + x * x;
float atan_1q = num / ( 1.f + bx_a + num );
// Restore the sign bit
uint32_t atan_2q = ux_s | (uint32_t &)atan_1q;
return (float &)atan_2q;
}
// Approximates atan2(y, x) normalized to the [0,4) range
// with a maximum error of 0.1620 degrees
float norm_atan2( float y, float x )
{
static const uint32_t sign_mask = 0x80000000;
static const float b = 0.596227f;
// Extract the sign bits
uint32_t ux_s = sign_mask & (uint32_t &)x;
uint32_t uy_s = sign_mask & (uint32_t &)y;
// Determine the quadrant offset
float q = (float)( ( ~ux_s & uy_s ) >> 29 | ux_s >> 30 );
// Calculate the arctangent in the first quadrant
float bxy_a = ::fabs( b * x * y );
float num = bxy_a + y * y;
float atan_1q = num / ( x * x + bxy_a + num );
// Translate it to the proper quadrant
uint32_t uatan_2q = (ux_s ^ uy_s) | (uint32_t &)atan_1q;
return q + (float &)uatan_2q;
}
In case you need more precision, there is a 3rd order rational function:
normalized_atan(x) ~ ( c x + x^2 + x^3) / ( 1 + (c + 1) x + (c + 1) x^2 + x^3)
where c = (1 + sqrt(17)) / 8
which has a maximum approximation error of 0.00811º
Maybe some kind of intelligent brute force like newton rapson.
So for solving asin() you go with steepest descent on sin()
Use a polynomial approximation. Least-squares fit is easiest (Microsoft Excel has it) and Chebyshev approximation is more accurate.
This question has been covered before: How do Trigonometric functions work?
Only continous functions are approximable by polynomials. And arcsin(x) is discontinous in point x=1.same arccos(x).But a range reduction to interval 1,sqrt(1/2) in that case avoid this situation. We have arcsin(x)=pi/2- arccos(x),arccos(x)=pi/2-arcsin(x).you can use matlab for minimax approximation.Aproximate only in range [0,sqrt(1/2)](if angle for that arcsin is request is bigger that sqrt(1/2) find cos(x).arctangent function only for x<1.arctan(x)=pi/2-arctan(1/x).