1 3b0f3d61 2020-01-22 neels /* Myers diff algorithm implementation, invented by Eugene W. Myers [1].
2 3b0f3d61 2020-01-22 neels * Implementations of both the Myers Divide Et Impera (using linear space)
3 3b0f3d61 2020-01-22 neels * and the canonical Myers algorithm (using quadratic space). */
5 3b0f3d61 2020-01-22 neels * Copyright (c) 2020 Neels Hofmeyr <neels@hofmeyr.de>
7 3b0f3d61 2020-01-22 neels * Permission to use, copy, modify, and distribute this software for any
8 3b0f3d61 2020-01-22 neels * purpose with or without fee is hereby granted, provided that the above
9 3b0f3d61 2020-01-22 neels * copyright notice and this permission notice appear in all copies.
11 3b0f3d61 2020-01-22 neels * THE SOFTWARE IS PROVIDED "AS IS" AND THE AUTHOR DISCLAIMS ALL WARRANTIES
12 3b0f3d61 2020-01-22 neels * WITH REGARD TO THIS SOFTWARE INCLUDING ALL IMPLIED WARRANTIES OF
13 3b0f3d61 2020-01-22 neels * MERCHANTABILITY AND FITNESS. IN NO EVENT SHALL THE AUTHOR BE LIABLE FOR
14 3b0f3d61 2020-01-22 neels * ANY SPECIAL, DIRECT, INDIRECT, OR CONSEQUENTIAL DAMAGES OR ANY DAMAGES
15 3b0f3d61 2020-01-22 neels * WHATSOEVER RESULTING FROM LOSS OF USE, DATA OR PROFITS, WHETHER IN AN
16 3b0f3d61 2020-01-22 neels * ACTION OF CONTRACT, NEGLIGENCE OR OTHER TORTIOUS ACTION, ARISING OUT OF
17 3b0f3d61 2020-01-22 neels * OR IN CONNECTION WITH THE USE OR PERFORMANCE OF THIS SOFTWARE.
20 e10a628a 2020-09-16 stsp #include <inttypes.h>
21 e10a628a 2020-09-16 stsp #include <stdbool.h>
22 e10a628a 2020-09-16 stsp #include <stdlib.h>
23 e10a628a 2020-09-16 stsp #include <string.h>
24 7a54ad3a 2020-09-20 stsp #include <stdio.h>
25 e10a628a 2020-09-16 stsp #include <errno.h>
27 1dfba055 2020-10-07 stsp #include <arraylist.h>
28 1dfba055 2020-10-07 stsp #include <diff_main.h>
30 85ab4559 2020-09-22 stsp #include "diff_internal.h"
31 2a1b94d0 2020-09-26 stsp #include "diff_debug.h"
33 3b0f3d61 2020-01-22 neels /* Myers' diff algorithm [1] is nicely explained in [2].
34 3b0f3d61 2020-01-22 neels * [1] http://www.xmailserver.org/diff2.pdf
35 3b0f3d61 2020-01-22 neels * [2] https://blog.jcoglan.com/2017/02/12/the-myers-diff-algorithm-part-1/ ff.
37 3b0f3d61 2020-01-22 neels * Myers approaches finding the smallest diff as a graph problem.
38 3b0f3d61 2020-01-22 neels * The crux is that the original algorithm requires quadratic amount of memory:
39 0d27172a 2020-05-06 neels * both sides' lengths added, and that squared. So if we're diffing lines of
40 0d27172a 2020-05-06 neels * text, two files with 1000 lines each would blow up to a matrix of about
41 0d27172a 2020-05-06 neels * 2000 * 2000 ints of state, about 16 Mb of RAM to figure out 2 kb of text.
42 0d27172a 2020-05-06 neels * The solution is using Myers' "divide and conquer" extension algorithm, which
43 0d27172a 2020-05-06 neels * does the original traversal from both ends of the files to reach a middle
44 0d27172a 2020-05-06 neels * where these "snakes" touch, hence does not need to backtrace the traversal,
45 0d27172a 2020-05-06 neels * and so gets away with only keeping a single column of that huge state matrix
46 0d27172a 2020-05-06 neels * in memory.
49 3b0f3d61 2020-01-22 neels struct diff_box {
50 3b0f3d61 2020-01-22 neels unsigned int left_start;
51 3b0f3d61 2020-01-22 neels unsigned int left_end;
52 3b0f3d61 2020-01-22 neels unsigned int right_start;
53 3b0f3d61 2020-01-22 neels unsigned int right_end;
56 3b0f3d61 2020-01-22 neels /* If the two contents of a file are A B C D E and X B C Y,
57 3b0f3d61 2020-01-22 neels * the Myers diff graph looks like:
61 3b0f3d61 2020-01-22 neels * k-1 0 1 2 3 4 5
62 3b0f3d61 2020-01-22 neels * \ A B C D E
63 3b0f3d61 2020-01-22 neels * 0 o-o-o-o-o-o
64 3b0f3d61 2020-01-22 neels * X | | | | | |
65 3b0f3d61 2020-01-22 neels * 1 o-o-o-o-o-o
66 3b0f3d61 2020-01-22 neels * B | |\| | | |
67 3b0f3d61 2020-01-22 neels * 2 o-o-o-o-o-o
68 3b0f3d61 2020-01-22 neels * C | | |\| | |
69 3b0f3d61 2020-01-22 neels * 3 o-o-o-o-o-o
70 3b0f3d61 2020-01-22 neels * Y | | | | | |\
71 3b0f3d61 2020-01-22 neels * 4 o-o-o-o-o-o c1
75 3b0f3d61 2020-01-22 neels * Moving right means delete an atom from the left-hand-side,
76 3b0f3d61 2020-01-22 neels * Moving down means add an atom from the right-hand-side.
77 0d27172a 2020-05-06 neels * Diagonals indicate identical atoms on both sides, the challenge is to use as
78 0d27172a 2020-05-06 neels * many diagonals as possible.
80 0d27172a 2020-05-06 neels * The original Myers algorithm walks all the way from the top left to the
81 0d27172a 2020-05-06 neels * bottom right, remembers all steps, and then backtraces to find the shortest
82 0d27172a 2020-05-06 neels * path. However, that requires keeping the entire graph in memory, which needs
83 3b0f3d61 2020-01-22 neels * quadratic space.
85 0d27172a 2020-05-06 neels * Myers adds a variant that uses linear space -- note, not linear time, only
86 0d27172a 2020-05-06 neels * linear space: walk forward and backward, find a meeting point in the middle,
87 0d27172a 2020-05-06 neels * and recurse on the two separate sections. This is called "divide and
88 0d27172a 2020-05-06 neels * conquer".
90 0d27172a 2020-05-06 neels * d: the step number, starting with 0, a.k.a. the distance from the starting
92 0d27172a 2020-05-06 neels * k: relative index in the state array for the forward scan, indicating on
93 0d27172a 2020-05-06 neels * which diagonal through the diff graph we currently are.
94 0d27172a 2020-05-06 neels * c: relative index in the state array for the backward scan, indicating the
95 0d27172a 2020-05-06 neels * diagonal number from the bottom up.
97 3b0f3d61 2020-01-22 neels * The "divide and conquer" traversal through the Myers graph looks like this:
99 3b0f3d61 2020-01-22 neels * | d= 0 1 2 3 2 1 0
100 3b0f3d61 2020-01-22 neels * ----+--------------------------------------------
104 3b0f3d61 2020-01-22 neels * 3 | 3,0 5,2 2
106 3b0f3d61 2020-01-22 neels * 2 | 2,0 5,3 1
108 3b0f3d61 2020-01-22 neels * 1 | 1,0 4,3 >= 4,3 5,4<-- 0
109 3b0f3d61 2020-01-22 neels * | / / \ /
110 3b0f3d61 2020-01-22 neels * 0 | -->0,0 3,3 4,4 -1
112 3b0f3d61 2020-01-22 neels * -1 | 0,1 1,2 3,4 -2
114 3b0f3d61 2020-01-22 neels * -2 | 0,2 -3
117 3b0f3d61 2020-01-22 neels * | forward-> <-backward
119 3b0f3d61 2020-01-22 neels * x,y pairs here are the coordinates in the Myers graph:
120 3b0f3d61 2020-01-22 neels * x = atom index in left-side source, y = atom index in the right-side source.
122 0d27172a 2020-05-06 neels * Only one forward column and one backward column are kept in mem, each need at
123 0d27172a 2020-05-06 neels * most left.len + 1 + right.len items. Note that each d step occupies either
124 0d27172a 2020-05-06 neels * the even or the odd items of a column: if e.g. the previous column is in the
125 0d27172a 2020-05-06 neels * odd items, the next column is formed in the even items, without overwriting
126 0d27172a 2020-05-06 neels * the previous column's results.
128 0d27172a 2020-05-06 neels * Also note that from the diagonal index k and the x coordinate, the y
129 0d27172a 2020-05-06 neels * coordinate can be derived:
130 3b0f3d61 2020-01-22 neels * y = x - k
131 0d27172a 2020-05-06 neels * Hence the state array only needs to keep the x coordinate, i.e. the position
132 0d27172a 2020-05-06 neels * in the left-hand file, and the y coordinate, i.e. position in the right-hand
133 0d27172a 2020-05-06 neels * file, is derived from the index in the state array.
135 0d27172a 2020-05-06 neels * The two traces meet at 4,3, the first step (here found in the forward
136 0d27172a 2020-05-06 neels * traversal) where a forward position is on or past a backward traced position
137 0d27172a 2020-05-06 neels * on the same diagonal.
139 3b0f3d61 2020-01-22 neels * This divides the problem space into:
141 3b0f3d61 2020-01-22 neels * 0 1 2 3 4 5
142 3b0f3d61 2020-01-22 neels * A B C D E
143 3b0f3d61 2020-01-22 neels * 0 o-o-o-o-o
144 3b0f3d61 2020-01-22 neels * X | | | | |
145 3b0f3d61 2020-01-22 neels * 1 o-o-o-o-o
146 3b0f3d61 2020-01-22 neels * B | |\| | |
147 3b0f3d61 2020-01-22 neels * 2 o-o-o-o-o
148 3b0f3d61 2020-01-22 neels * C | | |\| |
149 3b0f3d61 2020-01-22 neels * 3 o-o-o-o-*-o *: forward and backward meet here
153 3b0f3d61 2020-01-22 neels * Doing the same on each section lead to:
155 3b0f3d61 2020-01-22 neels * 0 1 2 3 4 5
156 3b0f3d61 2020-01-22 neels * A B C D E
159 3b0f3d61 2020-01-22 neels * 1 o-b b: backward d=1 first reaches here (sliding up the snake)
160 3b0f3d61 2020-01-22 neels * B \ f: then forward d=2 reaches here (sliding down the snake)
161 3b0f3d61 2020-01-22 neels * 2 o As result, the box from b to f is found to be identical;
162 0d27172a 2020-05-06 neels * C \ leaving a top box from 0,0 to 1,1 and a bottom trivial
163 0d27172a 2020-05-06 neels * 3 f-o tail 3,3 to 4,3.
167 3b0f3d61 2020-01-22 neels * 4 o *: forward and backward meet here
169 3b0f3d61 2020-01-22 neels * and solving the last top left box gives:
171 3b0f3d61 2020-01-22 neels * 0 1 2 3 4 5
172 3b0f3d61 2020-01-22 neels * A B C D E -A
173 3b0f3d61 2020-01-22 neels * 0 o-o +X
185 3b0f3d61 2020-01-22 neels #define xk_to_y(X, K) ((X) - (K))
186 3b0f3d61 2020-01-22 neels #define xc_to_y(X, C, DELTA) ((X) - (C) + (DELTA))
187 3b0f3d61 2020-01-22 neels #define k_to_c(K, DELTA) ((K) + (DELTA))
188 3b0f3d61 2020-01-22 neels #define c_to_k(C, DELTA) ((C) - (DELTA))
190 3b0f3d61 2020-01-22 neels /* Do one forwards step in the "divide and conquer" graph traversal.
191 3b0f3d61 2020-01-22 neels * left: the left side to diff.
192 3b0f3d61 2020-01-22 neels * right: the right side to diff against.
193 0d27172a 2020-05-06 neels * kd_forward: the traversal state for forwards traversal, modified by this
194 0d27172a 2020-05-06 neels * function.
195 3b0f3d61 2020-01-22 neels * This is carried over between invocations with increasing d.
196 0d27172a 2020-05-06 neels * kd_forward points at the center of the state array, allowing
197 0d27172a 2020-05-06 neels * negative indexes.
198 0d27172a 2020-05-06 neels * kd_backward: the traversal state for backwards traversal, to find a meeting
200 0d27172a 2020-05-06 neels * Since forwards is done first, kd_backward will be valid for d -
201 0d27172a 2020-05-06 neels * 1, not d.
202 0d27172a 2020-05-06 neels * kd_backward points at the center of the state array, allowing
203 0d27172a 2020-05-06 neels * negative indexes.
204 0d27172a 2020-05-06 neels * d: Step or distance counter, indicating for what value of d the kd_forward
205 0d27172a 2020-05-06 neels * should be populated.
206 0d27172a 2020-05-06 neels * For d == 0, kd_forward[0] is initialized, i.e. the first invocation should
207 0d27172a 2020-05-06 neels * be for d == 0.
208 3b0f3d61 2020-01-22 neels * meeting_snake: resulting meeting point, if any.
209 0d27172a 2020-05-06 neels * Return true when a meeting point has been identified.
211 ac2eeeff 2020-09-20 neels static int
212 ac2eeeff 2020-09-20 neels diff_divide_myers_forward(bool *found_midpoint,
213 ac2eeeff 2020-09-20 neels struct diff_data *left, struct diff_data *right,
214 61a7b578 2020-05-06 neels int *kd_forward, int *kd_backward, int d,
215 61a7b578 2020-05-06 neels struct diff_box *meeting_snake)
217 3b0f3d61 2020-01-22 neels int delta = (int)right->atoms.len - (int)left->atoms.len;
220 ac2eeeff 2020-09-20 neels *found_midpoint = false;
222 3b0f3d61 2020-01-22 neels debug("-- %s d=%d\n", __func__, d);
224 3b0f3d61 2020-01-22 neels for (k = d; k >= -d; k -= 2) {
225 3b0f3d61 2020-01-22 neels if (k < -(int)right->atoms.len || k > (int)left->atoms.len) {
226 0d27172a 2020-05-06 neels /* This diagonal is completely outside of the Myers
227 0d27172a 2020-05-06 neels * graph, don't calculate it. */
228 3b0f3d61 2020-01-22 neels if (k < -(int)right->atoms.len)
229 0d27172a 2020-05-06 neels debug(" %d k < -(int)right->atoms.len %d\n", k,
230 0d27172a 2020-05-06 neels -(int)right->atoms.len);
232 0d27172a 2020-05-06 neels debug(" %d k > left->atoms.len %d\n", k,
233 0d27172a 2020-05-06 neels left->atoms.len);
234 3b0f3d61 2020-01-22 neels if (k < 0) {
235 0d27172a 2020-05-06 neels /* We are traversing negatively, and already
236 0d27172a 2020-05-06 neels * below the entire graph, nothing will come of
237 0d27172a 2020-05-06 neels * this. */
238 3b0f3d61 2020-01-22 neels debug(" break");
241 3b0f3d61 2020-01-22 neels debug(" continue");
244 3b0f3d61 2020-01-22 neels debug("- k = %d\n", k);
245 3b0f3d61 2020-01-22 neels if (d == 0) {
246 0d27172a 2020-05-06 neels /* This is the initializing step. There is no prev_k
247 0d27172a 2020-05-06 neels * yet, get the initial x from the top left of the Myers
248 0d27172a 2020-05-06 neels * graph. */
251 0d27172a 2020-05-06 neels /* Favoring "-" lines first means favoring moving rightwards in
252 0d27172a 2020-05-06 neels * the Myers graph.
253 0d27172a 2020-05-06 neels * For this, all k should derive from k - 1, only the bottom
254 0d27172a 2020-05-06 neels * most k derive from k + 1:
256 3b0f3d61 2020-01-22 neels * | d= 0 1 2
257 3b0f3d61 2020-01-22 neels * ----+----------------
259 3b0f3d61 2020-01-22 neels * 2 | 2,0 <-- from prev_k = 2 - 1 = 1
263 3b0f3d61 2020-01-22 neels * 0 | -->0,0 3,3
265 0d27172a 2020-05-06 neels * -1 | 0,1 <-- bottom most for d=1 from
266 0d27172a 2020-05-06 neels * | \\ prev_k = -1 + 1 = 0
267 0d27172a 2020-05-06 neels * -2 | 0,2 <-- bottom most for d=2 from
268 0d27172a 2020-05-06 neels * prev_k = -2 + 1 = -1
270 0d27172a 2020-05-06 neels * Except when a k + 1 from a previous run already means a
271 0d27172a 2020-05-06 neels * further advancement in the graph.
272 3b0f3d61 2020-01-22 neels * If k == d, there is no k + 1 and k - 1 is the only option.
273 0d27172a 2020-05-06 neels * If k < d, use k + 1 in case that yields a larger x. Also use
274 0d27172a 2020-05-06 neels * k + 1 if k - 1 is outside the graph.
276 0d27172a 2020-05-06 neels else if (k > -d
277 0d27172a 2020-05-06 neels && (k == d
278 0d27172a 2020-05-06 neels || (k - 1 >= -(int)right->atoms.len
279 0d27172a 2020-05-06 neels && kd_forward[k - 1] >= kd_forward[k + 1]))) {
280 3b0f3d61 2020-01-22 neels /* Advance from k - 1.
281 0d27172a 2020-05-06 neels * From position prev_k, step to the right in the Myers
282 0d27172a 2020-05-06 neels * graph: x += 1.
284 3b0f3d61 2020-01-22 neels int prev_k = k - 1;
285 f71e8098 2020-05-05 neels int prev_x = kd_forward[prev_k];
286 3b0f3d61 2020-01-22 neels x = prev_x + 1;
288 3b0f3d61 2020-01-22 neels /* The bottom most one.
289 0d27172a 2020-05-06 neels * From position prev_k, step to the bottom in the Myers
290 0d27172a 2020-05-06 neels * graph: y += 1.
291 0d27172a 2020-05-06 neels * Incrementing y is achieved by decrementing k while
292 0d27172a 2020-05-06 neels * keeping the same x.
293 3b0f3d61 2020-01-22 neels * (since we're deriving y from y = x - k).
295 3b0f3d61 2020-01-22 neels int prev_k = k + 1;
296 f71e8098 2020-05-05 neels int prev_x = kd_forward[prev_k];
297 3b0f3d61 2020-01-22 neels x = prev_x;
300 f71e8098 2020-05-05 neels int x_before_slide = x;
301 3b0f3d61 2020-01-22 neels /* Slide down any snake that we might find here. */
302 b3fb4686 2020-09-20 neels while (x < left->atoms.len && xk_to_y(x, k) < right->atoms.len) {
303 b3fb4686 2020-09-20 neels bool same;
304 b3fb4686 2020-09-20 neels int r = diff_atom_same(&same,
305 b3fb4686 2020-09-20 neels &left->atoms.head[x],
306 b3fb4686 2020-09-20 neels &right->atoms.head[
307 b3fb4686 2020-09-20 neels xk_to_y(x, k)]);
310 b3fb4686 2020-09-20 neels if (!same)
314 3b0f3d61 2020-01-22 neels kd_forward[k] = x;
315 c5419a05 2020-05-05 neels if (x_before_slide != x) {
316 c5419a05 2020-05-05 neels debug(" down %d similar lines\n", x - x_before_slide);
319 3b0f3d61 2020-01-22 neels if (DEBUG) {
321 3b0f3d61 2020-01-22 neels for (fi = d; fi >= k; fi--) {
322 0d27172a 2020-05-06 neels debug("kd_forward[%d] = (%d, %d)\n", fi,
323 0d27172a 2020-05-06 neels kd_forward[fi], kd_forward[fi] - fi);
327 3b0f3d61 2020-01-22 neels if (x < 0 || x > left->atoms.len
328 3b0f3d61 2020-01-22 neels || xk_to_y(x, k) < 0 || xk_to_y(x, k) > right->atoms.len)
331 0d27172a 2020-05-06 neels /* Figured out a new forwards traversal, see if this has gone
332 0d27172a 2020-05-06 neels * onto or even past a preceding backwards traversal.
334 0d27172a 2020-05-06 neels * If the delta in length is odd, then d and backwards_d hit the
335 0d27172a 2020-05-06 neels * same state indexes:
336 3b0f3d61 2020-01-22 neels * | d= 0 1 2 1 0
337 3b0f3d61 2020-01-22 neels * ----+---------------- ----------------
343 3b0f3d61 2020-01-22 neels * 2 | 2,0====5,3 1
345 3b0f3d61 2020-01-22 neels * 1 | 1,0 5,4<-- 0
347 3b0f3d61 2020-01-22 neels * 0 | -->0,0 3,3====4,4 -1
349 3b0f3d61 2020-01-22 neels * -1 | 0,1 -2
351 3b0f3d61 2020-01-22 neels * -2 | 0,2 -3
354 0d27172a 2020-05-06 neels * If the delta is even, they end up off-by-one, i.e. on
355 0d27172a 2020-05-06 neels * different diagonals:
357 3b0f3d61 2020-01-22 neels * | d= 0 1 2 1 0
358 3b0f3d61 2020-01-22 neels * ----+---------------- ----------------
362 3b0f3d61 2020-01-22 neels * 2 | 2,0 off 2
364 3b0f3d61 2020-01-22 neels * 1 | 1,0 4,3 1
365 3b0f3d61 2020-01-22 neels * | / // \
366 3b0f3d61 2020-01-22 neels * 0 | -->0,0 3,3 4,4<-- 0
368 3b0f3d61 2020-01-22 neels * -1 | 0,1 3,4 -1
370 3b0f3d61 2020-01-22 neels * -2 | 0,2 -2
373 0d27172a 2020-05-06 neels * So in the forward path, we can only match up diagonals when
374 0d27172a 2020-05-06 neels * the delta is odd.
376 fd42ca98 2020-05-05 neels if ((delta & 1) == 0)
378 0d27172a 2020-05-06 neels /* Forwards is done first, so the backwards one was still at
379 0d27172a 2020-05-06 neels * d - 1. Can't do this for d == 0. */
380 3b0f3d61 2020-01-22 neels int backwards_d = d - 1;
381 fd42ca98 2020-05-05 neels if (backwards_d < 0)
384 fd42ca98 2020-05-05 neels debug("backwards_d = %d\n", backwards_d);
386 0d27172a 2020-05-06 neels /* If both sides have the same length, forward and backward
387 0d27172a 2020-05-06 neels * start on the same diagonal, meaning the backwards state index
389 0d27172a 2020-05-06 neels * As soon as the lengths are not the same, the backwards
390 0d27172a 2020-05-06 neels * traversal starts on a different diagonal, and c = k shifted
391 0d27172a 2020-05-06 neels * by the difference in length.
393 fd42ca98 2020-05-05 neels int c = k_to_c(k, delta);
395 0d27172a 2020-05-06 neels /* When the file sizes are very different, the traversal trees
396 0d27172a 2020-05-06 neels * start on far distant diagonals.
397 0d27172a 2020-05-06 neels * They don't necessarily meet straight on. See whether this
398 0d27172a 2020-05-06 neels * forward value is on a diagonal that is also valid in
399 0d27172a 2020-05-06 neels * kd_backward[], and match them if so. */
400 fd42ca98 2020-05-05 neels if (c >= -backwards_d && c <= backwards_d) {
401 0d27172a 2020-05-06 neels /* Current k is on a diagonal that exists in
402 0d27172a 2020-05-06 neels * kd_backward[]. If the two x positions have met or
403 0d27172a 2020-05-06 neels * passed (forward walked onto or past backward), then
404 0d27172a 2020-05-06 neels * we've found a midpoint / a mid-box.
406 0d27172a 2020-05-06 neels * When forwards and backwards traversals meet, the
407 0d27172a 2020-05-06 neels * endpoints of the mid-snake are not the two points in
408 0d27172a 2020-05-06 neels * kd_forward and kd_backward, but rather the section
409 0d27172a 2020-05-06 neels * that was slid (if any) of the current
410 0d27172a 2020-05-06 neels * forward/backward traversal only.
412 f71e8098 2020-05-05 neels * For example:
434 0d27172a 2020-05-06 neels * The forward traversal reached M from the top and slid
435 0d27172a 2020-05-06 neels * downwards to A. The backward traversal already
436 0d27172a 2020-05-06 neels * reached X, which is not a straight line from M
437 0d27172a 2020-05-06 neels * anymore, so picking a mid-snake from M to X would
438 0d27172a 2020-05-06 neels * yield a mistake.
440 0d27172a 2020-05-06 neels * The correct mid-snake is between M and A. M is where
441 0d27172a 2020-05-06 neels * the forward traversal hit the diagonal that the
442 0d27172a 2020-05-06 neels * backward traversal has already passed, and A is what
443 0d27172a 2020-05-06 neels * it reaches when sliding down identical lines.
445 fd42ca98 2020-05-05 neels int backward_x = kd_backward[c];
446 50198b5f 2020-05-05 neels debug("Compare: k=%d c=%d is (%d,%d) >= (%d,%d)?\n",
447 0d27172a 2020-05-06 neels k, c, x, xk_to_y(x, k), backward_x,
448 0d27172a 2020-05-06 neels xc_to_y(backward_x, c, delta));
449 f71e8098 2020-05-05 neels if (x >= backward_x) {
450 fd42ca98 2020-05-05 neels *meeting_snake = (struct diff_box){
451 f71e8098 2020-05-05 neels .left_start = x_before_slide,
452 fd42ca98 2020-05-05 neels .left_end = x,
453 0d27172a 2020-05-06 neels .right_start = xc_to_y(x_before_slide,
454 0d27172a 2020-05-06 neels c, delta),
455 fd42ca98 2020-05-05 neels .right_end = xk_to_y(x, k),
457 fd42ca98 2020-05-05 neels debug("HIT x=(%u,%u) - y=(%u,%u)\n",
458 fd42ca98 2020-05-05 neels meeting_snake->left_start,
459 fd42ca98 2020-05-05 neels meeting_snake->right_start,
460 fd42ca98 2020-05-05 neels meeting_snake->left_end,
461 fd42ca98 2020-05-05 neels meeting_snake->right_end);
462 0d27172a 2020-05-06 neels debug_dump_myers_graph(left, right, NULL,
463 0d27172a 2020-05-06 neels kd_forward, d,
464 0d27172a 2020-05-06 neels kd_backward, d-1);
465 ac2eeeff 2020-09-20 neels *found_midpoint = true;
471 0d27172a 2020-05-06 neels debug_dump_myers_graph(left, right, NULL, kd_forward, d,
472 0d27172a 2020-05-06 neels kd_backward, d-1);
476 3b0f3d61 2020-01-22 neels /* Do one backwards step in the "divide and conquer" graph traversal.
477 3b0f3d61 2020-01-22 neels * left: the left side to diff.
478 3b0f3d61 2020-01-22 neels * right: the right side to diff against.
479 0d27172a 2020-05-06 neels * kd_forward: the traversal state for forwards traversal, to find a meeting
481 0d27172a 2020-05-06 neels * Since forwards is done first, after this, both kd_forward and
482 0d27172a 2020-05-06 neels * kd_backward will be valid for d.
483 0d27172a 2020-05-06 neels * kd_forward points at the center of the state array, allowing
484 0d27172a 2020-05-06 neels * negative indexes.
485 0d27172a 2020-05-06 neels * kd_backward: the traversal state for backwards traversal, to find a meeting
487 3b0f3d61 2020-01-22 neels * This is carried over between invocations with increasing d.
488 0d27172a 2020-05-06 neels * kd_backward points at the center of the state array, allowing
489 0d27172a 2020-05-06 neels * negative indexes.
490 0d27172a 2020-05-06 neels * d: Step or distance counter, indicating for what value of d the kd_backward
491 0d27172a 2020-05-06 neels * should be populated.
492 0d27172a 2020-05-06 neels * Before the first invocation, kd_backward[0] shall point at the bottom
493 0d27172a 2020-05-06 neels * right of the Myers graph (left.len, right.len).
494 3b0f3d61 2020-01-22 neels * The first invocation will be for d == 1.
495 3b0f3d61 2020-01-22 neels * meeting_snake: resulting meeting point, if any.
496 0d27172a 2020-05-06 neels * Return true when a meeting point has been identified.
498 ac2eeeff 2020-09-20 neels static int
499 ac2eeeff 2020-09-20 neels diff_divide_myers_backward(bool *found_midpoint,
500 ac2eeeff 2020-09-20 neels struct diff_data *left, struct diff_data *right,
501 61a7b578 2020-05-06 neels int *kd_forward, int *kd_backward, int d,
502 61a7b578 2020-05-06 neels struct diff_box *meeting_snake)
504 3b0f3d61 2020-01-22 neels int delta = (int)right->atoms.len - (int)left->atoms.len;
508 ac2eeeff 2020-09-20 neels *found_midpoint = false;
510 3b0f3d61 2020-01-22 neels debug("-- %s d=%d\n", __func__, d);
512 3b0f3d61 2020-01-22 neels for (c = d; c >= -d; c -= 2) {
513 3b0f3d61 2020-01-22 neels if (c < -(int)left->atoms.len || c > (int)right->atoms.len) {
514 0d27172a 2020-05-06 neels /* This diagonal is completely outside of the Myers
515 0d27172a 2020-05-06 neels * graph, don't calculate it. */
516 3b0f3d61 2020-01-22 neels if (c < -(int)left->atoms.len)
517 0d27172a 2020-05-06 neels debug(" %d c < -(int)left->atoms.len %d\n", c,
518 0d27172a 2020-05-06 neels -(int)left->atoms.len);
520 0d27172a 2020-05-06 neels debug(" %d c > right->atoms.len %d\n", c,
521 0d27172a 2020-05-06 neels right->atoms.len);
522 3b0f3d61 2020-01-22 neels if (c < 0) {
523 0d27172a 2020-05-06 neels /* We are traversing negatively, and already
524 0d27172a 2020-05-06 neels * below the entire graph, nothing will come of
525 0d27172a 2020-05-06 neels * this. */
526 3b0f3d61 2020-01-22 neels debug(" break");
529 3b0f3d61 2020-01-22 neels debug(" continue");
532 3b0f3d61 2020-01-22 neels debug("- c = %d\n", c);
533 3b0f3d61 2020-01-22 neels if (d == 0) {
534 0d27172a 2020-05-06 neels /* This is the initializing step. There is no prev_c
535 0d27172a 2020-05-06 neels * yet, get the initial x from the bottom right of the
536 0d27172a 2020-05-06 neels * Myers graph. */
537 3b0f3d61 2020-01-22 neels x = left->atoms.len;
539 0d27172a 2020-05-06 neels /* Favoring "-" lines first means favoring moving rightwards in
540 0d27172a 2020-05-06 neels * the Myers graph.
541 0d27172a 2020-05-06 neels * For this, all c should derive from c - 1, only the bottom
542 0d27172a 2020-05-06 neels * most c derive from c + 1:
545 3b0f3d61 2020-01-22 neels * ---------------------------------------------------
549 3b0f3d61 2020-01-22 neels * from prev_c = c - 1 --> 5,2 2
553 3b0f3d61 2020-01-22 neels * 4,3 5,4<-- 0
555 3b0f3d61 2020-01-22 neels * bottom most for d=1 from c + 1 --> 4,4 -1
557 3b0f3d61 2020-01-22 neels * bottom most for d=2 --> 3,4 -2
559 0d27172a 2020-05-06 neels * Except when a c + 1 from a previous run already means a
560 0d27172a 2020-05-06 neels * further advancement in the graph.
561 3b0f3d61 2020-01-22 neels * If c == d, there is no c + 1 and c - 1 is the only option.
562 0d27172a 2020-05-06 neels * If c < d, use c + 1 in case that yields a larger x.
563 0d27172a 2020-05-06 neels * Also use c + 1 if c - 1 is outside the graph.
565 3b0f3d61 2020-01-22 neels else if (c > -d && (c == d
566 3b0f3d61 2020-01-22 neels || (c - 1 >= -(int)right->atoms.len
567 3b0f3d61 2020-01-22 neels && kd_backward[c - 1] <= kd_backward[c + 1]))) {
568 3b0f3d61 2020-01-22 neels /* A top one.
569 0d27172a 2020-05-06 neels * From position prev_c, step upwards in the Myers
570 0d27172a 2020-05-06 neels * graph: y -= 1.
571 0d27172a 2020-05-06 neels * Decrementing y is achieved by incrementing c while
572 0d27172a 2020-05-06 neels * keeping the same x. (since we're deriving y from
573 0d27172a 2020-05-06 neels * y = x - c + delta).
575 3b0f3d61 2020-01-22 neels int prev_c = c - 1;
576 f71e8098 2020-05-05 neels int prev_x = kd_backward[prev_c];
577 3b0f3d61 2020-01-22 neels x = prev_x;
579 3b0f3d61 2020-01-22 neels /* The bottom most one.
580 0d27172a 2020-05-06 neels * From position prev_c, step to the left in the Myers
581 0d27172a 2020-05-06 neels * graph: x -= 1.
583 3b0f3d61 2020-01-22 neels int prev_c = c + 1;
584 f71e8098 2020-05-05 neels int prev_x = kd_backward[prev_c];
585 3b0f3d61 2020-01-22 neels x = prev_x - 1;
588 0d27172a 2020-05-06 neels /* Slide up any snake that we might find here (sections of
589 0d27172a 2020-05-06 neels * identical lines on both sides). */
590 0d27172a 2020-05-06 neels debug("c=%d x-1=%d Yb-1=%d-1=%d\n", c, x-1, xc_to_y(x, c,
592 0d27172a 2020-05-06 neels xc_to_y(x, c, delta)-1);
593 3b0f3d61 2020-01-22 neels if (x > 0) {
594 0d27172a 2020-05-06 neels debug(" l=");
595 0d27172a 2020-05-06 neels debug_dump_atom(left, right, &left->atoms.head[x-1]);
597 3b0f3d61 2020-01-22 neels if (xc_to_y(x, c, delta) > 0) {
598 0d27172a 2020-05-06 neels debug(" r=");
599 0d27172a 2020-05-06 neels debug_dump_atom(right, left,
600 0d27172a 2020-05-06 neels &right->atoms.head[xc_to_y(x, c, delta)-1]);
602 f71e8098 2020-05-05 neels int x_before_slide = x;
603 b3fb4686 2020-09-20 neels while (x > 0 && xc_to_y(x, c, delta) > 0) {
604 b3fb4686 2020-09-20 neels bool same;
605 b3fb4686 2020-09-20 neels int r = diff_atom_same(&same,
606 b3fb4686 2020-09-20 neels &left->atoms.head[x-1],
607 b3fb4686 2020-09-20 neels &right->atoms.head[
608 b3fb4686 2020-09-20 neels xc_to_y(x, c, delta)-1]);
611 b3fb4686 2020-09-20 neels if (!same)
615 3b0f3d61 2020-01-22 neels kd_backward[c] = x;
616 c5419a05 2020-05-05 neels if (x_before_slide != x) {
617 c5419a05 2020-05-05 neels debug(" up %d similar lines\n", x_before_slide - x);
620 3b0f3d61 2020-01-22 neels if (DEBUG) {
622 3b0f3d61 2020-01-22 neels for (fi = d; fi >= c; fi--) {
623 0d27172a 2020-05-06 neels debug("kd_backward[%d] = (%d, %d)\n",
625 0d27172a 2020-05-06 neels kd_backward[fi],
626 3b0f3d61 2020-01-22 neels kd_backward[fi] - fi + delta);
630 3b0f3d61 2020-01-22 neels if (x < 0 || x > left->atoms.len
631 0d27172a 2020-05-06 neels || xc_to_y(x, c, delta) < 0
632 0d27172a 2020-05-06 neels || xc_to_y(x, c, delta) > right->atoms.len)
635 0d27172a 2020-05-06 neels /* Figured out a new backwards traversal, see if this has gone
636 0d27172a 2020-05-06 neels * onto or even past a preceding forwards traversal.
638 0d27172a 2020-05-06 neels * If the delta in length is even, then d and backwards_d hit
639 0d27172a 2020-05-06 neels * the same state indexes -- note how this is different from in
640 0d27172a 2020-05-06 neels * the forwards traversal, because now both d are the same:
642 3b0f3d61 2020-01-22 neels * | d= 0 1 2 2 1 0
643 3b0f3d61 2020-01-22 neels * ----+---------------- --------------------
649 3b0f3d61 2020-01-22 neels * 2 | 2,0====5,2 2
651 3b0f3d61 2020-01-22 neels * 1 | 1,0 5,3 1
653 3b0f3d61 2020-01-22 neels * 0 | -->0,0 3,3====4,3 5,4<-- 0
655 3b0f3d61 2020-01-22 neels * -1 | 0,1 4,4 -1
657 3b0f3d61 2020-01-22 neels * -2 | 0,2 -2
660 0d27172a 2020-05-06 neels * If the delta is odd, they end up off-by-one, i.e. on
661 0d27172a 2020-05-06 neels * different diagonals.
662 0d27172a 2020-05-06 neels * So in the backward path, we can only match up diagonals when
663 0d27172a 2020-05-06 neels * the delta is even.
665 0d27172a 2020-05-06 neels if ((delta & 1) != 0)
667 0d27172a 2020-05-06 neels /* Forwards was done first, now both d are the same. */
668 0d27172a 2020-05-06 neels int forwards_d = d;
670 0d27172a 2020-05-06 neels /* As soon as the lengths are not the same, the
671 0d27172a 2020-05-06 neels * backwards traversal starts on a different diagonal,
672 0d27172a 2020-05-06 neels * and c = k shifted by the difference in length.
674 0d27172a 2020-05-06 neels int k = c_to_k(c, delta);
676 0d27172a 2020-05-06 neels /* When the file sizes are very different, the traversal trees
677 0d27172a 2020-05-06 neels * start on far distant diagonals.
678 0d27172a 2020-05-06 neels * They don't necessarily meet straight on. See whether this
679 0d27172a 2020-05-06 neels * backward value is also on a valid diagonal in kd_forward[],
680 0d27172a 2020-05-06 neels * and match them if so. */
681 0d27172a 2020-05-06 neels if (k >= -forwards_d && k <= forwards_d) {
682 0d27172a 2020-05-06 neels /* Current c is on a diagonal that exists in
683 0d27172a 2020-05-06 neels * kd_forward[]. If the two x positions have met or
684 0d27172a 2020-05-06 neels * passed (backward walked onto or past forward), then
685 0d27172a 2020-05-06 neels * we've found a midpoint / a mid-box.
687 0d27172a 2020-05-06 neels * When forwards and backwards traversals meet, the
688 0d27172a 2020-05-06 neels * endpoints of the mid-snake are not the two points in
689 0d27172a 2020-05-06 neels * kd_forward and kd_backward, but rather the section
690 0d27172a 2020-05-06 neels * that was slid (if any) of the current
691 0d27172a 2020-05-06 neels * forward/backward traversal only.
693 0d27172a 2020-05-06 neels * For example:
713 0d27172a 2020-05-06 neels * The backward traversal reached M from the bottom and
714 0d27172a 2020-05-06 neels * slid upwards. The forward traversal already reached
715 0d27172a 2020-05-06 neels * X, which is not a straight line from M anymore, so
716 0d27172a 2020-05-06 neels * picking a mid-snake from M to X would yield a
717 0d27172a 2020-05-06 neels * mistake.
719 0d27172a 2020-05-06 neels * The correct mid-snake is between M and A. M is where
720 0d27172a 2020-05-06 neels * the backward traversal hit the diagonal that the
721 0d27172a 2020-05-06 neels * forwards traversal has already passed, and A is what
722 0d27172a 2020-05-06 neels * it reaches when sliding up identical lines.
725 0d27172a 2020-05-06 neels int forward_x = kd_forward[k];
726 0d27172a 2020-05-06 neels debug("Compare: k=%d c=%d is (%d,%d) >= (%d,%d)?\n",
727 0d27172a 2020-05-06 neels k, c, forward_x, xk_to_y(forward_x, k),
728 0d27172a 2020-05-06 neels x, xc_to_y(x, c, delta));
729 0d27172a 2020-05-06 neels if (forward_x >= x) {
730 0d27172a 2020-05-06 neels *meeting_snake = (struct diff_box){
731 0d27172a 2020-05-06 neels .left_start = x,
732 0d27172a 2020-05-06 neels .left_end = x_before_slide,
733 0d27172a 2020-05-06 neels .right_start = xc_to_y(x, c, delta),
734 0d27172a 2020-05-06 neels .right_end = xk_to_y(x_before_slide, k),
736 0d27172a 2020-05-06 neels debug("HIT x=%u,%u - y=%u,%u\n",
737 0d27172a 2020-05-06 neels meeting_snake->left_start,
738 0d27172a 2020-05-06 neels meeting_snake->right_start,
739 0d27172a 2020-05-06 neels meeting_snake->left_end,
740 0d27172a 2020-05-06 neels meeting_snake->right_end);
741 0d27172a 2020-05-06 neels debug_dump_myers_graph(left, right, NULL,
742 0d27172a 2020-05-06 neels kd_forward, d,
743 0d27172a 2020-05-06 neels kd_backward, d);
744 ac2eeeff 2020-09-20 neels *found_midpoint = true;
749 0d27172a 2020-05-06 neels debug_dump_myers_graph(left, right, NULL, kd_forward, d, kd_backward,
754 cd25827e 2020-09-20 neels /* Integer square root approximation */
755 cd25827e 2020-09-20 neels static int
756 cd25827e 2020-09-20 neels shift_sqrt(int val)
759 cd25827e 2020-09-20 neels for (i = 1; val > 0; val >>= 2)
764 0d27172a 2020-05-06 neels /* Myers "Divide et Impera": tracing forwards from the start and backwards from
765 0d27172a 2020-05-06 neels * the end to find a midpoint that divides the problem into smaller chunks.
766 0d27172a 2020-05-06 neels * Requires only linear amounts of memory. */
768 0d27172a 2020-05-06 neels diff_algo_myers_divide(const struct diff_algo_config *algo_config,
769 0d27172a 2020-05-06 neels struct diff_state *state)
771 3e6cba3a 2020-08-13 stsp int rc = ENOMEM;
772 3b0f3d61 2020-01-22 neels struct diff_data *left = &state->left;
773 3b0f3d61 2020-01-22 neels struct diff_data *right = &state->right;
775 3b0f3d61 2020-01-22 neels debug("\n** %s\n", __func__);
776 3b0f3d61 2020-01-22 neels debug("left:\n");
777 3b0f3d61 2020-01-22 neels debug_dump(left);
778 3b0f3d61 2020-01-22 neels debug("right:\n");
779 3b0f3d61 2020-01-22 neels debug_dump(right);
780 50198b5f 2020-05-05 neels debug_dump_myers_graph(left, right, NULL, NULL, 0, NULL, 0);
782 0d27172a 2020-05-06 neels /* Allocate two columns of a Myers graph, one for the forward and one
783 0d27172a 2020-05-06 neels * for the backward traversal. */
784 3b0f3d61 2020-01-22 neels unsigned int max = left->atoms.len + right->atoms.len;
785 3b0f3d61 2020-01-22 neels size_t kd_len = max + 1;
786 3b0f3d61 2020-01-22 neels size_t kd_buf_size = kd_len << 1;
787 3b0f3d61 2020-01-22 neels int *kd_buf = reallocarray(NULL, kd_buf_size, sizeof(int));
788 3b0f3d61 2020-01-22 neels if (!kd_buf)
789 3e6cba3a 2020-08-13 stsp return ENOMEM;
791 3b0f3d61 2020-01-22 neels for (i = 0; i < kd_buf_size; i++)
792 3b0f3d61 2020-01-22 neels kd_buf[i] = -1;
793 3b0f3d61 2020-01-22 neels int *kd_forward = kd_buf;
794 3b0f3d61 2020-01-22 neels int *kd_backward = kd_buf + kd_len;
795 cd25827e 2020-09-20 neels int max_effort = shift_sqrt(max/2);
797 0d27172a 2020-05-06 neels /* The 'k' axis in Myers spans positive and negative indexes, so point
798 0d27172a 2020-05-06 neels * the kd to the middle.
799 3b0f3d61 2020-01-22 neels * It is then possible to index from -max/2 .. max/2. */
800 3b0f3d61 2020-01-22 neels kd_forward += max/2;
801 3b0f3d61 2020-01-22 neels kd_backward += max/2;
804 3b0f3d61 2020-01-22 neels struct diff_box mid_snake = {};
805 a45330b1 2020-05-05 neels bool found_midpoint = false;
806 3b0f3d61 2020-01-22 neels for (d = 0; d <= (max/2); d++) {
808 3b0f3d61 2020-01-22 neels debug("-- d=%d\n", d);
809 ac2eeeff 2020-09-20 neels r = diff_divide_myers_forward(&found_midpoint, left, right,
810 ac2eeeff 2020-09-20 neels kd_forward, kd_backward, d,
811 ac2eeeff 2020-09-20 neels &mid_snake);
814 a45330b1 2020-05-05 neels if (found_midpoint)
816 ac2eeeff 2020-09-20 neels r = diff_divide_myers_backward(&found_midpoint, left, right,
817 ac2eeeff 2020-09-20 neels kd_forward, kd_backward, d,
818 ac2eeeff 2020-09-20 neels &mid_snake);
821 a45330b1 2020-05-05 neels if (found_midpoint)
824 cd25827e 2020-09-20 neels /* Limit the effort spent looking for a mid snake. If files have
825 cd25827e 2020-09-20 neels * very few lines in common, the effort spent to find nice mid
826 cd25827e 2020-09-20 neels * snakes is just not worth it, the diff result will still be
827 cd25827e 2020-09-20 neels * essentially minus everything on the left, plus everything on
828 cd25827e 2020-09-20 neels * the right, with a few useless matches here and there. */
829 cd25827e 2020-09-20 neels if (d > max_effort) {
830 cd25827e 2020-09-20 neels /* pick the furthest reaching point from
831 cd25827e 2020-09-20 neels * kd_forward and kd_backward, and use that as a
832 cd25827e 2020-09-20 neels * midpoint, to not step into another diff algo
833 cd25827e 2020-09-20 neels * recursion with unchanged box. */
834 cd25827e 2020-09-20 neels int delta = (int)right->atoms.len - (int)left->atoms.len;
838 cd25827e 2020-09-20 neels int best_forward_i = 0;
839 cd25827e 2020-09-20 neels int best_forward_distance = 0;
840 cd25827e 2020-09-20 neels int best_backward_i = 0;
841 cd25827e 2020-09-20 neels int best_backward_distance = 0;
842 cd25827e 2020-09-20 neels int distance;
843 cd25827e 2020-09-20 neels int best_forward_x;
844 cd25827e 2020-09-20 neels int best_forward_y;
845 cd25827e 2020-09-20 neels int best_backward_x;
846 cd25827e 2020-09-20 neels int best_backward_y;
848 cd25827e 2020-09-20 neels debug("~~~ d = %d > max_effort = %d\n", d, max_effort);
850 cd25827e 2020-09-20 neels for (i = d; i >= -d; i -= 2) {
851 cd25827e 2020-09-20 neels if (i >= -(int)right->atoms.len && i <= (int)left->atoms.len) {
852 cd25827e 2020-09-20 neels x = kd_forward[i];
853 cd25827e 2020-09-20 neels y = xk_to_y(x, i);
854 cd25827e 2020-09-20 neels distance = x + y;
855 cd25827e 2020-09-20 neels if (distance > best_forward_distance) {
856 cd25827e 2020-09-20 neels best_forward_distance = distance;
857 cd25827e 2020-09-20 neels best_forward_i = i;
861 cd25827e 2020-09-20 neels if (i >= -(int)left->atoms.len && i <= (int)right->atoms.len) {
862 cd25827e 2020-09-20 neels x = kd_backward[i];
863 cd25827e 2020-09-20 neels y = xc_to_y(x, i, delta);
864 cd25827e 2020-09-20 neels distance = (right->atoms.len - x)
865 cd25827e 2020-09-20 neels + (left->atoms.len - y);
866 cd25827e 2020-09-20 neels if (distance >= best_backward_distance) {
867 cd25827e 2020-09-20 neels best_backward_distance = distance;
868 cd25827e 2020-09-20 neels best_backward_i = i;
873 cd25827e 2020-09-20 neels /* The myers-divide didn't meet in the middle. We just
874 cd25827e 2020-09-20 neels * figured out the places where the forward path
875 cd25827e 2020-09-20 neels * advanced the most, and the backward path advanced the
876 cd25827e 2020-09-20 neels * most. Just divide at whichever one of those two is better.
884 cd25827e 2020-09-20 neels * F <-- cut here
888 cd25827e 2020-09-20 neels * or here --> B
896 cd25827e 2020-09-20 neels best_forward_x = kd_forward[best_forward_i];
897 cd25827e 2020-09-20 neels best_forward_y = xk_to_y(best_forward_x, best_forward_i);
898 cd25827e 2020-09-20 neels best_backward_x = kd_backward[best_backward_i];
899 cd25827e 2020-09-20 neels best_backward_y = xc_to_y(x, best_backward_i, delta);
901 cd25827e 2020-09-20 neels if (best_forward_distance >= best_backward_distance) {
902 cd25827e 2020-09-20 neels x = best_forward_x;
903 cd25827e 2020-09-20 neels y = best_forward_y;
905 cd25827e 2020-09-20 neels x = best_backward_x;
906 cd25827e 2020-09-20 neels y = best_backward_y;
909 cd25827e 2020-09-20 neels debug("max_effort cut at x=%d y=%d\n", x, y);
910 cd25827e 2020-09-20 neels if (x < 0 || y < 0
911 cd25827e 2020-09-20 neels || x > left->atoms.len || y > right->atoms.len)
914 cd25827e 2020-09-20 neels found_midpoint = true;
915 cd25827e 2020-09-20 neels mid_snake = (struct diff_box){
916 cd25827e 2020-09-20 neels .left_start = x,
917 cd25827e 2020-09-20 neels .left_end = x,
918 cd25827e 2020-09-20 neels .right_start = y,
919 cd25827e 2020-09-20 neels .right_end = y,
925 a45330b1 2020-05-05 neels if (!found_midpoint) {
926 0d27172a 2020-05-06 neels /* Divide and conquer failed to find a meeting point. Use the
927 0d27172a 2020-05-06 neels * fallback_algo defined in the algo_config (leave this to the
928 0d27172a 2020-05-06 neels * caller). This is just paranoia/sanity, we normally should
929 0d27172a 2020-05-06 neels * always find a midpoint.
931 3b0f3d61 2020-01-22 neels debug(" no midpoint \n");
932 3b0f3d61 2020-01-22 neels rc = DIFF_RC_USE_DIFF_ALGO_FALLBACK;
933 3b0f3d61 2020-01-22 neels goto return_rc;
935 3b0f3d61 2020-01-22 neels debug(" mid snake L: %u to %u of %u R: %u to %u of %u\n",
936 3b0f3d61 2020-01-22 neels mid_snake.left_start, mid_snake.left_end, left->atoms.len,
937 0d27172a 2020-05-06 neels mid_snake.right_start, mid_snake.right_end,
938 0d27172a 2020-05-06 neels right->atoms.len);
940 3b0f3d61 2020-01-22 neels /* Section before the mid-snake. */
941 3b0f3d61 2020-01-22 neels debug("Section before the mid-snake\n");
943 3b0f3d61 2020-01-22 neels struct diff_atom *left_atom = &left->atoms.head[0];
944 3b0f3d61 2020-01-22 neels unsigned int left_section_len = mid_snake.left_start;
945 3b0f3d61 2020-01-22 neels struct diff_atom *right_atom = &right->atoms.head[0];
946 3b0f3d61 2020-01-22 neels unsigned int right_section_len = mid_snake.right_start;
948 3b0f3d61 2020-01-22 neels if (left_section_len && right_section_len) {
949 0d27172a 2020-05-06 neels /* Record an unsolved chunk, the caller will apply
950 0d27172a 2020-05-06 neels * inner_algo() on this chunk. */
951 3b0f3d61 2020-01-22 neels if (!diff_state_add_chunk(state, false,
952 3b0f3d61 2020-01-22 neels left_atom, left_section_len,
953 0d27172a 2020-05-06 neels right_atom,
954 0d27172a 2020-05-06 neels right_section_len))
955 3b0f3d61 2020-01-22 neels goto return_rc;
956 3b0f3d61 2020-01-22 neels } else if (left_section_len && !right_section_len) {
957 0d27172a 2020-05-06 neels /* Only left atoms and none on the right, they form a
958 0d27172a 2020-05-06 neels * "minus" chunk, then. */
959 3b0f3d61 2020-01-22 neels if (!diff_state_add_chunk(state, true,
960 3b0f3d61 2020-01-22 neels left_atom, left_section_len,
961 3b0f3d61 2020-01-22 neels right_atom, 0))
962 3b0f3d61 2020-01-22 neels goto return_rc;
963 3b0f3d61 2020-01-22 neels } else if (!left_section_len && right_section_len) {
964 0d27172a 2020-05-06 neels /* No left atoms, only atoms on the right, they form a
965 0d27172a 2020-05-06 neels * "plus" chunk, then. */
966 3b0f3d61 2020-01-22 neels if (!diff_state_add_chunk(state, true,
967 3b0f3d61 2020-01-22 neels left_atom, 0,
968 0d27172a 2020-05-06 neels right_atom,
969 0d27172a 2020-05-06 neels right_section_len))
970 3b0f3d61 2020-01-22 neels goto return_rc;
972 0d27172a 2020-05-06 neels /* else: left_section_len == 0 and right_section_len == 0, i.e.
973 0d27172a 2020-05-06 neels * nothing before the mid-snake. */
975 cd25827e 2020-09-20 neels if (mid_snake.left_end > mid_snake.left_start) {
976 0d27172a 2020-05-06 neels /* The midpoint is a "snake", i.e. on a section of
977 0d27172a 2020-05-06 neels * identical data on both sides: that section
978 0d27172a 2020-05-06 neels * immediately becomes a solved diff chunk. */
979 a45330b1 2020-05-05 neels debug("the mid-snake\n");
980 a45330b1 2020-05-05 neels if (!diff_state_add_chunk(state, true,
981 0d27172a 2020-05-06 neels &left->atoms.head[mid_snake.left_start],
982 0d27172a 2020-05-06 neels mid_snake.left_end - mid_snake.left_start,
983 0d27172a 2020-05-06 neels &right->atoms.head[mid_snake.right_start],
984 0d27172a 2020-05-06 neels mid_snake.right_end - mid_snake.right_start))
985 a45330b1 2020-05-05 neels goto return_rc;
988 3b0f3d61 2020-01-22 neels /* Section after the mid-snake. */
989 3b0f3d61 2020-01-22 neels debug("Section after the mid-snake\n");
990 0d27172a 2020-05-06 neels debug(" left_end %u right_end %u\n",
991 0d27172a 2020-05-06 neels mid_snake.left_end, mid_snake.right_end);
992 0d27172a 2020-05-06 neels debug(" left_count %u right_count %u\n",
993 0d27172a 2020-05-06 neels left->atoms.len, right->atoms.len);
994 3b0f3d61 2020-01-22 neels left_atom = &left->atoms.head[mid_snake.left_end];
995 3b0f3d61 2020-01-22 neels left_section_len = left->atoms.len - mid_snake.left_end;
996 3b0f3d61 2020-01-22 neels right_atom = &right->atoms.head[mid_snake.right_end];
997 3b0f3d61 2020-01-22 neels right_section_len = right->atoms.len - mid_snake.right_end;
999 3b0f3d61 2020-01-22 neels if (left_section_len && right_section_len) {
1000 0d27172a 2020-05-06 neels /* Record an unsolved chunk, the caller will apply
1001 0d27172a 2020-05-06 neels * inner_algo() on this chunk. */
1002 3b0f3d61 2020-01-22 neels if (!diff_state_add_chunk(state, false,
1003 3b0f3d61 2020-01-22 neels left_atom, left_section_len,
1004 0d27172a 2020-05-06 neels right_atom,
1005 0d27172a 2020-05-06 neels right_section_len))
1006 3b0f3d61 2020-01-22 neels goto return_rc;
1007 3b0f3d61 2020-01-22 neels } else if (left_section_len && !right_section_len) {
1008 0d27172a 2020-05-06 neels /* Only left atoms and none on the right, they form a
1009 0d27172a 2020-05-06 neels * "minus" chunk, then. */
1010 3b0f3d61 2020-01-22 neels if (!diff_state_add_chunk(state, true,
1011 3b0f3d61 2020-01-22 neels left_atom, left_section_len,
1012 3b0f3d61 2020-01-22 neels right_atom, 0))
1013 3b0f3d61 2020-01-22 neels goto return_rc;
1014 3b0f3d61 2020-01-22 neels } else if (!left_section_len && right_section_len) {
1015 0d27172a 2020-05-06 neels /* No left atoms, only atoms on the right, they form a
1016 0d27172a 2020-05-06 neels * "plus" chunk, then. */
1017 3b0f3d61 2020-01-22 neels if (!diff_state_add_chunk(state, true,
1018 3b0f3d61 2020-01-22 neels left_atom, 0,
1019 0d27172a 2020-05-06 neels right_atom,
1020 0d27172a 2020-05-06 neels right_section_len))
1021 3b0f3d61 2020-01-22 neels goto return_rc;
1023 0d27172a 2020-05-06 neels /* else: left_section_len == 0 and right_section_len == 0, i.e.
1024 0d27172a 2020-05-06 neels * nothing after the mid-snake. */
1027 3b0f3d61 2020-01-22 neels rc = DIFF_RC_OK;
1029 3b0f3d61 2020-01-22 neels return_rc:
1030 3b0f3d61 2020-01-22 neels free(kd_buf);
1031 3b0f3d61 2020-01-22 neels debug("** END %s\n", __func__);
1032 3b0f3d61 2020-01-22 neels return rc;
1035 0d27172a 2020-05-06 neels /* Myers Diff tracing from the start all the way through to the end, requiring
1036 0d27172a 2020-05-06 neels * quadratic amounts of memory. This can fail if the required space surpasses
1037 0d27172a 2020-05-06 neels * algo_config->permitted_state_size. */
1039 0d27172a 2020-05-06 neels diff_algo_myers(const struct diff_algo_config *algo_config,
1040 0d27172a 2020-05-06 neels struct diff_state *state)
1042 0d27172a 2020-05-06 neels /* do a diff_divide_myers_forward() without a _backward(), so that it
1043 0d27172a 2020-05-06 neels * walks forward across the entire files to reach the end. Keep each
1044 0d27172a 2020-05-06 neels * run's state, and do a final backtrace. */
1045 3e6cba3a 2020-08-13 stsp int rc = ENOMEM;
1046 3b0f3d61 2020-01-22 neels struct diff_data *left = &state->left;
1047 3b0f3d61 2020-01-22 neels struct diff_data *right = &state->right;
1049 3b0f3d61 2020-01-22 neels debug("\n** %s\n", __func__);
1050 3b0f3d61 2020-01-22 neels debug("left:\n");
1051 3b0f3d61 2020-01-22 neels debug_dump(left);
1052 3b0f3d61 2020-01-22 neels debug("right:\n");
1053 3b0f3d61 2020-01-22 neels debug_dump(right);
1054 50198b5f 2020-05-05 neels debug_dump_myers_graph(left, right, NULL, NULL, 0, NULL, 0);
1056 0d27172a 2020-05-06 neels /* Allocate two columns of a Myers graph, one for the forward and one
1057 0d27172a 2020-05-06 neels * for the backward traversal. */
1058 3b0f3d61 2020-01-22 neels unsigned int max = left->atoms.len + right->atoms.len;
1059 3b0f3d61 2020-01-22 neels size_t kd_len = max + 1 + max;
1060 3b0f3d61 2020-01-22 neels size_t kd_buf_size = kd_len * kd_len;
1061 50198b5f 2020-05-05 neels size_t kd_state_size = kd_buf_size * sizeof(int);
1062 3b0f3d61 2020-01-22 neels debug("state size: %zu\n", kd_buf_size);
1063 3b0f3d61 2020-01-22 neels if (kd_buf_size < kd_len /* overflow? */
1064 50198b5f 2020-05-05 neels || kd_state_size > algo_config->permitted_state_size) {
1065 3b0f3d61 2020-01-22 neels debug("state size %zu > permitted_state_size %zu, use fallback_algo\n",
1066 50198b5f 2020-05-05 neels kd_state_size, algo_config->permitted_state_size);
1067 3b0f3d61 2020-01-22 neels return DIFF_RC_USE_DIFF_ALGO_FALLBACK;
1070 3b0f3d61 2020-01-22 neels int *kd_buf = reallocarray(NULL, kd_buf_size, sizeof(int));
1071 3b0f3d61 2020-01-22 neels if (!kd_buf)
1072 3e6cba3a 2020-08-13 stsp return ENOMEM;
1074 3b0f3d61 2020-01-22 neels for (i = 0; i < kd_buf_size; i++)
1075 3b0f3d61 2020-01-22 neels kd_buf[i] = -1;
1077 0d27172a 2020-05-06 neels /* The 'k' axis in Myers spans positive and negative indexes, so point
1078 0d27172a 2020-05-06 neels * the kd to the middle.
1079 3b0f3d61 2020-01-22 neels * It is then possible to index from -max .. max. */
1080 3b0f3d61 2020-01-22 neels int *kd_origin = kd_buf + max;
1081 3b0f3d61 2020-01-22 neels int *kd_column = kd_origin;
1084 3b0f3d61 2020-01-22 neels int backtrack_d = -1;
1085 3b0f3d61 2020-01-22 neels int backtrack_k = 0;
1087 3b0f3d61 2020-01-22 neels int x, y;
1088 3b0f3d61 2020-01-22 neels for (d = 0; d <= max; d++, kd_column += kd_len) {
1089 3b0f3d61 2020-01-22 neels debug("-- d=%d\n", d);
1091 3b0f3d61 2020-01-22 neels debug("-- %s d=%d\n", __func__, d);
1093 3b0f3d61 2020-01-22 neels for (k = d; k >= -d; k -= 2) {
1094 0d27172a 2020-05-06 neels if (k < -(int)right->atoms.len
1095 0d27172a 2020-05-06 neels || k > (int)left->atoms.len) {
1096 0d27172a 2020-05-06 neels /* This diagonal is completely outside of the
1097 0d27172a 2020-05-06 neels * Myers graph, don't calculate it. */
1098 3b0f3d61 2020-01-22 neels if (k < -(int)right->atoms.len)
1099 0d27172a 2020-05-06 neels debug(" %d k <"
1100 0d27172a 2020-05-06 neels " -(int)right->atoms.len %d\n",
1101 0d27172a 2020-05-06 neels k, -(int)right->atoms.len);
1103 0d27172a 2020-05-06 neels debug(" %d k > left->atoms.len %d\n", k,
1104 0d27172a 2020-05-06 neels left->atoms.len);
1105 3b0f3d61 2020-01-22 neels if (k < 0) {
1106 0d27172a 2020-05-06 neels /* We are traversing negatively, and
1107 0d27172a 2020-05-06 neels * already below the entire graph,
1108 0d27172a 2020-05-06 neels * nothing will come of this. */
1109 3b0f3d61 2020-01-22 neels debug(" break");
1112 3b0f3d61 2020-01-22 neels debug(" continue");
1113 3b0f3d61 2020-01-22 neels continue;
1116 3b0f3d61 2020-01-22 neels debug("- k = %d\n", k);
1117 3b0f3d61 2020-01-22 neels if (d == 0) {
1118 0d27172a 2020-05-06 neels /* This is the initializing step. There is no
1119 0d27172a 2020-05-06 neels * prev_k yet, get the initial x from the top
1120 0d27172a 2020-05-06 neels * left of the Myers graph. */
1123 3b0f3d61 2020-01-22 neels int *kd_prev_column = kd_column - kd_len;
1125 0d27172a 2020-05-06 neels /* Favoring "-" lines first means favoring
1126 0d27172a 2020-05-06 neels * moving rightwards in the Myers graph.
1127 0d27172a 2020-05-06 neels * For this, all k should derive from k - 1,
1128 0d27172a 2020-05-06 neels * only the bottom most k derive from k + 1:
1130 3b0f3d61 2020-01-22 neels * | d= 0 1 2
1131 3b0f3d61 2020-01-22 neels * ----+----------------
1133 0d27172a 2020-05-06 neels * 2 | 2,0 <-- from
1134 0d27172a 2020-05-06 neels * | / prev_k = 2 - 1 = 1
1135 3b0f3d61 2020-01-22 neels * 1 | 1,0
1137 3b0f3d61 2020-01-22 neels * 0 | -->0,0 3,3
1139 0d27172a 2020-05-06 neels * -1 | 0,1 <-- bottom most for d=1
1140 0d27172a 2020-05-06 neels * | \\ from prev_k = -1+1 = 0
1141 0d27172a 2020-05-06 neels * -2 | 0,2 <-- bottom most for
1142 0d27172a 2020-05-06 neels * d=2 from
1143 0d27172a 2020-05-06 neels * prev_k = -2+1 = -1
1145 0d27172a 2020-05-06 neels * Except when a k + 1 from a previous run
1146 0d27172a 2020-05-06 neels * already means a further advancement in the
1148 0d27172a 2020-05-06 neels * If k == d, there is no k + 1 and k - 1 is the
1149 0d27172a 2020-05-06 neels * only option.
1150 0d27172a 2020-05-06 neels * If k < d, use k + 1 in case that yields a
1151 0d27172a 2020-05-06 neels * larger x. Also use k + 1 if k - 1 is outside
1152 0d27172a 2020-05-06 neels * the graph.
1154 0d27172a 2020-05-06 neels if (k > -d
1155 0d27172a 2020-05-06 neels && (k == d
1156 0d27172a 2020-05-06 neels || (k - 1 >= -(int)right->atoms.len
1157 0d27172a 2020-05-06 neels && kd_prev_column[k - 1]
1158 0d27172a 2020-05-06 neels >= kd_prev_column[k + 1]))) {
1159 3b0f3d61 2020-01-22 neels /* Advance from k - 1.
1160 0d27172a 2020-05-06 neels * From position prev_k, step to the
1161 0d27172a 2020-05-06 neels * right in the Myers graph: x += 1.
1163 3b0f3d61 2020-01-22 neels int prev_k = k - 1;
1164 3b0f3d61 2020-01-22 neels int prev_x = kd_prev_column[prev_k];
1165 3b0f3d61 2020-01-22 neels x = prev_x + 1;
1167 3b0f3d61 2020-01-22 neels /* The bottom most one.
1168 0d27172a 2020-05-06 neels * From position prev_k, step to the
1169 0d27172a 2020-05-06 neels * bottom in the Myers graph: y += 1.
1170 0d27172a 2020-05-06 neels * Incrementing y is achieved by
1171 0d27172a 2020-05-06 neels * decrementing k while keeping the same
1172 0d27172a 2020-05-06 neels * x. (since we're deriving y from y =
1173 0d27172a 2020-05-06 neels * x - k).
1175 3b0f3d61 2020-01-22 neels int prev_k = k + 1;
1176 3b0f3d61 2020-01-22 neels int prev_x = kd_prev_column[prev_k];
1177 3b0f3d61 2020-01-22 neels x = prev_x;
1181 3b0f3d61 2020-01-22 neels /* Slide down any snake that we might find here. */
1182 0d27172a 2020-05-06 neels while (x < left->atoms.len
1183 b3fb4686 2020-09-20 neels && xk_to_y(x, k) < right->atoms.len) {
1184 b3fb4686 2020-09-20 neels bool same;
1185 b3fb4686 2020-09-20 neels int r = diff_atom_same(&same,
1186 b3fb4686 2020-09-20 neels &left->atoms.head[x],
1187 b3fb4686 2020-09-20 neels &right->atoms.head[
1188 b3fb4686 2020-09-20 neels xk_to_y(x, k)]);
1190 b3fb4686 2020-09-20 neels return r;
1191 b3fb4686 2020-09-20 neels if (!same)
1195 3b0f3d61 2020-01-22 neels kd_column[k] = x;
1197 3b0f3d61 2020-01-22 neels if (DEBUG) {
1199 3b0f3d61 2020-01-22 neels for (fi = d; fi >= k; fi-=2) {
1200 0d27172a 2020-05-06 neels debug("kd_column[%d] = (%d, %d)\n", fi,
1201 0d27172a 2020-05-06 neels kd_column[fi],
1202 0d27172a 2020-05-06 neels kd_column[fi] - fi);
1206 0d27172a 2020-05-06 neels if (x == left->atoms.len
1207 0d27172a 2020-05-06 neels && xk_to_y(x, k) == right->atoms.len) {
1208 3b0f3d61 2020-01-22 neels /* Found a path */
1209 3b0f3d61 2020-01-22 neels backtrack_d = d;
1210 3b0f3d61 2020-01-22 neels backtrack_k = k;
1211 3b0f3d61 2020-01-22 neels debug("Reached the end at d = %d, k = %d\n",
1212 3b0f3d61 2020-01-22 neels backtrack_d, backtrack_k);
1217 3b0f3d61 2020-01-22 neels if (backtrack_d >= 0)
1221 50198b5f 2020-05-05 neels debug_dump_myers_graph(left, right, kd_origin, NULL, 0, NULL, 0);
1223 3b0f3d61 2020-01-22 neels /* backtrack. A matrix spanning from start to end of the file is ready:
1225 3b0f3d61 2020-01-22 neels * | d= 0 1 2 3 4
1226 3b0f3d61 2020-01-22 neels * ----+---------------------------------
1230 3b0f3d61 2020-01-22 neels * 2 | 2,0
1232 3b0f3d61 2020-01-22 neels * 1 | 1,0 4,3
1233 3b0f3d61 2020-01-22 neels * | / / \
1234 3b0f3d61 2020-01-22 neels * 0 | -->0,0 3,3 4,4 --> backtrack_d = 4, backtrack_k = 0
1235 3b0f3d61 2020-01-22 neels * | \ / \
1236 3b0f3d61 2020-01-22 neels * -1 | 0,1 3,4
1238 3b0f3d61 2020-01-22 neels * -2 | 0,2
1241 3b0f3d61 2020-01-22 neels * From (4,4) backwards, find the previous position that is the largest, and remember it.
1244 3b0f3d61 2020-01-22 neels for (d = backtrack_d, k = backtrack_k; d >= 0; d--) {
1245 3b0f3d61 2020-01-22 neels x = kd_column[k];
1246 3b0f3d61 2020-01-22 neels y = xk_to_y(x, k);
1248 0d27172a 2020-05-06 neels /* When the best position is identified, remember it for that
1249 0d27172a 2020-05-06 neels * kd_column.
1250 0d27172a 2020-05-06 neels * That kd_column is no longer needed otherwise, so just
1251 0d27172a 2020-05-06 neels * re-purpose kd_column[0] = x and kd_column[1] = y,
1252 3b0f3d61 2020-01-22 neels * so that there is no need to allocate more memory.
1254 3b0f3d61 2020-01-22 neels kd_column[0] = x;
1255 3b0f3d61 2020-01-22 neels kd_column[1] = y;
1256 3b0f3d61 2020-01-22 neels debug("Backtrack d=%d: xy=(%d, %d)\n",
1257 3b0f3d61 2020-01-22 neels d, kd_column[0], kd_column[1]);
1259 3b0f3d61 2020-01-22 neels /* Don't access memory before kd_buf */
1260 3b0f3d61 2020-01-22 neels if (d == 0)
1262 3b0f3d61 2020-01-22 neels int *kd_prev_column = kd_column - kd_len;
1264 3b0f3d61 2020-01-22 neels /* When y == 0, backtracking downwards (k-1) is the only way.
1265 3b0f3d61 2020-01-22 neels * When x == 0, backtracking upwards (k+1) is the only way.
1267 3b0f3d61 2020-01-22 neels * | d= 0 1 2 3 4
1268 3b0f3d61 2020-01-22 neels * ----+---------------------------------
1271 3b0f3d61 2020-01-22 neels * | ..y == 0
1272 3b0f3d61 2020-01-22 neels * 2 | 2,0
1274 3b0f3d61 2020-01-22 neels * 1 | 1,0 4,3
1275 3b0f3d61 2020-01-22 neels * | / / \
1276 0d27172a 2020-05-06 neels * 0 | -->0,0 3,3 4,4 --> backtrack_d = 4,
1277 0d27172a 2020-05-06 neels * | \ / \ backtrack_k = 0
1278 3b0f3d61 2020-01-22 neels * -1 | 0,1 3,4
1280 3b0f3d61 2020-01-22 neels * -2 | 0,2__
1281 3b0f3d61 2020-01-22 neels * | x == 0
1283 3b0f3d61 2020-01-22 neels debug("prev[k-1] = %d,%d prev[k+1] = %d,%d\n",
1284 3b0f3d61 2020-01-22 neels kd_prev_column[k-1], xk_to_y(kd_prev_column[k-1],k-1),
1285 3b0f3d61 2020-01-22 neels kd_prev_column[k+1], xk_to_y(kd_prev_column[k+1],k+1));
1286 3b0f3d61 2020-01-22 neels if (y == 0
1287 0d27172a 2020-05-06 neels || (x > 0
1288 0d27172a 2020-05-06 neels && kd_prev_column[k - 1] >= kd_prev_column[k + 1])) {
1289 3b0f3d61 2020-01-22 neels k = k - 1;
1290 3b0f3d61 2020-01-22 neels debug("prev k=k-1=%d x=%d y=%d\n",
1291 0d27172a 2020-05-06 neels k, kd_prev_column[k],
1292 0d27172a 2020-05-06 neels xk_to_y(kd_prev_column[k], k));
1294 3b0f3d61 2020-01-22 neels k = k + 1;
1295 3b0f3d61 2020-01-22 neels debug("prev k=k+1=%d x=%d y=%d\n",
1296 0d27172a 2020-05-06 neels k, kd_prev_column[k],
1297 0d27172a 2020-05-06 neels xk_to_y(kd_prev_column[k], k));
1299 3b0f3d61 2020-01-22 neels kd_column = kd_prev_column;
1302 3b0f3d61 2020-01-22 neels /* Forwards again, this time recording the diff chunks.
1303 0d27172a 2020-05-06 neels * Definitely start from 0,0. kd_column[0] may actually point to the
1304 0d27172a 2020-05-06 neels * bottom of a snake starting at 0,0 */
1308 3b0f3d61 2020-01-22 neels kd_column = kd_origin;
1309 3b0f3d61 2020-01-22 neels for (d = 0; d <= backtrack_d; d++, kd_column += kd_len) {
1310 3b0f3d61 2020-01-22 neels int next_x = kd_column[0];
1311 3b0f3d61 2020-01-22 neels int next_y = kd_column[1];
1312 3b0f3d61 2020-01-22 neels debug("Forward track from xy(%d,%d) to xy(%d,%d)\n",
1313 3b0f3d61 2020-01-22 neels x, y, next_x, next_y);
1315 3b0f3d61 2020-01-22 neels struct diff_atom *left_atom = &left->atoms.head[x];
1316 3b0f3d61 2020-01-22 neels int left_section_len = next_x - x;
1317 3b0f3d61 2020-01-22 neels struct diff_atom *right_atom = &right->atoms.head[y];
1318 3b0f3d61 2020-01-22 neels int right_section_len = next_y - y;
1320 3e6cba3a 2020-08-13 stsp rc = ENOMEM;
1321 3b0f3d61 2020-01-22 neels if (left_section_len && right_section_len) {
1322 3b0f3d61 2020-01-22 neels /* This must be a snake slide.
1323 0d27172a 2020-05-06 neels * Snake slides have a straight line leading into them
1324 0d27172a 2020-05-06 neels * (except when starting at (0,0)). Find out whether the
1325 0d27172a 2020-05-06 neels * lead-in is horizontal or vertical:
1328 3b0f3d61 2020-01-22 neels * ---------->
1330 3b0f3d61 2020-01-22 neels * r| o-o o
1337 0d27172a 2020-05-06 neels * If left_section_len > right_section_len, the lead-in
1338 0d27172a 2020-05-06 neels * is horizontal, meaning first remove one atom from the
1339 0d27172a 2020-05-06 neels * left before sliding down the snake.
1340 0d27172a 2020-05-06 neels * If right_section_len > left_section_len, the lead-in
1341 0d27172a 2020-05-06 neels * is vetical, so add one atom from the right before
1342 0d27172a 2020-05-06 neels * sliding down the snake. */
1343 3b0f3d61 2020-01-22 neels if (left_section_len == right_section_len + 1) {
1344 3b0f3d61 2020-01-22 neels if (!diff_state_add_chunk(state, true,
1345 3b0f3d61 2020-01-22 neels left_atom, 1,
1346 3b0f3d61 2020-01-22 neels right_atom, 0))
1347 3b0f3d61 2020-01-22 neels goto return_rc;
1348 3b0f3d61 2020-01-22 neels left_atom++;
1349 3b0f3d61 2020-01-22 neels left_section_len--;
1350 3b0f3d61 2020-01-22 neels } else if (right_section_len == left_section_len + 1) {
1351 3b0f3d61 2020-01-22 neels if (!diff_state_add_chunk(state, true,
1352 3b0f3d61 2020-01-22 neels left_atom, 0,
1353 3b0f3d61 2020-01-22 neels right_atom, 1))
1354 3b0f3d61 2020-01-22 neels goto return_rc;
1355 3b0f3d61 2020-01-22 neels right_atom++;
1356 3b0f3d61 2020-01-22 neels right_section_len--;
1357 3b0f3d61 2020-01-22 neels } else if (left_section_len != right_section_len) {
1358 0d27172a 2020-05-06 neels /* The numbers are making no sense. Should never
1359 0d27172a 2020-05-06 neels * happen. */
1360 3b0f3d61 2020-01-22 neels rc = DIFF_RC_USE_DIFF_ALGO_FALLBACK;
1361 3b0f3d61 2020-01-22 neels goto return_rc;
1364 3b0f3d61 2020-01-22 neels if (!diff_state_add_chunk(state, true,
1365 3b0f3d61 2020-01-22 neels left_atom, left_section_len,
1366 0d27172a 2020-05-06 neels right_atom,
1367 0d27172a 2020-05-06 neels right_section_len))
1368 3b0f3d61 2020-01-22 neels goto return_rc;
1369 3b0f3d61 2020-01-22 neels } else if (left_section_len && !right_section_len) {
1370 0d27172a 2020-05-06 neels /* Only left atoms and none on the right, they form a
1371 0d27172a 2020-05-06 neels * "minus" chunk, then. */
1372 3b0f3d61 2020-01-22 neels if (!diff_state_add_chunk(state, true,
1373 3b0f3d61 2020-01-22 neels left_atom, left_section_len,
1374 3b0f3d61 2020-01-22 neels right_atom, 0))
1375 3b0f3d61 2020-01-22 neels goto return_rc;
1376 3b0f3d61 2020-01-22 neels } else if (!left_section_len && right_section_len) {
1377 0d27172a 2020-05-06 neels /* No left atoms, only atoms on the right, they form a
1378 0d27172a 2020-05-06 neels * "plus" chunk, then. */
1379 3b0f3d61 2020-01-22 neels if (!diff_state_add_chunk(state, true,
1380 3b0f3d61 2020-01-22 neels left_atom, 0,
1381 0d27172a 2020-05-06 neels right_atom,
1382 0d27172a 2020-05-06 neels right_section_len))
1383 3b0f3d61 2020-01-22 neels goto return_rc;
1386 3b0f3d61 2020-01-22 neels x = next_x;
1387 3b0f3d61 2020-01-22 neels y = next_y;
1390 3b0f3d61 2020-01-22 neels rc = DIFF_RC_OK;
1392 3b0f3d61 2020-01-22 neels return_rc:
1393 3b0f3d61 2020-01-22 neels free(kd_buf);
1394 3b0f3d61 2020-01-22 neels debug("** END %s rc=%d\n", __func__, rc);
1395 3b0f3d61 2020-01-22 neels return rc;
