Merge pull request #10 from e7appew/fix-typos

[wiggle/upstream.git] / diff.c

/*
 * wiggle - apply rejected patches
 *
 * Copyright (C) 2003 Neil Brown <neilb@cse.unsw.edu.au>
 * Copyright (C) 2011-2013 Neil Brown <neilb@suse.de>
 *
 *
 *    This program is free software; you can redistribute it and/or modify
 *    it under the terms of the GNU General Public License as published by
 *    the Free Software Foundation; either version 2 of the License, or
 *    (at your option) any later version.
 *
 *    This program is distributed in the hope that it will be useful,
 *    but WITHOUT ANY WARRANTY; without even the implied warranty of
 *    MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
 *    GNU General Public License for more details.
 *
 *    You should have received a copy of the GNU General Public License
 *    along with this program.
 *
 *    Author: Neil Brown
 *    Email: <neilb@suse.de>
 */

/*
 * Calculate longest common subsequence between two sequences
 *
 * Each sequence contains strings with
 *   hash start length
 * We produce a list of tripples: a b len
 * where A and B point to elements in the two sequences, and len is the number
 * of common elements there.  The list is terminated by an entry with len==0.
 *
 * This is roughly based on
 *   "An O(ND) Difference Algorithm and its Variations", Eugene Myers,
 *   Algorithmica Vol. 1 No. 2, 1986, pp. 251-266;
 * http://xmailserver.org/diff2.pdf
 *
 * However we don't run the basic algorithm both forward and backward until
 * we find an overlap as Myers suggests.  Rather we always run forwards, but
 * we record the location of the (possibly empty) snake that crosses the
 * midline.  When we finish, this recorded location for the best path shows
 * us where to divide and find further midpoints.
 *
 * In brief, the algorithm is as follows.
 *
 * Imagine a Cartesian Matrix where x co-ordinates correspond to symbols in
 * the first sequence (A, length a) and y co-ordinates correspond to symbols
 * in the second sequence (B, length b).  At the origin we have the first
 * sequence.
 * Movement in the x direction represents deleting the symbol as that point,
 * so from x=i-1 to x=i deletes symbol i from A.

 * Movement in the y direction represents adding the corresponding symbol
 * from B.  So to move from the origin 'a' spaces along X and then 'b' spaces
 * up Y will remove all of the first sequence and then add all of the second
 * sequence.  Similarly moving firstly up the Y axis, then along the X
 * direction will add the new sequence, then remove the old sequence.  Thus
 * the point a,b represents the second sequence and a part from 0,0 to a,b
 * represent an sequence of edits to change A into B.
 *
 * There are clearly many paths from 0,0 to a,b going through different
 * points in the matrix in different orders.  At some points in the matrix
 * the next symbol to be added from B is the same as the next symbol to be
 * removed from A.  At these points we can take a diagonal step to a new
 * point in the matrix without actually changing any symbol.  A sequence of
 * these diagonal steps is called a 'snake'.  The goal then is to find a path
 * of x-steps (removals), y-steps (additions) and diagonals (steps and
 * snakes) where the number of (non-diagonal) steps is minimal.
 *
 * i.e. we aim for as many long snakes as possible.
 * If the total number of 'steps' is called the 'cost', we aim to minimise
 * the cost.
 *
 * As storing the whole matrix in memory would be prohibitive with large
 * sequences we limit ourselves to linear storage proportional to a+b and
 * repeat the search at most log2(a+b) times building up the path as we go.
 * Specifically we perform a search on the full matrix and record where each
 * path crosses the half-way point. i.e. where x+y = (a+b)/2 (== mid).  This
 * tells us the mid point of the best path.  We then perform two searches,
 * one on each of the two halves and find the 1/4 and 3/4 way points.  This
 * continues recursively until we have all points.
 *
 * The storage is an array v of 'struct v'.  This is indexed by the
 * diagonal-number k = x-y.  Thus k can be negative and the array is
 * allocated to allow for that.  During the search there is an implicit value
 * 'c' which is the cost (length in steps) of all the paths currently under
 * consideration.
 * v[k] stores details of the longest reaching path of cost c that finishes
 *      on diagonal k.  "longest reaching" means "finishes closest to a,b".
 * Details are:
 *   The location of the end point.  'x' is stored. y = x - k.
 *   The diagonal of the midpoint crossing. md is stored. x = (mid + md)/2
 *                                                        y = (mid - md)/2
 *                                                          = x - md
 *   (note: md is a diagonal so md = x-y.  mid is an anti-diagonal: mid = x+y)
 *   The number of 'snakes' in the path (l).  This is used to allocate the
 *   array which will record the snakes and to terminate recursion.
 *
 * A path with an even cost (c is even) must end on an even diagonal (k is
 * even) and when c is odd, k must be odd.  So the v[] array is treated as
 * two sub arrays, the even part and the odd part.  One represents paths of
 * cost 'c', the other paths of cost c-1.
 *
 * Initially only v[0] is meaningful and there are no snakes.  We firstly
 * extend all paths under consideration with the longest possible snake on
 * that diagonal.
 *
 * Then we increment 'c' and calculate for each suitable 'k' whether the best
 * path to diagonal k of cost c comes from taking an x-step from the c-1 path
 * on diagonal k-1, or from taking a y-step from the c-1 path on diagonal
 * k+1.  Obviously we need to avoid stepping out of the matrix.  Finally we
 * check if the 'v' array can be extended or reduced at the boundaries.  If
 * we hit a border we must reduce.  If the best we could possibly do on that
 * diagonal is less than the worst result from the current leading path, then
 * we also reduce.  Otherwise we extend the range of 'k's we consider.
 *
 * We continue until we find a path has reached a,b.  This must be a minimal
 * cost path (cost==c).  At this point re-check the end of the snake at the
 * midpoint and report that.
 *
 * This all happens recursively for smaller and smaller subranges stopping
 * when we examine a submatrix and find that it contains no snakes.  As we
 * are usually dealing with sub-matrixes we are not walking from 0,0 to a,b
 * from alo,blo to ahi,bhi - low point to high point.  So the initial k is
 * alo-blo, not 0.
 *
 */

#include        "wiggle.h"
#include        <stdlib.h>

struct v {
        int x;  /* x location of furthest reaching path of current cost */
        int md; /* diagonal location of midline crossing */
        int l;  /* number of continuous common sequences found so far */
};

static int find_common(struct file *a, struct file *b,
                            int *alop, int *ahip,
                            int *blop, int *bhip,
                            struct v *v)
{
        /* Examine matrix from alo to ahi and blo to bhi.
         * i.e. including alo and blo, but less than ahi and bhi.
         * Finding longest subsequence and
         * return new {a,b}{lo,hi} either side of midline.
         * i.e. mid = ( (ahi-alo) + (bhi-blo) ) / 2
         *      alo+blo <= mid <= ahi+bhi
         *  and alo,blo to ahi,bhi is a common (possibly empty)
         *  subseq - a snake.
         *
         * v is scratch space which is indexable from
         * alo-bhi to ahi-blo inclusive.
         * i.e. even though there is no symbol at ahi or bhi, we do
         * consider paths that reach there as they simply cannot
         * go further in that direction.
         *
         * Return the number of snakes found.
         */

        int klo, khi;
        int alo = *alop;
        int ahi = *ahip;
        int blo = *blop;
        int bhi = *bhip;

        int mid = (ahi+bhi+alo+blo)/2;

        /* 'worst' is the worst-case extra cost that we need
         * to pay before reaching our destination.  It assumes
         * no more snakes in the furthest-reaching path so far.
         * We use this to know when we can trim the extreme
         * diagonals - when their best case does not improve on
         * the current worst case.
         */
        int worst = (ahi-alo)+(bhi-blo);

        klo = khi = alo-blo;
        v[klo].x = alo;
        v[klo].l = 0;

        while (1) {
                int x, y;
                int cost;
                int k;

                /* Find the longest snake extending on each current
                 * diagonal, and record if it crosses the midline.
                 * If we reach the end, return.
                 */
                for (k = klo ; k <= khi ; k += 2) {
                        int snake = 0;

                        x = v[k].x;
                        y = x-k;
                        if (y > bhi)
                                abort();

                        /* Follow any snake that is here */
                        while (x < ahi && y < bhi &&
                               match(&a->list[x], &b->list[y])
                                ) {
                                x++;
                                y++;
                                snake = 1;
                        }

                        /* Refine the worst-case remaining cost */
                        cost = (ahi-x)+(bhi-y);
                        if (cost < worst)
                                worst = cost;

                        /* Check for midline crossing */
                        if (x+y >= mid &&
                             v[k].x + v[k].x-k <= mid)
                                v[k].md = k;

                        v[k].x = x;
                        v[k].l += snake;

                        if (cost == 0) {
                                /* OK! We have arrived.
                                 * We crossed the midpoint on diagonal v[k].md
                                 */
                                if (x != ahi)
                                        abort();

                                /* The snake could start earlier than the
                                 * midline.  We cannot just search backwards
                                 * as that might find the wrong path - the
                                 * greediness of the diff algorithm is
                                 * asymmetric.
                                 * We could record the start of the snake in
                                 * 'v', but we will find the actual snake when
                                 * we recurse so there is no need.
                                 */
                                x = (v[k].md+mid)/2;
                                y = x-v[k].md;

                                *alop = x;
                                *blop = y;

                                /* Find the end of the snake using the same
                                 * greedy approach as when we first found the
                                 * snake
                                 */
                                while (x < ahi && y < bhi &&
                                       match(&a->list[x], &b->list[y])
                                        ) {
                                        x++;
                                        y++;
                                }
                                *ahip = x;
                                *bhip = y;

                                return v[k].l;
                        }
                }

                /* No success with previous cost, so increment cost (c) by 1
                 * and for each other diagonal, set from the end point of the
                 * diagonal on one side of it or the other.
                 */
                for (k = klo+1; k <= khi-1 ; k += 2) {
                        if (v[k-1].x+1 > ahi) {
                                /* cannot step to the right from previous
                                 * diagonal as there is no room.
                                 * So step up from next diagonal.
                                 */
                                v[k] = v[k+1];
                        } else if (v[k+1].x - k > bhi || v[k-1].x+1 >= v[k+1].x) {
                                /* Cannot step up from next diagonal as either
                                 * there is no room, or doing so wouldn't get us
                                 * as close to the endpoint.
                                 * So step to the right.
                                 */
                                v[k] = v[k-1];
                                v[k].x++;
                        } else {
                                /* There is room in both directions, but
                                 * stepping up from the next diagonal gets us
                                 * closer
                                 */
                                v[k] = v[k+1];
                        }
                }

                /* Now we need to either extend or contract klo and khi
                 * so they both change parity (odd vs even).
                 * If we extend we need to step up (for klo) or to the
                 * right (khi) from the adjacent diagonal.  This is
                 * not possible if we have hit the edge of the matrix, and
                 * not sensible if the new point has a best case remaining
                 * cost that is worse than our current worst case remaining
                 * cost.
                 * The best-case remaining cost is the absolute difference
                 * between the remaining number of additions and the remaining
                 * number of deletions - and assumes lots of snakes.
                 */
                /* new location if we step up from klo to klo-1*/
                x = v[klo].x; y = x - (klo-1);
                cost = abs((ahi-x)-(bhi-y));
                if (y <= bhi && cost <= worst) {
                        /* Looks acceptable - step up. */
                        v[klo-1] = v[klo];
                        klo--;
                } else
                        klo++;

                /* new location if we step to the right from khi to khi+1 */
                x = v[khi].x+1; y = x - (khi+1);
                cost = abs((ahi-x)-(bhi-y));
                if (x <= ahi && cost <= worst) {
                        /* Looks acceptable - step to the right */
                        v[khi+1] = v[khi];
                        v[khi+1].x++;
                        khi++;
                } else
                        khi--;
        }
}

static struct csl *lcsl(struct file *a, int alo, int ahi,
                        struct file *b, int blo, int bhi,
                        struct csl *csl,
                        struct v *v)
{
        /* lcsl == longest common sub-list.
         * This calls itself recursively as it finds the midpoint
         * of the best path.
         * On first call, 'csl' is NULL and will need to be allocated and
         * is returned.
         * On subsequence calls when 'csl' is not NULL, we add all the
         * snakes we find to csl, and return a pointer to the next
         * location where future snakes can be stored.
         */
        int len;
        int alo1 = alo;
        int ahi1 = ahi;
        int blo1 = blo;
        int bhi1 = bhi;
        struct csl *rv = NULL;

        if (ahi <= alo || bhi <= blo)
                return csl;

        len = find_common(a, b,
                          &alo1, &ahi1,
                          &blo1, &bhi1,
                          v);

        if (csl == NULL) {
                /* 'len+1' to hold a sentinel */
                rv = csl = xmalloc((len+1)*sizeof(*csl));
                csl->len = 0;
        }
        if (len) {
                /* There are more snakes to find - keep looking. */

                /* With depth-first recursion, this adds all the snakes
                 * before 'alo1' to 'csl'
                 */
                csl = lcsl(a, alo, alo1,
                           b, blo, blo1,
                           csl, v);

                if (ahi1 > alo1) {
                        /* need to add this common seq, possibly attach
                         * to last
                         */
                        if (csl->len &&
                            csl->a+csl->len == alo1 &&
                            csl->b+csl->len == blo1) {
                                csl->len += ahi1-alo1;
                        } else {
                                if (csl->len)
                                        csl++;
                                csl->len = ahi1-alo1;
                                csl->a = alo1;
                                csl->b = blo1;
                                csl[1].len = 0;
                        }
                }
                /* Now recurse to add all the snakes after ahi1 to csl */
                csl = lcsl(a, ahi1, ahi,
                           b, bhi1, bhi,
                           csl, v);
        }
        if (rv) {
                /* This was the first call.  Record the endpoint
                 * as a snake of length 0.  This might be extended.
                 * by 'fixup()' below.
                 */
                if (csl->len)
                        csl++;
                csl->a = ahi;
                csl->b = bhi;
#if 1
                if (rv+len != csl || csl->len != 0)
                        abort(); /* number of runs was wrong */
#endif
                return rv;
        } else
                /* intermediate call - return where we are up to */
                return csl;
}

/* If two common sequences are separated by only an add or remove,
 * and the first sequence ends the same as the middle text,
 * extend the second and contract the first in the hope that the
 * first might become empty.  This ameliorates against the greediness
 * of the 'diff' algorithm.
 * i.e. if we have:
 *   [ foo X ] X [ bar ]
 *   [ foo X ] [ bar ]
 * Then change it to:
 *   [ foo ] X [ X bar ]
 *   [ foo ] [ X bar ]
 * We treat the final zero-length 'csl' as a common sequence which
 * can be extended so we must make sure to add a new zero-length csl
 * to the end.
 * If this doesn't make the first sequence disappear, and (one of the)
 * X(s) was a newline, then move back so the newline is at the end
 * of the first sequence.  This encourages common sequences
 * to be whole-line units where possible.
 */
static void fixup(struct file *a, struct file *b, struct csl *list)
{
        struct csl *list1, *orig;
        int lasteol = -1;
        int found_end = 0;

        if (!list)
                return;

        /* 'list' and 'list1' are adjacent pointers into the csl.
         * If a match gets deleted, they might not be physically
         * adjacent any more.  Once we get to the end of the list
         * this will cease to matter - the list will be a bit
         * shorter is all.
         */
        orig = list;
        list1 = list+1;
        while (list->len) {
                if (list1->len == 0)
                        found_end = 1;

                /* If a single token is either inserted or deleted
                 * immediately after a matching token...
                 */
                if ((list->a+list->len == list1->a &&
                     list->b+list->len != list1->b &&
                     /* text at b inserted */
                     match(&b->list[list->b+list->len-1],
                           &b->list[list1->b-1])
                            )
                    ||
                    (list->b+list->len == list1->b &&
                     list->a+list->len != list1->a &&
                     /* text at a deleted */
                     match(&a->list[list->a+list->len-1],
                           &a->list[list1->a-1])
                            )
                        ) {
                        /* If the last common token is a simple end-of-line
                         * record where it is.  For a word-wise diff, this is
                         * any EOL.  For a line-wise diff this is a blank line.
                         * If we are looking at a deletion it must be deleting
                         * the eol, so record that deleted eol.
                         */
                        if (ends_line(a->list[list->a+list->len-1])
                            && a->list[list->a+list->len-1].len == 1
                            && lasteol == -1
                                ) {
                                lasteol = list1->a-1;
                        }
                        /* Expand the second match, shrink the first */
                        list1->a--;
                        list1->b--;
                        list1->len++;
                        list->len--;

                        /* If the first match has become empty, make it
                         * disappear.. (and forget about the eol).
                         */
                        if (list->len == 0) {
                                lasteol = -1;
                                if (found_end) {
                                        /* Deleting just before the last
                                         * entry */
                                        *list = *list1;
                                        list1->a += list1->len;
                                        list1->b += list1->len;
                                        list1->len = 0;
                                } else if (list > orig)
                                        /* Deleting in the  middle */
                                        list--;
                                else {
                                        /* deleting the first entry */
                                        *list = *list1++;
                                }
                        }
                } else {
                        /* Nothing interesting here, though if we
                         * shuffled back past an eol, shuffle
                         * forward to line up with that eol.
                         * This causes an eol to bind more strongly
                         * with the preceding line than the following.
                         */
                        if (lasteol >= 0) {
                                while (list1->a <= lasteol
                                       && (list1->len > 1 ||
                                           (found_end && list1->len > 0))) {
                                        list1->a++;
                                        list1->b++;
                                        list1->len--;
                                        list->len++;
                                }
                                lasteol = -1;
                        }
                        *++list = *list1;
                        if (found_end) {
                                list1->a += list1->len;
                                list1->b += list1->len;
                                list1->len = 0;
                        } else
                                list1++;
                }
                if (list->len && list1 == list)
                        abort();
        }
}

/* Main entry point - find the common-sub-list of files 'a' and 'b'.
 * The final element in the list will have 'len' == 0 and will point
 * beyond the end of the files.
 */
struct csl *diff(struct file a, struct file b)
{
        struct v *v;
        struct csl *csl;
        v = xmalloc(sizeof(struct v)*(a.elcnt+b.elcnt+2));
        v += b.elcnt+1;

        csl = lcsl(&a, 0, a.elcnt,
                   &b, 0, b.elcnt,
                   NULL, v);
        free(v-(b.elcnt+1));
        fixup(&a, &b, csl);
        if (!csl) {
                csl = xmalloc(sizeof(*csl));
                csl->len = 0;
                csl->a = a.elcnt;
                csl->b = b.elcnt;
        }
        return csl;
}

/* Alternate entry point - find the common-sub-list in two
 * subranges of files.
 */
struct csl *diff_partial(struct file a, struct file b,
                         int alo, int ahi, int blo, int bhi)
{
        struct v *v;
        struct csl *csl;
        v = xmalloc(sizeof(struct v)*(ahi-alo+bhi-blo+2));
        v += bhi-alo+1;

        csl = lcsl(&a, alo, ahi,
                   &b, blo, bhi,
                   NULL, v);
        free(v-(bhi-alo+1));
        fixup(&a, &b, csl);
        return csl;
}

struct csl *csl_join(struct csl *c1, struct csl *c2)
{
        int cnt1, cnt2;
        int offset = 0;
        if (c1 == NULL)
                return c2;
        if (c2 == NULL)
                return c1;

        for (cnt1 = 0; c1[cnt1].len; cnt1++)
                ;
        for (cnt2 = 0; c2[cnt2].len; cnt2++)
                ;
        if (cnt1 && cnt2 &&
            c1[cnt1-1].a + c1[cnt1-1].len == c2[0].a &&
            c1[cnt1-1].b + c1[cnt1-1].len == c2[0].b) {
                /* Merge these two */
                c1[cnt1-1].len += c2[0].len;
                offset = 1;
                cnt2--;
        }
        c1 = realloc(c1, (cnt1+cnt2+1)*sizeof(*c1));
        while (cnt2 >= 0) {
                c1[cnt1+cnt2] = c2[cnt2 + offset];
                cnt2--;
        }
        free(c2);
        return c1;
}

/* When rediffing a patch, we *must* make sure the hunk headers
 * line up.  So don't do a full diff, but rather find the hunk
 * headers and diff the bits between them.
 */
struct csl *diff_patch(struct file a, struct file b)
{
        int ap, bp;
        struct csl *csl = NULL;
        if (a.list[0].start[0] != '\0' ||
            b.list[0].start[0] != '\0')
                /* this is not a patch */
                return diff(a, b);

        ap = 0; bp = 0;
        while (ap < a.elcnt && bp < b.elcnt) {
                int alo = ap;
                int blo = bp;
                struct csl *cs;

                do
                        ap++;
                while (ap < a.elcnt &&
                       a.list[ap].start[0] != '\0');
                do
                        bp++;
                while (bp < b.elcnt &&
                       b.list[bp].start[0] != '\0');
                cs = diff_partial(a, b, alo, ap, blo, bp);
                csl = csl_join(csl, cs);
        }
        return csl;
}

#ifdef MAIN

main(int argc, char *argv[])
{
        struct file a, b;
        struct csl *csl;
        struct elmnt *lst = xmalloc(argc*sizeof(*lst));
        int arg;
        struct v *v;
        int ln;

        arg = 1;
        a.elcnt = 0;
        a.list = lst;
        while (argv[arg] && strcmp(argv[arg], "--")) {
                lst->hash = 0;
                lst->start = argv[arg];
                lst->len = strlen(argv[arg]);
                a.elcnt++;
                lst++;
                arg++;
        }
        if (!argv[arg]) {
                printf("AARGH\n");
                exit(1);
        }
        arg++;
        b.elcnt = 0;
        b.list = lst;
        while (argv[arg] && strcmp(argv[arg], "--")) {
                lst->hash = 0;
                lst->start = argv[arg];
                lst->len = strlen(argv[arg]);
                b.elcnt++;
                lst++;
                arg++;
        }

        csl = diff(a, b);
        fixup(&a, &b, csl);
        while (csl && csl->len) {
                int i;
                printf("%d,%d for %d:\n", csl->a, csl->b, csl->len);
                for (i = 0; i < csl->len; i++) {
                        printf("  %.*s (%.*s)\n",
                               a.list[csl->a+i].len, a.list[csl->a+i].start,
                               b.list[csl->b+i].len, b.list[csl->b+i].start);
                }
                csl++;
        }

        exit(0);
}

#endif