/*
 * Copyright (C) 2008-2009, Johannes E. Schindelin <johannes.schindelin@gmx.de>
 * Copyright (C) 2009, Johannes Schindelin <johannes.schindelin@gmx.de> and others
 *
 * This program and the accompanying materials are made available under the
 * terms of the Eclipse Distribution License v. 1.0 which is available at
 * https://www.eclipse.org/org/documents/edl-v10.php.
 *
 * SPDX-License-Identifier: BSD-3-Clause
 */

package org.eclipse.jgit.diff;

import java.text.MessageFormat;

import org.eclipse.jgit.errors.DiffInterruptedException;
import org.eclipse.jgit.internal.JGitText;
import org.eclipse.jgit.util.IntList;
import org.eclipse.jgit.util.LongList;

/**
 * Diff algorithm, based on "An O(ND) Difference Algorithm and its Variations",
 * by Eugene Myers.
 * <p>
 * The basic idea is to put the line numbers of text A as columns ("x") and the
 * lines of text B as rows ("y"). Now you try to find the shortest "edit path"
 * from the upper left corner to the lower right corner, where you can always go
 * horizontally or vertically, but diagonally from (x,y) to (x+1,y+1) only if
 * line x in text A is identical to line y in text B.
 * <p>
 * Myers' fundamental concept is the "furthest reaching D-path on diagonal k": a
 * D-path is an edit path starting at the upper left corner and containing
 * exactly D non-diagonal elements ("differences"). The furthest reaching D-path
 * on diagonal k is the one that contains the most (diagonal) elements which
 * ends on diagonal k (where k = y - x).
 * <p>
 * Example:
 *
 * <pre>
 *    H E L L O   W O R L D
 *    ____
 *  L     \___
 *  O         \___
 *  W             \________
 * </pre>
 * <p>
 * Since every D-path has exactly D horizontal or vertical elements, it can only
 * end on the diagonals -D, -D+2, ..., D-2, D.
 * <p>
 * Since every furthest reaching D-path contains at least one furthest reaching
 * (D-1)-path (except for D=0), we can construct them recursively.
 * <p>
 * Since we are really interested in the shortest edit path, we can start
 * looking for a 0-path, then a 1-path, and so on, until we find a path that
 * ends in the lower right corner.
 * <p>
 * To save space, we do not need to store all paths (which has quadratic space
 * requirements), but generate the D-paths simultaneously from both sides. When
 * the ends meet, we will have found "the middle" of the path. From the end
 * points of that diagonal part, we can generate the rest recursively.
 * <p>
 * This only requires linear space.
 * <p>
 * The overall (runtime) complexity is:
 *
 * <pre>
 *     O(N * D^2 + 2 * N/2 * (D/2)^2 + 4 * N/4 * (D/4)^2 + ...)
 *     = O(N * D^2 * 5 / 4) = O(N * D^2),
 * </pre>
 * <p>
 * (With each step, we have to find the middle parts of twice as many regions as
 * before, but the regions (as well as the D) are halved.)
 * <p>
 * So the overall runtime complexity stays the same with linear space, albeit
 * with a larger constant factor.
 *
 * @param <S>
 *            type of sequence.
 */
@SuppressWarnings("hiding")
public class MyersDiff<S extends Sequence> {
	/** Singleton instance of MyersDiff. */
	public static final DiffAlgorithm INSTANCE = new LowLevelDiffAlgorithm() {
		@SuppressWarnings("unused")
		@Override
		public <S extends Sequence> void diffNonCommon(EditList edits,
				HashedSequenceComparator<S> cmp, HashedSequence<S> a,
				HashedSequence<S> b, Edit region) {
			new MyersDiff<>(edits, cmp, a, b, region);
		}
	};

	/**
	 * The list of edits found during the last call to
	 * {@link #calculateEdits(Edit)}
	 */
	protected EditList edits;

	/** Comparison function for sequences. */
	protected HashedSequenceComparator<S> cmp;

	/**
	 * The first text to be compared. Referred to as "Text A" in the comments
	 */
	protected HashedSequence<S> a;

	/**
	 * The second text to be compared. Referred to as "Text B" in the comments
	 */
	protected HashedSequence<S> b;

	private MyersDiff(EditList edits, HashedSequenceComparator<S> cmp,
			HashedSequence<S> a, HashedSequence<S> b, Edit region) {
		this.edits = edits;
		this.cmp = cmp;
		this.a = a;
		this.b = b;
		calculateEdits(region);
	}

	// TODO: use ThreadLocal for future multi-threaded operations
	MiddleEdit middle = new MiddleEdit();

	/**
	 * Entrypoint into the algorithm this class is all about. This method triggers that the
	 * differences between A and B are calculated in form of a list of edits.
	 * @param r portion of the sequences to examine.
	 */
	private void calculateEdits(Edit r) {
		middle.initialize(r.beginA, r.endA, r.beginB, r.endB);
		if (middle.beginA >= middle.endA &&
				middle.beginB >= middle.endB)
			return;

		calculateEdits(middle.beginA, middle.endA,
				middle.beginB, middle.endB);
	}

	/**
	 * Calculates the differences between a given part of A against another
	 * given part of B
	 *
	 * @param beginA
	 *            start of the part of A which should be compared
	 *            (0&lt;=beginA&lt;sizeof(A))
	 * @param endA
	 *            end of the part of A which should be compared
	 *            (beginA&lt;=endA&lt;sizeof(A))
	 * @param beginB
	 *            start of the part of B which should be compared
	 *            (0&lt;=beginB&lt;sizeof(B))
	 * @param endB
	 *            end of the part of B which should be compared
	 *            (beginB&lt;=endB&lt;sizeof(B))
	 */
	protected void calculateEdits(int beginA, int endA,
			int beginB, int endB) {
		Edit edit = middle.calculate(beginA, endA, beginB, endB);

		if (beginA < edit.beginA || beginB < edit.beginB) {
			int k = edit.beginB - edit.beginA;
			int x = middle.backward.snake(k, edit.beginA);
			calculateEdits(beginA, x, beginB, k + x);
		}

		if (edit.getType() != Edit.Type.EMPTY)
			edits.add(edits.size(), edit);

		// after middle
		if (endA > edit.endA || endB > edit.endB) {
			int k = edit.endB - edit.endA;
			int x = middle.forward.snake(k, edit.endA);
			calculateEdits(x, endA, k + x, endB);
		}
	}

	/**
	 * A class to help bisecting the sequences a and b to find minimal
	 * edit paths.
	 *
	 * As the arrays are reused for space efficiency, you will need one
	 * instance per thread.
	 *
	 * The entry function is the calculate() method.
	 */
	class MiddleEdit {
		void initialize(int beginA, int endA, int beginB, int endB) {
			this.beginA = beginA; this.endA = endA;
			this.beginB = beginB; this.endB = endB;

			// strip common parts on either end
			int k = beginB - beginA;
			this.beginA = forward.snake(k, beginA);
			this.beginB = k + this.beginA;

			k = endB - endA;
			this.endA = backward.snake(k, endA);
			this.endB = k + this.endA;
		}

		/*
		 * This function calculates the "middle" Edit of the shortest
		 * edit path between the given subsequences of a and b.
		 *
		 * Once a forward path and a backward path meet, we found the
		 * middle part.  From the last snake end point on both of them,
		 * we construct the Edit.
		 *
		 * It is assumed that there is at least one edit in the range.
		 */
		// TODO: measure speed impact when this is synchronized
		Edit calculate(int beginA, int endA, int beginB, int endB) {
			if (beginA == endA || beginB == endB)
				return new Edit(beginA, endA, beginB, endB);
			this.beginA = beginA; this.endA = endA;
			this.beginB = beginB; this.endB = endB;

			/*
			 * Following the conventions in Myers' paper, "k" is
			 * the difference between the index into "b" and the
			 * index into "a".
			 */
			int minK = beginB - endA;
			int maxK = endB - beginA;

			forward.initialize(beginB - beginA, beginA, minK, maxK);
			backward.initialize(endB - endA, endA, minK, maxK);

			for (int d = 1; ; d++)
				if (forward.calculate(d) ||
						backward.calculate(d))
					return edit;
		}

		/*
		 * For each d, we need to hold the d-paths for the diagonals
		 * k = -d, -d + 2, ..., d - 2, d.  These are stored in the
		 * forward (and backward) array.
		 *
		 * As we allow subsequences, too, this needs some refinement:
		 * the forward paths start on the diagonal forwardK =
		 * beginB - beginA, and backward paths start on the diagonal
		 * backwardK = endB - endA.
		 *
		 * So, we need to hold the forward d-paths for the diagonals
		 * k = forwardK - d, forwardK - d + 2, ..., forwardK + d and
		 * the analogue for the backward d-paths.  This means that
		 * we can turn (k, d) into the forward array index using this
		 * formula:
		 *
		 *	i = (d + k - forwardK) / 2
		 *
		 * There is a further complication: the edit paths should not
		 * leave the specified subsequences, so k is bounded by
		 * minK = beginB - endA and maxK = endB - beginA.  However,
		 * (k - forwardK) _must_ be odd whenever d is odd, and it
		 * _must_ be even when d is even.
		 *
		 * The values in the "forward" and "backward" arrays are
		 * positions ("x") in the sequence a, to get the corresponding
		 * positions ("y") in the sequence b, you have to calculate
		 * the appropriate k and then y:
		 *
		 *	k = forwardK - d + i * 2
		 *	y = k + x
		 *
		 * (substitute backwardK for forwardK if you want to get the
		 * y position for an entry in the "backward" array.
		 */
		EditPaths forward = new ForwardEditPaths();
		EditPaths backward = new BackwardEditPaths();

		/* Some variables which are shared between methods */
		protected int beginA, endA, beginB, endB;
		protected Edit edit;

		abstract class EditPaths {
			private IntList x = new IntList();
			private LongList snake = new LongList();
			int beginK, endK, middleK;
			int prevBeginK, prevEndK;
			/* if we hit one end early, no need to look further */
			int minK, maxK; // TODO: better explanation

			final int getIndex(int d, int k) {
// TODO: remove
if (((d + k - middleK) % 2) != 0)
	throw new RuntimeException(MessageFormat.format(JGitText.get().unexpectedOddResult, Integer.valueOf(d), Integer.valueOf(k), Integer.valueOf(middleK)));
				return (d + k - middleK) / 2;
			}

			final int getX(int d, int k) {
// TODO: remove
if (k < beginK || k > endK)
	throw new RuntimeException(MessageFormat.format(JGitText.get().kNotInRange, Integer.valueOf(k), Integer.valueOf(beginK), Integer.valueOf(endK)));
				return x.get(getIndex(d, k));
			}

			final long getSnake(int d, int k) {
// TODO: remove
if (k < beginK || k > endK)
	throw new RuntimeException(MessageFormat.format(JGitText.get().kNotInRange, Integer.valueOf(k), Integer.valueOf(beginK), Integer.valueOf(endK)));
				return snake.get(getIndex(d, k));
			}

			private int forceKIntoRange(int k) {
				/* if k is odd, so must be the result */
				if (k < minK)
					return minK + ((k ^ minK) & 1);
				else if (k > maxK)
					return maxK - ((k ^ maxK) & 1);
				return k;
			}

			void initialize(int k, int x, int minK, int maxK) {
				this.minK = minK;
				this.maxK = maxK;
				beginK = endK = middleK = k;
				this.x.clear();
				this.x.add(x);
				snake.clear();
				snake.add(newSnake(k, x));
			}

			abstract int snake(int k, int x);
			abstract int getLeft(int x);
			abstract int getRight(int x);
			abstract boolean isBetter(int left, int right);
			abstract void adjustMinMaxK(int k, int x);
			abstract boolean meets(int d, int k, int x, long snake);

			final long newSnake(int k, int x) {
				long y = k + x;
				long ret = ((long) x) << 32;
				return ret | y;
			}

			final int snake2x(long snake) {
				return (int) (snake >>> 32);
			}

			final int snake2y(long snake) {
				return (int) snake;
			}

			final boolean makeEdit(long snake1, long snake2) {
				int x1 = snake2x(snake1), x2 = snake2x(snake2);
				int y1 = snake2y(snake1), y2 = snake2y(snake2);
				/*
				 * Check for incompatible partial edit paths:
				 * when there are ambiguities, we might have
				 * hit incompatible (i.e. non-overlapping)
				 * forward/backward paths.
				 *
				 * In that case, just pretend that we have
				 * an empty edit at the end of one snake; this
				 * will force a decision which path to take
				 * in the next recursion step.
				 */
				if (x1 > x2 || y1 > y2) {
					x1 = x2;
					y1 = y2;
				}
				edit = new Edit(x1, x2, y1, y2);
				return true;
			}

			boolean calculate(int d) {
				prevBeginK = beginK;
				prevEndK = endK;
				beginK = forceKIntoRange(middleK - d);
				endK = forceKIntoRange(middleK + d);
				// TODO: handle i more efficiently
				// TODO: walk snake(k, getX(d, k)) only once per (d, k)
				// TODO: move end points out of the loop to avoid conditionals inside the loop
				// go backwards so that we can avoid temp vars
				for (int k = endK; k >= beginK; k -= 2) {
					if (Thread.interrupted()) {
						throw new DiffInterruptedException();
					}
					int left = -1, right = -1;
					long leftSnake = -1L, rightSnake = -1L;
					// TODO: refactor into its own function
					if (k > prevBeginK) {
						int i = getIndex(d - 1, k - 1);
						left = x.get(i);
						int end = snake(k - 1, left);
						leftSnake = left != end ?
							newSnake(k - 1, end) :
							snake.get(i);
						if (meets(d, k - 1, end, leftSnake))
							return true;
						left = getLeft(end);
					}
					if (k < prevEndK) {
						int i = getIndex(d - 1, k + 1);
						right = x.get(i);
						int end = snake(k + 1, right);
						rightSnake = right != end ?
							newSnake(k + 1, end) :
							snake.get(i);
						if (meets(d, k + 1, end, rightSnake))
							return true;
						right = getRight(end);
					}
					int newX;
					long newSnake;
					if (k >= prevEndK ||
							(k > prevBeginK &&
							 isBetter(left, right))) {
						newX = left;
						newSnake = leftSnake;
					}
					else {
						newX = right;
						newSnake = rightSnake;
					}
					if (meets(d, k, newX, newSnake))
						return true;
					adjustMinMaxK(k, newX);
					int i = getIndex(d, k);
					x.set(i, newX);
					snake.set(i, newSnake);
				}
				return false;
			}
		}

		class ForwardEditPaths extends EditPaths {
			@Override
			final int snake(int k, int x) {
				for (; x < endA && k + x < endB; x++)
					if (!cmp.equals(a, x, b, k + x))
						break;
				return x;
			}

			@Override
			final int getLeft(int x) {
				return x;
			}

			@Override
			final int getRight(int x) {
				return x + 1;
			}

			@Override
			final boolean isBetter(int left, int right) {
				return left > right;
			}

			@Override
			final void adjustMinMaxK(int k, int x) {
				if (x >= endA || k + x >= endB) {
					if (k > backward.middleK)
						maxK = k;
					else
						minK = k;
				}
			}

			@Override
			final boolean meets(int d, int k, int x, long snake) {
				if (k < backward.beginK || k > backward.endK)
					return false;
				// TODO: move out of loop
				if (((d - 1 + k - backward.middleK) % 2) != 0)
					return false;
				if (x < backward.getX(d - 1, k))
					return false;
				makeEdit(snake, backward.getSnake(d - 1, k));
				return true;
			}
		}

		class BackwardEditPaths extends EditPaths {
			@Override
			final int snake(int k, int x) {
				for (; x > beginA && k + x > beginB; x--)
					if (!cmp.equals(a, x - 1, b, k + x - 1))
						break;
				return x;
			}

			@Override
			final int getLeft(int x) {
				return x - 1;
			}

			@Override
			final int getRight(int x) {
				return x;
			}

			@Override
			final boolean isBetter(int left, int right) {
				return left < right;
			}

			@Override
			final void adjustMinMaxK(int k, int x) {
				if (x <= beginA || k + x <= beginB) {
					if (k > forward.middleK)
						maxK = k;
					else
						minK = k;
				}
			}

			@Override
			final boolean meets(int d, int k, int x, long snake) {
				if (k < forward.beginK || k > forward.endK)
					return false;
				// TODO: move out of loop
				if (((d + k - forward.middleK) % 2) != 0)
					return false;
				if (x > forward.getX(d, k))
					return false;
				makeEdit(forward.getSnake(d, k), snake);
				return true;
			}
		}
	}

	/**
	 * Main method
	 *
	 * @param args
	 *            two filenames specifying the contents to be diffed
	 */
	public static void main(String[] args) {
		if (args.length != 2) {
			System.err.println(JGitText.get().need2Arguments);
			System.exit(1);
		}
		try {
			RawText a = new RawText(new java.io.File(args[0]));
			RawText b = new RawText(new java.io.File(args[1]));
			EditList r = INSTANCE.diff(RawTextComparator.DEFAULT, a, b);
			System.out.println(r.toString());
		} catch (Exception e) {
			e.printStackTrace();
		}
	}
}