public class MathUtil {
private static final LogHelper log = Application.getLogger();
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+
public static final double EPSILON = 0.00000001; // 10mm^3 in m^3
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+
/**
* The square of x (x^2). On Sun's JRE using this method is as fast as typing x*x.
* @param x x
public static double pow2(double x) {
return x * x;
}
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/**
* The cube of x (x^3).
* @param x x
public static double pow3(double x) {
return x * x * x;
}
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public static double pow4(double x) {
return (x * x) * (x * x);
}
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/**
* Clamps the value x to the range min - max.
* @param x Original value.
return max;
return x;
}
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public static float clamp(float x, float min, float max) {
if (x < min)
return min;
return max;
return x;
}
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public static int clamp(int x, int min, int max) {
if (x < min)
return min;
return max;
return x;
}
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+
+
/**
* Maps a value from one value range to another.
*
}
return (value - fromMin) / (fromMax - fromMin) * (toMax - toMin) + toMin;
}
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+
+
/**
* Maps a coordinate from one value range to another.
*
double a = (value - fromMin) / (fromMax - fromMin);
return toMax.multiply(a).add(toMin.multiply(1 - a));
}
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+
/**
* Compute the minimum of two values. This is performed by direct comparison.
* However, if one of the values is NaN and the other is not, the non-NaN value is
return x;
return (x < y) ? x : y;
}
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/**
* Compute the maximum of two values. This is performed by direct comparison.
* However, if one of the values is NaN and the other is not, the non-NaN value is
return y;
return (x < y) ? y : x;
}
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/**
* Compute the minimum of three values. This is performed by direct comparison.
* However, if one of the values is NaN and the other is not, the non-NaN value is
return min(y, z);
}
}
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+
/**
* Compute the minimum of three values. This is performed by direct comparison.
public static double min(double w, double x, double y, double z) {
return min(min(w, x), min(y, z));
}
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+
/**
* Compute the maximum of three values. This is performed by direct comparison.
* However, if one of the values is NaN and the other is not, the non-NaN value is
return max(y, z);
}
}
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/**
* Calculates the hypotenuse <code>sqrt(x^2+y^2)</code>. This method is SIGNIFICANTLY
* faster than <code>Math.hypot(x,y)</code>.
public static double hypot(double x, double y) {
return Math.sqrt(x * x + y * y);
}
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/**
* Reduce the angle x to the range 0 - 2*PI.
* @param x Original angle.
double d = Math.floor(x / (2 * Math.PI));
return x - d * 2 * Math.PI;
}
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/**
* Reduce the angle x to the range -PI - PI.
*
double d = Math.rint(x / (2 * Math.PI));
return x - d * 2 * Math.PI;
}
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+
/**
* Return the square root of a value. If the value is negative, zero is returned.
* This is safer in cases where rounding errors might make a value slightly negative.
}
return Math.sqrt(d);
}
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+
public static boolean equals(double a, double b) {
double absb = Math.abs(b);
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if (absb < EPSILON / 2) {
// Near zero
return Math.abs(a) < EPSILON / 2;
}
return Math.abs(a - b) < EPSILON * absb;
}
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+
/**
* Return the sign of the number. This corresponds to Math.signum, but ignores
* the special cases of zero and NaN. The value returned for those is arbitrary.
public static double sign(double x) {
return (x < 0) ? -1.0 : 1.0;
}
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/* Math.abs() is about 3x as fast as this:
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public static double abs(double x) {
return (x<0) ? -x : x;
}
- */
+ */
public static double average(Collection<? extends Number> values) {
if (values.isEmpty()) {
return Double.NaN;
}
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double avg = 0.0;
int count = 0;
for (Number n : values) {
}
return avg / count;
}
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public static double stddev(Collection<? extends Number> values) {
if (values.size() < 2) {
return Double.NaN;
}
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double avg = average(values);
double stddev = 0.0;
int count = 0;
stddev = Math.sqrt(stddev / (count - 1));
return stddev;
}
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public static double median(Collection<? extends Number> values) {
if (values.isEmpty()) {
return Double.NaN;
}
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List<Number> sorted = new ArrayList<Number>(values);
Collections.sort(sorted, new Comparator<Number>() {
@Override
return Double.compare(o1.doubleValue(), o2.doubleValue());
}
});
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int n = sorted.size();
if (n % 2 == 0) {
return (sorted.get(n / 2).doubleValue() + sorted.get(n / 2 - 1).doubleValue()) / 2;
}
}
+ /**
+ * Use interpolation to determine the value of the function at point t.
+ * Current implementation uses simple linear interpolation. The domain
+ * and range lists must include the same number of values, t must be within
+ * the domain, and the domain list must be sorted.
+ *
+ * @param domain list containing domain samples
+ * @param range list of corresponding range samples
+ * @param t domain value at which to interpolate
+ * @return returns Double.NaN if either list is null or empty or different size, or if t is outsize the domain.
+ */
+ public static double interpolate( List<Double> domain, List<Double> range, double t ) {
+
+ if ( domain == null || range == null || domain.size() != range.size() ) {
+ return Double.NaN;
+ }
+
+ int length = domain.size();
+ if ( length <= 1 || t < domain.get(0) || t > domain.get( length-1 ) ) {
+ return Double.NaN;
+ }
+
+ // Look for the index of the right end point.
+ int right = 1;
+ while( t > domain.get(right) ) {
+ right ++;
+ }
+ int left = right -1;
+
+ // Points are:
+
+ double deltax = domain.get(right) - domain.get(left);
+ double deltay = range.get(right) - range.get(left);
+
+ // For numerical stability, if deltax is small,
+ if ( Math.abs(deltax) < EPSILON ) {
+ if ( deltay < -1.0 * EPSILON ) {
+ // return neg infinity if deltay is negative
+ return Double.NEGATIVE_INFINITY;
+ }
+ else if ( deltay > EPSILON ) {
+ // return infinity if deltay is large
+ return Double.POSITIVE_INFINITY;
+ } else {
+ // otherwise return 0
+ return 0.0d;
+ }
+ }
+
+ return range.get(left) + ( t - domain.get(left) ) * deltay / deltax;
+
+ }
+
}