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 	*found_midpoint = false;
221 
222 	debug("-- %s d=%d\n", __func__, d);
223 
224 	for (k = d; k >= -d; k -= 2) {
225 		if (k < -(int)right->atoms.len || k > (int)left->atoms.len) {
226 			/* This diagonal is completely outside of the Myers
227 			 * graph, don't calculate it. */
228 			if (k < -(int)right->atoms.len)
229 				debug(" %d k < -(int)right->atoms.len %d\n", k,
230 				      -(int)right->atoms.len);
231 			else
232 				debug(" %d k > left->atoms.len %d\n", k,
233 				      left->atoms.len);
234 			if (k < 0) {
235 				/* We are traversing negatively, and already
236 				 * below the entire graph, nothing will come of
237 				 * this. */
238 				debug(" break");
239 				break;
240 			}
241 			debug(" continue");
242 			continue;
243 		}
244 		debug("- k = %d\n", k);
245 		if (d == 0) {
246 			/* This is the initializing step. There is no prev_k
247 			 * yet, get the initial x from the top left of the Myers
248 			 * graph. */
249 			x = 0;
250 		}
251 		/* Favoring "-" lines first means favoring moving rightwards in
252 		 * the Myers graph.
253 		 * For this, all k should derive from k - 1, only the bottom
254 		 * most k derive from k + 1:
255 		 *
256 		 *      | d=   0   1   2
257 		 *  ----+----------------
258 		 *  k=  |
259 		 *   2  |             2,0 <-- from prev_k = 2 - 1 = 1
260 		 *      |            /
261 		 *   1  |         1,0
262 		 *      |        /
263 		 *   0  |  -->0,0     3,3
264 		 *      |       \\   /
265 		 *  -1  |         0,1 <-- bottom most for d=1 from
266 		 *      |           \\    prev_k = -1 + 1 = 0
267 		 *  -2  |             0,2 <-- bottom most for d=2 from
268 		 *                            prev_k = -2 + 1 = -1
269 		 *
270 		 * Except when a k + 1 from a previous run already means a
271 		 * further advancement in the graph.
272 		 * If k == d, there is no k + 1 and k - 1 is the only option.
273 		 * If k < d, use k + 1 in case that yields a larger x. Also use
274 		 * k + 1 if k - 1 is outside the graph.
275 		 */
276 		else if (k > -d
277 			 && (k == d
278 			     || (k - 1 >= -(int)right->atoms.len
279 				 && kd_forward[k - 1] >= kd_forward[k + 1]))) {
280 			/* Advance from k - 1.
281 			 * From position prev_k, step to the right in the Myers
282 			 * graph: x += 1.
283 			 */
284 			int prev_k = k - 1;
285 			int prev_x = kd_forward[prev_k];
286 			x = prev_x + 1;
287 		} else {
288 			/* The bottom most one.
289 			 * From position prev_k, step to the bottom in the Myers
290 			 * graph: y += 1.
291 			 * Incrementing y is achieved by decrementing k while
292 			 * keeping the same x.
293 			 * (since we're deriving y from y = x - k).
294 			 */
295 			int prev_k = k + 1;
296 			int prev_x = kd_forward[prev_k];
297 			x = prev_x;
298 		}
299 
300 		int x_before_slide = x;
301 		/* Slide down any snake that we might find here. */
302 		while (x < left->atoms.len && xk_to_y(x, k) < right->atoms.len) {
303 			bool same;
304 			int r = diff_atom_same(&same,
305 					       &left->atoms.head[x],
306 					       &right->atoms.head[
307 						xk_to_y(x, k)]);
308 			if (r)
309 				return r;
310 			if (!same)
311 				break;
312 			x++;
313 		}
314 		kd_forward[k] = x;
315 		if (x_before_slide != x) {
316 			debug("  down %d similar lines\n", x - x_before_slide);
317 		}
318 
319 		if (DEBUG) {
320 			int fi;
321 			for (fi = d; fi >= k; fi--) {
322 				debug("kd_forward[%d] = (%d, %d)\n", fi,
323 				      kd_forward[fi], kd_forward[fi] - fi);
324 			}
325 		}
326 
327 		if (x < 0 || x > left->atoms.len
328 		    || xk_to_y(x, k) < 0 || xk_to_y(x, k) > right->atoms.len)
329 			continue;
330 
331 		/* Figured out a new forwards traversal, see if this has gone
332 		 * onto or even past a preceding backwards traversal.
333 		 *
334 		 * If the delta in length is odd, then d and backwards_d hit the
335 		 * same state indexes:
336 		 *      | d=   0   1   2      1   0
337 		 *  ----+----------------    ----------------
338 		 *  k=  |                              c=
339 		 *   4  |                               3
340 		 *      |
341 		 *   3  |                               2
342 		 *      |                same
343 		 *   2  |             2,0====5,3        1
344 		 *      |            /          \
345 		 *   1  |         1,0            5,4<-- 0
346 		 *      |        /              /
347 		 *   0  |  -->0,0     3,3====4,4       -1
348 		 *      |        \   /
349 		 *  -1  |         0,1                  -2
350 		 *      |            \
351 		 *  -2  |             0,2              -3
352 		 *      |
353 		 *
354 		 * If the delta is even, they end up off-by-one, i.e. on
355 		 * different diagonals:
356 		 *
357 		 *      | d=   0   1   2    1   0
358 		 *  ----+----------------  ----------------
359 		 *      |                            c=
360 		 *   3  |                             3
361 		 *      |
362 		 *   2  |             2,0 off         2
363 		 *      |            /   \\
364 		 *   1  |         1,0      4,3        1
365 		 *      |        /       //   \
366 		 *   0  |  -->0,0     3,3      4,4<-- 0
367 		 *      |        \   /        /
368 		 *  -1  |         0,1      3,4       -1
369 		 *      |            \   //
370 		 *  -2  |             0,2            -2
371 		 *      |
372 		 *
373 		 * So in the forward path, we can only match up diagonals when
374 		 * the delta is odd.
375 		 */
376 		if ((delta & 1) == 0)
377 			continue;
378 		 /* Forwards is done first, so the backwards one was still at
379 		  * d - 1. Can't do this for d == 0. */
380 		int backwards_d = d - 1;
381 		if (backwards_d < 0)
382 			continue;
383 
384 		debug("backwards_d = %d\n", backwards_d);
385 
386 		/* If both sides have the same length, forward and backward
387 		 * start on the same diagonal, meaning the backwards state index
388 		 * c == k.
389 		 * As soon as the lengths are not the same, the backwards
390 		 * traversal starts on a different diagonal, and c = k shifted
391 		 * by the difference in length.
392 		 */
393 		int c = k_to_c(k, delta);
394 
395 		/* When the file sizes are very different, the traversal trees
396 		 * start on far distant diagonals.
397 		 * They don't necessarily meet straight on. See whether this
398 		 * forward value is on a diagonal that is also valid in
399 		 * kd_backward[], and match them if so. */
400 		if (c >= -backwards_d && c <= backwards_d) {
401 			/* Current k is on a diagonal that exists in
402 			 * kd_backward[]. If the two x positions have met or
403 			 * passed (forward walked onto or past backward), then
404 			 * we've found a midpoint / a mid-box.
405 			 *
406 			 * When forwards and backwards traversals meet, the
407 			 * endpoints of the mid-snake are not the two points in
408 			 * kd_forward and kd_backward, but rather the section
409 			 * that was slid (if any) of the current
410 			 * forward/backward traversal only.
411 			 *
412 			 * For example:
413 			 *
414 			 *   o
415 			 *    \
416 			 *     o
417 			 *      \
418 			 *       o
419 			 *        \
420 			 *         o
421 			 *          \
422 			 *       X o o
423 			 *       | | |
424 			 *     o-o-o o
425 			 *          \|
426 			 *           M
427 			 *            \
428 			 *             o
429 			 *              \
430 			 *               A o
431 			 *               | |
432 			 *             o-o-o
433 			 *
434 			 * The forward traversal reached M from the top and slid
435 			 * downwards to A.  The backward traversal already
436 			 * reached X, which is not a straight line from M
437 			 * anymore, so picking a mid-snake from M to X would
438 			 * yield a mistake.
439 			 *
440 			 * The correct mid-snake is between M and A. M is where
441 			 * the forward traversal hit the diagonal that the
442 			 * backward traversal has already passed, and A is what
443 			 * it reaches when sliding down identical lines.
444 			 */
445 			int backward_x = kd_backward[c];
446 			debug("Compare:  k=%d c=%d  is (%d,%d) >= (%d,%d)?\n",
447 			      k, c, x, xk_to_y(x, k), backward_x,
448 			      xc_to_y(backward_x, c, delta));
449 			if (x >= backward_x) {
450 				*meeting_snake = (struct diff_box){
451 					.left_start = x_before_slide,
452 					.left_end = x,
453 					.right_start = xc_to_y(x_before_slide,
454 							       c, delta),
455 					.right_end = xk_to_y(x, k),
456 				};
457 				debug("HIT x=(%u,%u) - y=(%u,%u)\n",
458 				      meeting_snake->left_start,
459 				      meeting_snake->right_start,
460 				      meeting_snake->left_end,
461 				      meeting_snake->right_end);
462 				debug_dump_myers_graph(left, right, NULL,
463 						       kd_forward, d,
464 						       kd_backward, d-1);
465 				*found_midpoint = true;
466 				return 0;
467 			}
468 		}
469 	}
470 
471 	debug_dump_myers_graph(left, right, NULL, kd_forward, d,
472 			       kd_backward, d-1);
473 	return 0;
474 }
475 
476 /* Do one backwards step in the "divide and conquer" graph traversal.
477  * left: the left side to diff.
478  * right: the right side to diff against.
479  * kd_forward: the traversal state for forwards traversal, to find a meeting
480  *             point.
481  *             Since forwards is done first, after this, both kd_forward and
482  *             kd_backward will be valid for d.
483  *             kd_forward points at the center of the state array, allowing
484  *             negative indexes.
485  * kd_backward: the traversal state for backwards traversal, to find a meeting
486  *              point.
487  *              This is carried over between invocations with increasing d.
488  *              kd_backward points at the center of the state array, allowing
489  *              negative indexes.
490  * d: Step or distance counter, indicating for what value of d the kd_backward
491  *    should be populated.
492  *    Before the first invocation, kd_backward[0] shall point at the bottom
493  *    right of the Myers graph (left.len, right.len).
494  *    The first invocation will be for d == 1.
495  * meeting_snake: resulting meeting point, if any.
496  * Return true when a meeting point has been identified.
497  */
498 static int
499 diff_divide_myers_backward(bool *found_midpoint,
500 			   struct diff_data *left, struct diff_data *right,
501 			   int *kd_forward, int *kd_backward, int d,
502 			   struct diff_box *meeting_snake)
503 {
504 	int delta = (int)right->atoms.len - (int)left->atoms.len;
505 	int c;
506 	int x;
507 
508 	*found_midpoint = false;
509 
510 	debug("-- %s d=%d\n", __func__, d);
511 
512 	for (c = d; c >= -d; c -= 2) {
513 		if (c < -(int)left->atoms.len || c > (int)right->atoms.len) {
514 			/* This diagonal is completely outside of the Myers
515 			 * graph, don't calculate it. */
516 			if (c < -(int)left->atoms.len)
517 				debug(" %d c < -(int)left->atoms.len %d\n", c,
518 				      -(int)left->atoms.len);
519 			else
520 				debug(" %d c > right->atoms.len %d\n", c,
521 				      right->atoms.len);
522 			if (c < 0) {
523 				/* We are traversing negatively, and already
524 				 * below the entire graph, nothing will come of
525 				 * this. */
526 				debug(" break");
527 				break;
528 			}
529 			debug(" continue");
530 			continue;
531 		}
532 		debug("- c = %d\n", c);
533 		if (d == 0) {
534 			/* This is the initializing step. There is no prev_c
535 			 * yet, get the initial x from the bottom right of the
536 			 * Myers graph. */
537 			x = left->atoms.len;
538 		}
539 		/* Favoring "-" lines first means favoring moving rightwards in
540 		 * the Myers graph.
541 		 * For this, all c should derive from c - 1, only the bottom
542 		 * most c derive from c + 1:
543 		 *
544 		 *                                  2   1   0
545 		 *  ---------------------------------------------------
546 		 *                                               c=
547 		 *                                                3
548 		 *
549 		 *         from prev_c = c - 1 --> 5,2            2
550 		 *                                    \
551 		 *                                     5,3        1
552 		 *                                        \
553 		 *                                 4,3     5,4<-- 0
554 		 *                                    \   /
555 		 *  bottom most for d=1 from c + 1 --> 4,4       -1
556 		 *                                    /
557 		 *         bottom most for d=2 --> 3,4           -2
558 		 *
559 		 * Except when a c + 1 from a previous run already means a
560 		 * further advancement in the graph.
561 		 * If c == d, there is no c + 1 and c - 1 is the only option.
562 		 * If c < d, use c + 1 in case that yields a larger x.
563 		 * Also use c + 1 if c - 1 is outside the graph.
564 		 */
565 		else if (c > -d && (c == d
566 				    || (c - 1 >= -(int)right->atoms.len
567 					&& kd_backward[c - 1] <= kd_backward[c + 1]))) {
568 			/* A top one.
569 			 * From position prev_c, step upwards in the Myers
570 			 * graph: y -= 1.
571 			 * Decrementing y is achieved by incrementing c while
572 			 * keeping the same x. (since we're deriving y from
573 			 * y = x - c + delta).
574 			 */
575 			int prev_c = c - 1;
576 			int prev_x = kd_backward[prev_c];
577 			x = prev_x;
578 		} else {
579 			/* The bottom most one.
580 			 * From position prev_c, step to the left in the Myers
581 			 * graph: x -= 1.
582 			 */
583 			int prev_c = c + 1;
584 			int prev_x = kd_backward[prev_c];
585 			x = prev_x - 1;
586 		}
587 
588 		/* Slide up any snake that we might find here (sections of
589 		 * identical lines on both sides). */
590 		debug("c=%d x-1=%d Yb-1=%d-1=%d\n", c, x-1, xc_to_y(x, c,
591 								    delta),
592 		      xc_to_y(x, c, delta)-1);
593 		if (x > 0) {
594 			debug("  l=");
595 			debug_dump_atom(left, right, &left->atoms.head[x-1]);
596 		}
597 		if (xc_to_y(x, c, delta) > 0) {
598 			debug("  r=");
599 			debug_dump_atom(right, left,
600 				&right->atoms.head[xc_to_y(x, c, delta)-1]);
601 		}
602 		int x_before_slide = x;
603 		while (x > 0 && xc_to_y(x, c, delta) > 0) {
604 			bool same;
605 			int r = diff_atom_same(&same,
606 					       &left->atoms.head[x-1],
607 					       &right->atoms.head[
608 						xc_to_y(x, c, delta)-1]);
609 			if (r)
610 				return r;
611 			if (!same)
612 				break;
613 			x--;
614 		}
615 		kd_backward[c] = x;
616 		if (x_before_slide != x) {
617 			debug("  up %d similar lines\n", x_before_slide - x);
618 		}
619 
620 		if (DEBUG) {
621 			int fi;
622 			for (fi = d; fi >= c; fi--) {
623 				debug("kd_backward[%d] = (%d, %d)\n",
624 				      fi,
625 				      kd_backward[fi],
626 				      kd_backward[fi] - fi + delta);
627 			}
628 		}
629 
630 		if (x < 0 || x > left->atoms.len
631 		    || xc_to_y(x, c, delta) < 0
632 		    || xc_to_y(x, c, delta) > right->atoms.len)
633 			continue;
634 
635 		/* Figured out a new backwards traversal, see if this has gone
636 		 * onto or even past a preceding forwards traversal.
637 		 *
638 		 * If the delta in length is even, then d and backwards_d hit
639 		 * the same state indexes -- note how this is different from in
640 		 * the forwards traversal, because now both d are the same:
641 		 *
642 		 *      | d=   0   1   2      2   1   0
643 		 *  ----+----------------    --------------------
644 		 *  k=  |                                  c=
645 		 *   4  |
646 		 *      |
647 		 *   3  |                                   3
648 		 *      |                same
649 		 *   2  |             2,0====5,2            2
650 		 *      |            /          \
651 		 *   1  |         1,0            5,3        1
652 		 *      |        /              /  \
653 		 *   0  |  -->0,0     3,3====4,3    5,4<--  0
654 		 *      |        \   /             /
655 		 *  -1  |         0,1            4,4       -1
656 		 *      |            \
657 		 *  -2  |             0,2                  -2
658 		 *      |
659 		 *                                      -3
660 		 * If the delta is odd, they end up off-by-one, i.e. on
661 		 * different diagonals.
662 		 * So in the backward path, we can only match up diagonals when
663 		 * the delta is even.
664 		 */
665 		if ((delta & 1) != 0)
666 			continue;
667 		/* Forwards was done first, now both d are the same. */
668 		int forwards_d = d;
669 
670 		/* As soon as the lengths are not the same, the
671 		 * backwards traversal starts on a different diagonal,
672 		 * and c = k shifted by the difference in length.
673 		 */
674 		int k = c_to_k(c, delta);
675 
676 		/* When the file sizes are very different, the traversal trees
677 		 * start on far distant diagonals.
678 		 * They don't necessarily meet straight on. See whether this
679 		 * backward value is also on a valid diagonal in kd_forward[],
680 		 * and match them if so. */
681 		if (k >= -forwards_d && k <= forwards_d) {
682 			/* Current c is on a diagonal that exists in
683 			 * kd_forward[]. If the two x positions have met or
684 			 * passed (backward walked onto or past forward), then
685 			 * we've found a midpoint / a mid-box.
686 			 *
687 			 * When forwards and backwards traversals meet, the
688 			 * endpoints of the mid-snake are not the two points in
689 			 * kd_forward and kd_backward, but rather the section
690 			 * that was slid (if any) of the current
691 			 * forward/backward traversal only.
692 			 *
693 			 * For example:
694 			 *
695 			 *   o-o-o
696 			 *   | |
697 			 *   o A
698 			 *   |  \
699 			 *   o   o
700 			 *        \
701 			 *         M
702 			 *         |\
703 			 *         o o-o-o
704 			 *         | | |
705 			 *         o o X
706 			 *          \
707 			 *           o
708 			 *            \
709 			 *             o
710 			 *              \
711 			 *               o
712 			 *
713 			 * The backward traversal reached M from the bottom and
714 			 * slid upwards.  The forward traversal already reached
715 			 * X, which is not a straight line from M anymore, so
716 			 * picking a mid-snake from M to X would yield a
717 			 * mistake.
718 			 *
719 			 * The correct mid-snake is between M and A. M is where
720 			 * the backward traversal hit the diagonal that the
721 			 * forwards traversal has already passed, and A is what
722 			 * it reaches when sliding up identical lines.
723 			 */
724 
725 			int forward_x = kd_forward[k];
726 			debug("Compare:  k=%d c=%d  is (%d,%d) >= (%d,%d)?\n",
727 			      k, c, forward_x, xk_to_y(forward_x, k),
728 			      x, xc_to_y(x, c, delta));
729 			if (forward_x >= x) {
730 				*meeting_snake = (struct diff_box){
731 					.left_start = x,
732 					.left_end = x_before_slide,
733 					.right_start = xc_to_y(x, c, delta),
734 					.right_end = xk_to_y(x_before_slide, k),
735 				};
736 				debug("HIT x=%u,%u - y=%u,%u\n",
737 				      meeting_snake->left_start,
738 				      meeting_snake->right_start,
739 				      meeting_snake->left_end,
740 				      meeting_snake->right_end);
741 				debug_dump_myers_graph(left, right, NULL,
742 						       kd_forward, d,
743 						       kd_backward, d);
744 				*found_midpoint = true;
745 				return 0;
746 			}
747 		}
748 	}
749 	debug_dump_myers_graph(left, right, NULL, kd_forward, d, kd_backward,
750 			       d);
751 	return 0;
752 }
753 
754 /* Integer square root approximation */
755 static int
756 shift_sqrt(int val)
757 {
758 	int i;
759         for (i = 1; val > 0; val >>= 2)
760 		i <<= 1;
761         return i;
762 }
763 
764 /* Myers "Divide et Impera": tracing forwards from the start and backwards from
765  * the end to find a midpoint that divides the problem into smaller chunks.
766  * Requires only linear amounts of memory. */
767 int
768 diff_algo_myers_divide(const struct diff_algo_config *algo_config,
769 		       struct diff_state *state)
770 {
771 	int rc = ENOMEM;
772 	struct diff_data *left = &state->left;
773 	struct diff_data *right = &state->right;
774 
775 	debug("\n** %s\n", __func__);
776 	debug("left:\n");
777 	debug_dump(left);
778 	debug("right:\n");
779 	debug_dump(right);
780 	debug_dump_myers_graph(left, right, NULL, NULL, 0, NULL, 0);
781 
782 	/* Allocate two columns of a Myers graph, one for the forward and one
783 	 * for the backward traversal. */
784 	unsigned int max = left->atoms.len + right->atoms.len;
785 	size_t kd_len = max + 1;
786 	size_t kd_buf_size = kd_len << 1;
787 	int *kd_buf = reallocarray(NULL, kd_buf_size, sizeof(int));
788 	if (!kd_buf)
789 		return ENOMEM;
790 	int i;
791 	for (i = 0; i < kd_buf_size; i++)
792 		kd_buf[i] = -1;
793 	int *kd_forward = kd_buf;
794 	int *kd_backward = kd_buf + kd_len;
795 	int max_effort = shift_sqrt(max/2);
796 
797 	/* The 'k' axis in Myers spans positive and negative indexes, so point
798 	 * the kd to the middle.
799 	 * It is then possible to index from -max/2 .. max/2. */
800 	kd_forward += max/2;
801 	kd_backward += max/2;
802 
803 	int d;
804 	struct diff_box mid_snake = {};
805 	bool found_midpoint = false;
806 	for (d = 0; d <= (max/2); d++) {
807 		int r;
808 		debug("-- d=%d\n", d);
809 		r = diff_divide_myers_forward(&found_midpoint, left, right,
810 					      kd_forward, kd_backward, d,
811 					      &mid_snake);
812 		if (r)
813 			return r;
814 		if (found_midpoint)
815 			break;
816 		r = diff_divide_myers_backward(&found_midpoint, left, right,
817 					       kd_forward, kd_backward, d,
818 					       &mid_snake);
819 		if (r)
820 			return r;
821 		if (found_midpoint)
822 			break;
823 
824 		/* Limit the effort spent looking for a mid snake. If files have
825 		 * very few lines in common, the effort spent to find nice mid
826 		 * snakes is just not worth it, the diff result will still be
827 		 * essentially minus everything on the left, plus everything on
828 		 * the right, with a few useless matches here and there. */
829 		if (d > max_effort) {
830 			/* pick the furthest reaching point from
831 			 * kd_forward and kd_backward, and use that as a
832 			 * midpoint, to not step into another diff algo
833 			 * recursion with unchanged box. */
834 			int delta = (int)right->atoms.len - (int)left->atoms.len;
835 			int x;
836 			int y;
837 			int i;
838 			int best_forward_i = 0;
839 			int best_forward_distance = 0;
840 			int best_backward_i = 0;
841 			int best_backward_distance = 0;
842 			int distance;
843 			int best_forward_x;
844 			int best_forward_y;
845 			int best_backward_x;
846 			int best_backward_y;
847 
848 			debug("~~~ d = %d > max_effort = %d\n", d, max_effort);
849 
850 			for (i = d; i >= -d; i -= 2) {
851 				if (i >= -(int)right->atoms.len && i <= (int)left->atoms.len) {
852 					x = kd_forward[i];
853 					y = xk_to_y(x, i);
854 					distance = x + y;
855 					if (distance > best_forward_distance) {
856 						best_forward_distance = distance;
857 						best_forward_i = i;
858 					}
859 				}
860 
861 				if (i >= -(int)left->atoms.len && i <= (int)right->atoms.len) {
862 					x = kd_backward[i];
863 					y = xc_to_y(x, i, delta);
864 					distance = (right->atoms.len - x)
865 						+ (left->atoms.len - y);
866 					if (distance >= best_backward_distance) {
867 						best_backward_distance = distance;
868 						best_backward_i = i;
869 					}
870 				}
871 			}
872 
873 			/* The myers-divide didn't meet in the middle. We just
874 			 * figured out the places where the forward path
875 			 * advanced the most, and the backward path advanced the
876 			 * most. Just divide at whichever one of those two is better.
877 			 *
878 			 *   o-o
879 			 *     |
880 			 *     o
881 			 *      \
882 			 *       o
883 			 *        \
884 			 *         F <-- cut here
885 			 *
886 			 *
887 			 *
888 			 *           or here --> B
889 			 *                        \
890 			 *                         o
891 			 *                          \
892 			 *                           o
893 			 *                           |
894 			 *                           o-o
895 			 */
896 			best_forward_x = kd_forward[best_forward_i];
897 			best_forward_y = xk_to_y(best_forward_x, best_forward_i);
898 			best_backward_x = kd_backward[best_backward_i];
899 			best_backward_y = xc_to_y(x, best_backward_i, delta);
900 
901 			if (best_forward_distance >= best_backward_distance) {
902 				x = best_forward_x;
903 				y = best_forward_y;
904 			} else {
905 				x = best_backward_x;
906 				y = best_backward_y;
907 			}
908 
909 			debug("max_effort cut at x=%d y=%d\n", x, y);
910 			if (x < 0 || y < 0
911 			    || x > left->atoms.len || y > right->atoms.len)
912 				break;
913 
914 			found_midpoint = true;
915 			mid_snake = (struct diff_box){
916 				.left_start = x,
917 				.left_end = x,
918 				.right_start = y,
919 				.right_end = y,
920 			};
921 			break;
922 		}
923 	}
924 
925 	if (!found_midpoint) {
926 		/* Divide and conquer failed to find a meeting point. Use the
927 		 * fallback_algo defined in the algo_config (leave this to the
928 		 * caller). This is just paranoia/sanity, we normally should
929 		 * always find a midpoint.
930 		 */
931 		debug(" no midpoint \n");
932 		rc = DIFF_RC_USE_DIFF_ALGO_FALLBACK;
933 		goto return_rc;
934 	} else {
935 		debug(" mid snake L: %u to %u of %u   R: %u to %u of %u\n",
936 		      mid_snake.left_start, mid_snake.left_end, left->atoms.len,
937 		      mid_snake.right_start, mid_snake.right_end,
938 		      right->atoms.len);
939 
940 		/* Section before the mid-snake.  */
941 		debug("Section before the mid-snake\n");
942 
943 		struct diff_atom *left_atom = &left->atoms.head[0];
944 		unsigned int left_section_len = mid_snake.left_start;
945 		struct diff_atom *right_atom = &right->atoms.head[0];
946 		unsigned int right_section_len = mid_snake.right_start;
947 
948 		if (left_section_len && right_section_len) {
949 			/* Record an unsolved chunk, the caller will apply
950 			 * inner_algo() on this chunk. */
951 			if (!diff_state_add_chunk(state, false,
952 						  left_atom, left_section_len,
953 						  right_atom,
954 						  right_section_len))
955 				goto return_rc;
956 		} else if (left_section_len && !right_section_len) {
957 			/* Only left atoms and none on the right, they form a
958 			 * "minus" chunk, then. */
959 			if (!diff_state_add_chunk(state, true,
960 						  left_atom, left_section_len,
961 						  right_atom, 0))
962 				goto return_rc;
963 		} else if (!left_section_len && right_section_len) {
964 			/* No left atoms, only atoms on the right, they form a
965 			 * "plus" chunk, then. */
966 			if (!diff_state_add_chunk(state, true,
967 						  left_atom, 0,
968 						  right_atom,
969 						  right_section_len))
970 				goto return_rc;
971 		}
972 		/* else: left_section_len == 0 and right_section_len == 0, i.e.
973 		 * nothing before the mid-snake. */
974 
975 		if (mid_snake.left_end > mid_snake.left_start) {
976 			/* The midpoint is a "snake", i.e. on a section of
977 			 * identical data on both sides: that section
978 			 * immediately becomes a solved diff chunk. */
979 			debug("the mid-snake\n");
980 			if (!diff_state_add_chunk(state, true,
981 				  &left->atoms.head[mid_snake.left_start],
982 				  mid_snake.left_end - mid_snake.left_start,
983 				  &right->atoms.head[mid_snake.right_start],
984 				  mid_snake.right_end - mid_snake.right_start))
985 				goto return_rc;
986 		}
987 
988 		/* Section after the mid-snake. */
989 		debug("Section after the mid-snake\n");
990 		debug("  left_end %u  right_end %u\n",
991 		      mid_snake.left_end, mid_snake.right_end);
992 		debug("  left_count %u  right_count %u\n",
993 		      left->atoms.len, right->atoms.len);
994 		left_atom = &left->atoms.head[mid_snake.left_end];
995 		left_section_len = left->atoms.len - mid_snake.left_end;
996 		right_atom = &right->atoms.head[mid_snake.right_end];
997 		right_section_len = right->atoms.len - mid_snake.right_end;
998 
999 		if (left_section_len && right_section_len) {
1000 			/* Record an unsolved chunk, the caller will apply
1001 			 * inner_algo() on this chunk. */
1002 			if (!diff_state_add_chunk(state, false,
1003 						  left_atom, left_section_len,
1004 						  right_atom,
1005 						  right_section_len))
1006 				goto return_rc;
1007 		} else if (left_section_len && !right_section_len) {
1008 			/* Only left atoms and none on the right, they form a
1009 			 * "minus" chunk, then. */
1010 			if (!diff_state_add_chunk(state, true,
1011 						  left_atom, left_section_len,
1012 						  right_atom, 0))
1013 				goto return_rc;
1014 		} else if (!left_section_len && right_section_len) {
1015 			/* No left atoms, only atoms on the right, they form a
1016 			 * "plus" chunk, then. */
1017 			if (!diff_state_add_chunk(state, true,
1018 						  left_atom, 0,
1019 						  right_atom,
1020 						  right_section_len))
1021 				goto return_rc;
1022 		}
1023 		/* else: left_section_len == 0 and right_section_len == 0, i.e.
1024 		 * nothing after the mid-snake. */
1025 	}
1026 
1027 	rc = DIFF_RC_OK;
1028 
1029 return_rc:
1030 	free(kd_buf);
1031 	debug("** END %s\n", __func__);
1032 	return rc;
1033 }
1034 
1035 /* Myers Diff tracing from the start all the way through to the end, requiring
1036  * quadratic amounts of memory. This can fail if the required space surpasses
1037  * algo_config->permitted_state_size. */
1038 int
1039 diff_algo_myers(const struct diff_algo_config *algo_config,
1040 		struct diff_state *state)
1041 {
1042 	/* do a diff_divide_myers_forward() without a _backward(), so that it
1043 	 * walks forward across the entire files to reach the end. Keep each
1044 	 * run's state, and do a final backtrace. */
1045 	int rc = ENOMEM;
1046 	struct diff_data *left = &state->left;
1047 	struct diff_data *right = &state->right;
1048 
1049 	debug("\n** %s\n", __func__);
1050 	debug("left:\n");
1051 	debug_dump(left);
1052 	debug("right:\n");
1053 	debug_dump(right);
1054 	debug_dump_myers_graph(left, right, NULL, NULL, 0, NULL, 0);
1055 
1056 	/* Allocate two columns of a Myers graph, one for the forward and one
1057 	 * for the backward traversal. */
1058 	unsigned int max = left->atoms.len + right->atoms.len;
1059 	size_t kd_len = max + 1 + max;
1060 	size_t kd_buf_size = kd_len * kd_len;
1061 	size_t kd_state_size = kd_buf_size * sizeof(int);
1062 	debug("state size: %zu\n", kd_buf_size);
1063 	if (kd_buf_size < kd_len /* overflow? */
1064 	    || kd_state_size > algo_config->permitted_state_size) {
1065 		debug("state size %zu > permitted_state_size %zu, use fallback_algo\n",
1066 		      kd_state_size, algo_config->permitted_state_size);
1067 		return DIFF_RC_USE_DIFF_ALGO_FALLBACK;
1068 	}
1069 
1070 	int *kd_buf = reallocarray(NULL, kd_buf_size, sizeof(int));
1071 	if (!kd_buf)
1072 		return ENOMEM;
1073 	int i;
1074 	for (i = 0; i < kd_buf_size; i++)
1075 		kd_buf[i] = -1;
1076 
1077 	/* The 'k' axis in Myers spans positive and negative indexes, so point
1078 	 * the kd to the middle.
1079 	 * It is then possible to index from -max .. max. */
1080 	int *kd_origin = kd_buf + max;
1081 	int *kd_column = kd_origin;
1082 
1083 	int d;
1084 	int backtrack_d = -1;
1085 	int backtrack_k = 0;
1086 	int k;
1087 	int x, y;
1088 	for (d = 0; d <= max; d++, kd_column += kd_len) {
1089 		debug("-- d=%d\n", d);
1090 
1091 		debug("-- %s d=%d\n", __func__, d);
1092 
1093 		for (k = d; k >= -d; k -= 2) {
1094 			if (k < -(int)right->atoms.len
1095 			    || k > (int)left->atoms.len) {
1096 				/* This diagonal is completely outside of the
1097 				 * Myers graph, don't calculate it. */
1098 				if (k < -(int)right->atoms.len)
1099 					debug(" %d k <"
1100 					      " -(int)right->atoms.len %d\n",
1101 					      k, -(int)right->atoms.len);
1102 				else
1103 					debug(" %d k > left->atoms.len %d\n", k,
1104 					      left->atoms.len);
1105 				if (k < 0) {
1106 					/* We are traversing negatively, and
1107 					 * already below the entire graph,
1108 					 * nothing will come of this. */
1109 					debug(" break");
1110 					break;
1111 				}
1112 				debug(" continue");
1113 				continue;
1114 			}
1115 
1116 			debug("- k = %d\n", k);
1117 			if (d == 0) {
1118 				/* This is the initializing step. There is no
1119 				 * prev_k yet, get the initial x from the top
1120 				 * left of the Myers graph. */
1121 				x = 0;
1122 			} else {
1123 				int *kd_prev_column = kd_column - kd_len;
1124 
1125 				/* Favoring "-" lines first means favoring
1126 				 * moving rightwards in the Myers graph.
1127 				 * For this, all k should derive from k - 1,
1128 				 * only the bottom most k derive from k + 1:
1129 				 *
1130 				 *      | d=   0   1   2
1131 				 *  ----+----------------
1132 				 *  k=  |
1133 				 *   2  |             2,0 <-- from
1134 				 *      |            /        prev_k = 2 - 1 = 1
1135 				 *   1  |         1,0
1136 				 *      |        /
1137 				 *   0  |  -->0,0     3,3
1138 				 *      |       \\   /
1139 				 *  -1  |         0,1 <-- bottom most for d=1
1140 				 *      |           \\    from prev_k = -1+1 = 0
1141 				 *  -2  |             0,2 <-- bottom most for
1142 				 *                            d=2 from
1143 				 *                            prev_k = -2+1 = -1
1144 				 *
1145 				 * Except when a k + 1 from a previous run
1146 				 * already means a further advancement in the
1147 				 * graph.
1148 				 * If k == d, there is no k + 1 and k - 1 is the
1149 				 * only option.
1150 				 * If k < d, use k + 1 in case that yields a
1151 				 * larger x. Also use k + 1 if k - 1 is outside
1152 				 * the graph.
1153 				 */
1154 				if (k > -d
1155 				    && (k == d
1156 					|| (k - 1 >= -(int)right->atoms.len
1157 					    && kd_prev_column[k - 1]
1158 					       >= kd_prev_column[k + 1]))) {
1159 					/* Advance from k - 1.
1160 					 * From position prev_k, step to the
1161 					 * right in the Myers graph: x += 1.
1162 					 */
1163 					int prev_k = k - 1;
1164 					int prev_x = kd_prev_column[prev_k];
1165 					x = prev_x + 1;
1166 				} else {
1167 					/* The bottom most one.
1168 					 * From position prev_k, step to the
1169 					 * bottom in the Myers graph: y += 1.
1170 					 * Incrementing y is achieved by
1171 					 * decrementing k while keeping the same
1172 					 * x. (since we're deriving y from y =
1173 					 * x - k).
1174 					 */
1175 					int prev_k = k + 1;
1176 					int prev_x = kd_prev_column[prev_k];
1177 					x = prev_x;
1178 				}
1179 			}
1180 
1181 			/* Slide down any snake that we might find here. */
1182 			while (x < left->atoms.len
1183 			       && xk_to_y(x, k) < right->atoms.len) {
1184 				bool same;
1185 				int r = diff_atom_same(&same,
1186 						       &left->atoms.head[x],
1187 						       &right->atoms.head[
1188 							xk_to_y(x, k)]);
1189 				if (r)
1190 					return r;
1191 				if (!same)
1192 					break;
1193 				x++;
1194 			}
1195 			kd_column[k] = x;
1196 
1197 			if (DEBUG) {
1198 				int fi;
1199 				for (fi = d; fi >= k; fi-=2) {
1200 					debug("kd_column[%d] = (%d, %d)\n", fi,
1201 					      kd_column[fi],
1202 					      kd_column[fi] - fi);
1203 				}
1204 			}
1205 
1206 			if (x == left->atoms.len
1207 			    && xk_to_y(x, k) == right->atoms.len) {
1208 				/* Found a path */
1209 				backtrack_d = d;
1210 				backtrack_k = k;
1211 				debug("Reached the end at d = %d, k = %d\n",
1212 				      backtrack_d, backtrack_k);
1213 				break;
1214 			}
1215 		}
1216 
1217 		if (backtrack_d >= 0)
1218 			break;
1219 	}
1220 
1221 	debug_dump_myers_graph(left, right, kd_origin, NULL, 0, NULL, 0);
1222 
1223 	/* backtrack. A matrix spanning from start to end of the file is ready:
1224 	 *
1225 	 *      | d=   0   1   2   3   4
1226 	 *  ----+---------------------------------
1227 	 *  k=  |
1228 	 *   3  |
1229 	 *      |
1230 	 *   2  |             2,0
1231 	 *      |            /
1232 	 *   1  |         1,0     4,3
1233 	 *      |        /       /   \
1234 	 *   0  |  -->0,0     3,3     4,4 --> backtrack_d = 4, backtrack_k = 0
1235 	 *      |        \   /   \
1236 	 *  -1  |         0,1     3,4
1237 	 *      |            \
1238 	 *  -2  |             0,2
1239 	 *      |
1240 	 *
1241 	 * From (4,4) backwards, find the previous position that is the largest, and remember it.
1242 	 *
1243 	 */
1244 	for (d = backtrack_d, k = backtrack_k; d >= 0; d--) {
1245 		x = kd_column[k];
1246 		y = xk_to_y(x, k);
1247 
1248 		/* When the best position is identified, remember it for that
1249 		 * kd_column.
1250 		 * That kd_column is no longer needed otherwise, so just
1251 		 * re-purpose kd_column[0] = x and kd_column[1] = y,
1252 		 * so that there is no need to allocate more memory.
1253 		 */
1254 		kd_column[0] = x;
1255 		kd_column[1] = y;
1256 		debug("Backtrack d=%d: xy=(%d, %d)\n",
1257 		      d, kd_column[0], kd_column[1]);
1258 
1259 		/* Don't access memory before kd_buf */
1260 		if (d == 0)
1261 			break;
1262 		int *kd_prev_column = kd_column - kd_len;
1263 
1264 		/* When y == 0, backtracking downwards (k-1) is the only way.
1265 		 * When x == 0, backtracking upwards (k+1) is the only way.
1266 		 *
1267 		 *      | d=   0   1   2   3   4
1268 		 *  ----+---------------------------------
1269 		 *  k=  |
1270 		 *   3  |
1271 		 *      |                ..y == 0
1272 		 *   2  |             2,0
1273 		 *      |            /
1274 		 *   1  |         1,0     4,3
1275 		 *      |        /       /   \
1276 		 *   0  |  -->0,0     3,3     4,4 --> backtrack_d = 4,
1277 		 *      |        \   /   \            backtrack_k = 0
1278 		 *  -1  |         0,1     3,4
1279 		 *      |            \
1280 		 *  -2  |             0,2__
1281 		 *      |                  x == 0
1282 		 */
1283 		debug("prev[k-1] = %d,%d  prev[k+1] = %d,%d\n",
1284 		      kd_prev_column[k-1], xk_to_y(kd_prev_column[k-1],k-1),
1285 		      kd_prev_column[k+1], xk_to_y(kd_prev_column[k+1],k+1));
1286 		if (y == 0
1287 		    || (x > 0
1288 			&& kd_prev_column[k - 1] >= kd_prev_column[k + 1])) {
1289 			k = k - 1;
1290 			debug("prev k=k-1=%d x=%d y=%d\n",
1291 			      k, kd_prev_column[k],
1292 			      xk_to_y(kd_prev_column[k], k));
1293 		} else {
1294 			k = k + 1;
1295 			debug("prev k=k+1=%d x=%d y=%d\n",
1296 			      k, kd_prev_column[k],
1297 			      xk_to_y(kd_prev_column[k], k));
1298 		}
1299 		kd_column = kd_prev_column;
1300 	}
1301 
1302 	/* Forwards again, this time recording the diff chunks.
1303 	 * Definitely start from 0,0. kd_column[0] may actually point to the
1304 	 * bottom of a snake starting at 0,0 */
1305 	x = 0;
1306 	y = 0;
1307 
1308 	kd_column = kd_origin;
1309 	for (d = 0; d <= backtrack_d; d++, kd_column += kd_len) {
1310 		int next_x = kd_column[0];
1311 		int next_y = kd_column[1];
1312 		debug("Forward track from xy(%d,%d) to xy(%d,%d)\n",
1313 		      x, y, next_x, next_y);
1314 
1315 		struct diff_atom *left_atom = &left->atoms.head[x];
1316 		int left_section_len = next_x - x;
1317 		struct diff_atom *right_atom = &right->atoms.head[y];
1318 		int right_section_len = next_y - y;
1319 
1320 		rc = ENOMEM;
1321 		if (left_section_len && right_section_len) {
1322 			/* This must be a snake slide.
1323 			 * Snake slides have a straight line leading into them
1324 			 * (except when starting at (0,0)). Find out whether the
1325 			 * lead-in is horizontal or vertical:
1326 			 *
1327 			 *     left
1328 			 *  ---------->
1329 			 *  |
1330 			 * r|   o-o        o
1331 			 * i|      \       |
1332 			 * g|       o      o
1333 			 * h|        \      \
1334 			 * t|         o      o
1335 			 *  v
1336 			 *
1337 			 * If left_section_len > right_section_len, the lead-in
1338 			 * is horizontal, meaning first remove one atom from the
1339 			 * left before sliding down the snake.
1340 			 * If right_section_len > left_section_len, the lead-in
1341 			 * is vetical, so add one atom from the right before
1342 			 * sliding down the snake. */
1343 			if (left_section_len == right_section_len + 1) {
1344 				if (!diff_state_add_chunk(state, true,
1345 							  left_atom, 1,
1346 							  right_atom, 0))
1347 					goto return_rc;
1348 				left_atom++;
1349 				left_section_len--;
1350 			} else if (right_section_len == left_section_len + 1) {
1351 				if (!diff_state_add_chunk(state, true,
1352 							  left_atom, 0,
1353 							  right_atom, 1))
1354 					goto return_rc;
1355 				right_atom++;
1356 				right_section_len--;
1357 			} else if (left_section_len != right_section_len) {
1358 				/* The numbers are making no sense. Should never
1359 				 * happen. */
1360 				rc = DIFF_RC_USE_DIFF_ALGO_FALLBACK;
1361 				goto return_rc;
1362 			}
1363 
1364 			if (!diff_state_add_chunk(state, true,
1365 						  left_atom, left_section_len,
1366 						  right_atom,
1367 						  right_section_len))
1368 				goto return_rc;
1369 		} else if (left_section_len && !right_section_len) {
1370 			/* Only left atoms and none on the right, they form a
1371 			 * "minus" chunk, then. */
1372 			if (!diff_state_add_chunk(state, true,
1373 						  left_atom, left_section_len,
1374 						  right_atom, 0))
1375 				goto return_rc;
1376 		} else if (!left_section_len && right_section_len) {
1377 			/* No left atoms, only atoms on the right, they form a
1378 			 * "plus" chunk, then. */
1379 			if (!diff_state_add_chunk(state, true,
1380 						  left_atom, 0,
1381 						  right_atom,
1382 						  right_section_len))
1383 				goto return_rc;
1384 		}
1385 
1386 		x = next_x;
1387 		y = next_y;
1388 	}
1389 
1390 	rc = DIFF_RC_OK;
1391 
1392 return_rc:
1393 	free(kd_buf);
1394 	debug("** END %s rc=%d\n", __func__, rc);
1395 	return rc;
1396 }