/*
 * Copyright (C) 2017 The Android Open Source Project
 *
 * Licensed under the Apache License, Version 2.0 (the "License");
 * you may not use this file except in compliance with the License.
 * You may obtain a copy of the License at
 *
 *      http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 */

package com.android.dialer.common.backoff;

import com.android.dialer.common.Assert;

/**
 * Given an initial delay, D, a maximum total backoff time, T, and a maximum number of backoffs, N,
 * this class calculates a base multiplier, b, and a scaling factor, g, such that the initial
 * backoff is g*b = D and the sum of the scaled backoffs is T = g*b + g*b^2 + g*b^3 + ... g*b^N
 *
 * <p>T/D = (1 - b^N)/(1 - b) but this cannot be written as a simple equation for b in terms of T, N
 * and D so instead use Newton's method (https://en.wikipedia.org/wiki/Newton%27s_method) to find an
 * approximate value for b.
 *
 * <p>Example usage using the {@code ExponentialBackoff} would be:
 *
 * <pre>
 *   // Retry with exponential backoff for up to 2 minutes, with an initial delay of 100 millis
 *   // and a maximum of 10 retries
 *   long initialDelayMillis = 100;
 *   int maxTries = 10;
 *   double base = ExponentialBaseCalculator.findBase(
 *       initialDelayMillis,
 *       TimeUnit.MINUTES.toMillis(2),
 *       maxTries);
 *   ExponentialBackoff backoff = new ExponentialBackoff(initialDelayMillis, base, maxTries);
 *   while (backoff.isInRange()) {
 *     ...
 *     long delay = backoff.getNextBackoff();
 *     // Wait for the indicated time...
 *   }
 * </pre>
 */
public final class ExponentialBaseCalculator {
  private static final int MAX_STEPS = 1000;
  private static final double DEFAULT_TOLERANCE_MILLIS = 1;

  /**
   * Calculate an exponential backoff base multiplier such that the first backoff delay will be as
   * specified and the sum of the delays after doing the indicated maximum number of backoffs will
   * be as specified.
   *
   * @throws IllegalArgumentException if the initial delay is greater than the total backoff time
   * @throws IllegalArgumentException if the maximum number of backoffs is not greater than 1
   * @throws IllegalStateException if it fails to find an acceptable base multiplier
   */
  public static double findBase(
      long initialDelayMillis, long totalBackoffTimeMillis, int maximumBackoffs) {
    Assert.checkArgument(initialDelayMillis < totalBackoffTimeMillis);
    Assert.checkArgument(maximumBackoffs > 1);
    long scaledTotalTime = Math.round(((double) totalBackoffTimeMillis) / initialDelayMillis);
    double scaledTolerance = DEFAULT_TOLERANCE_MILLIS / initialDelayMillis;
    return getBaseImpl(scaledTotalTime, maximumBackoffs, scaledTolerance);
  }

  /**
   * T/D = (1 - b^N)/(1 - b) but this cannot be written as a simple equation for b in terms of T, D
   * and N so instead we use Newtons method to find an approximate value for b.
   *
   * <p>Let f(b) = (1 - b^N)/(1 - b) - T/D then we want to find b* such that f(b*) = 0, or more
   * precisely |f(b*)| < tolerance
   *
   * <p>Using Newton's method we can interatively find b* as follows: b1 = b0 - f(b0)/f'(b0), where
   * b0 is the current best guess for b* and b1 is the next best guess.
   *
   * <p>f'(b) = (f(b) + T/D - N*b^(N - 1))/(1 - b)
   *
   * <p>so
   *
   * <p>b1 = b0 - f(b0)(1 - b0)/(f(b0) + T/D - N*b0^(N - 1))
   */
  private static double getBaseImpl(long t, int n, double tolerance) {
    double b0 = 2; // Initial guess for b*
    double b0n = Math.pow(b0, n);
    double fb0 = f(b0, t, b0n);
    if (Math.abs(fb0) < tolerance) {
      // Initial guess was pretty good
      return b0;
    }

    for (int i = 0; i < MAX_STEPS; i++) {
      double fpb0 = fp(b0, t, n, fb0, b0n);
      double b1 = b0 - fb0 / fpb0;
      double b1n = Math.pow(b1, n);
      double fb1 = f(b1, t, b1n);

      if (Math.abs(fb1) < tolerance) {
        // Found an acceptable value
        return b1;
      }

      b0 = b1;
      b0n = b1n;
      fb0 = fb1;
    }

    throw new IllegalStateException("Failed to find base. Too many iterations.");
  }

  // Evaluate f(b), the function we are trying to find the zero for.
  // Note: passing b^N as a parameter so it only has to be calculated once
  private static double f(double b, long t, double bn) {
    return (1 - bn) / (1 - b) - t;
  }

  // Evaluate f'(b), the derivative of the function we are trying to find the zero for.
  // Note: passing f(b) and b^N as parameters for efficiency
  private static double fp(double b, long t, int n, double fb, double bn) {
    return (fb + t - n * bn / b) / (1 - b);
  }
}