/*

 bdiff.c - efficient binary diff extension for Mercurial

 Copyright 2005, 2006 Matt Mackall <mpm@selenic.com>

 This software may be used and distributed according to the terms of

 the GNU General Public License, incorporated herein by reference.

 Based roughly on Python difflib

*/

#include <stdlib.h>

#include <string.h>

#include <limits.h>

#include "compat.h"

#include "bitmanipulation.h"

#include "bdiff.h"

struct pos {

	int pos, len;

};

int bdiff_splitlines(const char *a, ssize_t len, struct bdiff_line **lr)

{

	unsigned hash;

	int i;

	const char *p, *b = a;

	const char * const plast = a + len - 1;

	struct bdiff_line *l;

	/* count the lines */

	i = 1; /* extra line for sentinel */

			i++;

	*lr = l = (struct bdiff_line *)malloc(sizeof(struct bdiff_line) * i);

	if (!l)

		return -1;

	/* build the line array and calculate hashes */

	hash = 0;

	for (p = a; p < a + len; p++) {

		/* Leonid Yuriev's hash */

		hash = (hash * 1664525) + (unsigned char)*p + 1013904223;

		if (*p == '\n' || p == plast) {

			l->hash = hash;

			hash = 0;

			l->len = p - b + 1;

			l->l = b;

			l->n = INT_MAX;

			l++;

			b = p + 1;

		}

	}

	/* set up a sentinel */

	l->hash = 0;

	l->len = 0;

	l->l = a + len;

	return i - 1;

}

static inline int cmp(struct bdiff_line *a, struct bdiff_line *b)

{

	return a->hash != b->hash || a->len != b->len || memcmp(a->l, b->l, a->len);

}

static int equatelines(struct bdiff_line *a, int an, struct bdiff_line *b,

	int bn)

{

	int i, j, buckets = 1, t, scale;

	struct pos *h = NULL;

	/* build a hash table of the next highest power of 2 */

	while (buckets < bn + 1)

		buckets *= 2;

	/* try to allocate a large hash table to avoid collisions */

	for (scale = 4; scale; scale /= 2) {

		h = (struct pos *)malloc(scale * buckets * sizeof(struct pos));

		if (h)

			break;

	}

	if (!h)

		return 0;

	buckets = buckets * scale - 1;

	/* clear the hash table */

	for (i = 0; i <= buckets; i++) {

		h[i].pos = -1;

		h[i].len = 0;

	}

	/* add lines to the hash table chains */

	for (i = 0; i < bn; i++) {

		/* find the equivalence class */

		for (j = b[i].hash & buckets; h[j].pos != -1;

		     j = (j + 1) & buckets)

			if (!cmp(b + i, b + h[j].pos))

				break;

		/* add to the head of the equivalence class */

		b[i].n = h[j].pos;

		b[i].e = j;

		h[j].pos = i;

		h[j].len++; /* keep track of popularity */

	}

	/* compute popularity threshold */

	t = (bn >= 31000) ? bn / 1000 : 1000000 / (bn + 1);

	/* match items in a to their equivalence class in b */

	for (i = 0; i < an; i++) {

		/* find the equivalence class */

		for (j = a[i].hash & buckets; h[j].pos != -1;

		     j = (j + 1) & buckets)

			if (!cmp(a + i, b + h[j].pos))

				break;

		a[i].e = j; /* use equivalence class for quick compare */

		if (h[j].len <= t)

			a[i].n = h[j].pos; /* point to head of match list */

		else

			a[i].n = -1; /* too popular */

	}

	/* discard hash tables */

	free(h);

	return 1;

}

static int longest_match(struct bdiff_line *a, struct bdiff_line *b,

			struct pos *pos,

			 int a1, int a2, int b1, int b2, int *omi, int *omj)

{

	int mi = a1, mj = b1, mk = 0, i, j, k, half;

	/* window our search on large regions to better bound

	   worst-case performance. by choosing a window at the end, we

	   reduce skipping overhead on the b chains. */

	if (a2 - a1 > 30000)

		a1 = a2 - 30000;

	half = (a1 + a2) / 2;

	for (i = a1; i < a2; i++) {

		/* skip all lines in b after the current block */

		for (j = a[i].n; j >= b2; j = b[j].n)

			;

		/* loop through all lines match a[i] in b */

		for (; j >= b1; j = b[j].n) {

			/* does this extend an earlier match? */

			for (k = 1; j - k >= b1 && i - k >= a1; k++) {

				/* reached an earlier match? */

				if (pos[j - k].pos == i - k) {

					k += pos[j - k].len;

					break;

				}

				/* previous line mismatch? */

				if (a[i - k].e != b[j - k].e)

					break;

			}

			pos[j].pos = i;

			pos[j].len = k;

			/* best match so far? we prefer matches closer

			   to the middle to balance recursion */

			if (k > mk || (k == mk && (i <= mi || i < half))) {

				mi = i;

				mj = j;

				mk = k;

			}

		}

	}

	if (mk) {

		mi = mi - mk + 1;

		mj = mj - mk + 1;

	}

	/* expand match to include subsequent popular lines */

	while (mi + mk < a2 && mj + mk < b2 &&

	       a[mi + mk].e == b[mj + mk].e)

		mk++;

	*omi = mi;

	*omj = mj;

	return mk;

}

static struct bdiff_hunk *recurse(struct bdiff_line *a, struct bdiff_line *b,

				struct pos *pos,

			    int a1, int a2, int b1, int b2, struct bdiff_hunk *l)

{

	int i, j, k;

	while (1) {

		/* find the longest match in this chunk */

		k = longest_match(a, b, pos, a1, a2, b1, b2, &i, &j);

		if (!k)

			return l;

		/* and recurse on the remaining chunks on either side */

		l = recurse(a, b, pos, a1, i, b1, j, l);

		if (!l)

			return NULL;

		l->next = (struct bdiff_hunk *)malloc(sizeof(struct bdiff_hunk));

		if (!l->next)

			return NULL;

		l = l->next;

		l->a1 = i;

		l->a2 = i + k;

		l->b1 = j;

		l->b2 = j + k;

		l->next = NULL;

		/* tail-recursion didn't happen, so do equivalent iteration */

		a1 = i + k;

		b1 = j + k;

	}

}

int bdiff_diff(struct bdiff_line *a, int an, struct bdiff_line *b,

		int bn, struct bdiff_hunk *base)

{

	struct bdiff_hunk *curr;

	struct pos *pos;

	int t, count = 0;

	/* allocate and fill arrays */

	t = equatelines(a, an, b, bn);

	pos = (struct pos *)calloc(bn ? bn : 1, sizeof(struct pos));

	if (pos && t) {

		/* generate the matching block list */

		curr = recurse(a, b, pos, 0, an, 0, bn, base);

		if (!curr)

			return -1;

		/* sentinel end hunk */

		curr->next = (struct bdiff_hunk *)malloc(sizeof(struct bdiff_hunk));

		if (!curr->next)

			return -1;

		curr = curr->next;

		curr->a1 = curr->a2 = an;

		curr->b1 = curr->b2 = bn;

		curr->next = NULL;

	}

	free(pos);

	/* normalize the hunk list, try to push each hunk towards the end */

	for (curr = base->next; curr; curr = curr->next) {

		struct bdiff_hunk *next = curr->next;

		if (!next)

			break;

		if (curr->a2 == next->a1 || curr->b2 == next->b1)

			while (curr->a2 < an && curr->b2 < bn

			       && next->a1 < next->a2

			       && next->b1 < next->b2

			       && !cmp(a + curr->a2, b + curr->b2)) {

				curr->a2++;

				next->a1++;

				curr->b2++;

				next->b1++;

			}

	}

	for (curr = base->next; curr; curr = curr->next)

		count++;

	return count;

}

void bdiff_freehunks(struct bdiff_hunk *l)

{

	struct bdiff_hunk *n;

	for (; l; l = n) {

		n = l->next;

		free(l);

	}

}