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 	for (k = d; k >= -d; k -= 2) {
226 		if (k < -(int)right->atoms.len || k > (int)left->atoms.len) {
227 			/* This diagonal is completely outside of the Myers
228 			 * graph, don't calculate it. */
229 			if (k < 0) {
230 				/* We are traversing negatively, and already
231 				 * below the entire graph, nothing will come of
232 				 * this. */
233 				debug(" break\n");
234 				break;
235 			}
236 			debug(" continue\n");
237 			continue;
238 		}
239 		if (d == 0) {
240 			/* This is the initializing step. There is no prev_k
241 			 * yet, get the initial x from the top left of the Myers
242 			 * graph. */
243 			x = 0;
244 			prev_x = x;
245 			prev_y = xk_to_y(x, k);
246 		}
247 		/* Favoring "-" lines first means favoring moving rightwards in
248 		 * the Myers graph.
249 		 * For this, all k should derive from k - 1, only the bottom
250 		 * most k derive from k + 1:
251 		 *
252 		 *      | d=   0   1   2
253 		 *  ----+----------------
254 		 *  k=  |
255 		 *   2  |             2,0 <-- from prev_k = 2 - 1 = 1
256 		 *      |            /
257 		 *   1  |         1,0
258 		 *      |        /
259 		 *   0  |  -->0,0     3,3
260 		 *      |       \\   /
261 		 *  -1  |         0,1 <-- bottom most for d=1 from
262 		 *      |           \\    prev_k = -1 + 1 = 0
263 		 *  -2  |             0,2 <-- bottom most for d=2 from
264 		 *                            prev_k = -2 + 1 = -1
265 		 *
266 		 * Except when a k + 1 from a previous run already means a
267 		 * further advancement in the graph.
268 		 * If k == d, there is no k + 1 and k - 1 is the only option.
269 		 * If k < d, use k + 1 in case that yields a larger x. Also use
270 		 * k + 1 if k - 1 is outside the graph.
271 		 */
272 		else if (k > -d
273 			 && (k == d
274 			     || (k - 1 >= -(int)right->atoms.len
275 				 && kd_forward[k - 1] >= kd_forward[k + 1]))) {
276 			/* Advance from k - 1.
277 			 * From position prev_k, step to the right in the Myers
278 			 * graph: x += 1.
279 			 */
280 			int prev_k = k - 1;
281 			prev_x = kd_forward[prev_k];
282 			prev_y = xk_to_y(prev_x, prev_k);
283 			x = prev_x + 1;
284 		} else {
285 			/* The bottom most one.
286 			 * From position prev_k, step to the bottom in the Myers
287 			 * graph: y += 1.
288 			 * Incrementing y is achieved by decrementing k while
289 			 * keeping the same x.
290 			 * (since we're deriving y from y = x - k).
291 			 */
292 			int prev_k = k + 1;
293 			prev_x = kd_forward[prev_k];
294 			prev_y = xk_to_y(prev_x, prev_k);
295 			x = prev_x;
296 		}
297 
298 		x_before_slide = x;
299 		/* Slide down any snake that we might find here. */
300 		while (x < left->atoms.len && xk_to_y(x, k) < right->atoms.len) {
301 			bool same;
302 			int r = diff_atom_same(&same,
303 					       &left->atoms.head[x],
304 					       &right->atoms.head[
305 						xk_to_y(x, k)]);
306 			if (r)
307 				return r;
308 			if (!same)
309 				break;
310 			x++;
311 		}
312 		kd_forward[k] = x;
313 #if 0
314 		if (x_before_slide != x) {
315 			debug("  down %d similar lines\n", x - x_before_slide);
316 		}
317 
318 #if DEBUG
319 		{
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 #endif
327 #endif
328 
329 		if (x < 0 || x > left->atoms.len
330 		    || xk_to_y(x, k) < 0 || xk_to_y(x, k) > right->atoms.len)
331 			continue;
332 
333 		/* Figured out a new forwards traversal, see if this has gone
334 		 * onto or even past a preceding backwards traversal.
335 		 *
336 		 * If the delta in length is odd, then d and backwards_d hit the
337 		 * same state indexes:
338 		 *      | d=   0   1   2      1   0
339 		 *  ----+----------------    ----------------
340 		 *  k=  |                              c=
341 		 *   4  |                               3
342 		 *      |
343 		 *   3  |                               2
344 		 *      |                same
345 		 *   2  |             2,0====5,3        1
346 		 *      |            /          \
347 		 *   1  |         1,0            5,4<-- 0
348 		 *      |        /              /
349 		 *   0  |  -->0,0     3,3====4,4       -1
350 		 *      |        \   /
351 		 *  -1  |         0,1                  -2
352 		 *      |            \
353 		 *  -2  |             0,2              -3
354 		 *      |
355 		 *
356 		 * If the delta is even, they end up off-by-one, i.e. on
357 		 * different diagonals:
358 		 *
359 		 *      | d=   0   1   2    1   0
360 		 *  ----+----------------  ----------------
361 		 *      |                            c=
362 		 *   3  |                             3
363 		 *      |
364 		 *   2  |             2,0 off         2
365 		 *      |            /   \\
366 		 *   1  |         1,0      4,3        1
367 		 *      |        /       //   \
368 		 *   0  |  -->0,0     3,3      4,4<-- 0
369 		 *      |        \   /        /
370 		 *  -1  |         0,1      3,4       -1
371 		 *      |            \   //
372 		 *  -2  |             0,2            -2
373 		 *      |
374 		 *
375 		 * So in the forward path, we can only match up diagonals when
376 		 * the delta is odd.
377 		 */
378 		if ((delta & 1) == 0)
379 			continue;
380 		 /* Forwards is done first, so the backwards one was still at
381 		  * d - 1. Can't do this for d == 0. */
382 		int backwards_d = d - 1;
383 		if (backwards_d < 0)
384 			continue;
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 			if (x >= backward_x) {
447 				if (x_before_slide != x) {
448 					/* met after sliding up a mid-snake */
449 					*meeting_snake = (struct diff_box){
450 						.left_start = x_before_slide,
451 						.left_end = x,
452 						.right_start = xc_to_y(x_before_slide,
453 								       c, delta),
454 						.right_end = xk_to_y(x, k),
455 					};
456 				} else {
457 					/* met after a side step, non-identical
458 					 * line. Mark that as box divider
459 					 * instead. This makes sure that
460 					 * myers_divide never returns the same
461 					 * box that came as input, avoiding
462 					 * "infinite" looping. */
463 					*meeting_snake = (struct diff_box){
464 						.left_start = prev_x,
465 						.left_end = x,
466 						.right_start = prev_y,
467 						.right_end = xk_to_y(x, k),
468 					};
469 				}
470 				debug("HIT x=(%u,%u) - y=(%u,%u)\n",
471 				      meeting_snake->left_start,
472 				      meeting_snake->right_start,
473 				      meeting_snake->left_end,
474 				      meeting_snake->right_end);
475 				debug_dump_myers_graph(left, right, NULL,
476 						       kd_forward, d,
477 						       kd_backward, d-1);
478 				*found_midpoint = true;
479 				return 0;
480 			}
481 		}
482 	}
483 
484 	return 0;
485 }
486 
487 /* Do one backwards step in the "divide and conquer" graph traversal.
488  * left: the left side to diff.
489  * right: the right side to diff against.
490  * kd_forward: the traversal state for forwards traversal, to find a meeting
491  *             point.
492  *             Since forwards is done first, after this, both kd_forward and
493  *             kd_backward will be valid for d.
494  *             kd_forward points at the center of the state array, allowing
495  *             negative indexes.
496  * kd_backward: the traversal state for backwards traversal, to find a meeting
497  *              point.
498  *              This is carried over between invocations with increasing d.
499  *              kd_backward points at the center of the state array, allowing
500  *              negative indexes.
501  * d: Step or distance counter, indicating for what value of d the kd_backward
502  *    should be populated.
503  *    Before the first invocation, kd_backward[0] shall point at the bottom
504  *    right of the Myers graph (left.len, right.len).
505  *    The first invocation will be for d == 1.
506  * meeting_snake: resulting meeting point, if any.
507  * Return true when a meeting point has been identified.
508  */
509 static int
510 diff_divide_myers_backward(bool *found_midpoint,
511 			   struct diff_data *left, struct diff_data *right,
512 			   int *kd_forward, int *kd_backward, int d,
513 			   struct diff_box *meeting_snake)
514 {
515 	int delta = (int)right->atoms.len - (int)left->atoms.len;
516 	int c;
517 	int x;
518 	int prev_x;
519 	int prev_y;
520 	int x_before_slide;
521 
522 	*found_midpoint = false;
523 
524 	for (c = d; c >= -d; c -= 2) {
525 		if (c < -(int)left->atoms.len || c > (int)right->atoms.len) {
526 			/* This diagonal is completely outside of the Myers
527 			 * graph, don't calculate it. */
528 			if (c < 0) {
529 				/* We are traversing negatively, and already
530 				 * below the entire graph, nothing will come of
531 				 * this. */
532 				break;
533 			}
534 			continue;
535 		}
536 		if (d == 0) {
537 			/* This is the initializing step. There is no prev_c
538 			 * yet, get the initial x from the bottom right of the
539 			 * Myers graph. */
540 			x = left->atoms.len;
541 			prev_x = x;
542 			prev_y = xc_to_y(x, c, delta);
543 		}
544 		/* Favoring "-" lines first means favoring moving rightwards in
545 		 * the Myers graph.
546 		 * For this, all c should derive from c - 1, only the bottom
547 		 * most c derive from c + 1:
548 		 *
549 		 *                                  2   1   0
550 		 *  ---------------------------------------------------
551 		 *                                               c=
552 		 *                                                3
553 		 *
554 		 *         from prev_c = c - 1 --> 5,2            2
555 		 *                                    \
556 		 *                                     5,3        1
557 		 *                                        \
558 		 *                                 4,3     5,4<-- 0
559 		 *                                    \   /
560 		 *  bottom most for d=1 from c + 1 --> 4,4       -1
561 		 *                                    /
562 		 *         bottom most for d=2 --> 3,4           -2
563 		 *
564 		 * Except when a c + 1 from a previous run already means a
565 		 * further advancement in the graph.
566 		 * If c == d, there is no c + 1 and c - 1 is the only option.
567 		 * If c < d, use c + 1 in case that yields a larger x.
568 		 * Also use c + 1 if c - 1 is outside the graph.
569 		 */
570 		else if (c > -d && (c == d
571 				    || (c - 1 >= -(int)right->atoms.len
572 					&& kd_backward[c - 1] <= kd_backward[c + 1]))) {
573 			/* A top one.
574 			 * From position prev_c, step upwards in the Myers
575 			 * graph: y -= 1.
576 			 * Decrementing y is achieved by incrementing c while
577 			 * keeping the same x. (since we're deriving y from
578 			 * y = x - c + delta).
579 			 */
580 			int prev_c = c - 1;
581 			prev_x = kd_backward[prev_c];
582 			prev_y = xc_to_y(prev_x, prev_c, delta);
583 			x = prev_x;
584 		} else {
585 			/* The bottom most one.
586 			 * From position prev_c, step to the left in the Myers
587 			 * graph: x -= 1.
588 			 */
589 			int prev_c = c + 1;
590 			prev_x = kd_backward[prev_c];
591 			prev_y = xc_to_y(prev_x, prev_c, delta);
592 			x = prev_x - 1;
593 		}
594 
595 		/* Slide up any snake that we might find here (sections of
596 		 * identical lines on both sides). */
597 #if 0
598 		debug("c=%d x-1=%d Yb-1=%d-1=%d\n", c, x-1, xc_to_y(x, c,
599 								    delta),
600 		      xc_to_y(x, c, delta)-1);
601 		if (x > 0) {
602 			debug("  l=");
603 			debug_dump_atom(left, right, &left->atoms.head[x-1]);
604 		}
605 		if (xc_to_y(x, c, delta) > 0) {
606 			debug("  r=");
607 			debug_dump_atom(right, left,
608 				&right->atoms.head[xc_to_y(x, c, delta)-1]);
609 		}
610 #endif
611 		x_before_slide = x;
612 		while (x > 0 && xc_to_y(x, c, delta) > 0) {
613 			bool same;
614 			int r = diff_atom_same(&same,
615 					       &left->atoms.head[x-1],
616 					       &right->atoms.head[
617 						xc_to_y(x, c, delta)-1]);
618 			if (r)
619 				return r;
620 			if (!same)
621 				break;
622 			x--;
623 		}
624 		kd_backward[c] = x;
625 #if 0
626 		if (x_before_slide != x) {
627 			debug("  up %d similar lines\n", x_before_slide - x);
628 		}
629 
630 		if (DEBUG) {
631 			int fi;
632 			for (fi = d; fi >= c; fi--) {
633 				debug("kd_backward[%d] = (%d, %d)\n",
634 				      fi,
635 				      kd_backward[fi],
636 				      kd_backward[fi] - fi + delta);
637 			}
638 		}
639 #endif
640 
641 		if (x < 0 || x > left->atoms.len
642 		    || xc_to_y(x, c, delta) < 0
643 		    || xc_to_y(x, c, delta) > right->atoms.len)
644 			continue;
645 
646 		/* Figured out a new backwards traversal, see if this has gone
647 		 * onto or even past a preceding forwards traversal.
648 		 *
649 		 * If the delta in length is even, then d and backwards_d hit
650 		 * the same state indexes -- note how this is different from in
651 		 * the forwards traversal, because now both d are the same:
652 		 *
653 		 *      | d=   0   1   2      2   1   0
654 		 *  ----+----------------    --------------------
655 		 *  k=  |                                  c=
656 		 *   4  |
657 		 *      |
658 		 *   3  |                                   3
659 		 *      |                same
660 		 *   2  |             2,0====5,2            2
661 		 *      |            /          \
662 		 *   1  |         1,0            5,3        1
663 		 *      |        /              /  \
664 		 *   0  |  -->0,0     3,3====4,3    5,4<--  0
665 		 *      |        \   /             /
666 		 *  -1  |         0,1            4,4       -1
667 		 *      |            \
668 		 *  -2  |             0,2                  -2
669 		 *      |
670 		 *                                      -3
671 		 * If the delta is odd, they end up off-by-one, i.e. on
672 		 * different diagonals.
673 		 * So in the backward path, we can only match up diagonals when
674 		 * the delta is even.
675 		 */
676 		if ((delta & 1) != 0)
677 			continue;
678 		/* Forwards was done first, now both d are the same. */
679 		int forwards_d = d;
680 
681 		/* As soon as the lengths are not the same, the
682 		 * backwards traversal starts on a different diagonal,
683 		 * and c = k shifted by the difference in length.
684 		 */
685 		int k = c_to_k(c, delta);
686 
687 		/* When the file sizes are very different, the traversal trees
688 		 * start on far distant diagonals.
689 		 * They don't necessarily meet straight on. See whether this
690 		 * backward value is also on a valid diagonal in kd_forward[],
691 		 * and match them if so. */
692 		if (k >= -forwards_d && k <= forwards_d) {
693 			/* Current c is on a diagonal that exists in
694 			 * kd_forward[]. If the two x positions have met or
695 			 * passed (backward walked onto or past forward), then
696 			 * we've found a midpoint / a mid-box.
697 			 *
698 			 * When forwards and backwards traversals meet, the
699 			 * endpoints of the mid-snake are not the two points in
700 			 * kd_forward and kd_backward, but rather the section
701 			 * that was slid (if any) of the current
702 			 * forward/backward traversal only.
703 			 *
704 			 * For example:
705 			 *
706 			 *   o-o-o
707 			 *   | |
708 			 *   o A
709 			 *   |  \
710 			 *   o   o
711 			 *        \
712 			 *         M
713 			 *         |\
714 			 *         o o-o-o
715 			 *         | | |
716 			 *         o o X
717 			 *          \
718 			 *           o
719 			 *            \
720 			 *             o
721 			 *              \
722 			 *               o
723 			 *
724 			 * The backward traversal reached M from the bottom and
725 			 * slid upwards.  The forward traversal already reached
726 			 * X, which is not a straight line from M anymore, so
727 			 * picking a mid-snake from M to X would yield a
728 			 * mistake.
729 			 *
730 			 * The correct mid-snake is between M and A. M is where
731 			 * the backward traversal hit the diagonal that the
732 			 * forwards traversal has already passed, and A is what
733 			 * it reaches when sliding up identical lines.
734 			 */
735 
736 			int forward_x = kd_forward[k];
737 			if (forward_x >= x) {
738 				if (x_before_slide != x) {
739 					/* met after sliding down a mid-snake */
740 					*meeting_snake = (struct diff_box){
741 						.left_start = x,
742 						.left_end = x_before_slide,
743 						.right_start = xc_to_y(x, c, delta),
744 						.right_end = xk_to_y(x_before_slide, k),
745 					};
746 				} else {
747 					/* met after a side step, non-identical
748 					 * line. Mark that as box divider
749 					 * instead. This makes sure that
750 					 * myers_divide never returns the same
751 					 * box that came as input, avoiding
752 					 * "infinite" looping. */
753 					*meeting_snake = (struct diff_box){
754 						.left_start = x,
755 						.left_end = prev_x,
756 						.right_start = xc_to_y(x, c, delta),
757 						.right_end = prev_y,
758 					};
759 				}
760 				debug("HIT x=%u,%u - y=%u,%u\n",
761 				      meeting_snake->left_start,
762 				      meeting_snake->right_start,
763 				      meeting_snake->left_end,
764 				      meeting_snake->right_end);
765 				debug_dump_myers_graph(left, right, NULL,
766 						       kd_forward, d,
767 						       kd_backward, d);
768 				*found_midpoint = true;
769 				return 0;
770 			}
771 		}
772 	}
773 	return 0;
774 }
775 
776 /* Integer square root approximation */
777 static int
778 shift_sqrt(int val)
779 {
780 	int i;
781         for (i = 1; val > 0; val >>= 2)
782 		i <<= 1;
783         return i;
784 }
785 
786 /* Myers "Divide et Impera": tracing forwards from the start and backwards from
787  * the end to find a midpoint that divides the problem into smaller chunks.
788  * Requires only linear amounts of memory. */
789 int
790 diff_algo_myers_divide(const struct diff_algo_config *algo_config,
791 		       struct diff_state *state)
792 {
793 	int rc = ENOMEM;
794 	struct diff_data *left = &state->left;
795 	struct diff_data *right = &state->right;
796 	int *kd_buf;
797 
798 	debug("\n** %s\n", __func__);
799 	debug("left:\n");
800 	debug_dump(left);
801 	debug("right:\n");
802 	debug_dump(right);
803 
804 	/* Allocate two columns of a Myers graph, one for the forward and one
805 	 * for the backward traversal. */
806 	unsigned int max = left->atoms.len + right->atoms.len;
807 	size_t kd_len = max + 1;
808 	size_t kd_buf_size = kd_len << 1;
809 
810 	if (state->kd_buf_size < kd_buf_size) {
811 		kd_buf = reallocarray(state->kd_buf, kd_buf_size,
812 		    sizeof(int));
813 		if (!kd_buf)
814 			return ENOMEM;
815 		state->kd_buf = kd_buf;
816 		state->kd_buf_size = kd_buf_size;
817 	} else
818 		kd_buf = state->kd_buf;
819 	int i;
820 	for (i = 0; i < kd_buf_size; i++)
821 		kd_buf[i] = -1;
822 	int *kd_forward = kd_buf;
823 	int *kd_backward = kd_buf + kd_len;
824 	int max_effort = shift_sqrt(max/2);
825 
826 	/* The 'k' axis in Myers spans positive and negative indexes, so point
827 	 * the kd to the middle.
828 	 * It is then possible to index from -max/2 .. max/2. */
829 	kd_forward += max/2;
830 	kd_backward += max/2;
831 
832 	int d;
833 	struct diff_box mid_snake = {};
834 	bool found_midpoint = false;
835 	for (d = 0; d <= (max/2); d++) {
836 		int r;
837 		r = diff_divide_myers_forward(&found_midpoint, left, right,
838 					      kd_forward, kd_backward, d,
839 					      &mid_snake);
840 		if (r)
841 			return r;
842 		if (found_midpoint)
843 			break;
844 		r = diff_divide_myers_backward(&found_midpoint, left, right,
845 					       kd_forward, kd_backward, d,
846 					       &mid_snake);
847 		if (r)
848 			return r;
849 		if (found_midpoint)
850 			break;
851 
852 		/* Limit the effort spent looking for a mid snake. If files have
853 		 * very few lines in common, the effort spent to find nice mid
854 		 * snakes is just not worth it, the diff result will still be
855 		 * essentially minus everything on the left, plus everything on
856 		 * the right, with a few useless matches here and there. */
857 		if (d > max_effort) {
858 			/* pick the furthest reaching point from
859 			 * kd_forward and kd_backward, and use that as a
860 			 * midpoint, to not step into another diff algo
861 			 * recursion with unchanged box. */
862 			int delta = (int)right->atoms.len - (int)left->atoms.len;
863 			int x = 0;
864 			int y;
865 			int i;
866 			int best_forward_i = 0;
867 			int best_forward_distance = 0;
868 			int best_backward_i = 0;
869 			int best_backward_distance = 0;
870 			int distance;
871 			int best_forward_x;
872 			int best_forward_y;
873 			int best_backward_x;
874 			int best_backward_y;
875 
876 			debug("~~~ HIT d = %d > max_effort = %d\n", d, max_effort);
877 			debug_dump_myers_graph(left, right, NULL,
878 					       kd_forward, d,
879 					       kd_backward, d);
880 
881 			for (i = d; i >= -d; i -= 2) {
882 				if (i >= -(int)right->atoms.len && i <= (int)left->atoms.len) {
883 					x = kd_forward[i];
884 					y = xk_to_y(x, i);
885 					distance = x + y;
886 					if (distance > best_forward_distance) {
887 						best_forward_distance = distance;
888 						best_forward_i = i;
889 					}
890 				}
891 
892 				if (i >= -(int)left->atoms.len && i <= (int)right->atoms.len) {
893 					x = kd_backward[i];
894 					y = xc_to_y(x, i, delta);
895 					distance = (right->atoms.len - x)
896 						+ (left->atoms.len - y);
897 					if (distance >= best_backward_distance) {
898 						best_backward_distance = distance;
899 						best_backward_i = i;
900 					}
901 				}
902 			}
903 
904 			/* The myers-divide didn't meet in the middle. We just
905 			 * figured out the places where the forward path
906 			 * advanced the most, and the backward path advanced the
907 			 * most. Just divide at whichever one of those two is better.
908 			 *
909 			 *   o-o
910 			 *     |
911 			 *     o
912 			 *      \
913 			 *       o
914 			 *        \
915 			 *         F <-- cut here
916 			 *
917 			 *
918 			 *
919 			 *           or here --> B
920 			 *                        \
921 			 *                         o
922 			 *                          \
923 			 *                           o
924 			 *                           |
925 			 *                           o-o
926 			 */
927 			best_forward_x = kd_forward[best_forward_i];
928 			best_forward_y = xk_to_y(best_forward_x, best_forward_i);
929 			best_backward_x = kd_backward[best_backward_i];
930 			best_backward_y = xc_to_y(best_backward_x, best_backward_i, delta);
931 
932 			if (best_forward_distance >= best_backward_distance) {
933 				x = best_forward_x;
934 				y = best_forward_y;
935 			} else {
936 				x = best_backward_x;
937 				y = best_backward_y;
938 			}
939 
940 			debug("max_effort cut at x=%d y=%d\n", x, y);
941 			if (x < 0 || y < 0
942 			    || x > left->atoms.len || y > right->atoms.len)
943 				break;
944 
945 			found_midpoint = true;
946 			mid_snake = (struct diff_box){
947 				.left_start = x,
948 				.left_end = x,
949 				.right_start = y,
950 				.right_end = y,
951 			};
952 			break;
953 		}
954 	}
955 
956 	if (!found_midpoint) {
957 		/* Divide and conquer failed to find a meeting point. Use the
958 		 * fallback_algo defined in the algo_config (leave this to the
959 		 * caller). This is just paranoia/sanity, we normally should
960 		 * always find a midpoint.
961 		 */
962 		debug(" no midpoint \n");
963 		rc = DIFF_RC_USE_DIFF_ALGO_FALLBACK;
964 		goto return_rc;
965 	} else {
966 		debug(" mid snake L: %u to %u of %u   R: %u to %u of %u\n",
967 		      mid_snake.left_start, mid_snake.left_end, left->atoms.len,
968 		      mid_snake.right_start, mid_snake.right_end,
969 		      right->atoms.len);
970 
971 		/* Section before the mid-snake.  */
972 		debug("Section before the mid-snake\n");
973 
974 		struct diff_atom *left_atom = &left->atoms.head[0];
975 		unsigned int left_section_len = mid_snake.left_start;
976 		struct diff_atom *right_atom = &right->atoms.head[0];
977 		unsigned int right_section_len = mid_snake.right_start;
978 
979 		if (left_section_len && right_section_len) {
980 			/* Record an unsolved chunk, the caller will apply
981 			 * inner_algo() on this chunk. */
982 			if (!diff_state_add_chunk(state, false,
983 						  left_atom, left_section_len,
984 						  right_atom,
985 						  right_section_len))
986 				goto return_rc;
987 		} else if (left_section_len && !right_section_len) {
988 			/* Only left atoms and none on the right, they form a
989 			 * "minus" chunk, then. */
990 			if (!diff_state_add_chunk(state, true,
991 						  left_atom, left_section_len,
992 						  right_atom, 0))
993 				goto return_rc;
994 		} else if (!left_section_len && right_section_len) {
995 			/* No left atoms, only atoms on the right, they form a
996 			 * "plus" chunk, then. */
997 			if (!diff_state_add_chunk(state, true,
998 						  left_atom, 0,
999 						  right_atom,
1000 						  right_section_len))
1001 				goto return_rc;
1002 		}
1003 		/* else: left_section_len == 0 and right_section_len == 0, i.e.
1004 		 * nothing before the mid-snake. */
1005 
1006 		if (mid_snake.left_end > mid_snake.left_start
1007 		    || mid_snake.right_end > mid_snake.right_start) {
1008 			/* The midpoint is a section of identical data on both
1009 			 * sides, or a certain differing line: that section
1010 			 * immediately becomes a solved chunk. */
1011 			debug("the mid-snake\n");
1012 			if (!diff_state_add_chunk(state, true,
1013 				  &left->atoms.head[mid_snake.left_start],
1014 				  mid_snake.left_end - mid_snake.left_start,
1015 				  &right->atoms.head[mid_snake.right_start],
1016 				  mid_snake.right_end - mid_snake.right_start))
1017 				goto return_rc;
1018 		}
1019 
1020 		/* Section after the mid-snake. */
1021 		debug("Section after the mid-snake\n");
1022 		debug("  left_end %u  right_end %u\n",
1023 		      mid_snake.left_end, mid_snake.right_end);
1024 		debug("  left_count %u  right_count %u\n",
1025 		      left->atoms.len, right->atoms.len);
1026 		left_atom = &left->atoms.head[mid_snake.left_end];
1027 		left_section_len = left->atoms.len - mid_snake.left_end;
1028 		right_atom = &right->atoms.head[mid_snake.right_end];
1029 		right_section_len = right->atoms.len - mid_snake.right_end;
1030 
1031 		if (left_section_len && right_section_len) {
1032 			/* Record an unsolved chunk, the caller will apply
1033 			 * inner_algo() on this chunk. */
1034 			if (!diff_state_add_chunk(state, false,
1035 						  left_atom, left_section_len,
1036 						  right_atom,
1037 						  right_section_len))
1038 				goto return_rc;
1039 		} else if (left_section_len && !right_section_len) {
1040 			/* Only left atoms and none on the right, they form a
1041 			 * "minus" chunk, then. */
1042 			if (!diff_state_add_chunk(state, true,
1043 						  left_atom, left_section_len,
1044 						  right_atom, 0))
1045 				goto return_rc;
1046 		} else if (!left_section_len && right_section_len) {
1047 			/* No left atoms, only atoms on the right, they form a
1048 			 * "plus" chunk, then. */
1049 			if (!diff_state_add_chunk(state, true,
1050 						  left_atom, 0,
1051 						  right_atom,
1052 						  right_section_len))
1053 				goto return_rc;
1054 		}
1055 		/* else: left_section_len == 0 and right_section_len == 0, i.e.
1056 		 * nothing after the mid-snake. */
1057 	}
1058 
1059 	rc = DIFF_RC_OK;
1060 
1061 return_rc:
1062 	debug("** END %s\n", __func__);
1063 	return rc;
1064 }
1065 
1066 /* Myers Diff tracing from the start all the way through to the end, requiring
1067  * quadratic amounts of memory. This can fail if the required space surpasses
1068  * algo_config->permitted_state_size. */
1069 int
1070 diff_algo_myers(const struct diff_algo_config *algo_config,
1071 		struct diff_state *state)
1072 {
1073 	/* do a diff_divide_myers_forward() without a _backward(), so that it
1074 	 * walks forward across the entire files to reach the end. Keep each
1075 	 * run's state, and do a final backtrace. */
1076 	int rc = ENOMEM;
1077 	struct diff_data *left = &state->left;
1078 	struct diff_data *right = &state->right;
1079 	int *kd_buf;
1080 
1081 	debug("\n** %s\n", __func__);
1082 	debug("left:\n");
1083 	debug_dump(left);
1084 	debug("right:\n");
1085 	debug_dump(right);
1086 	debug_dump_myers_graph(left, right, NULL, NULL, 0, NULL, 0);
1087 
1088 	/* Allocate two columns of a Myers graph, one for the forward and one
1089 	 * for the backward traversal. */
1090 	unsigned int max = left->atoms.len + right->atoms.len;
1091 	size_t kd_len = max + 1 + max;
1092 	size_t kd_buf_size = kd_len * kd_len;
1093 	size_t kd_state_size = kd_buf_size * sizeof(int);
1094 	debug("state size: %zu\n", kd_state_size);
1095 	if (kd_buf_size < kd_len /* overflow? */
1096 	    || kd_state_size > algo_config->permitted_state_size) {
1097 		debug("state size %zu > permitted_state_size %zu, use fallback_algo\n",
1098 		      kd_state_size, algo_config->permitted_state_size);
1099 		return DIFF_RC_USE_DIFF_ALGO_FALLBACK;
1100 	}
1101 
1102 	if (state->kd_buf_size < kd_buf_size) {
1103 		kd_buf = reallocarray(state->kd_buf, kd_buf_size,
1104 		    sizeof(int));
1105 		if (!kd_buf)
1106 			return ENOMEM;
1107 		state->kd_buf = kd_buf;
1108 		state->kd_buf_size = kd_buf_size;
1109 	} else
1110 		kd_buf = state->kd_buf;
1111 
1112 	int i;
1113 	for (i = 0; i < kd_buf_size; i++)
1114 		kd_buf[i] = -1;
1115 
1116 	/* The 'k' axis in Myers spans positive and negative indexes, so point
1117 	 * the kd to the middle.
1118 	 * It is then possible to index from -max .. max. */
1119 	int *kd_origin = kd_buf + max;
1120 	int *kd_column = kd_origin;
1121 
1122 	int d;
1123 	int backtrack_d = -1;
1124 	int backtrack_k = 0;
1125 	int k;
1126 	int x, y;
1127 	for (d = 0; d <= max; d++, kd_column += kd_len) {
1128 		debug("-- %s d=%d\n", __func__, d);
1129 
1130 		for (k = d; k >= -d; k -= 2) {
1131 			if (k < -(int)right->atoms.len
1132 			    || k > (int)left->atoms.len) {
1133 				/* This diagonal is completely outside of the
1134 				 * Myers graph, don't calculate it. */
1135 				if (k < -(int)right->atoms.len)
1136 					debug(" %d k <"
1137 					      " -(int)right->atoms.len %d\n",
1138 					      k, -(int)right->atoms.len);
1139 				else
1140 					debug(" %d k > left->atoms.len %d\n", k,
1141 					      left->atoms.len);
1142 				if (k < 0) {
1143 					/* We are traversing negatively, and
1144 					 * already below the entire graph,
1145 					 * nothing will come of this. */
1146 					debug(" break\n");
1147 					break;
1148 				}
1149 				debug(" continue\n");
1150 				continue;
1151 			}
1152 
1153 			if (d == 0) {
1154 				/* This is the initializing step. There is no
1155 				 * prev_k yet, get the initial x from the top
1156 				 * left of the Myers graph. */
1157 				x = 0;
1158 			} else {
1159 				int *kd_prev_column = kd_column - kd_len;
1160 
1161 				/* Favoring "-" lines first means favoring
1162 				 * moving rightwards in the Myers graph.
1163 				 * For this, all k should derive from k - 1,
1164 				 * only the bottom most k derive from k + 1:
1165 				 *
1166 				 *      | d=   0   1   2
1167 				 *  ----+----------------
1168 				 *  k=  |
1169 				 *   2  |             2,0 <-- from
1170 				 *      |            /        prev_k = 2 - 1 = 1
1171 				 *   1  |         1,0
1172 				 *      |        /
1173 				 *   0  |  -->0,0     3,3
1174 				 *      |       \\   /
1175 				 *  -1  |         0,1 <-- bottom most for d=1
1176 				 *      |           \\    from prev_k = -1+1 = 0
1177 				 *  -2  |             0,2 <-- bottom most for
1178 				 *                            d=2 from
1179 				 *                            prev_k = -2+1 = -1
1180 				 *
1181 				 * Except when a k + 1 from a previous run
1182 				 * already means a further advancement in the
1183 				 * graph.
1184 				 * If k == d, there is no k + 1 and k - 1 is the
1185 				 * only option.
1186 				 * If k < d, use k + 1 in case that yields a
1187 				 * larger x. Also use k + 1 if k - 1 is outside
1188 				 * the graph.
1189 				 */
1190 				if (k > -d
1191 				    && (k == d
1192 					|| (k - 1 >= -(int)right->atoms.len
1193 					    && kd_prev_column[k - 1]
1194 					       >= kd_prev_column[k + 1]))) {
1195 					/* Advance from k - 1.
1196 					 * From position prev_k, step to the
1197 					 * right in the Myers graph: x += 1.
1198 					 */
1199 					int prev_k = k - 1;
1200 					int prev_x = kd_prev_column[prev_k];
1201 					x = prev_x + 1;
1202 				} else {
1203 					/* The bottom most one.
1204 					 * From position prev_k, step to the
1205 					 * bottom in the Myers graph: y += 1.
1206 					 * Incrementing y is achieved by
1207 					 * decrementing k while keeping the same
1208 					 * x. (since we're deriving y from y =
1209 					 * x - k).
1210 					 */
1211 					int prev_k = k + 1;
1212 					int prev_x = kd_prev_column[prev_k];
1213 					x = prev_x;
1214 				}
1215 			}
1216 
1217 			/* Slide down any snake that we might find here. */
1218 			while (x < left->atoms.len
1219 			       && xk_to_y(x, k) < right->atoms.len) {
1220 				bool same;
1221 				int r = diff_atom_same(&same,
1222 						       &left->atoms.head[x],
1223 						       &right->atoms.head[
1224 							xk_to_y(x, k)]);
1225 				if (r)
1226 					return r;
1227 				if (!same)
1228 					break;
1229 				x++;
1230 			}
1231 			kd_column[k] = x;
1232 
1233 			if (x == left->atoms.len
1234 			    && xk_to_y(x, k) == right->atoms.len) {
1235 				/* Found a path */
1236 				backtrack_d = d;
1237 				backtrack_k = k;
1238 				debug("Reached the end at d = %d, k = %d\n",
1239 				      backtrack_d, backtrack_k);
1240 				break;
1241 			}
1242 		}
1243 
1244 		if (backtrack_d >= 0)
1245 			break;
1246 	}
1247 
1248 	debug_dump_myers_graph(left, right, kd_origin, NULL, 0, NULL, 0);
1249 
1250 	/* backtrack. A matrix spanning from start to end of the file is ready:
1251 	 *
1252 	 *      | d=   0   1   2   3   4
1253 	 *  ----+---------------------------------
1254 	 *  k=  |
1255 	 *   3  |
1256 	 *      |
1257 	 *   2  |             2,0
1258 	 *      |            /
1259 	 *   1  |         1,0     4,3
1260 	 *      |        /       /   \
1261 	 *   0  |  -->0,0     3,3     4,4 --> backtrack_d = 4, backtrack_k = 0
1262 	 *      |        \   /   \
1263 	 *  -1  |         0,1     3,4
1264 	 *      |            \
1265 	 *  -2  |             0,2
1266 	 *      |
1267 	 *
1268 	 * From (4,4) backwards, find the previous position that is the largest, and remember it.
1269 	 *
1270 	 */
1271 	for (d = backtrack_d, k = backtrack_k; d >= 0; d--) {
1272 		x = kd_column[k];
1273 		y = xk_to_y(x, k);
1274 
1275 		/* When the best position is identified, remember it for that
1276 		 * kd_column.
1277 		 * That kd_column is no longer needed otherwise, so just
1278 		 * re-purpose kd_column[0] = x and kd_column[1] = y,
1279 		 * so that there is no need to allocate more memory.
1280 		 */
1281 		kd_column[0] = x;
1282 		kd_column[1] = y;
1283 		debug("Backtrack d=%d: xy=(%d, %d)\n",
1284 		      d, kd_column[0], kd_column[1]);
1285 
1286 		/* Don't access memory before kd_buf */
1287 		if (d == 0)
1288 			break;
1289 		int *kd_prev_column = kd_column - kd_len;
1290 
1291 		/* When y == 0, backtracking downwards (k-1) is the only way.
1292 		 * When x == 0, backtracking upwards (k+1) is the only way.
1293 		 *
1294 		 *      | d=   0   1   2   3   4
1295 		 *  ----+---------------------------------
1296 		 *  k=  |
1297 		 *   3  |
1298 		 *      |                ..y == 0
1299 		 *   2  |             2,0
1300 		 *      |            /
1301 		 *   1  |         1,0     4,3
1302 		 *      |        /       /   \
1303 		 *   0  |  -->0,0     3,3     4,4 --> backtrack_d = 4,
1304 		 *      |        \   /   \            backtrack_k = 0
1305 		 *  -1  |         0,1     3,4
1306 		 *      |            \
1307 		 *  -2  |             0,2__
1308 		 *      |                  x == 0
1309 		 */
1310 		if (y == 0
1311 		    || (x > 0
1312 			&& kd_prev_column[k - 1] >= kd_prev_column[k + 1])) {
1313 			k = k - 1;
1314 			debug("prev k=k-1=%d x=%d y=%d\n",
1315 			      k, kd_prev_column[k],
1316 			      xk_to_y(kd_prev_column[k], k));
1317 		} else {
1318 			k = k + 1;
1319 			debug("prev k=k+1=%d x=%d y=%d\n",
1320 			      k, kd_prev_column[k],
1321 			      xk_to_y(kd_prev_column[k], k));
1322 		}
1323 		kd_column = kd_prev_column;
1324 	}
1325 
1326 	/* Forwards again, this time recording the diff chunks.
1327 	 * Definitely start from 0,0. kd_column[0] may actually point to the
1328 	 * bottom of a snake starting at 0,0 */
1329 	x = 0;
1330 	y = 0;
1331 
1332 	kd_column = kd_origin;
1333 	for (d = 0; d <= backtrack_d; d++, kd_column += kd_len) {
1334 		int next_x = kd_column[0];
1335 		int next_y = kd_column[1];
1336 		debug("Forward track from xy(%d,%d) to xy(%d,%d)\n",
1337 		      x, y, next_x, next_y);
1338 
1339 		struct diff_atom *left_atom = &left->atoms.head[x];
1340 		int left_section_len = next_x - x;
1341 		struct diff_atom *right_atom = &right->atoms.head[y];
1342 		int right_section_len = next_y - y;
1343 
1344 		rc = ENOMEM;
1345 		if (left_section_len && right_section_len) {
1346 			/* This must be a snake slide.
1347 			 * Snake slides have a straight line leading into them
1348 			 * (except when starting at (0,0)). Find out whether the
1349 			 * lead-in is horizontal or vertical:
1350 			 *
1351 			 *     left
1352 			 *  ---------->
1353 			 *  |
1354 			 * r|   o-o        o
1355 			 * i|      \       |
1356 			 * g|       o      o
1357 			 * h|        \      \
1358 			 * t|         o      o
1359 			 *  v
1360 			 *
1361 			 * If left_section_len > right_section_len, the lead-in
1362 			 * is horizontal, meaning first remove one atom from the
1363 			 * left before sliding down the snake.
1364 			 * If right_section_len > left_section_len, the lead-in
1365 			 * is vetical, so add one atom from the right before
1366 			 * sliding down the snake. */
1367 			if (left_section_len == right_section_len + 1) {
1368 				if (!diff_state_add_chunk(state, true,
1369 							  left_atom, 1,
1370 							  right_atom, 0))
1371 					goto return_rc;
1372 				left_atom++;
1373 				left_section_len--;
1374 			} else if (right_section_len == left_section_len + 1) {
1375 				if (!diff_state_add_chunk(state, true,
1376 							  left_atom, 0,
1377 							  right_atom, 1))
1378 					goto return_rc;
1379 				right_atom++;
1380 				right_section_len--;
1381 			} else if (left_section_len != right_section_len) {
1382 				/* The numbers are making no sense. Should never
1383 				 * happen. */
1384 				rc = DIFF_RC_USE_DIFF_ALGO_FALLBACK;
1385 				goto return_rc;
1386 			}
1387 
1388 			if (!diff_state_add_chunk(state, true,
1389 						  left_atom, left_section_len,
1390 						  right_atom,
1391 						  right_section_len))
1392 				goto return_rc;
1393 		} else if (left_section_len && !right_section_len) {
1394 			/* Only left atoms and none on the right, they form a
1395 			 * "minus" chunk, then. */
1396 			if (!diff_state_add_chunk(state, true,
1397 						  left_atom, left_section_len,
1398 						  right_atom, 0))
1399 				goto return_rc;
1400 		} else if (!left_section_len && right_section_len) {
1401 			/* No left atoms, only atoms on the right, they form a
1402 			 * "plus" chunk, then. */
1403 			if (!diff_state_add_chunk(state, true,
1404 						  left_atom, 0,
1405 						  right_atom,
1406 						  right_section_len))
1407 				goto return_rc;
1408 		}
1409 
1410 		x = next_x;
1411 		y = next_y;
1412 	}
1413 
1414 	rc = DIFF_RC_OK;
1415 
1416 return_rc:
1417 	debug("** END %s rc=%d\n", __func__, rc);
1418 	return rc;
1419 }