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package edu.stanford.nlp.math;
import java.util.Collection;
/**
* The class SloppyMath contains methods for performing basic
* numeric operations. In some cases, such as max and min, they cut a few
* corners in
* the implementation for the sake of efficiency. In particular, they may
* not handle special notions like NaN and -0.0 correctly. This was the
* origin of the class name, but many other methods are just useful
* math additions, such as logSum. This class just has static math methds.
*
* @author Christopher Manning
* @version 2003/01/02
*/
public final class SloppyMath {
private SloppyMath() {} // this class is just static methods.
/** Round a double to the nearest integer, via conventional rules
* (.5 rounds up, .49 rounds down), and return the result, still as a double.
*
* @param x What to round
* @return The rounded value
*/
public static double round(double x) {
return Math.floor(x + 0.5d);
}
/** Round a double to the given number of decimal places,
* rounding to the nearest value via conventional rules (5 rounds up, 49
* rounds down).
* E.g. round(3.1416, 2) == 3.14, round(431.5, -2) == 400,
* round(431.5, 0) = 432
*/
public static double round(double x, int precision) {
double power = Math.pow(10.0, precision);
return round(x * power) / power;
}
/**
* max() that works on three integers. Like many of the other max() functions in this class,
* doesn't perform special checks like NaN or -0.0f to save time.
*
* @return The maximum of three int values.
*/
public static int max(int a, int b, int c) {
int ma;
ma = a;
if (b > ma) {
ma = b;
}
if (c > ma) {
ma = c;
}
return ma;
}
public static int max(Collection vals) {
if (vals.isEmpty()) { throw new RuntimeException(); }
int max = Integer.MIN_VALUE;
for (int i : vals) {
if (i > max) { max = i; }
}
return max;
}
/**
* Returns the greater of two int values. That
* is, the result is the argument closer to positive infinity. If
* the arguments have the same value, the result is that same
* value. Does none of the special checks for NaN or -0.0f that
* Math.max does.
*
* @param a an argument.
* @param b another argument.
* @return the larger of a and b.
*/
public static int max(int a, int b) {
return (a >= b) ? a : b;
}
/**
* Returns the greater of two float values. That is,
* the result is the argument closer to positive infinity. If the
* arguments have the same value, the result is that same
* value. Does none of the special checks for NaN or -0.0f that
* Math.max does.
*
* @param a an argument.
* @param b another argument.
* @return the larger of a and b.
*/
public static float max(float a, float b) {
return (a >= b) ? a : b;
}
/**
* Returns the greater of two double values. That
* is, the result is the argument closer to positive infinity. If
* the arguments have the same value, the result is that same
* value. Does none of the special checks for NaN or -0.0f that
* Math.max does.
*
* @param a an argument.
* @param b another argument.
* @return the larger of a and b.
*/
public static double max(double a, double b) {
return (a >= b) ? a : b;
}
/**
* Returns the minimum of three int values.
*/
public static int min(int a, int b, int c) {
int mi;
mi = a;
if (b < mi) {
mi = b;
}
if (c < mi) {
mi = c;
}
return mi;
}
/**
* Returns the smaller of two float values. That is,
* the result is the value closer to negative infinity. If the
* arguments have the same value, the result is that same
* value. Does none of the special checks for NaN or -0.0f that
* Math.max does.
*
* @param a an argument.
* @param b another argument.
* @return the smaller of a and b.
*/
public static float min(float a, float b) {
return (a <= b) ? a : b;
}
/**
* Returns the smaller of two double values. That
* is, the result is the value closer to negative infinity. If the
* arguments have the same value, the result is that same
* value. Does none of the special checks for NaN or -0.0f that
* Math.max does.
*
* @param a an argument.
* @param b another argument.
* @return the smaller of a and b.
*/
public static double min(double a, double b) {
return (a <= b) ? a : b;
}
/**
* @return an approximation of the log of the Gamma function of x. Laczos Approximation
* Reference: Numerical Recipes in C
* http://www.library.cornell.edu/nr/cbookcpdf.html
* from www.cs.berkeley.edu/~milch/blog/versions/blog-0.1.3/blog/distrib
*/
public static double lgamma(double x) {
double[] cof = {76.18009172947146, -86.50532032941677,
24.01409824083091,-1.231739572450155,
0.1208650973866179e-2,-0.5395239384953e-5};
double xxx = x;
double tmp = x + 5.5;
tmp -= ((x + 0.5) * Math.log(tmp));
double ser = 1.000000000190015;
for (int j = 0; j < 6; j++) {
xxx++;
ser += cof[j] / xxx;
}
return -tmp + Math.log(2.5066282746310005*ser / x);
}
/**
* Returns true if the argument is a "dangerous" double to have
* around, namely one that is infinite, NaN or zero.
*/
public static boolean isDangerous(double d) {
return Double.isInfinite(d) || Double.isNaN(d) || d == 0.0;
}
/**
* Returns true if the argument is a "very dangerous" double to have
* around, namely one that is infinite or NaN.
*/
public static boolean isVeryDangerous(double d) {
return Double.isInfinite(d) || Double.isNaN(d);
}
public static boolean isCloseTo(double a, double b) {
if (a>b) {
return (a-b)<1e-4;
} else {
return (b-a)<1e-4;
}
}
/**
* If a difference is bigger than this in log terms, then the sum or
* difference of them will just be the larger (to 12 or so decimal
* places for double, and 7 or 8 for float).
*/
static final double LOGTOLERANCE = 30.0;
static final float LOGTOLERANCE_F = 20.0f;
public static double gamma(double n) {
return Math.sqrt(2.0*Math.PI/n) * Math.pow((n/Math.E)*Math.sqrt(n*Math.sinh((1.0/n)+(1/810*Math.pow(n,6)))),n);
}
/**
* Convenience method for log to a different base
*/
public static double log(double num, double base) {
return Math.log(num)/Math.log(base);
}
/**
* Returns the log of the sum of two numbers, which are
* themselves input in log form. This uses natural logarithms.
* Reasonable care is taken to do this as efficiently as possible
* (under the assumption that the numbers might differ greatly in
* magnitude), with high accuracy, and without numerical overflow.
* Also, handle correctly the case of arguments being -Inf (e.g.,
* probability 0).
*
* @param lx First number, in log form
* @param ly Second number, in log form
* @return log(exp(lx) + exp(ly))
*/
public static float logAdd(float lx, float ly) {
float max, negDiff;
if (lx > ly) {
max = lx;
negDiff = ly - lx;
} else {
max = ly;
negDiff = lx - ly;
}
if (max == Double.NEGATIVE_INFINITY) {
return max;
} else if (negDiff < -LOGTOLERANCE_F) {
return max;
} else {
return max + (float) Math.log(1.0 + Math.exp(negDiff));
}
}
/**
* Returns the log of the sum of two numbers, which are
* themselves input in log form. This uses natural logarithms.
* Reasonable care is taken to do this as efficiently as possible
* (under the assumption that the numbers might differ greatly in
* magnitude), with high accuracy, and without numerical overflow.
* Also, handle correctly the case of arguments being -Inf (e.g.,
* probability 0).
*
* @param lx First number, in log form
* @param ly Second number, in log form
* @return log(exp(lx) + exp(ly))
*/
public static double logAdd(double lx, double ly) {
double max, negDiff;
if (lx > ly) {
max = lx;
negDiff = ly - lx;
} else {
max = ly;
negDiff = lx - ly;
}
if (max == Double.NEGATIVE_INFINITY) {
return max;
} else if (negDiff < -LOGTOLERANCE) {
return max;
} else {
return max + Math.log(1.0 + Math.exp(negDiff));
}
}
/**
* Computes n choose k in an efficient way. Works with
* k == 0 or k == n but undefined if k < 0 or k > n
*
* @return fact(n) / fact(k) * fact(n-k)
*/
public static int nChooseK(int n, int k) {
k = Math.min(k, n - k);
if (k == 0) {
return 1;
}
int accum = n;
for (int i = 1; i < k; i++) {
accum *= (n - i);
accum /= i;
}
return accum / k;
}
/**
* Returns an approximation to Math.pow(a,b) that is ~27x faster
* with a margin of error possibly around ~10%. From
* http://martin.ankerl.com/2007/10/04/optimized-pow-approximation-for-java-and-c-c/
*/
public static double pow(final double a, final double b) {
final int x = (int) (Double.doubleToLongBits(a) >> 32);
final int y = (int) (b * (x - 1072632447) + 1072632447);
return Double.longBitsToDouble(((long) y) << 32);
}
/**
* Exponentiation like we learned in grade school:
* multiply b by itself e times. Uses power of two trick.
* e must be nonnegative!!! no checking!!! For e <= 0,
* the exponent is treated as 0, and 1 is returned. 0^0 also
* returns 1.
* Biased to do quickly small exponents, like the CRF needs.
*
* @param b base
* @param e exponent
* @return b^e
*/
public static int intPow(int b, int e) {
switch (e) {
case 0:
return 1;
case 1:
return b;
case 2:
return b * b;
default:
int result = 1;
int currPow = b;
while (e > 0) {
if ((e & 1) != 0) {
result *= currPow;
}
currPow *= currPow;
e >>= 1;
}
return result;
}
}
/**
* Exponentiation like we learned in grade school:
* multiply b by itself e times. Uses power of two trick.
* e must be nonnegative!!! no checking!!!
*
* @param b base
* @param e exponent
* @return b^e
*/
public static float intPow(float b, int e) {
float result = 1.0f;
float currPow = b;
while (e > 0) {
if ((e & 1) != 0) {
result *= currPow;
}
currPow *= currPow;
e >>= 1;
}
return result;
}
/**
* Exponentiation like we learned in grade school:
* multiply b by itself e times. Uses power of two trick.
* e must be nonnegative!!! no checking!!!
* @param b base
* @param e exponent
* @return b^e
*/
public static double intPow(double b, int e) {
double result = 1.0;
double currPow = b;
while (e > 0) {
if ((e & 1) != 0) {
result *= currPow;
}
currPow *= currPow;
e >>= 1;
}
return result;
}
/**
* Find a hypergeometric distribution. This uses exact math, trying
* fairly hard to avoid numeric overflow by interleaving
* multiplications and divisions.
* (To do: make it even better at avoiding overflow, by using loops
* that will do either a multiple or divide based on the size of the
* intermediate result.)
*
* @param k The number of black balls drawn
* @param n The total number of balls
* @param r The number of black balls
* @param m The number of balls drawn
* @return The hypergeometric value
*/
public static double hypergeometric(int k, int n, int r, int m) {
if (k < 0 || r > n || m > n || n <= 0 || m < 0 || r < 0) {
throw new IllegalArgumentException("Invalid hypergeometric");
}
// exploit symmetry of problem
if (m > n / 2) {
m = n - m;
k = r - k;
}
if (r > n / 2) {
r = n - r;
k = m - k;
}
if (m > r) {
int temp = m;
m = r;
r = temp;
}
// now we have that k <= m <= r <= n/2
if (k < (m + r) - n || k > m) {
return 0.0;
}
// Do limit cases explicitly
// It's unclear whether this is a good idea. I put it in fearing
// numerical errors when the numbers seemed off, but actually there
// was a bug in the Fisher's exact routine.
if (r == n) {
if (k == m) {
return 1.0;
} else {
return 0.0;
}
} else if (r == n - 1) {
if (k == m) {
return (n - m) / (double) n;
} else if (k == m - 1) {
return m / (double) n;
} else {
return 0.0;
}
} else if (m == 1) {
if (k == 0) {
return (n - r) / (double) n;
} else if (k == 1) {
return r / (double) n;
} else {
return 0.0;
}
} else if (m == 0) {
if (k == 0) {
return 1.0;
} else {
return 0.0;
}
} else if (k == 0) {
double ans = 1.0;
for (int m0 = 0; m0 < m; m0++) {
ans *= ((n - r) - m0);
ans /= (n - m0);
}
return ans;
}
double ans = 1.0;
// do (n-r)x...x((n-r)-((m-k)-1))/n x...x (n-((m-k-1)))
// leaving rest of denominator to get to multimply by (n-(m-1))
// that's k things which goes into next loop
for (int nr = n - r, n0 = n; nr > (n - r) - (m - k); nr--, n0--) {
// System.out.println("Multiplying by " + nr);
ans *= nr;
// System.out.println("Dividing by " + n0);
ans /= n0;
}
// System.out.println("Done phase 1");
for (int k0 = 0; k0 < k; k0++) {
ans *= (m - k0);
// System.out.println("Multiplying by " + (m-k0));
ans /= ((n - (m - k0)) + 1);
// System.out.println("Dividing by " + ((n-(m+k0)+1)));
ans *= (r - k0);
// System.out.println("Multiplying by " + (r-k0));
ans /= (k0 + 1);
// System.out.println("Dividing by " + (k0+1));
}
return ans;
}
/**
* Find a one tailed exact binomial test probability. Finds the chance
* of this or a higher result
*
* @param k number of successes
* @param n Number of trials
* @param p Probability of a success
*/
public static double exactBinomial(int k, int n, double p) {
double total = 0.0;
for (int m = k; m <= n; m++) {
double nChooseM = 1.0;
for (int r = 1; r <= m; r++) {
nChooseM *= (n - r) + 1;
nChooseM /= r;
}
// System.out.println(n + " choose " + m + " is " + nChooseM);
// System.out.println("prob contribution is " +
// (nChooseM * Math.pow(p, m) * Math.pow(1.0-p, n - m)));
total += nChooseM * Math.pow(p, m) * Math.pow(1.0 - p, n - m);
}
return total;
}
/**
* Find a one-tailed Fisher's exact probability. Chance of having seen
* this or a more extreme departure from what you would have expected
* given independence. I.e., k >= the value passed in.
* Warning: this was done just for collocations, where you are
* concerned with the case of k being larger than predicted. It doesn't
* correctly handle other cases, such as k being smaller than expected.
*
* @param k The number of black balls drawn
* @param n The total number of balls
* @param r The number of black balls
* @param m The number of balls drawn
* @return The Fisher's exact p-value
*/
public static double oneTailedFishersExact(int k, int n, int r, int m) {
if (k < 0 || k < (m + r) - n || k > r || k > m || r > n || m > n) {
throw new IllegalArgumentException("Invalid Fisher's exact: " + "k=" + k + " n=" + n + " r=" + r + " m=" + m + " k<0=" + (k < 0) + " k<(m+r)-n=" + (k < (m + r) - n) + " k>r=" + (k > r) + " k>m=" + (k > m) + " r>n=" + (r > n) + "m>n=" + (m > n));
}
// exploit symmetry of problem
if (m > n / 2) {
m = n - m;
k = r - k;
}
if (r > n / 2) {
r = n - r;
k = m - k;
}
if (m > r) {
int temp = m;
m = r;
r = temp;
}
// now we have that k <= m <= r <= n/2
double total = 0.0;
if (k > m / 2) {
// sum from k to m
for (int k0 = k; k0 <= m; k0++) {
// System.out.println("Calling hypg(" + k0 + "; " + n +
// ", " + r + ", " + m + ")");
total += SloppyMath.hypergeometric(k0, n, r, m);
}
} else {
// sum from max(0, (m+r)-n) to k-1, and then subtract from 1
int min = Math.max(0, (m + r) - n);
for (int k0 = min; k0 < k; k0++) {
// System.out.println("Calling hypg(" + k0 + "; " + n +
// ", " + r + ", " + m + ")");
total += SloppyMath.hypergeometric(k0, n, r, m);
}
total = 1.0 - total;
}
return total;
}
/**
* Find a 2x2 chi-square value.
* Note: could do this more neatly using simplified formula for 2x2 case.
*
* @param k The number of black balls drawn
* @param n The total number of balls
* @param r The number of black balls
* @param m The number of balls drawn
* @return The Fisher's exact p-value
*/
public static double chiSquare2by2(int k, int n, int r, int m) {
int[][] cg = {{k, r - k}, {m - k, n - (k + (r - k) + (m - k))}};
int[] cgr = {r, n - r};
int[] cgc = {m, n - m};
double total = 0.0;
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
double exp = (double) cgr[i] * cgc[j] / n;
total += (cg[i][j] - exp) * (cg[i][j] - exp) / exp;
}
}
return total;
}
/**
* Compute the sigmoid function with mean zero.
* Care is taken to compute an accurate answer without
* numerical overflow. (Added by rajatr)
*
* @param x Point to compute sigmoid at.
* @return Value of the sigmoid, given by 1/(1+exp(-x))
*/
public static double sigmoid(double x) {
if (x<0) {
double num = Math.exp(x);
return num / (1.0 + num);
}
else {
double den = 1.0 + Math.exp(-x);
return 1.0 / den;
}
}
public static double poisson(int x, double lambda) {
if (x<0 || lambda<=0.0) throw new RuntimeException("Bad arguments: " + x + " and " + lambda);
double p = (Math.exp(-lambda) * Math.pow(lambda, x)) / factorial(x);
if (Double.isInfinite(p) || p<=0.0) throw new RuntimeException(Math.exp(-lambda) +" "+ Math.pow(lambda, x) +" "+ factorial(x));
return p;
}
/**
* Uses floating point so that it can represent the really big numbers that come up.
* @param x Argumet to take factorial of
* @return Factorial of argument
*/
public static double factorial(int x) {
double result = 1.0;
for (int i=x; i>1; i--) {
result *= i;
}
return result;
}
/**
* Tests the hypergeometric distribution code, or other functions
* provided in this module.
*
* @param args Either none, and the log add rountines are tested, or the
* following 4 arguments: k (cell), n (total), r (row), m (col)
*/
public static void main(String[] args) {
if (args.length == 0) {
System.err.println("Usage: java edu.stanford.nlp.math.SloppyMath " + "[-logAdd|-fishers k n r m|-binomial r n p");
} else if (args[0].equals("-logAdd")) {
System.out.println("Log adds of neg infinity numbers, etc.");
System.out.println("(logs) -Inf + -Inf = " + logAdd(Double.NEGATIVE_INFINITY, Double.NEGATIVE_INFINITY));
System.out.println("(logs) -Inf + -7 = " + logAdd(Double.NEGATIVE_INFINITY, -7.0));
System.out.println("(logs) -7 + -Inf = " + logAdd(-7.0, Double.NEGATIVE_INFINITY));
System.out.println("(logs) -50 + -7 = " + logAdd(-50.0, -7.0));
System.out.println("(logs) -11 + -7 = " + logAdd(-11.0, -7.0));
System.out.println("(logs) -7 + -11 = " + logAdd(-7.0, -11.0));
System.out.println("real 1/2 + 1/2 = " + logAdd(Math.log(0.5), Math.log(0.5)));
} else if (args[0].equals("-fishers")) {
int k = Integer.parseInt(args[1]);
int n = Integer.parseInt(args[2]);
int r = Integer.parseInt(args[3]);
int m = Integer.parseInt(args[4]);
double ans = SloppyMath.hypergeometric(k, n, r, m);
System.out.println("hypg(" + k + "; " + n + ", " + r + ", " + m + ") = " + ans);
ans = SloppyMath.oneTailedFishersExact(k, n, r, m);
System.out.println("1-tailed Fisher's exact(" + k + "; " + n + ", " + r + ", " + m + ") = " + ans);
double ansChi = SloppyMath.chiSquare2by2(k, n, r, m);
System.out.println("chiSquare(" + k + "; " + n + ", " + r + ", " + m + ") = " + ansChi);
System.out.println("Swapping arguments should give same hypg:");
ans = SloppyMath.hypergeometric(k, n, r, m);
System.out.println("hypg(" + k + "; " + n + ", " + m + ", " + r + ") = " + ans);
int othrow = n - m;
int othcol = n - r;
int cell12 = m - k;
int cell21 = r - k;
int cell22 = othrow - (r - k);
ans = SloppyMath.hypergeometric(cell12, n, othcol, m);
System.out.println("hypg(" + cell12 + "; " + n + ", " + othcol + ", " + m + ") = " + ans);
ans = SloppyMath.hypergeometric(cell21, n, r, othrow);
System.out.println("hypg(" + cell21 + "; " + n + ", " + r + ", " + othrow + ") = " + ans);
ans = SloppyMath.hypergeometric(cell22, n, othcol, othrow);
System.out.println("hypg(" + cell22 + "; " + n + ", " + othcol + ", " + othrow + ") = " + ans);
} else if (args[0].equals("-binomial")) {
int k = Integer.parseInt(args[1]);
int n = Integer.parseInt(args[2]);
double p = Double.parseDouble(args[3]);
double ans = SloppyMath.exactBinomial(k, n, p);
System.out.println("Binomial p(X >= " + k + "; " + n + ", " + p + ") = " + ans);
} else {
System.err.println("Unknown option: " + args[0]);
}
}
}
