/*******************************************************************************
 * Copyright (c) 2000, 2014 IBM Corporation and others.
 * All rights reserved. This program and the accompanying materials
 * are made available under the terms of the Eclipse Public License v1.0
 * which accompanies this distribution, and is available at
 * http://www.eclipse.org/legal/epl-v10.html
 *
 * Contributors:
 *     IBM Corporation - initial API and implementation
 *******************************************************************************/
package org.eclipse.compare.internal.core;

import org.eclipse.core.runtime.OperationCanceledException;
import org.eclipse.core.runtime.SubMonitor;


/* Used to determine the change set responsible for each line */
public abstract class LCS {
	public static final double TOO_LONG = 100000000.0;	// 10^8, the value of N*M when
														// to start binding the
														// run time

	private static final double POW_LIMIT = 1.5; // limit the time to
													// D^POW_LIMIT

	private int max_differences; // the maximum number of differences from
									// each end to consider

	private int length;

	/**
	 * Myers' algorithm for longest common subsequence. O((M + N)D) worst case
	 * time, O(M + N + D^2) expected time, O(M + N) space
	 * (http://citeseer.ist.psu.edu/myers86ond.html)
	 *
	 * Note: Beyond implementing the algorithm as described in the paper I have
	 * added diagonal range compression which helps when finding the LCS of a
	 * very long and a very short sequence, also bound the running time to (N +
	 * M)^1.5 when both sequences are very long.
	 *
	 * After this method is called, the longest common subsequence is available
	 * by calling getResult() where result[0] is composed of
	 * entries from l1 and result[1] is composed of entries from l2
	 * @param subMonitor
	 */
	public void longestCommonSubsequence(SubMonitor subMonitor) {
		int length1 = getLength1();
		int length2 = getLength2();
		if (length1 == 0 || length2 == 0) {
			this.length = 0;
			return;
		}

		this.max_differences = (length1 + length2 + 1) / 2; // ceil((N+M)/2)
		if (!isCappingDisabled() && (double) length1 * (double) length2 > TOO_LONG) {
			// limit complexity to D^POW_LIMIT for long sequences
			this.max_differences = (int) Math.pow(this.max_differences, POW_LIMIT - 1.0);
		}

		initializeLcs(length1);

		subMonitor.beginTask(null, length1);

		/*
		 * The common prefixes and suffixes are always part of some LCS, include
		 * them now to reduce our search space
		 */
		int forwardBound;
		int max = Math.min(length1, length2);
		for (forwardBound = 0; forwardBound < max
				&& isRangeEqual(forwardBound, forwardBound); forwardBound++) {
			setLcs(forwardBound, forwardBound);
			worked(subMonitor, 1);
		}

		int backBoundL1 = length1 - 1;
		int backBoundL2 = length2 - 1;

		while (backBoundL1 >= forwardBound && backBoundL2 >= forwardBound
				&& isRangeEqual(backBoundL1, backBoundL2)) {
			setLcs(backBoundL1, backBoundL2);
			backBoundL1--;
			backBoundL2--;
			worked(subMonitor, 1);
		}

		this.length = forwardBound
				+ length1
				- backBoundL1
				- 1
				+ lcs_rec(forwardBound, backBoundL1, forwardBound,
						backBoundL2, new int[2][length1 + length2 + 1],
						new int[3], subMonitor);

	}

	private boolean isCappingDisabled() {
		return ComparePlugin.getDefault().isCappingDisabled();
	}

	/**
	 * The recursive helper function for Myers' LCS. Computes the LCS of
	 * l1[bottoml1 .. topl1] and l2[bottoml2 .. topl2] fills in the appropriate
	 * location in lcs and returns the length
	 *
	 * @param l1 The 1st sequence
	 * @param bottoml1 Index in the 1st sequence to start from (inclusive)
	 * @param topl1 Index in the 1st sequence to end on (inclusive)
	 * @param l2 The 2nd sequence
	 * @param bottoml2 Index in the 2nd sequence to start from (inclusive)
	 * @param topl2 Index in the 2nd sequence to end on (inclusive)
	 * @param V should be allocated as int[2][l1.length + l2.length + 1], used
	 *            to store furthest reaching D-paths
	 * @param snake should be allocated as int[3], used to store the beginning
	 *            x, y coordinates and the length of the latest snake traversed
	 * @param subMonitor
	 * @param lcs should be allocated as TextLine[2][l1.length], used to store
	 *            the common points found to be part of the LCS where lcs[0]
	 *            references lines of l1 and lcs[1] references lines of l2.
	 *
	 * @return the length of the LCS
	 */
	private int lcs_rec(
			int bottoml1, int topl1,
			int bottoml2, int topl2,
			int[][] V, int[] snake, SubMonitor subMonitor) {

		// check that both sequences are non-empty
		if (bottoml1 > topl1 || bottoml2 > topl2) {
			return 0;
		}

		int d = find_middle_snake(bottoml1, topl1, bottoml2, topl2, V, snake, subMonitor);
		// System.out.println(snake[0] + " " + snake[1] + " " + snake[2]);

		// need to store these so we don't lose them when they're overwritten by
		// the recursion
		int len = snake[2];
		int startx = snake[0];
		int starty = snake[1];

		// the middle snake is part of the LCS, store it
		for (int i = 0; i < len; i++) {
			setLcs(startx + i, starty + i);
			worked(subMonitor, 1);
		}

		if (d > 1) {
			return len
					+ lcs_rec(bottoml1, startx - 1, bottoml2, starty - 1, V, snake, subMonitor)
					+ lcs_rec(startx + len, topl1, starty + len, topl2, V, snake, subMonitor);
		} else if (d == 1) {
			/*
			 * In this case the sequences differ by exactly 1 line. We have
			 * already saved all the lines after the difference in the for loop
			 * above, now we need to save all the lines before the difference.
			 */
			int max = Math.min(startx - bottoml1, starty - bottoml2);
			for (int i = 0; i < max; i++) {
				setLcs(bottoml1 + i, bottoml2 + i);
				worked(subMonitor, 1);
			}
			return max + len;
		}

		return len;
	}

	private void worked(SubMonitor subMonitor, int work) {
		if (subMonitor.isCanceled())
			throw new OperationCanceledException();
		subMonitor.worked(work);
	}

	/**
	 * Helper function for Myers' LCS algorithm to find the middle snake for
	 * l1[bottoml1..topl1] and l2[bottoml2..topl2] The x, y coodrdinates of the
	 * start of the middle snake are saved in snake[0], snake[1] respectively
	 * and the length of the snake is saved in s[2].
	 *
	 * @param l1 The 1st sequence
	 * @param bottoml1 Index in the 1st sequence to start from (inclusive)
	 * @param topl1 Index in the 1st sequence to end on (inclusive)
	 * @param l2 The 2nd sequence
	 * @param bottoml2 Index in the 2nd sequence to start from (inclusive)
	 * @param topl2 Index in the 2nd sequence to end on (inclusive)
	 * @param V should be allocated as int[2][l1.length + l2.length + 1], used
	 *            to store furthest reaching D-paths
	 * @param snake should be allocated as int[3], used to store the beginning
	 *            x, y coordinates and the length of the middle snake
	 * @subMonitor subMonitor
	 *
	 * @return The number of differences (SES) between l1[bottoml1..topl1] and
	 *         l2[bottoml2..topl2]
	 */
	private int find_middle_snake(
			int bottoml1, int topl1,
			int bottoml2, int topl2,
			int[][] V, int[] snake,
			SubMonitor subMonitor) {
		int N = topl1 - bottoml1 + 1;
		int M = topl2 - bottoml2 + 1;
		// System.out.println("N: " + N + " M: " + M + " bottom: " + bottoml1 +
		// ", " +
		// bottoml2 + " top: " + topl1 + ", " + topl2);
		int delta = N - M;
		boolean isEven;
		if ((delta & 1) == 1) {
			isEven = false;
		} else {
			isEven = true;
		}

		int limit = Math.min(this.max_differences, (N + M + 1) / 2); // ceil((N+M)/2)

		int value_to_add_forward; // a 0 or 1 that we add to the start offset
									// to make it odd/even
		if ((M & 1) == 1) {
			value_to_add_forward = 1;
		} else {
			value_to_add_forward = 0;
		}

		int value_to_add_backward;
		if ((N & 1) == 1) {
			value_to_add_backward = 1;
		} else {
			value_to_add_backward = 0;
		}

		int start_forward = -M;
		int end_forward = N;
		int start_backward = -N;
		int end_backward = M;

		V[0][limit + 1] = 0;
		V[1][limit - 1] = N;
		for (int d = 0; d <= limit; d++) {

			int start_diag = Math.max(value_to_add_forward + start_forward, -d);
			int end_diag = Math.min(end_forward, d);
			value_to_add_forward = 1 - value_to_add_forward;

			// compute forward furthest reaching paths
			for (int k = start_diag; k <= end_diag; k += 2) {
				int x;
				if (k == -d
						|| (k < d && V[0][limit + k - 1] < V[0][limit + k + 1])) {
					x = V[0][limit + k + 1];
				} else {
					x = V[0][limit + k - 1] + 1;
				}

				int y = x - k;

				snake[0] = x + bottoml1;
				snake[1] = y + bottoml2;
				snake[2] = 0;
				// System.out.println("1 x: " + x + " y: " + y + " k: " + k + "
				// d: " + d );
				while (x < N && y < M
						&& isRangeEqual(x + bottoml1, y + bottoml2)) {
					x++;
					y++;
					snake[2]++;
				}
				V[0][limit + k] = x;
				// System.out.println(x + " " + V[1][limit+k -delta] + " " + k +
				// " " + delta);
				if (!isEven && k >= delta - d + 1 && k <= delta + d - 1
						&& x >= V[1][limit + k - delta]) {
					// System.out.println("Returning: " + (2*d-1));
					return 2 * d - 1;
				}

				// check to see if we can cut down the diagonal range
				if (x >= N && end_forward > k - 1) {
					end_forward = k - 1;
				} else if (y >= M) {
					start_forward = k + 1;
					value_to_add_forward = 0;
				}
			}

			start_diag = Math.max(value_to_add_backward + start_backward, -d);
			end_diag = Math.min(end_backward, d);
			value_to_add_backward = 1 - value_to_add_backward;

			// compute backward furthest reaching paths
			for (int k = start_diag; k <= end_diag; k += 2) {
				int x;
				if (k == d
						|| (k != -d && V[1][limit + k - 1] < V[1][limit + k + 1])) {
					x = V[1][limit + k - 1];
				} else {
					x = V[1][limit + k + 1] - 1;
				}

				int y = x - k - delta;

				snake[2] = 0;
				// System.out.println("2 x: " + x + " y: " + y + " k: " + k + "
				// d: " + d);
				while (x > 0 && y > 0
						&& isRangeEqual(x - 1 + bottoml1, y - 1 + bottoml2)) {
					x--;
					y--;
					snake[2]++;
				}
				V[1][limit + k] = x;

				if (isEven && k >= -delta - d && k <= d - delta
						&& x <= V[0][limit + k + delta]) {
					// System.out.println("Returning: " + 2*d);
					snake[0] = bottoml1 + x;
					snake[1] = bottoml2 + y;

					return 2 * d;
				}

				// check to see if we can cut down our diagonal range
				if (x <= 0) {
					start_backward = k + 1;
					value_to_add_backward = 0;
				} else if (y <= 0 && end_backward > k - 1) {
					end_backward = k - 1;
				}
			}
			worked(subMonitor, 1);
		}

		/*
		 * computing the true LCS is too expensive, instead find the diagonal
		 * with the most progress and pretend a midle snake of length 0 occurs
		 * there.
		 */

		int[] most_progress = findMostProgress(M, N, limit, V);

		snake[0] = bottoml1 + most_progress[0];
		snake[1] = bottoml2 + most_progress[1];
		snake[2] = 0;
		return 5; /*
					 * HACK: since we didn't really finish the LCS computation
					 * we don't really know the length of the SES. We don't do
					 * anything with the result anyway, unless it's <=1. We know
					 * for a fact SES > 1 so 5 is as good a number as any to
					 * return here
					 */
	}

	/**
	 * Takes the array with furthest reaching D-paths from an LCS computation
	 * and returns the x,y coordinates and progress made in the middle diagonal
	 * among those with maximum progress, both from the front and from the back.
	 *
	 * @param M the length of the 1st sequence for which LCS is being computed
	 * @param N the length of the 2nd sequence for which LCS is being computed
	 * @param limit the number of steps made in an attempt to find the LCS from
	 *            the front and back
	 * @param V the array storing the furthest reaching D-paths for the LCS
	 *            computation
	 * @return The result as an array of 3 integers where result[0] is the x
	 *         coordinate of the current location in the diagonal with the most
	 *         progress, result[1] is the y coordinate of the current location
	 *         in the diagonal with the most progress and result[2] is the
	 *         amount of progress made in that diagonal
	 */
	private static int[] findMostProgress(int M, int N, int limit, int[][] V) {
		int delta = N - M;

		int forward_start_diag;
		if ((M & 1) == (limit & 1)) {
			forward_start_diag = Math.max(-M, -limit);
		} else {
			forward_start_diag = Math.max(1 - M, -limit);
		}

		int forward_end_diag = Math.min(N, limit);

		int backward_start_diag;
		if ((N & 1) == (limit & 1)) {
			backward_start_diag = Math.max(-N, -limit);
		} else {
			backward_start_diag = Math.max(1 - N, -limit);
		}

		int backward_end_diag = Math.min(M, limit);

		int[][] max_progress = new int[Math.max(forward_end_diag
				- forward_start_diag, backward_end_diag - backward_start_diag) / 2 + 1][3];
		int num_progress = 0; // the 1st entry is current, it is initialized
								// with 0s

		// first search the forward diagonals
		for (int k = forward_start_diag; k <= forward_end_diag; k += 2) {
			int x = V[0][limit + k];
			int y = x - k;
			if (x > N || y > M) {
				continue;
			}

			int progress = x + y;
			if (progress > max_progress[0][2]) {
				num_progress = 0;
				max_progress[0][0] = x;
				max_progress[0][1] = y;
				max_progress[0][2] = progress;
			} else if (progress == max_progress[0][2]) {
				num_progress++;
				max_progress[num_progress][0] = x;
				max_progress[num_progress][1] = y;
				max_progress[num_progress][2] = progress;
			}
		}

		boolean max_progress_forward = true; // initially the maximum
												// progress is in the forward
												// direction

		// now search the backward diagonals
		for (int k = backward_start_diag; k <= backward_end_diag; k += 2) {
			int x = V[1][limit + k];
			int y = x - k - delta;
			if (x < 0 || y < 0) {
				continue;
			}

			int progress = N - x + M - y;
			if (progress > max_progress[0][2]) {
				num_progress = 0;
				max_progress_forward = false;
				max_progress[0][0] = x;
				max_progress[0][1] = y;
				max_progress[0][2] = progress;
			} else if (progress == max_progress[0][2] && !max_progress_forward) {
				num_progress++;
				max_progress[num_progress][0] = x;
				max_progress[num_progress][1] = y;
				max_progress[num_progress][2] = progress;
			}
		}

		// return the middle diagonal with maximal progress.
		return max_progress[num_progress / 2];
	}

	protected abstract int getLength2();

	protected abstract int getLength1();

	protected abstract boolean isRangeEqual(int i1, int i2);

	protected abstract void setLcs(int sl1, int sl2);

	protected abstract void initializeLcs(int lcsLength);

	public int getLength() {
		return this.length;
	}
}
