mathutils.java

Apache的common math数学软件包

     * Double.MAX_VALUE, Double.POSITIVE_INFINITY is returned</li>
     * </ul>
     * </p>
     * 
     * @param n argument
     * @return <code>n!</code>
     * @throws IllegalArgumentException if n < 0
     */
    public static double factorialDouble(final int n) {
        if (n < 0) {
            throw new IllegalArgumentException("must have n >= 0 for n!");
        }
        return Math.floor(Math.exp(factorialLog(n)) + 0.5);
    }

    /**
     * Returns the natural logarithm of n!.
     * <p>
     * <Strong>Preconditions</strong>:
     * <ul>
     * <li> <code>n >= 0</code> (otherwise
     * <code>IllegalArgumentException</code> is thrown)</li>
     * </ul></p>
     * 
     * @param n argument
     * @return <code>n!</code>
     * @throws IllegalArgumentException if preconditions are not met.
     */
    public static double factorialLog(final int n) {
        if (n < 0) {
            throw new IllegalArgumentException("must have n > 0 for n!");
        }
        double logSum = 0;
        for (int i = 2; i <= n; i++) {
            logSum += Math.log((double)i);
        }
        return logSum;
    }

    /**
     * <p>
     * Gets the greatest common divisor of the absolute value of two numbers,
     * using the "binary gcd" method which avoids division and modulo
     * operations. See Knuth 4.5.2 algorithm B. This algorithm is due to Josef
     * Stein (1961).
     * </p>
     * 
     * @param u a non-zero number
     * @param v a non-zero number
     * @return the greatest common divisor, never zero
     * @since 1.1
     */
    public static int gcd(int u, int v) {
        if (u * v == 0) {
            return (Math.abs(u) + Math.abs(v));
        }
        // keep u and v negative, as negative integers range down to
        // -2^31, while positive numbers can only be as large as 2^31-1
        // (i.e. we can't necessarily negate a negative number without
        // overflow)
        /* assert u!=0 && v!=0; */
        if (u > 0) {
            u = -u;
        } // make u negative
        if (v > 0) {
            v = -v;
        } // make v negative
        // B1. [Find power of 2]
        int k = 0;
        while ((u & 1) == 0 && (v & 1) == 0 && k < 31) { // while u and v are
                                                            // both even...
            u /= 2;
            v /= 2;
            k++; // cast out twos.
        }
        if (k == 31) {
            throw new ArithmeticException("overflow: gcd is 2^31");
        }
        // B2. Initialize: u and v have been divided by 2^k and at least
        // one is odd.
        int t = ((u & 1) == 1) ? v : -(u / 2)/* B3 */;
        // t negative: u was odd, v may be even (t replaces v)
        // t positive: u was even, v is odd (t replaces u)
        do {
            /* assert u<0 && v<0; */
            // B4/B3: cast out twos from t.
            while ((t & 1) == 0) { // while t is even..
                t /= 2; // cast out twos
            }
            // B5 [reset max(u,v)]
            if (t > 0) {
                u = -t;
            } else {
                v = t;
            }
            // B6/B3. at this point both u and v should be odd.
            t = (v - u) / 2;
            // |u| larger: t positive (replace u)
            // |v| larger: t negative (replace v)
        } while (t != 0);
        return -u * (1 << k); // gcd is u*2^k
    }

    /**
     * Returns an integer hash code representing the given double value.
     * 
     * @param value the value to be hashed
     * @return the hash code
     */
    public static int hash(double value) {
        long bits = Double.doubleToLongBits(value);
        return (int)(bits ^ (bits >>> 32));
    }

    /**
     * Returns an integer hash code representing the given double array value.
     * 
     * @param value the value to be hashed (may be null)
     * @return the hash code
     * @since 1.2
     */
    public static int hash(double[] value) {
        if (value == null) {
            return 0;
        }
        int result = value.length;
        for (int i = 0; i < value.length; ++i) {
            result = result * 31 + hash(value[i]);
        }
        return result;
    }

    /**
     * For a byte value x, this method returns (byte)(+1) if x >= 0 and
     * (byte)(-1) if x < 0.
     * 
     * @param x the value, a byte
     * @return (byte)(+1) or (byte)(-1), depending on the sign of x
     */
    public static byte indicator(final byte x) {
        return (x >= ZB) ? PB : NB;
    }

    /**
     * For a double precision value x, this method returns +1.0 if x >= 0 and
     * -1.0 if x < 0. Returns <code>NaN</code> if <code>x</code> is
     * <code>NaN</code>.
     * 
     * @param x the value, a double
     * @return +1.0 or -1.0, depending on the sign of x
     */
    public static double indicator(final double x) {
        if (Double.isNaN(x)) {
            return Double.NaN;
        }
        return (x >= 0.0) ? 1.0 : -1.0;
    }

    /**
     * For a float value x, this method returns +1.0F if x >= 0 and -1.0F if x <
     * 0. Returns <code>NaN</code> if <code>x</code> is <code>NaN</code>.
     * 
     * @param x the value, a float
     * @return +1.0F or -1.0F, depending on the sign of x
     */
    public static float indicator(final float x) {
        if (Float.isNaN(x)) {
            return Float.NaN;
        }
        return (x >= 0.0F) ? 1.0F : -1.0F;
    }

    /**
     * For an int value x, this method returns +1 if x >= 0 and -1 if x < 0.
     * 
     * @param x the value, an int
     * @return +1 or -1, depending on the sign of x
     */
    public static int indicator(final int x) {
        return (x >= 0) ? 1 : -1;
    }

    /**
     * For a long value x, this method returns +1L if x >= 0 and -1L if x < 0.
     * 
     * @param x the value, a long
     * @return +1L or -1L, depending on the sign of x
     */
    public static long indicator(final long x) {
        return (x >= 0L) ? 1L : -1L;
    }

    /**
     * For a short value x, this method returns (short)(+1) if x >= 0 and
     * (short)(-1) if x < 0.
     * 
     * @param x the value, a short
     * @return (short)(+1) or (short)(-1), depending on the sign of x
     */
    public static short indicator(final short x) {
        return (x >= ZS) ? PS : NS;
    }

    /**
     * Returns the least common multiple between two integer values.
     * 
     * @param a the first integer value.
     * @param b the second integer value.
     * @return the least common multiple between a and b.
     * @throws ArithmeticException if the lcm is too large to store as an int
     * @since 1.1
     */
    public static int lcm(int a, int b) {
        return Math.abs(mulAndCheck(a / gcd(a, b), b));
    }

    /** 
     * <p>Returns the 
     * <a href="http://mathworld.wolfram.com/Logarithm.html">logarithm</a>
     * for base <code>b</code> of <code>x</code>.
     * </p>
     * <p>Returns <code>NaN<code> if either argument is negative.  If 
     * <code>base</code> is 0 and <code>x</code> is positive, 0 is returned.
     * If <code>base</code> is positive and <code>x</code> is 0, 
     * <code>Double.NEGATIVE_INFINITY</code> is returned.  If both arguments
     * are 0, the result is <code>NaN</code>.</p>
     * 
     * @param base the base of the logarithm, must be greater than 0
     * @param x argument, must be greater than 0
     * @return the value of the logarithm - the number y such that base^y = x.
     * @since 1.2
     */ 
    public static double log(double base, double x) {
        return Math.log(x)/Math.log(base);
    }

    /**
     * Multiply two integers, checking for overflow.
     * 
     * @param x a factor
     * @param y a factor
     * @return the product <code>x*y</code>
     * @throws ArithmeticException if the result can not be represented as an
     *         int
     * @since 1.1
     */
    public static int mulAndCheck(int x, int y) {
        long m = ((long)x) * ((long)y);
        if (m < Integer.MIN_VALUE || m > Integer.MAX_VALUE) {
            throw new ArithmeticException("overflow: mul");
        }
        return (int)m;
    }

    /**
     * Multiply two long integers, checking for overflow.
     * 
     * @param a first value
     * @param b second value
     * @return the product <code>a * b</code>
     * @throws ArithmeticException if the result can not be represented as an
     *         long
     * @since 1.2
     */
    public static long mulAndCheck(long a, long b) {
        long ret;
        String msg = "overflow: multiply";
        if (a > b) {
            // use symmetry to reduce boundry cases
            ret = mulAndCheck(b, a);
        } else {
            if (a < 0) {
                if (b < 0) {
                    // check for positive overflow with negative a, negative b
                    if (a >= Long.MAX_VALUE / b) {
                        ret = a * b;
                    } else {
                        throw new ArithmeticException(msg);
                    }
                } else if (b > 0) {
                    // check for negative overflow with negative a, positive b
                    if (Long.MIN_VALUE / b <= a) {
                        ret = a * b;
                    } else {
                        throw new ArithmeticException(msg);
                        
                    }
                } else {
                    // assert b == 0
                    ret = 0;
                }
            } else if (a > 0) {
                // assert a > 0
                // assert b > 0
                
                // check for positive overflow with positive a, positive b
                if (a <= Long.MAX_VALUE / b) {
                    ret = a * b;
                } else {
                    throw new ArithmeticException(msg);
                }
            } else {
                // assert a == 0
                ret = 0;
            }
        }
        return ret;
    }

    /**
     * Get the next machine representable number after a number, moving
     * in the direction of another number.
     * <p>
     * If <code>direction</code> is greater than or equal to<code>d</code>,
     * the smallest machine representable number strictly greater than
     * <code>d</code> is returned; otherwise the largest representable number
     * strictly less than <code>d</code> is returned.</p>
     * <p>
     * If <code>d</code> is NaN or Infinite, it is returned unchanged.</p>
     * 
     * @param d base number
     * @param direction (the only important thing is whether
     * direction is greater or smaller than d)
     * @return the next machine representable number in the specified direction
     * @since 1.2
     */
    public static double nextAfter(double d, double direction) {

        // handling of some important special cases
        if (Double.isNaN(d) || Double.isInfinite(d)) {
                return d;
        } else if (d == 0) {
                return (direction < 0) ? -Double.MIN_VALUE : Double.MIN_VALUE;
        }
        // special cases MAX_VALUE to infinity and  MIN_VALUE to 0
        // are handled just as normal numbers

        // split the double in raw components
        long bits     = Double.doubleToLongBits(d);
        long sign     = bits & 0x8000000000000000L;
        long exponent = bits & 0x7ff0000000000000L;
        long mantissa = bits & 0x000fffffffffffffL;

        if (d * (direction - d) >= 0) {
                // we should increase the mantissa
                if (mantissa == 0x000fffffffffffffL) {
                        return Double.longBitsToDouble(sign |
                                        (exponent + 0x0010000000000000L));