Smoothing algorithm that doesn't miss the bell curve - android

I am writing an android app which measure the Altitude on location change.
I want to see if the user is going up hill or down hill (on a lift or mountain biking). naturally when you are descending you will have small rises and when you are going up hill you can descend slightly.
I have a smoothing algorithm, which takes the average of the previous ten altitude readings and the average of the next ten then compares the two for an increase or decrease.
This roughly has the effect I am looking for apart from it misses the bell curve and there are still some areas where there is a dip in a general rise which I don't want to see.
statistics is not my strong point but is there a better way to smooth this data?
here is my code
qu="SELECT ID,SPEED,ALTITUDE,ISCLIMB from trip_data where tripid="+Tripid+" order by gmttimestamp;";
c= db.rawQuery(qu, null);
if(c!=null && c.moveToFirst())
{
int av=10;
for(int i=av;i<c.getCount()-av;i++)
{
double prevAlt=0;
double nxtAlt=0;
for(int b=0;b<av;b++)
{
c.moveToPosition(i-b);
prevAlt+=c.getDouble(2);
}
prevAlt/=av;
lastAlt=curAlt;
c.moveToPosition(i);
int id=c.getInt(0);
curSpeed=c.getDouble(1);
curAlt=c.getDouble(2);
for(int b=1;b<av+1;b++)
{
c.moveToPosition(i+b);
nxtAlt+=c.getDouble(2);
}
nxtAlt/=av;
int isC=0;
Log.i("corrections", "preivous ="+prevAlt+" and the next is "+nxtAlt);
db.execSQL("UPDATE TRIP_DATA set PREVALT ="+prevAlt+", NEXTALT="+nxtAlt+", DALT="+(curAlt-lastAlt)+" where id="+id+"");
if(nxtAlt>prevAlt)
{
isC=1;
}else
{
isC=0;
}
String ins="UPDATE trip_data set ISCLIMB="+isC+" where ID="+id+";";
db.execSQL(ins);
Log.i("corrections", ins);
}

Take a look at Savitsky-Golay filters. They give a weight to each point.
You can even use them to calculate the smoothed 1st derative directly.
For example to get the derative from point i using a 5 point quadratic filter (all point in an array):
// coefficients for 5 point 1ste derative
// -2, -1, 0, 1, 2
// factor = 10
double derative = (point[x - 2] * -2 + point[x - 1] * -1 + point[x] * 0 + point[x + 1] * 1 + point[x + 2] * 2) / 10;

Related

Cross correlation to find sonar echoes

I'm trying to detect echoes of my chirp in my sound recording on Android and it seems cross correlation is the most appropriate way of finding where the FFTs of the two signals are similar and from there I can identify peaks in the cross correlated array which will correspond to distances.
From my understanding, I have come up with the following cross correlation function. Is this correct? I wasn't sure whether to add zeros to the beginning as and start a few elements back?
public double[] xcorr1(double[] recording, double[] chirp) {
double[] recordingZeroPadded = new double[recording.length + chirp.length];
for (int i = recording.length; i < recording.length + chirp.length; ++i)
recordingZeroPadded[i] = 0;
for (int i = 0; i < recording.length; ++i)
recordingZeroPadded[i] = recording[i];
double[] result = new double[recording.length + chirp.length - 1];
for (int offset = 0; offset < recordingZeroPadded.length - chirp.length; ++offset)
for (int i = 0; i < chirp.length; ++i)
result[offset] += chirp[i] * recordingZeroPadded[offset + i];
return result;
}
Secondary question:
According to this answer, it can also be calculated like
corr(a, b) = ifft(fft(a_and_zeros) * fft(b_and_zeros[reversed]))
which I don't understand at all but seems easy enough to implement. That said I have failed (assuming my xcorr1 is correct). I feel like I've completely misunderstood this?
public double[] xcorr2(double[] recording, double[] chirp) {
// assume same length arguments for now
DoubleFFT_1D fft = new DoubleFFT_1D(recording.length);
fft.realForward(recording);
reverse(chirp);
fft.realForward(chirp);
double[] result = new double[recording.length];
for (int i = 0; i < result.length; ++i)
result [i] = recording[i] * chirp[i];
fft.realInverse(result, true);
return result;
}
Assuming I got both working, which function would be most appropriate given that the arrays will contain a few thousand elements?
EDIT: Btw, I have tried adding zeros to both ends of both arrays for the FFT version.
EDIT after SleuthEye's response:
Can you just verify that, because I'm dealing with 'actual' data, I need only do half the computations (the real parts) by doing a real transform?
From your code, it looks as though the odd numbered elements in the array returned by the REAL transform are imaginary. What's going on here?
How am I going from an array of real numbers to complex? Or is this the purpose of a transform; to move real numbers into the complex domain? (but the real numbers are just a subset of the complex numbers and so wouldn't they already be in this domain?)
If realForward is in fact returning imaginary/complex numbers, how does it differ to complexForward? And how do I interpret the results? The magnitude of the complex number?
I apologise for my lack of understanding with regard to transforms, I have only so far studied fourier series.
Thanks for the code. Here is 'my' working implementation:
public double[] xcorr2(double[] recording, double[] chirp) {
// pad to power of 2 for optimisation
int y = 1;
while (Math.pow(2,y) < recording.length + chirp.length)
++y;
int paddedLength = (int)Math.pow(2,y);
double[] paddedRecording = new double[paddedLength];
double[] paddedChirp = new double[paddedLength];
for (int i = 0; i < recording.length; ++i)
paddedRecording[i] = recording[i];
for (int i = recording.length; i < paddedLength; ++i)
paddedRecording[i] = 0;
for (int i = 0; i < chirp.length; ++i)
paddedChirp[i] = chirp[i];
for (int i = chirp.length; i < paddedLength; ++i)
paddedChirp[i] = 0;
reverse(chirp);
DoubleFFT_1D fft = new DoubleFFT_1D(paddedLength);
fft.realForward(paddedRecording);
fft.realForward(paddedChirp);
double[] result = new double[paddedLength];
result[0] = paddedRecording[0] * paddedChirp[0]; // value at f=0Hz is real-valued
result[1] = paddedRecording[1] * paddedChirp[1]; // value at f=fs/2 is real-valued and packed at index 1
for (int i = 1; i < result.length / 2; ++i) {
double a = paddedRecording[2*i];
double b = paddedRecording[2*i + 1];
double c = paddedChirp[2*i];
double d = paddedChirp[2*i + 1];
// (a+b*j)*(c-d*j) = (a*c+b*d) + (b*c-a*d)*j
result[2*i] = a*c + b*d;
result[2*i + 1] = b*c - a*d;
}
fft.realInverse(result, true);
// discard trailing zeros
double[] result2 = new double[recording.length + chirp.length - 1];
for (int i = 0; i < result2.length; ++i)
result2[i] = result[i];
return result2;
}
However, until about 5000 elements each, xcorr1 seems to be quicker. Am I doing anything particularly slow (perhaps the constant 'new'ing of memory -- maybe I should cast to an ArrayList)? Or the arbitrary way in which I generated the arrays to test them? Or should I do the conjugates instead of reversing it? That said, performance isn't really an issue so unless there's something obvious you needn't bother pointing out optimisations.
Your implementation of xcorr1 does correspond to the standard signal-processing definition of cross-correlation.
Relative to your interrogation with respect to adding zeros at the beginning: adding chirp.length-1 zeros would make index 0 of the result correspond to the start of transmission. Note however that the peak of the correlation output occurs chirp.length-1 samples after the start of echoes (the chirp has to be aligned with the full received echo). Using the peak index to obtain echo delays, you would then have to adjust for that correlator delay either by subtracting the delay or by discarding the first chirp.length-1 output results. Noting that the additional zeros correspond to that many extra outputs at the beginning, you'd probably be better off not adding those zeros at the beginning in the first place.
For xcorr2 however, a few things need to be addressed. First, if the recording and chirp inputs are not already zero-padded to at least chirp+recording data length you would need to do so (preferably to a power of 2 length for performance reasons). As you are aware, they would both need to be padded to the same length.
Second, you didn't take into account that the multiplication indicated in the posted reference answer, correspond in fact to complex multiplications (whereas DoubleFFT_1D.realForward API uses doubles). Now if you are going to implement something such as a complex multiplication with the chirp's FFT, you might as well actually implement the multiplication with the complex conjugate of the chirp's FFT (the alternate implementation indicated in the reference answer), removing the need to reverse the time-domain values.
Also accounting for DoubleFFT_1D.realForward packing order for even length transforms, you would get:
// [...]
fft.realForward(paddedRecording);
fft.realForward(paddedChirp);
result[0] = paddedRecording[0]*paddedChirp[0]; // value at f=0Hz is real-valued
result[1] = paddedRecording[1]*paddedChirp[1]; // value at f=fs/2 is real-valued and packed at index 1
for (int i = 1; i < result.length/2; ++i) {
double a = paddedRecording[2*i];
double b = paddedRecording[2*i+1];
double c = paddedChirp[2*i];
double d = paddedChirp[2*i+1];
// (a+b*j)*(c-d*j) = (a*c+b*d) + (b*c-a*d)*j
result[2*i] = a*c + b*d;
result[2*i+1] = b*c - a*d;
}
fft.realInverse(result, true);
// [...]
Note that the result array would be of the same size as paddedRecording and paddedChirp, but only the first recording.length+chirp.length-1 should be kept.
Finally, relative to which function is the most appropriate for arrays of a few thousand elements, the FFT version xcorr2 is likely going to be much faster (provided you restrict array lengths to powers of 2).
The direct version doesn't require zero-padding first. You just take recording of length M and chirp of length N and calculate result of length N+M-1. Work through a tiny example by hand to grok the steps:
recording = [1, 2, 3]
chirp = [4, 5]
1 2 3
4 5
1 2 3
4 5
1 2 3
4 5
1 2 3
4 5
result = [1*5, 1*4 + 2*5, 2*4 + 3*5, 3*4] = [5, 14, 23, 4]
The FFT method is much faster if you have long arrays. In this case you have to zero-pad each input to size M+N-1 so that both input arrays are the same size before taking the FFT.
Also, the FFT output is complex numbers, so you need to use complex multiplication. (1+2j)*(3+4j) is -5+10j, not 3+8j. I don't know how your complex numbers are arranged or handled, but make sure this is right.
Or is this the purpose of a transform; to move real numbers into the complex domain?
No, the Fourier transform transforms from the time domain to the frequency domain. The time domain data can be either real or complex, and the frequency domain data can be either real or complex. In most cases you have real data with a complex spectrum. You need to read up on the Fourier transform.
If realForward is in fact returning imaginary/complex numbers, how does it differ to complexForward?
The real FFT takes a real input, while the complex FFT takes a complex input. Both transforms produce complex numbers as their output. That's what the DFT does. The only time a DFT produces real output is if the input data is symmetrical (in which case you can use the DCT to save even more time).

Detect whether a polygon is well formed or not (Google Map Android) [duplicate]

From the man page for XFillPolygon:
If shape is Complex, the path may self-intersect. Note that contiguous coincident points in the path are not treated as self-intersection.
If shape is Convex, for every pair of points inside the polygon, the line segment connecting them does not intersect the path. If known by the client, specifying Convex can improve performance. If you specify Convex for a path that is not convex, the graphics results are undefined.
If shape is Nonconvex, the path does not self-intersect, but the shape is not wholly convex. If known by the client, specifying Nonconvex instead of Complex may improve performance. If you specify Nonconvex for a self-intersecting path, the graphics results are undefined.
I am having performance problems with fill XFillPolygon and, as the man page suggests, the first step I want to take is to specify the correct shape of the polygon. I am currently using Complex to be on the safe side.
Is there an efficient algorithm to determine if a polygon (defined by a series of coordinates) is convex, non-convex or complex?
You can make things a lot easier than the Gift-Wrapping Algorithm... that's a good answer when you have a set of points w/o any particular boundary and need to find the convex hull.
In contrast, consider the case where the polygon is not self-intersecting, and it consists of a set of points in a list where the consecutive points form the boundary. In this case it is much easier to figure out whether a polygon is convex or not (and you don't have to calculate any angles, either):
For each consecutive pair of edges of the polygon (each triplet of points), compute the z-component of the cross product of the vectors defined by the edges pointing towards the points in increasing order. Take the cross product of these vectors:
given p[k], p[k+1], p[k+2] each with coordinates x, y:
dx1 = x[k+1]-x[k]
dy1 = y[k+1]-y[k]
dx2 = x[k+2]-x[k+1]
dy2 = y[k+2]-y[k+1]
zcrossproduct = dx1*dy2 - dy1*dx2
The polygon is convex if the z-components of the cross products are either all positive or all negative. Otherwise the polygon is nonconvex.
If there are N points, make sure you calculate N cross products, e.g. be sure to use the triplets (p[N-2],p[N-1],p[0]) and (p[N-1],p[0],p[1]).
If the polygon is self-intersecting, then it fails the technical definition of convexity even if its directed angles are all in the same direction, in which case the above approach would not produce the correct result.
This question is now the first item in either Bing or Google when you search for "determine convex polygon." However, none of the answers are good enough.
The (now deleted) answer by #EugeneYokota works by checking whether an unordered set of points can be made into a convex polygon, but that's not what the OP asked for. He asked for a method to check whether a given polygon is convex or not. (A "polygon" in computer science is usually defined [as in the XFillPolygon documentation] as an ordered array of 2D points, with consecutive points joined with a side as well as the last point to the first.) Also, the gift wrapping algorithm in this case would have the time-complexity of O(n^2) for n points - which is much larger than actually needed to solve this problem, while the question asks for an efficient algorithm.
#JasonS's answer, along with the other answers that follow his idea, accepts star polygons such as a pentagram or the one in #zenna's comment, but star polygons are not considered to be convex. As
#plasmacel notes in a comment, this is a good approach to use if you have prior knowledge that the polygon is not self-intersecting, but it can fail if you do not have that knowledge.
#Sekhat's answer is correct but it also has the time-complexity of O(n^2) and thus is inefficient.
#LorenPechtel's added answer after her edit is the best one here but it is vague.
A correct algorithm with optimal complexity
The algorithm I present here has the time-complexity of O(n), correctly tests whether a polygon is convex or not, and passes all the tests I have thrown at it. The idea is to traverse the sides of the polygon, noting the direction of each side and the signed change of direction between consecutive sides. "Signed" here means left-ward is positive and right-ward is negative (or the reverse) and straight-ahead is zero. Those angles are normalized to be between minus-pi (exclusive) and pi (inclusive). Summing all these direction-change angles (a.k.a the deflection angles) together will result in plus-or-minus one turn (i.e. 360 degrees) for a convex polygon, while a star-like polygon (or a self-intersecting loop) will have a different sum ( n * 360 degrees, for n turns overall, for polygons where all the deflection angles are of the same sign). So we must check that the sum of the direction-change angles is plus-or-minus one turn. We also check that the direction-change angles are all positive or all negative and not reverses (pi radians), all points are actual 2D points, and that no consecutive vertices are identical. (That last point is debatable--you may want to allow repeated vertices but I prefer to prohibit them.) The combination of those checks catches all convex and non-convex polygons.
Here is code for Python 3 that implements the algorithm and includes some minor efficiencies. The code looks longer than it really is due to the the comment lines and the bookkeeping involved in avoiding repeated point accesses.
TWO_PI = 2 * pi
def is_convex_polygon(polygon):
"""Return True if the polynomial defined by the sequence of 2D
points is 'strictly convex': points are valid, side lengths non-
zero, interior angles are strictly between zero and a straight
angle, and the polygon does not intersect itself.
NOTES: 1. Algorithm: the signed changes of the direction angles
from one side to the next side must be all positive or
all negative, and their sum must equal plus-or-minus
one full turn (2 pi radians). Also check for too few,
invalid, or repeated points.
2. No check is explicitly done for zero internal angles
(180 degree direction-change angle) as this is covered
in other ways, including the `n < 3` check.
"""
try: # needed for any bad points or direction changes
# Check for too few points
if len(polygon) < 3:
return False
# Get starting information
old_x, old_y = polygon[-2]
new_x, new_y = polygon[-1]
new_direction = atan2(new_y - old_y, new_x - old_x)
angle_sum = 0.0
# Check each point (the side ending there, its angle) and accum. angles
for ndx, newpoint in enumerate(polygon):
# Update point coordinates and side directions, check side length
old_x, old_y, old_direction = new_x, new_y, new_direction
new_x, new_y = newpoint
new_direction = atan2(new_y - old_y, new_x - old_x)
if old_x == new_x and old_y == new_y:
return False # repeated consecutive points
# Calculate & check the normalized direction-change angle
angle = new_direction - old_direction
if angle <= -pi:
angle += TWO_PI # make it in half-open interval (-Pi, Pi]
elif angle > pi:
angle -= TWO_PI
if ndx == 0: # if first time through loop, initialize orientation
if angle == 0.0:
return False
orientation = 1.0 if angle > 0.0 else -1.0
else: # if other time through loop, check orientation is stable
if orientation * angle <= 0.0: # not both pos. or both neg.
return False
# Accumulate the direction-change angle
angle_sum += angle
# Check that the total number of full turns is plus-or-minus 1
return abs(round(angle_sum / TWO_PI)) == 1
except (ArithmeticError, TypeError, ValueError):
return False # any exception means not a proper convex polygon
The following Java function/method is an implementation of the algorithm described in this answer.
public boolean isConvex()
{
if (_vertices.size() < 4)
return true;
boolean sign = false;
int n = _vertices.size();
for(int i = 0; i < n; i++)
{
double dx1 = _vertices.get((i + 2) % n).X - _vertices.get((i + 1) % n).X;
double dy1 = _vertices.get((i + 2) % n).Y - _vertices.get((i + 1) % n).Y;
double dx2 = _vertices.get(i).X - _vertices.get((i + 1) % n).X;
double dy2 = _vertices.get(i).Y - _vertices.get((i + 1) % n).Y;
double zcrossproduct = dx1 * dy2 - dy1 * dx2;
if (i == 0)
sign = zcrossproduct > 0;
else if (sign != (zcrossproduct > 0))
return false;
}
return true;
}
The algorithm is guaranteed to work as long as the vertices are ordered (either clockwise or counter-clockwise), and you don't have self-intersecting edges (i.e. it only works for simple polygons).
Here's a test to check if a polygon is convex.
Consider each set of three points along the polygon--a vertex, the vertex before, the vertex after. If every angle is 180 degrees or less you have a convex polygon. When you figure out each angle, also keep a running total of (180 - angle). For a convex polygon, this will total 360.
This test runs in O(n) time.
Note, also, that in most cases this calculation is something you can do once and save — most of the time you have a set of polygons to work with that don't go changing all the time.
To test if a polygon is convex, every point of the polygon should be level with or behind each line.
Here's an example picture:
The answer by #RoryDaulton
seems the best to me, but what if one of the angles is exactly 0?
Some may want such an edge case to return True, in which case, change "<=" to "<" in the line :
if orientation * angle < 0.0: # not both pos. or both neg.
Here are my test cases which highlight the issue :
# A square
assert is_convex_polygon( ((0,0), (1,0), (1,1), (0,1)) )
# This LOOKS like a square, but it has an extra point on one of the edges.
assert is_convex_polygon( ((0,0), (0.5,0), (1,0), (1,1), (0,1)) )
The 2nd assert fails in the original answer. Should it?
For my use case, I would prefer it didn't.
This method would work on simple polygons (no self intersecting edges) assuming that the vertices are ordered (either clockwise or counter)
For an array of vertices:
vertices = [(0,0),(1,0),(1,1),(0,1)]
The following python implementation checks whether the z component of all the cross products have the same sign
def zCrossProduct(a,b,c):
return (a[0]-b[0])*(b[1]-c[1])-(a[1]-b[1])*(b[0]-c[0])
def isConvex(vertices):
if len(vertices)<4:
return True
signs= [zCrossProduct(a,b,c)>0 for a,b,c in zip(vertices[2:],vertices[1:],vertices)]
return all(signs) or not any(signs)
I implemented both algorithms: the one posted by #UriGoren (with a small improvement - only integer math) and the one from #RoryDaulton, in Java. I had some problems because my polygon is closed, so both algorithms were considering the second as concave, when it was convex. So i changed it to prevent such situation. My methods also uses a base index (which can be or not 0).
These are my test vertices:
// concave
int []x = {0,100,200,200,100,0,0};
int []y = {50,0,50,200,50,200,50};
// convex
int []x = {0,100,200,100,0,0};
int []y = {50,0,50,200,200,50};
And now the algorithms:
private boolean isConvex1(int[] x, int[] y, int base, int n) // Rory Daulton
{
final double TWO_PI = 2 * Math.PI;
// points is 'strictly convex': points are valid, side lengths non-zero, interior angles are strictly between zero and a straight
// angle, and the polygon does not intersect itself.
// NOTES: 1. Algorithm: the signed changes of the direction angles from one side to the next side must be all positive or
// all negative, and their sum must equal plus-or-minus one full turn (2 pi radians). Also check for too few,
// invalid, or repeated points.
// 2. No check is explicitly done for zero internal angles(180 degree direction-change angle) as this is covered
// in other ways, including the `n < 3` check.
// needed for any bad points or direction changes
// Check for too few points
if (n <= 3) return true;
if (x[base] == x[n-1] && y[base] == y[n-1]) // if its a closed polygon, ignore last vertex
n--;
// Get starting information
int old_x = x[n-2], old_y = y[n-2];
int new_x = x[n-1], new_y = y[n-1];
double new_direction = Math.atan2(new_y - old_y, new_x - old_x), old_direction;
double angle_sum = 0.0, orientation=0;
// Check each point (the side ending there, its angle) and accum. angles for ndx, newpoint in enumerate(polygon):
for (int i = 0; i < n; i++)
{
// Update point coordinates and side directions, check side length
old_x = new_x; old_y = new_y; old_direction = new_direction;
int p = base++;
new_x = x[p]; new_y = y[p];
new_direction = Math.atan2(new_y - old_y, new_x - old_x);
if (old_x == new_x && old_y == new_y)
return false; // repeated consecutive points
// Calculate & check the normalized direction-change angle
double angle = new_direction - old_direction;
if (angle <= -Math.PI)
angle += TWO_PI; // make it in half-open interval (-Pi, Pi]
else if (angle > Math.PI)
angle -= TWO_PI;
if (i == 0) // if first time through loop, initialize orientation
{
if (angle == 0.0) return false;
orientation = angle > 0 ? 1 : -1;
}
else // if other time through loop, check orientation is stable
if (orientation * angle <= 0) // not both pos. or both neg.
return false;
// Accumulate the direction-change angle
angle_sum += angle;
// Check that the total number of full turns is plus-or-minus 1
}
return Math.abs(Math.round(angle_sum / TWO_PI)) == 1;
}
And now from Uri Goren
private boolean isConvex2(int[] x, int[] y, int base, int n)
{
if (n < 4)
return true;
boolean sign = false;
if (x[base] == x[n-1] && y[base] == y[n-1]) // if its a closed polygon, ignore last vertex
n--;
for(int p=0; p < n; p++)
{
int i = base++;
int i1 = i+1; if (i1 >= n) i1 = base + i1-n;
int i2 = i+2; if (i2 >= n) i2 = base + i2-n;
int dx1 = x[i1] - x[i];
int dy1 = y[i1] - y[i];
int dx2 = x[i2] - x[i1];
int dy2 = y[i2] - y[i1];
int crossproduct = dx1*dy2 - dy1*dx2;
if (i == base)
sign = crossproduct > 0;
else
if (sign != (crossproduct > 0))
return false;
}
return true;
}
For a non complex (intersecting) polygon to be convex, vector frames obtained from any two connected linearly independent lines a,b must be point-convex otherwise the polygon is concave.
For example the lines a,b are convex to the point p and concave to it below for each case i.e. above: p exists inside a,b and below: p exists outside a,b
Similarly for each polygon below, if each line pair making up a sharp edge is point-convex to the centroid c then the polygon is convex otherwise it’s concave.
blunt edges (wronged green) are to be ignored
N.B
This approach would require you compute the centroid of your polygon beforehand since it doesn’t employ angles but vector algebra/transformations
Adapted Uri's code into matlab. Hope this may help.
Be aware that Uri's algorithm only works for simple polygons! So, be sure to test if the polygon is simple first!
% M [ x1 x2 x3 ...
% y1 y2 y3 ...]
% test if a polygon is convex
function ret = isConvex(M)
N = size(M,2);
if (N<4)
ret = 1;
return;
end
x0 = M(1, 1:end);
x1 = [x0(2:end), x0(1)];
x2 = [x0(3:end), x0(1:2)];
y0 = M(2, 1:end);
y1 = [y0(2:end), y0(1)];
y2 = [y0(3:end), y0(1:2)];
dx1 = x2 - x1;
dy1 = y2 - y1;
dx2 = x0 - x1;
dy2 = y0 - y1;
zcrossproduct = dx1 .* dy2 - dy1 .* dx2;
% equality allows two consecutive edges to be parallel
t1 = sum(zcrossproduct >= 0);
t2 = sum(zcrossproduct <= 0);
ret = t1 == N || t2 == N;
end

Android Floating point multiplication off by a factor 0.004739

Can anyone explain to me why the speed calculation is off?
It is a floating point calculation. I show (below) in the log print all the factors involved.
According to my hand calculator:
length: sqrt(3.497924^2 + (-1.134711)^2) = 3.67736881
speed: 1.05 * 3.67736881 = 3.86123725
this is off by a factor 4.739 10^-3 from the printed result 3.879624
The code:
float factor = yTouch(paddle, ballBounds);
float speed = SPEED_TRANSFER[level] * PointF.length(mSpeed.x, mSpeed.y);
double alpha = Math.PI/2;
double beta = Math.atan2(mSpeed.y, mSpeed.x);
double tau = -(2*alpha+beta);
Log.v("debug", String.format("speed(%f) mSpeed(%f,%f) transfer(%2.2f)
length(%f) Original tau(%3.2f)", speed, mSpeed.x, mSpeed.y,
SPEED_TRANSFER[level], PointF.length(mSpeed.x, mSpeed.y), tau/Math.PI));
The Log:
03-24 10:48:53.421: V/debug(3236): speed(3.879624) mSpeed(3.497924,-1.134711) transfer(1.05) length(3.677369) Original tau(-0.90)
Edit
I tried this, the direct multiplication is right.
Log.v("debug", "Direct multiplication:" + (1.05f * 3.677369f) );
03-24 11:49:08.687: V/debug(7644): Direct multiplication:3.8612373
Can you print more digits of SPEED_TRANSFER[level] in your log line? You're looking at an error in the 3rd decimal place, but only showing 2 places of one of the factors. If SPEED_TRANSFER[level] == 1.055 for instance the result is correct (although I would have expected this to print as 1.06 in your Log line, but that's another matter).

Calculate distance between two points directly in SQLite

In my web/MySQL application I have something like this to get distance between two points:
6371 * acos(cos(radians(-19.83996)) * cos(radians(lat)) * cos(radians(-43.94910) - radians(lng)) + sin(radians(-19.83996)) * sin(radians(lat)))
But I tested in SQLite and these mathematical functions (acos, cos, radians, sin) do not exist. Is there something equivalent for me to calculate the distance directly in the database?
However, I have an iPhone application that uses this method to calculate. Works perfectly, but now I need to perform this same search in the database in an Android application.
Thanks in advance.
UPDATE
I have 9000 points to calculate distance and obtain the 5 nearby locations of a given point.
Here is what I would do:
Take your given point. Measure a (that is an arbitrary value to be refined) ~2km wide square around it and take the values for the east/west/north/south bounds.
Make a query for elements inside this square. This one is easy, you just have to
select * from points where lat between ? and ? and lon between ? and ?
Count your result. Not enough result (less than 5, obviously, but I would say twice that to be sure), retry with a larger radius. Too much (say, more than 100), try again with a smaller radius.
Once you have enough, load them, make sure all 5 elements you need are not only in the Xkm wide square, but also in the Xkm radius circle (to avoid having a potential closer element not detected by the previous approximation).
Another approach
Valid only if your given point is relatively close to those you are searching.
Measure a local approximation of a flat earth. Close to your point, you can consider a linear relation between lat, lon, and distance. That allows you to make a request sorted by a simple calculus. (multiplication and addition). Again, select a little more points in order to make the proper calculation after the SQLite request.
Insert cos_lat_rad,sin_lat_rad,cos_lon_rad,sin_lon_rad in to your table
contentValues.put("cos_lat_rad", Math.cos(deg2rad(latitude)));
contentValues.put("sin_lat_rad", Math.sin(deg2rad(latitude)));
contentValues.put("cos_lon_rad", Math.cos(deg2rad(longitude)));
contentValues.put("sin_lon_rad", Math.sin(deg2rad(longitude)));
degree to radian
public static double deg2rad(double deg) {
return (deg * Math.PI / 180.0);
}
query , distance in km
Cursor c=database.dis(String.valueOf(Math.cos((double) distance / (double) 6380)), Math.cos(deg2rad(latitude)), Math.sin(deg2rad(latitude)), Math.cos(deg2rad(longitude)), Math.sin(deg2rad(longitude)));
query
public Cursor dis(String dis, double cos_lat_rad, double sin_lat_rad, double cos_lon_rad, double sin_lon_rad) {
Cursor cursor = sqLiteDatabase.rawQuery("SELECT * ,(" + sin_lat_rad + "*\"sin_lat_rad\"+" + cos_lat_rad + "*\"cos_lat_rad\"*(" + sin_lon_rad + "*\"sin_lon_rad\"+" + cos_lon_rad + "*\"cos_lon_rad\")) AS \"distance_acos\" FROM parish WHERE ("+sin_lat_rad+" * \"sin_lat_rad\" +"+ cos_lat_rad +"* \"cos_lat_rad\" * (+"+sin_lon_rad +"* \"sin_lon_rad\" + "+cos_lon_rad +"* \"cos_lon_rad\")) >"+dis+ " ORDER BY \"distance_acos\" DESC ", null);
return cursor;
}
convert distance_acos to km
if(c.moveToFirst())
do {
double distance_acos= c.getDouble(c.getColumnIndex("distance_acos"));
String Distance=String.valueOf(Math.acos(distance_acos) * 6380);
}while (c.moveToNext());
Reply of Angel does not solve the problem for SQLite since it still includes ACOS function which does not exist in SQLite
What I have concluded after a thorough research is that one should pick a rough estimation to use in SQLite with the formula below:
Distance estimation for km:
SQRT(SQUARE((TO_LAT-FROM_LAT)*110)+
SQUARE((TO_LONG-FROM_LONG)*COS(TO_LAT)*111))
For windows:
Install minGw full options. modify environment variable: system variable path, including
c:\mingw\bin
test functionality command:
g++ --version
copy files: extension-functions.c, sqlite3.h, sqlite3ext.h in sqlite3 program directory. Go to sqlite3 directory and compile:
gcc -shared -I "path" -o libsqlitefunctions.so extension-functions.c
(path = path of sqlite3ext.h; i.e. C:\sqlite3)
If the program is built so that loading extensions is permitted, the following will work:
sqlite> SELECT load_extension('./libsqlitefunctions.so');
sqlite> select cos(radians(45));
0.707106781186548
SQLite Distance implementation:
From: https://www.movable-type.co.uk/scripts/latlong.html https://en.wikipedia.org/wiki/Haversine_formula
Distance
This uses the ‘haversine’ formula to calculate the great-circle distance between two points – that is, the shortest distance over the earth’s surface – giving an ‘as-the-crow-flies’ distance between the points (ignoring any hills they fly over, of course!).
Haversine
Formula:
a = sin²(Δφ/2) + cos φ1 * cos φ2 * sin²(Δλ/2)
c = 2 * atan2( √a, √(1−a) )
c = 2 *
d = R * c
where φ is latitude, λ is longitude, R is earth’s radius (mean radius = 6,371km);
note that angles need to be in radians to pass to trig functions!
JavaScript:
const R = 6378136.6 ; // meters equatorial radius
const φ1 = lat1 * Math.PI/180; // φ, λ in radians
const φ2 = lat2 * Math.PI/180;
const Δφ = (lat2-lat1) * Math.PI/180;
const Δλ = (lon2-lon1) * Math.PI/180;
const a = Math.sin(Δφ/2) * Math.sin(Δφ/2) +
Math.cos(φ1) * Math.cos(φ2) *
Math.sin(Δλ/2) * Math.sin(Δλ/2);
const c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
const c = 2 * Math.asen(MIN (1, Math.sqrt(a))); //sqlite implementation
const d = R * c; // in metres
Distance = 2 * R * ASIN( MIN (1, SQRT( SIN( (RADIANS(lat1)-RADIANS(lat2))/2 )^2 + COS( RADIANS(lat1) )*COS( RADIANS(lat2) )*SIN( (RADIANS(long1)-RADIANS(long2))/2 )^2 )))
Physical Properties of Earth https://en.wikipedia.org/wiki/Earth_ellipsoid :Ecuatorial radius: 6378.1366 Kms. Average radius: 6367 Kms
Constant = 2 * 6378136.6 = 12756273.2
SQLite query command with coordinates taken from table PAR:
ROUND (
12756273.2 * ASIN(
MIN (1 ,
SQRT(
POWER( SIN(RADIANS(PAR.Lat1 - PAR.Lat2)/2) , 2) +
COS(RADIANS(PAR.Lat1)) * COS(RADIANS(PAR.Lat2)) * POWER ( SIN(RADIANS(PAR.Long1 - PAR.Long2)/2) , 2)
)
)
)
, 0) AS Distance
In android we have a Location class, we need initialize the location class and in it we have a method called distanceBetween() which gives the distance between two geopoints.
Refer this LINK

Android audio FFT to display fundamental frequency

I have been working on an Android project for awhile that displays the fundamental frequency of an input signal (to act as a tuner). I have successfully implemented the AudioRecord class and am getting data from it. However, I am having a hard time performing an FFT on this data to get the fundamental frequency of the input signal. I have been looking at the post here, and am using FFT in Java and Complex class to go with it.
I have successfully used the FFT function found in FFT in Java, but I am not sure if I am obtaining the correct results. For the magnitude of the FFT (sqrt[rere+imim]) I am getting values that start high, around 15000 Hz, and then slowly diminish to about 300 Hz. Doesn't seem right.
Also, as far as the raw data from the mic goes, the data seems fine, except that the first 50 values or so are always the number 3, unless I hit the tuning button again while still in the application and then I only get about 15. Is that normal?
Here is a bit of my code.
First of all, I convert the short data (obtained from the microphone) to a double using the following code which is from the post I have been looking at. This snippet of code I do not completely understand, but I think it works.
//Conversion from short to double
double[] micBufferData = new double[bufferSizeInBytes];//size may need to change
final int bytesPerSample = 2; // As it is 16bit PCM
final double amplification = 1.0; // choose a number as you like
for (int index = 0, floatIndex = 0; index < bufferSizeInBytes - bytesPerSample + 1; index += bytesPerSample, floatIndex++) {
double sample = 0;
for (int b = 0; b < bytesPerSample; b++) {
int v = audioData[index + b];
if (b < bytesPerSample - 1 || bytesPerSample == 1) {
v &= 0xFF;
}
sample += v << (b * 8);
}
double sample32 = amplification * (sample / 32768.0);
micBufferData[floatIndex] = sample32;
}
The code then continues as follows:
//Create Complex array for use in FFT
Complex[] fftTempArray = new Complex[bufferSizeInBytes];
for (int i=0; i<bufferSizeInBytes; i++)
{
fftTempArray[i] = new Complex(micBufferData[i], 0);
}
//Obtain array of FFT data
final Complex[] fftArray = FFT.fft(fftTempArray);
final Complex[] fftInverse = FFT.ifft(fftTempArray);
//Create an array of magnitude of fftArray
double[] magnitude = new double[fftArray.length];
for (int i=0; i<fftArray.length; i++){
magnitude[i]= fftArray[i].abs();
}
fft.setTextColor(Color.GREEN);
fft.setText("fftArray is "+ fftArray[500] +" and fftTempArray is "+fftTempArray[500] + " and fftInverse is "+fftInverse[500]+" and audioData is "+audioData[500]+ " and magnitude is "+ magnitude[1] + ", "+magnitude[500]+", "+magnitude[1000]+" Good job!");
for(int i = 2; i < samples; i++){
fft.append(" " + magnitude[i] + " Hz");
}
That last bit is just to check what values I am getting (and to keep me sane!). In the post referred to above, it talks about needing the sampling frequency and gives this code:
private double ComputeFrequency(int arrayIndex) {
return ((1.0 * sampleRate) / (1.0 * fftOutWindowSize)) * arrayIndex;
}
How do I implement this code? I don't realy understand where fftOutWindowSize and arrayIndex comes from?
Any help is greatly appreciated!
Dustin
Recently I'm working on a project which requires almost the same. Probably you don't need any help anymore but I will give my thoughts anyway. Maybe someone need this in the future.
I'm not sure whether the short to double function works, I don't understand that snippet of code neither. It is wrote for byte to double conversion.
In the code: "double[] micBufferData = new double[bufferSizeInBytes];" I think the size of micBufferData should be "bufferSizeInBytes / 2", since every sample takes two bytes and the size of micBufferData should be the sample number.
FFT algorithms do require a FFT window size, and it has to be a number which is the power of 2. However many algorithms can receive an arbitrary of number as input and it will do the rest. In the document of those algorithms should have the requirements of input. In your case, the size of the Complex array can be the input of FFT algorithms. And I don't really know the detail of the FFT algorithm but I think the inverse one is not needed.
To use the code you gave at last, you should firstly find the peak index in the sample array. I used double array as input instead of Complex, so in my case it is something like: double maxVal = -1;int maxIndex = -1;
for( int j=0; j < mFftSize / 2; ++j ) {
double v = fftResult[2*j] * fftResult[2*j] + fftResult[2*j+1] * fftResult[2*j+1];
if( v > maxVal ) {
maxVal = v;
maxIndex = j;
}
}
2*j is the real part and 2*j+1 is the imaginary part. maxIndex is the index of the peak magnitude you want (More detail here), and use it as input to the ComputeFrequency function. The return value is the frequency of the sample array you want.
Hopefully it can help someone.
You should pick an FFT window size depending on your time versus frequency resolution requirements, and not just use the audio buffer size when creating your FFT temp array.
The array index is your int i, as used in your magnitude[i] print statement.
The fundamental pitch frequency for music is often different from FFT peak magnitude, so you may want to research some pitch estimation algorithms.
I suspect that the strange results you're getting are because you might need to unpack the FFT. How this is done will depend on the library that you're using (see here for docs on how it's packed in GSL, for example). The packing may mean that the real and imaginary components are not in the positions in the array that you expect.
For your other questions about window size and resolution, if you're creating a tuner then I'd suggest trying a window size of about 20ms (eg 1024 samples at 44.1kHz). For a tuner you need quite high resolution, so you could try zero-padding by a factor of 8 or 16 which will give you a resolution of 3-6Hz.

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