/*
 * Copyright (c) 2016 Martin Davis.
 *
 * All rights reserved. This program and the accompanying materials
 * are made available under the terms of the Eclipse Public License 2.0
 * and Eclipse Distribution License v. 1.0 which accompanies this distribution.
 * The Eclipse Public License is available at http://www.eclipse.org/legal/epl-v20.html
 * and the Eclipse Distribution License is available at
 *
 * http://www.eclipse.org/org/documents/edl-v10.php.
 */
package org.locationtech.jts.math;

import java.util.Random;

/**
 * Various utility functions for mathematical and numerical operations.
 * 
 * @author mbdavis
 *
 */
public class MathUtil 
{
  /**
   * Clamps a <tt>double</tt> value to a given range.
   * @param x the value to clamp
   * @param min the minimum value of the range
   * @param max the maximum value of the range
   * @return the clamped value
   */
  public static double clamp(double x, double min, double max)
  {
    if (x < min) return min;
    if (x > max) return max;
    return x;
  }
  
  /**
   * Clamps an <tt>int</tt> value to a given range.
   * @param x the value to clamp
   * @param min the minimum value of the range
   * @param max the maximum value of the range
   * @return the clamped value
   */
  public static int clamp(int x, int min, int max)
  {
    if (x < min) return min;
    if (x > max) return max;
    return x;
  }
  
  /**
   * Clamps an integer to a given maximum limit.
   * 
   * @param x the value to clamp
   * @param max the maximum value
   * @return the clamped value
   */
  public static int clampMax(int x, int max)
  {
    if (x > max) return max;
    return x;
  }
  
  /**
   * Computes the ceiling function of the dividend of two integers.
   * 
   * @param num the numerator
   * @param denom the denominator
   * @return the ceiling of num / denom
   */
  public static int ceil(int num, int denom) {
    int div = num / denom;
    return div * denom >= num ? div : div + 1;
  }
  
  private static final double LOG_10 = Math.log(10);
  
  /**
   * Computes the base-10 logarithm of a <tt>double</tt> value.
   * <ul>
   * <li>If the argument is NaN or less than zero, then the result is NaN.
   * <li>If the argument is positive infinity, then the result is positive infinity.
   * <li>If the argument is positive zero or negative zero, then the result is negative infinity.
   * </ul>
   *   
   * @param x a positive number
   * @return the value log a, the base-10 logarithm of the input value
   */
  public static double log10(double x)
  {
    double ln = Math.log(x);
    if (Double.isInfinite(ln)) return ln;
    if (Double.isNaN(ln)) return ln;
    return ln / LOG_10;
  }
  
  /**
   * Computes an index which wraps around a given maximum value.
   * For values &gt;= 0, this is equals to <tt>val % max</tt>.
   * For values &lt; 0, this is equal to <tt>max - (-val) % max</tt>
   * 
   * @param index the value to wrap
   * @param max the maximum value (or modulus)
   * @return the wrapped index
   */
  public static int wrap(int index, int max)
  {
    if (index < 0) {
      return max - ((-index) % max);
    }
    return index % max;
  }

  /**
   * Computes the average of two numbers.
   * 
   * @param x1 a number
   * @param x2 a number
   * @return the average of the inputs
   */
  public static double average(double x1, double x2)
  {
    return (x1 + x2) / 2.0;
  }
  
  public static double max(double v1, double v2, double v3)
  {
    double max = v1;
    if (v2 > max) max = v2;
    if (v3 > max) max = v3;
    return max;
  }
  
  public static double max(double v1, double v2, double v3, double v4)
  {
    double max = v1;
    if (v2 > max) max = v2;
    if (v3 > max) max = v3;
    if (v4 > max) max = v4;
    return max;
  }
  
  public static double min(double v1, double v2, double v3, double v4)
  {
    double min = v1;
    if (v2 < min) min = v2;
    if (v3 < min) min = v3;
    if (v4 < min) min = v4;
    return min;
  }
  
  /**
   * The inverse of the Golden Ratio phi.
   */
  public static final double PHI_INV = (Math.sqrt(5) - 1.0) / 2.0;

  /**
   * Generates a quasi-random sequence of numbers in the range [0,1].
   * They are produced by an additive recurrence with 1/&phi; as the constant.
   * This produces a low-discrepancy sequence which is more evenly
   * distribute than random numbers.
   * <p>
   * See <a href='https://en.wikipedia.org/wiki/Low-discrepancy_sequence#Additive_recurrence'>Wikipedia: Low-discrepancy Sequences - Additive Recurrence</a>.
   * <p>
   * The sequence is initialized by calling it 
   * with any positive fractional number; 0 works well for most uses.
   * 
   * @param curr the current number in the sequence
   * @return the next value in the sequence
   */
  public static double quasirandom( double curr) {
    return quasirandom(curr, PHI_INV);
  }
  
  /**
   * Generates a quasi-random sequence of numbers in the range [0,1].
   * They are produced by an additive recurrence with constant &alpha;.
   * <pre>
   *     R(&alpha;) :  t<sub>n</sub> = { t<sub>0</sub> + n&alpha; },  n = 1,2,3,...   
   * </pre>
   * When &alpha; is irrational this produces a 
   * <a href='https://en.wikipedia.org/wiki/Low-discrepancy_sequence#Additive_recurrence'>Low discrepancy sequence</a>
   *  which is more evenly
   * distributed than random numbers.
   * <p>
   * The sequence is initialized by calling it 
   * with any positive fractional number. 0 works well for most uses.
   * 
   * @param curr the current number in the sequence
   * @param alpha the sequence additive constant
   * @return the next value in the sequence
   */
  public static double quasirandom( double curr, double alpha) {
    double next = curr + alpha;
    if (next < 1) return next;
    return next - Math.floor(next);
  }
  
  /**
   * Generates a randomly-shuffled list of the integers from [0..n-1].
   * <p>
   * One use is to randomize points inserted into a {@link KDtree}.
   * 
   * @param n the number of integers to shuffle
   * @return the shuffled array
   */
  public static int[] shuffle(int n) {
    final Random rnd = new Random(13);
    int[] ints = new int[n];
    for (int i = 0; i < n; i++) {
      ints[i] = i;
    }
    for (int i = n - 1; i >= 1; i--) {
      int j = rnd.nextInt(i + 1);
      int last = ints[i];
      ints[i] = ints[j];
      ints[j] = last;
    }
    return ints;
  }
}
