1 /* Myers diff algorithm implementation, invented by Eugene W. Myers [1].
2  * Implementations of both the Myers Divide Et Impera (using linear space)
3  * and the canonical Myers algorithm (using quadratic space). */
4 /*
5  * Copyright (c) 2020 Neels Hofmeyr <neels@hofmeyr.de>
6  *
7  * Permission to use, copy, modify, and distribute this software for any
8  * purpose with or without fee is hereby granted, provided that the above
9  * copyright notice and this permission notice appear in all copies.
10  *
11  * THE SOFTWARE IS PROVIDED "AS IS" AND THE AUTHOR DISCLAIMS ALL WARRANTIES
12  * WITH REGARD TO THIS SOFTWARE INCLUDING ALL IMPLIED WARRANTIES OF
13  * MERCHANTABILITY AND FITNESS. IN NO EVENT SHALL THE AUTHOR BE LIABLE FOR
14  * ANY SPECIAL, DIRECT, INDIRECT, OR CONSEQUENTIAL DAMAGES OR ANY DAMAGES
15  * WHATSOEVER RESULTING FROM LOSS OF USE, DATA OR PROFITS, WHETHER IN AN
16  * ACTION OF CONTRACT, NEGLIGENCE OR OTHER TORTIOUS ACTION, ARISING OUT OF
17  * OR IN CONNECTION WITH THE USE OR PERFORMANCE OF THIS SOFTWARE.
18  */
19 
20 #include <inttypes.h>
21 #include <stdbool.h>
22 #include <stdlib.h>
23 #include <string.h>
24 #include <stdio.h>
25 #include <errno.h>
26 
27 #include <arraylist.h>
28 #include <diff_main.h>
29 
30 #include "diff_internal.h"
31 #include "diff_debug.h"
32 
33 /* Myers' diff algorithm [1] is nicely explained in [2].
34  * [1] http://www.xmailserver.org/diff2.pdf
35  * [2] https://blog.jcoglan.com/2017/02/12/the-myers-diff-algorithm-part-1/ ff.
36  *
37  * Myers approaches finding the smallest diff as a graph problem.
38  * The crux is that the original algorithm requires quadratic amount of memory:
39  * both sides' lengths added, and that squared. So if we're diffing lines of
40  * text, two files with 1000 lines each would blow up to a matrix of about
41  * 2000 * 2000 ints of state, about 16 Mb of RAM to figure out 2 kb of text.
42  * The solution is using Myers' "divide and conquer" extension algorithm, which
43  * does the original traversal from both ends of the files to reach a middle
44  * where these "snakes" touch, hence does not need to backtrace the traversal,
45  * and so gets away with only keeping a single column of that huge state matrix
46  * in memory.
47  */
48 
49 struct diff_box {
50 	unsigned int left_start;
51 	unsigned int left_end;
52 	unsigned int right_start;
53 	unsigned int right_end;
54 };
55 
56 /* If the two contents of a file are A B C D E and X B C Y,
57  * the Myers diff graph looks like:
58  *
59  *   k0  k1
60  *    \   \
61  * k-1     0 1 2 3 4 5
62  *   \      A B C D E
63  *     0   o-o-o-o-o-o
64  *      X  | | | | | |
65  *     1   o-o-o-o-o-o
66  *      B  | |\| | | |
67  *     2   o-o-o-o-o-o
68  *      C  | | |\| | |
69  *     3   o-o-o-o-o-o
70  *      Y  | | | | | |\
71  *     4   o-o-o-o-o-o c1
72  *                  \ \
73  *                 c-1 c0
74  *
75  * Moving right means delete an atom from the left-hand-side,
76  * Moving down means add an atom from the right-hand-side.
77  * Diagonals indicate identical atoms on both sides, the challenge is to use as
78  * many diagonals as possible.
79  *
80  * The original Myers algorithm walks all the way from the top left to the
81  * bottom right, remembers all steps, and then backtraces to find the shortest
82  * path. However, that requires keeping the entire graph in memory, which needs
83  * quadratic space.
84  *
85  * Myers adds a variant that uses linear space -- note, not linear time, only
86  * linear space: walk forward and backward, find a meeting point in the middle,
87  * and recurse on the two separate sections. This is called "divide and
88  * conquer".
89  *
90  * d: the step number, starting with 0, a.k.a. the distance from the starting
91  *    point.
92  * k: relative index in the state array for the forward scan, indicating on
93  *    which diagonal through the diff graph we currently are.
94  * c: relative index in the state array for the backward scan, indicating the
95  *    diagonal number from the bottom up.
96  *
97  * The "divide and conquer" traversal through the Myers graph looks like this:
98  *
99  *      | d=   0   1   2   3      2   1   0
100  *  ----+--------------------------------------------
101  *  k=  |                                      c=
102  *   4  |                                       3
103  *      |
104  *   3  |                 3,0    5,2            2
105  *      |                /          \
106  *   2  |             2,0            5,3        1
107  *      |            /                 \
108  *   1  |         1,0     4,3 >= 4,3    5,4<--  0
109  *      |        /       /          \  /
110  *   0  |  -->0,0     3,3            4,4       -1
111  *      |        \   /              /
112  *  -1  |         0,1     1,2    3,4           -2
113  *      |            \   /
114  *  -2  |             0,2                      -3
115  *      |                \
116  *      |                 0,3
117  *      |  forward->                 <-backward
118  *
119  * x,y pairs here are the coordinates in the Myers graph:
120  * x = atom index in left-side source, y = atom index in the right-side source.
121  *
122  * Only one forward column and one backward column are kept in mem, each need at
123  * most left.len + 1 + right.len items.  Note that each d step occupies either
124  * the even or the odd items of a column: if e.g. the previous column is in the
125  * odd items, the next column is formed in the even items, without overwriting
126  * the previous column's results.
127  *
128  * Also note that from the diagonal index k and the x coordinate, the y
129  * coordinate can be derived:
130  *    y = x - k
131  * Hence the state array only needs to keep the x coordinate, i.e. the position
132  * in the left-hand file, and the y coordinate, i.e. position in the right-hand
133  * file, is derived from the index in the state array.
134  *
135  * The two traces meet at 4,3, the first step (here found in the forward
136  * traversal) where a forward position is on or past a backward traced position
137  * on the same diagonal.
138  *
139  * This divides the problem space into:
140  *
141  *         0 1 2 3 4 5
142  *          A B C D E
143  *     0   o-o-o-o-o
144  *      X  | | | | |
145  *     1   o-o-o-o-o
146  *      B  | |\| | |
147  *     2   o-o-o-o-o
148  *      C  | | |\| |
149  *     3   o-o-o-o-*-o   *: forward and backward meet here
150  *      Y          | |
151  *     4           o-o
152  *
153  * Doing the same on each section lead to:
154  *
155  *         0 1 2 3 4 5
156  *          A B C D E
157  *     0   o-o
158  *      X  | |
159  *     1   o-b    b: backward d=1 first reaches here (sliding up the snake)
160  *      B     \   f: then forward d=2 reaches here (sliding down the snake)
161  *     2       o     As result, the box from b to f is found to be identical;
162  *      C       \    leaving a top box from 0,0 to 1,1 and a bottom trivial
163  *     3         f-o tail 3,3 to 4,3.
164  *
165  *     3           o-*
166  *      Y            |
167  *     4             o   *: forward and backward meet here
168  *
169  * and solving the last top left box gives:
170  *
171  *         0 1 2 3 4 5
172  *          A B C D E           -A
173  *     0   o-o                  +X
174  *      X    |                   B
175  *     1     o                   C
176  *      B     \                 -D
177  *     2       o                -E
178  *      C       \               +Y
179  *     3         o-o-o
180  *      Y            |
181  *     4             o
182  *
183  */
184 
185 #define xk_to_y(X, K) ((X) - (K))
186 #define xc_to_y(X, C, DELTA) ((X) - (C) + (DELTA))
187 #define k_to_c(K, DELTA) ((K) + (DELTA))
188 #define c_to_k(C, DELTA) ((C) - (DELTA))
189 
190 /* Do one forwards step in the "divide and conquer" graph traversal.
191  * left: the left side to diff.
192  * right: the right side to diff against.
193  * kd_forward: the traversal state for forwards traversal, modified by this
194  *             function.
195  *             This is carried over between invocations with increasing d.
196  *             kd_forward points at the center of the state array, allowing
197  *             negative indexes.
198  * kd_backward: the traversal state for backwards traversal, to find a meeting
199  *              point.
200  *              Since forwards is done first, kd_backward will be valid for d -
201  *              1, not d.
202  *              kd_backward points at the center of the state array, allowing
203  *              negative indexes.
204  * d: Step or distance counter, indicating for what value of d the kd_forward
205  *    should be populated.
206  *    For d == 0, kd_forward[0] is initialized, i.e. the first invocation should
207  *    be for d == 0.
208  * meeting_snake: resulting meeting point, if any.
209  * Return true when a meeting point has been identified.
210  */
211 static int
212 diff_divide_myers_forward(bool *found_midpoint,
213 			  struct diff_data *left, struct diff_data *right,
214 			  int *kd_forward, int *kd_backward, int d,
215 			  struct diff_box *meeting_snake)
216 {
217 	int delta = (int)right->atoms.len - (int)left->atoms.len;
218 	int k;
219 	int x;
220 	int prev_x;
221 	int prev_y;
222 	int x_before_slide;
223 	*found_midpoint = false;
224 
225 	debug("-- %s d=%d\n", __func__, d);
226 
227 	for (k = d; k >= -d; k -= 2) {
228 		if (k < -(int)right->atoms.len || k > (int)left->atoms.len) {
229 			/* This diagonal is completely outside of the Myers
230 			 * graph, don't calculate it. */
231 			if (k < -(int)right->atoms.len)
232 				debug(" %d k < -(int)right->atoms.len %d\n", k,
233 				      -(int)right->atoms.len);
234 			else
235 				debug(" %d k > left->atoms.len %d\n", k,
236 				      left->atoms.len);
237 			if (k < 0) {
238 				/* We are traversing negatively, and already
239 				 * below the entire graph, nothing will come of
240 				 * this. */
241 				debug(" break");
242 				break;
243 			}
244 			debug(" continue");
245 			continue;
246 		}
247 		debug("- k = %d\n", k);
248 		if (d == 0) {
249 			/* This is the initializing step. There is no prev_k
250 			 * yet, get the initial x from the top left of the Myers
251 			 * graph. */
252 			x = 0;
253 		}
254 		/* Favoring "-" lines first means favoring moving rightwards in
255 		 * the Myers graph.
256 		 * For this, all k should derive from k - 1, only the bottom
257 		 * most k derive from k + 1:
258 		 *
259 		 *      | d=   0   1   2
260 		 *  ----+----------------
261 		 *  k=  |
262 		 *   2  |             2,0 <-- from prev_k = 2 - 1 = 1
263 		 *      |            /
264 		 *   1  |         1,0
265 		 *      |        /
266 		 *   0  |  -->0,0     3,3
267 		 *      |       \\   /
268 		 *  -1  |         0,1 <-- bottom most for d=1 from
269 		 *      |           \\    prev_k = -1 + 1 = 0
270 		 *  -2  |             0,2 <-- bottom most for d=2 from
271 		 *                            prev_k = -2 + 1 = -1
272 		 *
273 		 * Except when a k + 1 from a previous run already means a
274 		 * further advancement in the graph.
275 		 * If k == d, there is no k + 1 and k - 1 is the only option.
276 		 * If k < d, use k + 1 in case that yields a larger x. Also use
277 		 * k + 1 if k - 1 is outside the graph.
278 		 */
279 		else if (k > -d
280 			 && (k == d
281 			     || (k - 1 >= -(int)right->atoms.len
282 				 && kd_forward[k - 1] >= kd_forward[k + 1]))) {
283 			/* Advance from k - 1.
284 			 * From position prev_k, step to the right in the Myers
285 			 * graph: x += 1.
286 			 */
287 			int prev_k = k - 1;
288 			prev_x = kd_forward[prev_k];
289 			prev_y = xk_to_y(prev_x, prev_k);
290 			x = prev_x + 1;
291 		} else {
292 			/* The bottom most one.
293 			 * From position prev_k, step to the bottom in the Myers
294 			 * graph: y += 1.
295 			 * Incrementing y is achieved by decrementing k while
296 			 * keeping the same x.
297 			 * (since we're deriving y from y = x - k).
298 			 */
299 			int prev_k = k + 1;
300 			prev_x = kd_forward[prev_k];
301 			prev_y = xk_to_y(prev_x, prev_k);
302 			x = prev_x;
303 		}
304 
305 		x_before_slide = x;
306 		/* Slide down any snake that we might find here. */
307 		while (x < left->atoms.len && xk_to_y(x, k) < right->atoms.len) {
308 			bool same;
309 			int r = diff_atom_same(&same,
310 					       &left->atoms.head[x],
311 					       &right->atoms.head[
312 						xk_to_y(x, k)]);
313 			if (r)
314 				return r;
315 			if (!same)
316 				break;
317 			x++;
318 		}
319 		kd_forward[k] = x;
320 		if (x_before_slide != x) {
321 			debug("  down %d similar lines\n", x - x_before_slide);
322 		}
323 
324 #if DEBUG
325 		{
326 			int fi;
327 			for (fi = d; fi >= k; fi--) {
328 				debug("kd_forward[%d] = (%d, %d)\n", fi,
329 				      kd_forward[fi], kd_forward[fi] - fi);
330 			}
331 		}
332 #endif
333 
334 		if (x < 0 || x > left->atoms.len
335 		    || xk_to_y(x, k) < 0 || xk_to_y(x, k) > right->atoms.len)
336 			continue;
337 
338 		/* Figured out a new forwards traversal, see if this has gone
339 		 * onto or even past a preceding backwards traversal.
340 		 *
341 		 * If the delta in length is odd, then d and backwards_d hit the
342 		 * same state indexes:
343 		 *      | d=   0   1   2      1   0
344 		 *  ----+----------------    ----------------
345 		 *  k=  |                              c=
346 		 *   4  |                               3
347 		 *      |
348 		 *   3  |                               2
349 		 *      |                same
350 		 *   2  |             2,0====5,3        1
351 		 *      |            /          \
352 		 *   1  |         1,0            5,4<-- 0
353 		 *      |        /              /
354 		 *   0  |  -->0,0     3,3====4,4       -1
355 		 *      |        \   /
356 		 *  -1  |         0,1                  -2
357 		 *      |            \
358 		 *  -2  |             0,2              -3
359 		 *      |
360 		 *
361 		 * If the delta is even, they end up off-by-one, i.e. on
362 		 * different diagonals:
363 		 *
364 		 *      | d=   0   1   2    1   0
365 		 *  ----+----------------  ----------------
366 		 *      |                            c=
367 		 *   3  |                             3
368 		 *      |
369 		 *   2  |             2,0 off         2
370 		 *      |            /   \\
371 		 *   1  |         1,0      4,3        1
372 		 *      |        /       //   \
373 		 *   0  |  -->0,0     3,3      4,4<-- 0
374 		 *      |        \   /        /
375 		 *  -1  |         0,1      3,4       -1
376 		 *      |            \   //
377 		 *  -2  |             0,2            -2
378 		 *      |
379 		 *
380 		 * So in the forward path, we can only match up diagonals when
381 		 * the delta is odd.
382 		 */
383 		if ((delta & 1) == 0)
384 			continue;
385 		 /* Forwards is done first, so the backwards one was still at
386 		  * d - 1. Can't do this for d == 0. */
387 		int backwards_d = d - 1;
388 		if (backwards_d < 0)
389 			continue;
390 
391 		debug("backwards_d = %d\n", backwards_d);
392 
393 		/* If both sides have the same length, forward and backward
394 		 * start on the same diagonal, meaning the backwards state index
395 		 * c == k.
396 		 * As soon as the lengths are not the same, the backwards
397 		 * traversal starts on a different diagonal, and c = k shifted
398 		 * by the difference in length.
399 		 */
400 		int c = k_to_c(k, delta);
401 
402 		/* When the file sizes are very different, the traversal trees
403 		 * start on far distant diagonals.
404 		 * They don't necessarily meet straight on. See whether this
405 		 * forward value is on a diagonal that is also valid in
406 		 * kd_backward[], and match them if so. */
407 		if (c >= -backwards_d && c <= backwards_d) {
408 			/* Current k is on a diagonal that exists in
409 			 * kd_backward[]. If the two x positions have met or
410 			 * passed (forward walked onto or past backward), then
411 			 * we've found a midpoint / a mid-box.
412 			 *
413 			 * When forwards and backwards traversals meet, the
414 			 * endpoints of the mid-snake are not the two points in
415 			 * kd_forward and kd_backward, but rather the section
416 			 * that was slid (if any) of the current
417 			 * forward/backward traversal only.
418 			 *
419 			 * For example:
420 			 *
421 			 *   o
422 			 *    \
423 			 *     o
424 			 *      \
425 			 *       o
426 			 *        \
427 			 *         o
428 			 *          \
429 			 *       X o o
430 			 *       | | |
431 			 *     o-o-o o
432 			 *          \|
433 			 *           M
434 			 *            \
435 			 *             o
436 			 *              \
437 			 *               A o
438 			 *               | |
439 			 *             o-o-o
440 			 *
441 			 * The forward traversal reached M from the top and slid
442 			 * downwards to A.  The backward traversal already
443 			 * reached X, which is not a straight line from M
444 			 * anymore, so picking a mid-snake from M to X would
445 			 * yield a mistake.
446 			 *
447 			 * The correct mid-snake is between M and A. M is where
448 			 * the forward traversal hit the diagonal that the
449 			 * backward traversal has already passed, and A is what
450 			 * it reaches when sliding down identical lines.
451 			 */
452 			int backward_x = kd_backward[c];
453 			debug("Compare:  k=%d c=%d  is (%d,%d) >= (%d,%d)?\n",
454 			      k, c, x, xk_to_y(x, k), backward_x,
455 			      xc_to_y(backward_x, c, delta));
456 			if (x >= backward_x) {
457 				if (x_before_slide != x) {
458 					/* met after sliding up a mid-snake */
459 					*meeting_snake = (struct diff_box){
460 						.left_start = x_before_slide,
461 						.left_end = x,
462 						.right_start = xc_to_y(x_before_slide,
463 								       c, delta),
464 						.right_end = xk_to_y(x, k),
465 					};
466 				} else {
467 					/* met after a side step, non-identical
468 					 * line. Mark that as box divider
469 					 * instead. This makes sure that
470 					 * myers_divide never returns the same
471 					 * box that came as input, avoiding
472 					 * "infinite" looping. */
473 					*meeting_snake = (struct diff_box){
474 						.left_start = prev_x,
475 						.left_end = x,
476 						.right_start = prev_y,
477 						.right_end = xk_to_y(x, k),
478 					};
479 				}
480 				debug("HIT x=(%u,%u) - y=(%u,%u)\n",
481 				      meeting_snake->left_start,
482 				      meeting_snake->right_start,
483 				      meeting_snake->left_end,
484 				      meeting_snake->right_end);
485 				debug_dump_myers_graph(left, right, NULL,
486 						       kd_forward, d,
487 						       kd_backward, d-1);
488 				*found_midpoint = true;
489 				return 0;
490 			}
491 		}
492 	}
493 
494 	debug_dump_myers_graph(left, right, NULL, kd_forward, d,
495 			       kd_backward, d-1);
496 	return 0;
497 }
498 
499 /* Do one backwards step in the "divide and conquer" graph traversal.
500  * left: the left side to diff.
501  * right: the right side to diff against.
502  * kd_forward: the traversal state for forwards traversal, to find a meeting
503  *             point.
504  *             Since forwards is done first, after this, both kd_forward and
505  *             kd_backward will be valid for d.
506  *             kd_forward points at the center of the state array, allowing
507  *             negative indexes.
508  * kd_backward: the traversal state for backwards traversal, to find a meeting
509  *              point.
510  *              This is carried over between invocations with increasing d.
511  *              kd_backward points at the center of the state array, allowing
512  *              negative indexes.
513  * d: Step or distance counter, indicating for what value of d the kd_backward
514  *    should be populated.
515  *    Before the first invocation, kd_backward[0] shall point at the bottom
516  *    right of the Myers graph (left.len, right.len).
517  *    The first invocation will be for d == 1.
518  * meeting_snake: resulting meeting point, if any.
519  * Return true when a meeting point has been identified.
520  */
521 static int
522 diff_divide_myers_backward(bool *found_midpoint,
523 			   struct diff_data *left, struct diff_data *right,
524 			   int *kd_forward, int *kd_backward, int d,
525 			   struct diff_box *meeting_snake)
526 {
527 	int delta = (int)right->atoms.len - (int)left->atoms.len;
528 	int c;
529 	int x;
530 	int prev_x;
531 	int prev_y;
532 	int x_before_slide;
533 
534 	*found_midpoint = false;
535 
536 	debug("-- %s d=%d\n", __func__, d);
537 
538 	for (c = d; c >= -d; c -= 2) {
539 		if (c < -(int)left->atoms.len || c > (int)right->atoms.len) {
540 			/* This diagonal is completely outside of the Myers
541 			 * graph, don't calculate it. */
542 			if (c < -(int)left->atoms.len)
543 				debug(" %d c < -(int)left->atoms.len %d\n", c,
544 				      -(int)left->atoms.len);
545 			else
546 				debug(" %d c > right->atoms.len %d\n", c,
547 				      right->atoms.len);
548 			if (c < 0) {
549 				/* We are traversing negatively, and already
550 				 * below the entire graph, nothing will come of
551 				 * this. */
552 				debug(" break");
553 				break;
554 			}
555 			debug(" continue");
556 			continue;
557 		}
558 		debug("- c = %d\n", c);
559 		if (d == 0) {
560 			/* This is the initializing step. There is no prev_c
561 			 * yet, get the initial x from the bottom right of the
562 			 * Myers graph. */
563 			x = left->atoms.len;
564 		}
565 		/* Favoring "-" lines first means favoring moving rightwards in
566 		 * the Myers graph.
567 		 * For this, all c should derive from c - 1, only the bottom
568 		 * most c derive from c + 1:
569 		 *
570 		 *                                  2   1   0
571 		 *  ---------------------------------------------------
572 		 *                                               c=
573 		 *                                                3
574 		 *
575 		 *         from prev_c = c - 1 --> 5,2            2
576 		 *                                    \
577 		 *                                     5,3        1
578 		 *                                        \
579 		 *                                 4,3     5,4<-- 0
580 		 *                                    \   /
581 		 *  bottom most for d=1 from c + 1 --> 4,4       -1
582 		 *                                    /
583 		 *         bottom most for d=2 --> 3,4           -2
584 		 *
585 		 * Except when a c + 1 from a previous run already means a
586 		 * further advancement in the graph.
587 		 * If c == d, there is no c + 1 and c - 1 is the only option.
588 		 * If c < d, use c + 1 in case that yields a larger x.
589 		 * Also use c + 1 if c - 1 is outside the graph.
590 		 */
591 		else if (c > -d && (c == d
592 				    || (c - 1 >= -(int)right->atoms.len
593 					&& kd_backward[c - 1] <= kd_backward[c + 1]))) {
594 			/* A top one.
595 			 * From position prev_c, step upwards in the Myers
596 			 * graph: y -= 1.
597 			 * Decrementing y is achieved by incrementing c while
598 			 * keeping the same x. (since we're deriving y from
599 			 * y = x - c + delta).
600 			 */
601 			int prev_c = c - 1;
602 			prev_x = kd_backward[prev_c];
603 			prev_y = xc_to_y(prev_x, prev_c, delta);
604 			x = prev_x;
605 		} else {
606 			/* The bottom most one.
607 			 * From position prev_c, step to the left in the Myers
608 			 * graph: x -= 1.
609 			 */
610 			int prev_c = c + 1;
611 			prev_x = kd_backward[prev_c];
612 			prev_y = xc_to_y(prev_x, prev_c, delta);
613 			x = prev_x - 1;
614 		}
615 
616 		/* Slide up any snake that we might find here (sections of
617 		 * identical lines on both sides). */
618 		debug("c=%d x-1=%d Yb-1=%d-1=%d\n", c, x-1, xc_to_y(x, c,
619 								    delta),
620 		      xc_to_y(x, c, delta)-1);
621 		if (x > 0) {
622 			debug("  l=");
623 			debug_dump_atom(left, right, &left->atoms.head[x-1]);
624 		}
625 		if (xc_to_y(x, c, delta) > 0) {
626 			debug("  r=");
627 			debug_dump_atom(right, left,
628 				&right->atoms.head[xc_to_y(x, c, delta)-1]);
629 		}
630 		x_before_slide = x;
631 		while (x > 0 && xc_to_y(x, c, delta) > 0) {
632 			bool same;
633 			int r = diff_atom_same(&same,
634 					       &left->atoms.head[x-1],
635 					       &right->atoms.head[
636 						xc_to_y(x, c, delta)-1]);
637 			if (r)
638 				return r;
639 			if (!same)
640 				break;
641 			x--;
642 		}
643 		kd_backward[c] = x;
644 		if (x_before_slide != x) {
645 			debug("  up %d similar lines\n", x_before_slide - x);
646 		}
647 
648 		if (DEBUG) {
649 			int fi;
650 			for (fi = d; fi >= c; fi--) {
651 				debug("kd_backward[%d] = (%d, %d)\n",
652 				      fi,
653 				      kd_backward[fi],
654 				      kd_backward[fi] - fi + delta);
655 			}
656 		}
657 
658 		if (x < 0 || x > left->atoms.len
659 		    || xc_to_y(x, c, delta) < 0
660 		    || xc_to_y(x, c, delta) > right->atoms.len)
661 			continue;
662 
663 		/* Figured out a new backwards traversal, see if this has gone
664 		 * onto or even past a preceding forwards traversal.
665 		 *
666 		 * If the delta in length is even, then d and backwards_d hit
667 		 * the same state indexes -- note how this is different from in
668 		 * the forwards traversal, because now both d are the same:
669 		 *
670 		 *      | d=   0   1   2      2   1   0
671 		 *  ----+----------------    --------------------
672 		 *  k=  |                                  c=
673 		 *   4  |
674 		 *      |
675 		 *   3  |                                   3
676 		 *      |                same
677 		 *   2  |             2,0====5,2            2
678 		 *      |            /          \
679 		 *   1  |         1,0            5,3        1
680 		 *      |        /              /  \
681 		 *   0  |  -->0,0     3,3====4,3    5,4<--  0
682 		 *      |        \   /             /
683 		 *  -1  |         0,1            4,4       -1
684 		 *      |            \
685 		 *  -2  |             0,2                  -2
686 		 *      |
687 		 *                                      -3
688 		 * If the delta is odd, they end up off-by-one, i.e. on
689 		 * different diagonals.
690 		 * So in the backward path, we can only match up diagonals when
691 		 * the delta is even.
692 		 */
693 		if ((delta & 1) != 0)
694 			continue;
695 		/* Forwards was done first, now both d are the same. */
696 		int forwards_d = d;
697 
698 		/* As soon as the lengths are not the same, the
699 		 * backwards traversal starts on a different diagonal,
700 		 * and c = k shifted by the difference in length.
701 		 */
702 		int k = c_to_k(c, delta);
703 
704 		/* When the file sizes are very different, the traversal trees
705 		 * start on far distant diagonals.
706 		 * They don't necessarily meet straight on. See whether this
707 		 * backward value is also on a valid diagonal in kd_forward[],
708 		 * and match them if so. */
709 		if (k >= -forwards_d && k <= forwards_d) {
710 			/* Current c is on a diagonal that exists in
711 			 * kd_forward[]. If the two x positions have met or
712 			 * passed (backward walked onto or past forward), then
713 			 * we've found a midpoint / a mid-box.
714 			 *
715 			 * When forwards and backwards traversals meet, the
716 			 * endpoints of the mid-snake are not the two points in
717 			 * kd_forward and kd_backward, but rather the section
718 			 * that was slid (if any) of the current
719 			 * forward/backward traversal only.
720 			 *
721 			 * For example:
722 			 *
723 			 *   o-o-o
724 			 *   | |
725 			 *   o A
726 			 *   |  \
727 			 *   o   o
728 			 *        \
729 			 *         M
730 			 *         |\
731 			 *         o o-o-o
732 			 *         | | |
733 			 *         o o X
734 			 *          \
735 			 *           o
736 			 *            \
737 			 *             o
738 			 *              \
739 			 *               o
740 			 *
741 			 * The backward traversal reached M from the bottom and
742 			 * slid upwards.  The forward traversal already reached
743 			 * X, which is not a straight line from M anymore, so
744 			 * picking a mid-snake from M to X would yield a
745 			 * mistake.
746 			 *
747 			 * The correct mid-snake is between M and A. M is where
748 			 * the backward traversal hit the diagonal that the
749 			 * forwards traversal has already passed, and A is what
750 			 * it reaches when sliding up identical lines.
751 			 */
752 
753 			int forward_x = kd_forward[k];
754 			debug("Compare:  k=%d c=%d  is (%d,%d) >= (%d,%d)?\n",
755 			      k, c, forward_x, xk_to_y(forward_x, k),
756 			      x, xc_to_y(x, c, delta));
757 			if (forward_x >= x) {
758 				if (x_before_slide != x) {
759 					/* met after sliding down a mid-snake */
760 					*meeting_snake = (struct diff_box){
761 						.left_start = x,
762 						.left_end = x_before_slide,
763 						.right_start = xc_to_y(x, c, delta),
764 						.right_end = xk_to_y(x_before_slide, k),
765 					};
766 				} else {
767 					/* met after a side step, non-identical
768 					 * line. Mark that as box divider
769 					 * instead. This makes sure that
770 					 * myers_divide never returns the same
771 					 * box that came as input, avoiding
772 					 * "infinite" looping. */
773 					*meeting_snake = (struct diff_box){
774 						.left_start = x,
775 						.left_end = prev_x,
776 						.right_start = xc_to_y(x, c, delta),
777 						.right_end = prev_y,
778 					};
779 				}
780 				debug("HIT x=%u,%u - y=%u,%u\n",
781 				      meeting_snake->left_start,
782 				      meeting_snake->right_start,
783 				      meeting_snake->left_end,
784 				      meeting_snake->right_end);
785 				debug_dump_myers_graph(left, right, NULL,
786 						       kd_forward, d,
787 						       kd_backward, d);
788 				*found_midpoint = true;
789 				return 0;
790 			}
791 		}
792 	}
793 	debug_dump_myers_graph(left, right, NULL, kd_forward, d, kd_backward,
794 			       d);
795 	return 0;
796 }
797 
798 /* Integer square root approximation */
799 static int
800 shift_sqrt(int val)
801 {
802 	int i;
803         for (i = 1; val > 0; val >>= 2)
804 		i <<= 1;
805         return i;
806 }
807 
808 /* Myers "Divide et Impera": tracing forwards from the start and backwards from
809  * the end to find a midpoint that divides the problem into smaller chunks.
810  * Requires only linear amounts of memory. */
811 int
812 diff_algo_myers_divide(const struct diff_algo_config *algo_config,
813 		       struct diff_state *state)
814 {
815 	int rc = ENOMEM;
816 	struct diff_data *left = &state->left;
817 	struct diff_data *right = &state->right;
818 
819 	debug("\n** %s\n", __func__);
820 	debug("left:\n");
821 	debug_dump(left);
822 	debug("right:\n");
823 	debug_dump(right);
824 	debug_dump_myers_graph(left, right, NULL, NULL, 0, NULL, 0);
825 
826 	/* Allocate two columns of a Myers graph, one for the forward and one
827 	 * for the backward traversal. */
828 	unsigned int max = left->atoms.len + right->atoms.len;
829 	size_t kd_len = max + 1;
830 	size_t kd_buf_size = kd_len << 1;
831 	int *kd_buf = reallocarray(NULL, kd_buf_size, sizeof(int));
832 	if (!kd_buf)
833 		return ENOMEM;
834 	int i;
835 	for (i = 0; i < kd_buf_size; i++)
836 		kd_buf[i] = -1;
837 	int *kd_forward = kd_buf;
838 	int *kd_backward = kd_buf + kd_len;
839 	int max_effort = shift_sqrt(max/2);
840 
841 	/* The 'k' axis in Myers spans positive and negative indexes, so point
842 	 * the kd to the middle.
843 	 * It is then possible to index from -max/2 .. max/2. */
844 	kd_forward += max/2;
845 	kd_backward += max/2;
846 
847 	int d;
848 	struct diff_box mid_snake = {};
849 	bool found_midpoint = false;
850 	for (d = 0; d <= (max/2); d++) {
851 		int r;
852 		debug("-- d=%d\n", d);
853 		r = diff_divide_myers_forward(&found_midpoint, left, right,
854 					      kd_forward, kd_backward, d,
855 					      &mid_snake);
856 		if (r)
857 			return r;
858 		if (found_midpoint)
859 			break;
860 		r = diff_divide_myers_backward(&found_midpoint, left, right,
861 					       kd_forward, kd_backward, d,
862 					       &mid_snake);
863 		if (r)
864 			return r;
865 		if (found_midpoint)
866 			break;
867 
868 		/* Limit the effort spent looking for a mid snake. If files have
869 		 * very few lines in common, the effort spent to find nice mid
870 		 * snakes is just not worth it, the diff result will still be
871 		 * essentially minus everything on the left, plus everything on
872 		 * the right, with a few useless matches here and there. */
873 		if (d > max_effort) {
874 			/* pick the furthest reaching point from
875 			 * kd_forward and kd_backward, and use that as a
876 			 * midpoint, to not step into another diff algo
877 			 * recursion with unchanged box. */
878 			int delta = (int)right->atoms.len - (int)left->atoms.len;
879 			int x;
880 			int y;
881 			int i;
882 			int best_forward_i = 0;
883 			int best_forward_distance = 0;
884 			int best_backward_i = 0;
885 			int best_backward_distance = 0;
886 			int distance;
887 			int best_forward_x;
888 			int best_forward_y;
889 			int best_backward_x;
890 			int best_backward_y;
891 
892 			debug("~~~ d = %d > max_effort = %d\n", d, max_effort);
893 
894 			for (i = d; i >= -d; i -= 2) {
895 				if (i >= -(int)right->atoms.len && i <= (int)left->atoms.len) {
896 					x = kd_forward[i];
897 					y = xk_to_y(x, i);
898 					distance = x + y;
899 					if (distance > best_forward_distance) {
900 						best_forward_distance = distance;
901 						best_forward_i = i;
902 					}
903 				}
904 
905 				if (i >= -(int)left->atoms.len && i <= (int)right->atoms.len) {
906 					x = kd_backward[i];
907 					y = xc_to_y(x, i, delta);
908 					distance = (right->atoms.len - x)
909 						+ (left->atoms.len - y);
910 					if (distance >= best_backward_distance) {
911 						best_backward_distance = distance;
912 						best_backward_i = i;
913 					}
914 				}
915 			}
916 
917 			/* The myers-divide didn't meet in the middle. We just
918 			 * figured out the places where the forward path
919 			 * advanced the most, and the backward path advanced the
920 			 * most. Just divide at whichever one of those two is better.
921 			 *
922 			 *   o-o
923 			 *     |
924 			 *     o
925 			 *      \
926 			 *       o
927 			 *        \
928 			 *         F <-- cut here
929 			 *
930 			 *
931 			 *
932 			 *           or here --> B
933 			 *                        \
934 			 *                         o
935 			 *                          \
936 			 *                           o
937 			 *                           |
938 			 *                           o-o
939 			 */
940 			best_forward_x = kd_forward[best_forward_i];
941 			best_forward_y = xk_to_y(best_forward_x, best_forward_i);
942 			best_backward_x = kd_backward[best_backward_i];
943 			best_backward_y = xc_to_y(x, best_backward_i, delta);
944 
945 			if (best_forward_distance >= best_backward_distance) {
946 				x = best_forward_x;
947 				y = best_forward_y;
948 			} else {
949 				x = best_backward_x;
950 				y = best_backward_y;
951 			}
952 
953 			debug("max_effort cut at x=%d y=%d\n", x, y);
954 			if (x < 0 || y < 0
955 			    || x > left->atoms.len || y > right->atoms.len)
956 				break;
957 
958 			found_midpoint = true;
959 			mid_snake = (struct diff_box){
960 				.left_start = x,
961 				.left_end = x,
962 				.right_start = y,
963 				.right_end = y,
964 			};
965 			break;
966 		}
967 	}
968 
969 	if (!found_midpoint) {
970 		/* Divide and conquer failed to find a meeting point. Use the
971 		 * fallback_algo defined in the algo_config (leave this to the
972 		 * caller). This is just paranoia/sanity, we normally should
973 		 * always find a midpoint.
974 		 */
975 		debug(" no midpoint \n");
976 		rc = DIFF_RC_USE_DIFF_ALGO_FALLBACK;
977 		goto return_rc;
978 	} else {
979 		debug(" mid snake L: %u to %u of %u   R: %u to %u of %u\n",
980 		      mid_snake.left_start, mid_snake.left_end, left->atoms.len,
981 		      mid_snake.right_start, mid_snake.right_end,
982 		      right->atoms.len);
983 
984 		/* Section before the mid-snake.  */
985 		debug("Section before the mid-snake\n");
986 
987 		struct diff_atom *left_atom = &left->atoms.head[0];
988 		unsigned int left_section_len = mid_snake.left_start;
989 		struct diff_atom *right_atom = &right->atoms.head[0];
990 		unsigned int right_section_len = mid_snake.right_start;
991 
992 		if (left_section_len && right_section_len) {
993 			/* Record an unsolved chunk, the caller will apply
994 			 * inner_algo() on this chunk. */
995 			if (!diff_state_add_chunk(state, false,
996 						  left_atom, left_section_len,
997 						  right_atom,
998 						  right_section_len))
999 				goto return_rc;
1000 		} else if (left_section_len && !right_section_len) {
1001 			/* Only left atoms and none on the right, they form a
1002 			 * "minus" chunk, then. */
1003 			if (!diff_state_add_chunk(state, true,
1004 						  left_atom, left_section_len,
1005 						  right_atom, 0))
1006 				goto return_rc;
1007 		} else if (!left_section_len && right_section_len) {
1008 			/* No left atoms, only atoms on the right, they form a
1009 			 * "plus" chunk, then. */
1010 			if (!diff_state_add_chunk(state, true,
1011 						  left_atom, 0,
1012 						  right_atom,
1013 						  right_section_len))
1014 				goto return_rc;
1015 		}
1016 		/* else: left_section_len == 0 and right_section_len == 0, i.e.
1017 		 * nothing before the mid-snake. */
1018 
1019 		if (mid_snake.left_end > mid_snake.left_start
1020 		    || mid_snake.right_end > mid_snake.right_start) {
1021 			/* The midpoint is a section of identical data on both
1022 			 * sides, or a certain differing line: that section
1023 			 * immediately becomes a solved chunk. */
1024 			debug("the mid-snake\n");
1025 			if (!diff_state_add_chunk(state, true,
1026 				  &left->atoms.head[mid_snake.left_start],
1027 				  mid_snake.left_end - mid_snake.left_start,
1028 				  &right->atoms.head[mid_snake.right_start],
1029 				  mid_snake.right_end - mid_snake.right_start))
1030 				goto return_rc;
1031 		}
1032 
1033 		/* Section after the mid-snake. */
1034 		debug("Section after the mid-snake\n");
1035 		debug("  left_end %u  right_end %u\n",
1036 		      mid_snake.left_end, mid_snake.right_end);
1037 		debug("  left_count %u  right_count %u\n",
1038 		      left->atoms.len, right->atoms.len);
1039 		left_atom = &left->atoms.head[mid_snake.left_end];
1040 		left_section_len = left->atoms.len - mid_snake.left_end;
1041 		right_atom = &right->atoms.head[mid_snake.right_end];
1042 		right_section_len = right->atoms.len - mid_snake.right_end;
1043 
1044 		if (left_section_len && right_section_len) {
1045 			/* Record an unsolved chunk, the caller will apply
1046 			 * inner_algo() on this chunk. */
1047 			if (!diff_state_add_chunk(state, false,
1048 						  left_atom, left_section_len,
1049 						  right_atom,
1050 						  right_section_len))
1051 				goto return_rc;
1052 		} else if (left_section_len && !right_section_len) {
1053 			/* Only left atoms and none on the right, they form a
1054 			 * "minus" chunk, then. */
1055 			if (!diff_state_add_chunk(state, true,
1056 						  left_atom, left_section_len,
1057 						  right_atom, 0))
1058 				goto return_rc;
1059 		} else if (!left_section_len && right_section_len) {
1060 			/* No left atoms, only atoms on the right, they form a
1061 			 * "plus" chunk, then. */
1062 			if (!diff_state_add_chunk(state, true,
1063 						  left_atom, 0,
1064 						  right_atom,
1065 						  right_section_len))
1066 				goto return_rc;
1067 		}
1068 		/* else: left_section_len == 0 and right_section_len == 0, i.e.
1069 		 * nothing after the mid-snake. */
1070 	}
1071 
1072 	rc = DIFF_RC_OK;
1073 
1074 return_rc:
1075 	free(kd_buf);
1076 	debug("** END %s\n", __func__);
1077 	return rc;
1078 }
1079 
1080 /* Myers Diff tracing from the start all the way through to the end, requiring
1081  * quadratic amounts of memory. This can fail if the required space surpasses
1082  * algo_config->permitted_state_size. */
1083 int
1084 diff_algo_myers(const struct diff_algo_config *algo_config,
1085 		struct diff_state *state)
1086 {
1087 	/* do a diff_divide_myers_forward() without a _backward(), so that it
1088 	 * walks forward across the entire files to reach the end. Keep each
1089 	 * run's state, and do a final backtrace. */
1090 	int rc = ENOMEM;
1091 	struct diff_data *left = &state->left;
1092 	struct diff_data *right = &state->right;
1093 
1094 	debug("\n** %s\n", __func__);
1095 	debug("left:\n");
1096 	debug_dump(left);
1097 	debug("right:\n");
1098 	debug_dump(right);
1099 	debug_dump_myers_graph(left, right, NULL, NULL, 0, NULL, 0);
1100 
1101 	/* Allocate two columns of a Myers graph, one for the forward and one
1102 	 * for the backward traversal. */
1103 	unsigned int max = left->atoms.len + right->atoms.len;
1104 	size_t kd_len = max + 1 + max;
1105 	size_t kd_buf_size = kd_len * kd_len;
1106 	size_t kd_state_size = kd_buf_size * sizeof(int);
1107 	debug("state size: %zu\n", kd_state_size);
1108 	if (kd_buf_size < kd_len /* overflow? */
1109 	    || kd_state_size > algo_config->permitted_state_size) {
1110 		debug("state size %zu > permitted_state_size %zu, use fallback_algo\n",
1111 		      kd_state_size, algo_config->permitted_state_size);
1112 		return DIFF_RC_USE_DIFF_ALGO_FALLBACK;
1113 	}
1114 
1115 	int *kd_buf = reallocarray(NULL, kd_buf_size, sizeof(int));
1116 	if (!kd_buf)
1117 		return ENOMEM;
1118 	int i;
1119 	for (i = 0; i < kd_buf_size; i++)
1120 		kd_buf[i] = -1;
1121 
1122 	/* The 'k' axis in Myers spans positive and negative indexes, so point
1123 	 * the kd to the middle.
1124 	 * It is then possible to index from -max .. max. */
1125 	int *kd_origin = kd_buf + max;
1126 	int *kd_column = kd_origin;
1127 
1128 	int d;
1129 	int backtrack_d = -1;
1130 	int backtrack_k = 0;
1131 	int k;
1132 	int x, y;
1133 	for (d = 0; d <= max; d++, kd_column += kd_len) {
1134 		debug("-- d=%d\n", d);
1135 
1136 		debug("-- %s d=%d\n", __func__, d);
1137 
1138 		for (k = d; k >= -d; k -= 2) {
1139 			if (k < -(int)right->atoms.len
1140 			    || k > (int)left->atoms.len) {
1141 				/* This diagonal is completely outside of the
1142 				 * Myers graph, don't calculate it. */
1143 				if (k < -(int)right->atoms.len)
1144 					debug(" %d k <"
1145 					      " -(int)right->atoms.len %d\n",
1146 					      k, -(int)right->atoms.len);
1147 				else
1148 					debug(" %d k > left->atoms.len %d\n", k,
1149 					      left->atoms.len);
1150 				if (k < 0) {
1151 					/* We are traversing negatively, and
1152 					 * already below the entire graph,
1153 					 * nothing will come of this. */
1154 					debug(" break");
1155 					break;
1156 				}
1157 				debug(" continue");
1158 				continue;
1159 			}
1160 
1161 			debug("- k = %d\n", k);
1162 			if (d == 0) {
1163 				/* This is the initializing step. There is no
1164 				 * prev_k yet, get the initial x from the top
1165 				 * left of the Myers graph. */
1166 				x = 0;
1167 			} else {
1168 				int *kd_prev_column = kd_column - kd_len;
1169 
1170 				/* Favoring "-" lines first means favoring
1171 				 * moving rightwards in the Myers graph.
1172 				 * For this, all k should derive from k - 1,
1173 				 * only the bottom most k derive from k + 1:
1174 				 *
1175 				 *      | d=   0   1   2
1176 				 *  ----+----------------
1177 				 *  k=  |
1178 				 *   2  |             2,0 <-- from
1179 				 *      |            /        prev_k = 2 - 1 = 1
1180 				 *   1  |         1,0
1181 				 *      |        /
1182 				 *   0  |  -->0,0     3,3
1183 				 *      |       \\   /
1184 				 *  -1  |         0,1 <-- bottom most for d=1
1185 				 *      |           \\    from prev_k = -1+1 = 0
1186 				 *  -2  |             0,2 <-- bottom most for
1187 				 *                            d=2 from
1188 				 *                            prev_k = -2+1 = -1
1189 				 *
1190 				 * Except when a k + 1 from a previous run
1191 				 * already means a further advancement in the
1192 				 * graph.
1193 				 * If k == d, there is no k + 1 and k - 1 is the
1194 				 * only option.
1195 				 * If k < d, use k + 1 in case that yields a
1196 				 * larger x. Also use k + 1 if k - 1 is outside
1197 				 * the graph.
1198 				 */
1199 				if (k > -d
1200 				    && (k == d
1201 					|| (k - 1 >= -(int)right->atoms.len
1202 					    && kd_prev_column[k - 1]
1203 					       >= kd_prev_column[k + 1]))) {
1204 					/* Advance from k - 1.
1205 					 * From position prev_k, step to the
1206 					 * right in the Myers graph: x += 1.
1207 					 */
1208 					int prev_k = k - 1;
1209 					int prev_x = kd_prev_column[prev_k];
1210 					x = prev_x + 1;
1211 				} else {
1212 					/* The bottom most one.
1213 					 * From position prev_k, step to the
1214 					 * bottom in the Myers graph: y += 1.
1215 					 * Incrementing y is achieved by
1216 					 * decrementing k while keeping the same
1217 					 * x. (since we're deriving y from y =
1218 					 * x - k).
1219 					 */
1220 					int prev_k = k + 1;
1221 					int prev_x = kd_prev_column[prev_k];
1222 					x = prev_x;
1223 				}
1224 			}
1225 
1226 			/* Slide down any snake that we might find here. */
1227 			while (x < left->atoms.len
1228 			       && xk_to_y(x, k) < right->atoms.len) {
1229 				bool same;
1230 				int r = diff_atom_same(&same,
1231 						       &left->atoms.head[x],
1232 						       &right->atoms.head[
1233 							xk_to_y(x, k)]);
1234 				if (r)
1235 					return r;
1236 				if (!same)
1237 					break;
1238 				x++;
1239 			}
1240 			kd_column[k] = x;
1241 
1242 			if (DEBUG) {
1243 				int fi;
1244 				for (fi = d; fi >= k; fi-=2) {
1245 					debug("kd_column[%d] = (%d, %d)\n", fi,
1246 					      kd_column[fi],
1247 					      kd_column[fi] - fi);
1248 				}
1249 			}
1250 
1251 			if (x == left->atoms.len
1252 			    && xk_to_y(x, k) == right->atoms.len) {
1253 				/* Found a path */
1254 				backtrack_d = d;
1255 				backtrack_k = k;
1256 				debug("Reached the end at d = %d, k = %d\n",
1257 				      backtrack_d, backtrack_k);
1258 				break;
1259 			}
1260 		}
1261 
1262 		if (backtrack_d >= 0)
1263 			break;
1264 	}
1265 
1266 	debug_dump_myers_graph(left, right, kd_origin, NULL, 0, NULL, 0);
1267 
1268 	/* backtrack. A matrix spanning from start to end of the file is ready:
1269 	 *
1270 	 *      | d=   0   1   2   3   4
1271 	 *  ----+---------------------------------
1272 	 *  k=  |
1273 	 *   3  |
1274 	 *      |
1275 	 *   2  |             2,0
1276 	 *      |            /
1277 	 *   1  |         1,0     4,3
1278 	 *      |        /       /   \
1279 	 *   0  |  -->0,0     3,3     4,4 --> backtrack_d = 4, backtrack_k = 0
1280 	 *      |        \   /   \
1281 	 *  -1  |         0,1     3,4
1282 	 *      |            \
1283 	 *  -2  |             0,2
1284 	 *      |
1285 	 *
1286 	 * From (4,4) backwards, find the previous position that is the largest, and remember it.
1287 	 *
1288 	 */
1289 	for (d = backtrack_d, k = backtrack_k; d >= 0; d--) {
1290 		x = kd_column[k];
1291 		y = xk_to_y(x, k);
1292 
1293 		/* When the best position is identified, remember it for that
1294 		 * kd_column.
1295 		 * That kd_column is no longer needed otherwise, so just
1296 		 * re-purpose kd_column[0] = x and kd_column[1] = y,
1297 		 * so that there is no need to allocate more memory.
1298 		 */
1299 		kd_column[0] = x;
1300 		kd_column[1] = y;
1301 		debug("Backtrack d=%d: xy=(%d, %d)\n",
1302 		      d, kd_column[0], kd_column[1]);
1303 
1304 		/* Don't access memory before kd_buf */
1305 		if (d == 0)
1306 			break;
1307 		int *kd_prev_column = kd_column - kd_len;
1308 
1309 		/* When y == 0, backtracking downwards (k-1) is the only way.
1310 		 * When x == 0, backtracking upwards (k+1) is the only way.
1311 		 *
1312 		 *      | d=   0   1   2   3   4
1313 		 *  ----+---------------------------------
1314 		 *  k=  |
1315 		 *   3  |
1316 		 *      |                ..y == 0
1317 		 *   2  |             2,0
1318 		 *      |            /
1319 		 *   1  |         1,0     4,3
1320 		 *      |        /       /   \
1321 		 *   0  |  -->0,0     3,3     4,4 --> backtrack_d = 4,
1322 		 *      |        \   /   \            backtrack_k = 0
1323 		 *  -1  |         0,1     3,4
1324 		 *      |            \
1325 		 *  -2  |             0,2__
1326 		 *      |                  x == 0
1327 		 */
1328 		if (y == 0
1329 		    || (x > 0
1330 			&& kd_prev_column[k - 1] >= kd_prev_column[k + 1])) {
1331 			k = k - 1;
1332 			debug("prev k=k-1=%d x=%d y=%d\n",
1333 			      k, kd_prev_column[k],
1334 			      xk_to_y(kd_prev_column[k], k));
1335 		} else {
1336 			k = k + 1;
1337 			debug("prev k=k+1=%d x=%d y=%d\n",
1338 			      k, kd_prev_column[k],
1339 			      xk_to_y(kd_prev_column[k], k));
1340 		}
1341 		kd_column = kd_prev_column;
1342 	}
1343 
1344 	/* Forwards again, this time recording the diff chunks.
1345 	 * Definitely start from 0,0. kd_column[0] may actually point to the
1346 	 * bottom of a snake starting at 0,0 */
1347 	x = 0;
1348 	y = 0;
1349 
1350 	kd_column = kd_origin;
1351 	for (d = 0; d <= backtrack_d; d++, kd_column += kd_len) {
1352 		int next_x = kd_column[0];
1353 		int next_y = kd_column[1];
1354 		debug("Forward track from xy(%d,%d) to xy(%d,%d)\n",
1355 		      x, y, next_x, next_y);
1356 
1357 		struct diff_atom *left_atom = &left->atoms.head[x];
1358 		int left_section_len = next_x - x;
1359 		struct diff_atom *right_atom = &right->atoms.head[y];
1360 		int right_section_len = next_y - y;
1361 
1362 		rc = ENOMEM;
1363 		if (left_section_len && right_section_len) {
1364 			/* This must be a snake slide.
1365 			 * Snake slides have a straight line leading into them
1366 			 * (except when starting at (0,0)). Find out whether the
1367 			 * lead-in is horizontal or vertical:
1368 			 *
1369 			 *     left
1370 			 *  ---------->
1371 			 *  |
1372 			 * r|   o-o        o
1373 			 * i|      \       |
1374 			 * g|       o      o
1375 			 * h|        \      \
1376 			 * t|         o      o
1377 			 *  v
1378 			 *
1379 			 * If left_section_len > right_section_len, the lead-in
1380 			 * is horizontal, meaning first remove one atom from the
1381 			 * left before sliding down the snake.
1382 			 * If right_section_len > left_section_len, the lead-in
1383 			 * is vetical, so add one atom from the right before
1384 			 * sliding down the snake. */
1385 			if (left_section_len == right_section_len + 1) {
1386 				if (!diff_state_add_chunk(state, true,
1387 							  left_atom, 1,
1388 							  right_atom, 0))
1389 					goto return_rc;
1390 				left_atom++;
1391 				left_section_len--;
1392 			} else if (right_section_len == left_section_len + 1) {
1393 				if (!diff_state_add_chunk(state, true,
1394 							  left_atom, 0,
1395 							  right_atom, 1))
1396 					goto return_rc;
1397 				right_atom++;
1398 				right_section_len--;
1399 			} else if (left_section_len != right_section_len) {
1400 				/* The numbers are making no sense. Should never
1401 				 * happen. */
1402 				rc = DIFF_RC_USE_DIFF_ALGO_FALLBACK;
1403 				goto return_rc;
1404 			}
1405 
1406 			if (!diff_state_add_chunk(state, true,
1407 						  left_atom, left_section_len,
1408 						  right_atom,
1409 						  right_section_len))
1410 				goto return_rc;
1411 		} else if (left_section_len && !right_section_len) {
1412 			/* Only left atoms and none on the right, they form a
1413 			 * "minus" chunk, then. */
1414 			if (!diff_state_add_chunk(state, true,
1415 						  left_atom, left_section_len,
1416 						  right_atom, 0))
1417 				goto return_rc;
1418 		} else if (!left_section_len && right_section_len) {
1419 			/* No left atoms, only atoms on the right, they form a
1420 			 * "plus" chunk, then. */
1421 			if (!diff_state_add_chunk(state, true,
1422 						  left_atom, 0,
1423 						  right_atom,
1424 						  right_section_len))
1425 				goto return_rc;
1426 		}
1427 
1428 		x = next_x;
1429 		y = next_y;
1430 	}
1431 
1432 	rc = DIFF_RC_OK;
1433 
1434 return_rc:
1435 	free(kd_buf);
1436 	debug("** END %s rc=%d\n", __func__, rc);
1437 	return rc;
1438 }