forked from github/putty
-
Notifications
You must be signed in to change notification settings - Fork 0
/
Copy pathtree234.c
1638 lines (1508 loc) · 52.9 KB
/
tree234.c
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
541
542
543
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
575
576
577
578
579
580
581
582
583
584
585
586
587
588
589
590
591
592
593
594
595
596
597
598
599
600
601
602
603
604
605
606
607
608
609
610
611
612
613
614
615
616
617
618
619
620
621
622
623
624
625
626
627
628
629
630
631
632
633
634
635
636
637
638
639
640
641
642
643
644
645
646
647
648
649
650
651
652
653
654
655
656
657
658
659
660
661
662
663
664
665
666
667
668
669
670
671
672
673
674
675
676
677
678
679
680
681
682
683
684
685
686
687
688
689
690
691
692
693
694
695
696
697
698
699
700
701
702
703
704
705
706
707
708
709
710
711
712
713
714
715
716
717
718
719
720
721
722
723
724
725
726
727
728
729
730
731
732
733
734
735
736
737
738
739
740
741
742
743
744
745
746
747
748
749
750
751
752
753
754
755
756
757
758
759
760
761
762
763
764
765
766
767
768
769
770
771
772
773
774
775
776
777
778
779
780
781
782
783
784
785
786
787
788
789
790
791
792
793
794
795
796
797
798
799
800
801
802
803
804
805
806
807
808
809
810
811
812
813
814
815
816
817
818
819
820
821
822
823
824
825
826
827
828
829
830
831
832
833
834
835
836
837
838
839
840
841
842
843
844
845
846
847
848
849
850
851
852
853
854
855
856
857
858
859
860
861
862
863
864
865
866
867
868
869
870
871
872
873
874
875
876
877
878
879
880
881
882
883
884
885
886
887
888
889
890
891
892
893
894
895
896
897
898
899
900
901
902
903
904
905
906
907
908
909
910
911
912
913
914
915
916
917
918
919
920
921
922
923
924
925
926
927
928
929
930
931
932
933
934
935
936
937
938
939
940
941
942
943
944
945
946
947
948
949
950
951
952
953
954
955
956
957
958
959
960
961
962
963
964
965
966
967
968
969
970
971
972
973
974
975
976
977
978
979
980
981
982
983
984
985
986
987
988
989
990
991
992
993
994
995
996
997
998
999
1000
/*
* tree234.c: reasonably generic counted 2-3-4 tree routines.
*
* This file is copyright 1999-2001 Simon Tatham.
*
* Permission is hereby granted, free of charge, to any person
* obtaining a copy of this software and associated documentation
* files (the "Software"), to deal in the Software without
* restriction, including without limitation the rights to use,
* copy, modify, merge, publish, distribute, sublicense, and/or
* sell copies of the Software, and to permit persons to whom the
* Software is furnished to do so, subject to the following
* conditions:
*
* The above copyright notice and this permission notice shall be
* included in all copies or substantial portions of the Software.
*
* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
* EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES
* OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND
* NONINFRINGEMENT. IN NO EVENT SHALL SIMON TATHAM BE LIABLE FOR
* ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF
* CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN
* CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
* SOFTWARE.
*/
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include "defs.h"
#include "tree234.h"
#include "puttymem.h"
#ifdef TEST
static int verbose = 0;
#define LOG(x) do \
{ \
if (verbose > 2) \
printf x; \
} while (0)
#else
#define LOG(x)
#endif
typedef struct node234_Tag node234;
struct tree234_Tag {
node234 *root;
cmpfn234 cmp;
};
struct node234_Tag {
node234 *parent;
node234 *kids[4];
int counts[4];
void *elems[3];
};
/*
* Create a 2-3-4 tree.
*/
tree234 *newtree234(cmpfn234 cmp)
{
tree234 *ret = snew(tree234);
LOG(("created tree %p\n", ret));
ret->root = NULL;
ret->cmp = cmp;
return ret;
}
/*
* Free a 2-3-4 tree (not including freeing the elements).
*/
static void freenode234(node234 * n)
{
if (!n)
return;
freenode234(n->kids[0]);
freenode234(n->kids[1]);
freenode234(n->kids[2]);
freenode234(n->kids[3]);
sfree(n);
}
void freetree234(tree234 * t)
{
freenode234(t->root);
sfree(t);
}
/*
* Internal function to count a node.
*/
static int countnode234(node234 * n)
{
int count = 0;
int i;
if (!n)
return 0;
for (i = 0; i < 4; i++)
count += n->counts[i];
for (i = 0; i < 3; i++)
if (n->elems[i])
count++;
return count;
}
/*
* Internal function to return the number of elements in a node.
*/
static int elements234(node234 *n)
{
int i;
for (i = 0; i < 3; i++)
if (!n->elems[i])
break;
return i;
}
/*
* Count the elements in a tree.
*/
int count234(tree234 * t)
{
if (t->root)
return countnode234(t->root);
else
return 0;
}
/*
* Add an element e to a 2-3-4 tree t. Returns e on success, or if
* an existing element compares equal, returns that.
*/
static void *add234_internal(tree234 * t, void *e, int index)
{
node234 *n, **np, *left, *right;
void *orig_e = e;
int c, lcount, rcount;
LOG(("adding node %p to tree %p\n", e, t));
if (t->root == NULL) {
t->root = snew(node234);
t->root->elems[1] = t->root->elems[2] = NULL;
t->root->kids[0] = t->root->kids[1] = NULL;
t->root->kids[2] = t->root->kids[3] = NULL;
t->root->counts[0] = t->root->counts[1] = 0;
t->root->counts[2] = t->root->counts[3] = 0;
t->root->parent = NULL;
t->root->elems[0] = e;
LOG((" created root %p\n", t->root));
return orig_e;
}
n = NULL; /* placate gcc; will always be set below since t->root != NULL */
np = &t->root;
while (*np) {
int childnum;
n = *np;
LOG((" node %p: %p/%d [%p] %p/%d [%p] %p/%d [%p] %p/%d\n",
n,
n->kids[0], n->counts[0], n->elems[0],
n->kids[1], n->counts[1], n->elems[1],
n->kids[2], n->counts[2], n->elems[2],
n->kids[3], n->counts[3]));
if (index >= 0) {
if (!n->kids[0]) {
/*
* Leaf node. We want to insert at kid position
* equal to the index:
*
* 0 A 1 B 2 C 3
*/
childnum = index;
} else {
/*
* Internal node. We always descend through it (add
* always starts at the bottom, never in the
* middle).
*/
do { /* this is a do ... while (0) to allow `break' */
if (index <= n->counts[0]) {
childnum = 0;
break;
}
index -= n->counts[0] + 1;
if (index <= n->counts[1]) {
childnum = 1;
break;
}
index -= n->counts[1] + 1;
if (index <= n->counts[2]) {
childnum = 2;
break;
}
index -= n->counts[2] + 1;
if (index <= n->counts[3]) {
childnum = 3;
break;
}
return NULL; /* error: index out of range */
} while (0);
}
} else {
if ((c = t->cmp(e, n->elems[0])) < 0)
childnum = 0;
else if (c == 0)
return n->elems[0]; /* already exists */
else if (n->elems[1] == NULL
|| (c = t->cmp(e, n->elems[1])) < 0) childnum = 1;
else if (c == 0)
return n->elems[1]; /* already exists */
else if (n->elems[2] == NULL
|| (c = t->cmp(e, n->elems[2])) < 0) childnum = 2;
else if (c == 0)
return n->elems[2]; /* already exists */
else
childnum = 3;
}
np = &n->kids[childnum];
LOG((" moving to child %d (%p)\n", childnum, *np));
}
/*
* We need to insert the new element in n at position np.
*/
left = NULL;
lcount = 0;
right = NULL;
rcount = 0;
while (n) {
LOG((" at %p: %p/%d [%p] %p/%d [%p] %p/%d [%p] %p/%d\n",
n,
n->kids[0], n->counts[0], n->elems[0],
n->kids[1], n->counts[1], n->elems[1],
n->kids[2], n->counts[2], n->elems[2],
n->kids[3], n->counts[3]));
LOG((" need to insert %p/%d [%p] %p/%d at position %d\n",
left, lcount, e, right, rcount, (int)(np - n->kids)));
if (n->elems[1] == NULL) {
/*
* Insert in a 2-node; simple.
*/
if (np == &n->kids[0]) {
LOG((" inserting on left of 2-node\n"));
n->kids[2] = n->kids[1];
n->counts[2] = n->counts[1];
n->elems[1] = n->elems[0];
n->kids[1] = right;
n->counts[1] = rcount;
n->elems[0] = e;
n->kids[0] = left;
n->counts[0] = lcount;
} else { /* np == &n->kids[1] */
LOG((" inserting on right of 2-node\n"));
n->kids[2] = right;
n->counts[2] = rcount;
n->elems[1] = e;
n->kids[1] = left;
n->counts[1] = lcount;
}
if (n->kids[0])
n->kids[0]->parent = n;
if (n->kids[1])
n->kids[1]->parent = n;
if (n->kids[2])
n->kids[2]->parent = n;
LOG((" done\n"));
break;
} else if (n->elems[2] == NULL) {
/*
* Insert in a 3-node; simple.
*/
if (np == &n->kids[0]) {
LOG((" inserting on left of 3-node\n"));
n->kids[3] = n->kids[2];
n->counts[3] = n->counts[2];
n->elems[2] = n->elems[1];
n->kids[2] = n->kids[1];
n->counts[2] = n->counts[1];
n->elems[1] = n->elems[0];
n->kids[1] = right;
n->counts[1] = rcount;
n->elems[0] = e;
n->kids[0] = left;
n->counts[0] = lcount;
} else if (np == &n->kids[1]) {
LOG((" inserting in middle of 3-node\n"));
n->kids[3] = n->kids[2];
n->counts[3] = n->counts[2];
n->elems[2] = n->elems[1];
n->kids[2] = right;
n->counts[2] = rcount;
n->elems[1] = e;
n->kids[1] = left;
n->counts[1] = lcount;
} else { /* np == &n->kids[2] */
LOG((" inserting on right of 3-node\n"));
n->kids[3] = right;
n->counts[3] = rcount;
n->elems[2] = e;
n->kids[2] = left;
n->counts[2] = lcount;
}
if (n->kids[0])
n->kids[0]->parent = n;
if (n->kids[1])
n->kids[1]->parent = n;
if (n->kids[2])
n->kids[2]->parent = n;
if (n->kids[3])
n->kids[3]->parent = n;
LOG((" done\n"));
break;
} else {
node234 *m = snew(node234);
m->parent = n->parent;
LOG((" splitting a 4-node; created new node %p\n", m));
/*
* Insert in a 4-node; split into a 2-node and a
* 3-node, and move focus up a level.
*
* I don't think it matters which way round we put the
* 2 and the 3. For simplicity, we'll put the 3 first
* always.
*/
if (np == &n->kids[0]) {
m->kids[0] = left;
m->counts[0] = lcount;
m->elems[0] = e;
m->kids[1] = right;
m->counts[1] = rcount;
m->elems[1] = n->elems[0];
m->kids[2] = n->kids[1];
m->counts[2] = n->counts[1];
e = n->elems[1];
n->kids[0] = n->kids[2];
n->counts[0] = n->counts[2];
n->elems[0] = n->elems[2];
n->kids[1] = n->kids[3];
n->counts[1] = n->counts[3];
} else if (np == &n->kids[1]) {
m->kids[0] = n->kids[0];
m->counts[0] = n->counts[0];
m->elems[0] = n->elems[0];
m->kids[1] = left;
m->counts[1] = lcount;
m->elems[1] = e;
m->kids[2] = right;
m->counts[2] = rcount;
e = n->elems[1];
n->kids[0] = n->kids[2];
n->counts[0] = n->counts[2];
n->elems[0] = n->elems[2];
n->kids[1] = n->kids[3];
n->counts[1] = n->counts[3];
} else if (np == &n->kids[2]) {
m->kids[0] = n->kids[0];
m->counts[0] = n->counts[0];
m->elems[0] = n->elems[0];
m->kids[1] = n->kids[1];
m->counts[1] = n->counts[1];
m->elems[1] = n->elems[1];
m->kids[2] = left;
m->counts[2] = lcount;
/* e = e; */
n->kids[0] = right;
n->counts[0] = rcount;
n->elems[0] = n->elems[2];
n->kids[1] = n->kids[3];
n->counts[1] = n->counts[3];
} else { /* np == &n->kids[3] */
m->kids[0] = n->kids[0];
m->counts[0] = n->counts[0];
m->elems[0] = n->elems[0];
m->kids[1] = n->kids[1];
m->counts[1] = n->counts[1];
m->elems[1] = n->elems[1];
m->kids[2] = n->kids[2];
m->counts[2] = n->counts[2];
n->kids[0] = left;
n->counts[0] = lcount;
n->elems[0] = e;
n->kids[1] = right;
n->counts[1] = rcount;
e = n->elems[2];
}
m->kids[3] = n->kids[3] = n->kids[2] = NULL;
m->counts[3] = n->counts[3] = n->counts[2] = 0;
m->elems[2] = n->elems[2] = n->elems[1] = NULL;
if (m->kids[0])
m->kids[0]->parent = m;
if (m->kids[1])
m->kids[1]->parent = m;
if (m->kids[2])
m->kids[2]->parent = m;
if (n->kids[0])
n->kids[0]->parent = n;
if (n->kids[1])
n->kids[1]->parent = n;
LOG((" left (%p): %p/%d [%p] %p/%d [%p] %p/%d\n", m,
m->kids[0], m->counts[0], m->elems[0],
m->kids[1], m->counts[1], m->elems[1],
m->kids[2], m->counts[2]));
LOG((" right (%p): %p/%d [%p] %p/%d\n", n,
n->kids[0], n->counts[0], n->elems[0],
n->kids[1], n->counts[1]));
left = m;
lcount = countnode234(left);
right = n;
rcount = countnode234(right);
}
if (n->parent)
np = (n->parent->kids[0] == n ? &n->parent->kids[0] :
n->parent->kids[1] == n ? &n->parent->kids[1] :
n->parent->kids[2] == n ? &n->parent->kids[2] :
&n->parent->kids[3]);
n = n->parent;
}
/*
* If we've come out of here by `break', n will still be
* non-NULL and all we need to do is go back up the tree
* updating counts. If we've come here because n is NULL, we
* need to create a new root for the tree because the old one
* has just split into two. */
if (n) {
while (n->parent) {
int count = countnode234(n);
int childnum;
childnum = (n->parent->kids[0] == n ? 0 :
n->parent->kids[1] == n ? 1 :
n->parent->kids[2] == n ? 2 : 3);
n->parent->counts[childnum] = count;
n = n->parent;
}
} else {
LOG((" root is overloaded, split into two\n"));
t->root = snew(node234);
t->root->kids[0] = left;
t->root->counts[0] = lcount;
t->root->elems[0] = e;
t->root->kids[1] = right;
t->root->counts[1] = rcount;
t->root->elems[1] = NULL;
t->root->kids[2] = NULL;
t->root->counts[2] = 0;
t->root->elems[2] = NULL;
t->root->kids[3] = NULL;
t->root->counts[3] = 0;
t->root->parent = NULL;
if (t->root->kids[0])
t->root->kids[0]->parent = t->root;
if (t->root->kids[1])
t->root->kids[1]->parent = t->root;
LOG((" new root is %p/%d [%p] %p/%d\n",
t->root->kids[0], t->root->counts[0],
t->root->elems[0], t->root->kids[1], t->root->counts[1]));
}
return orig_e;
}
void *add234(tree234 * t, void *e)
{
if (!t->cmp) /* tree is unsorted */
return NULL;
return add234_internal(t, e, -1);
}
void *addpos234(tree234 * t, void *e, int index)
{
if (index < 0 || /* index out of range */
t->cmp) /* tree is sorted */
return NULL; /* return failure */
return add234_internal(t, e, index); /* this checks the upper bound */
}
/*
* Look up the element at a given numeric index in a 2-3-4 tree.
* Returns NULL if the index is out of range.
*/
void *index234(tree234 * t, int index)
{
node234 *n;
if (!t->root)
return NULL; /* tree is empty */
if (index < 0 || index >= countnode234(t->root))
return NULL; /* out of range */
n = t->root;
while (n) {
if (index < n->counts[0])
n = n->kids[0];
else if (index -= n->counts[0] + 1, index < 0)
return n->elems[0];
else if (index < n->counts[1])
n = n->kids[1];
else if (index -= n->counts[1] + 1, index < 0)
return n->elems[1];
else if (index < n->counts[2])
n = n->kids[2];
else if (index -= n->counts[2] + 1, index < 0)
return n->elems[2];
else
n = n->kids[3];
}
/* We shouldn't ever get here. I wonder how we did. */
return NULL;
}
/*
* Find an element e in a sorted 2-3-4 tree t. Returns NULL if not
* found. e is always passed as the first argument to cmp, so cmp
* can be an asymmetric function if desired. cmp can also be passed
* as NULL, in which case the compare function from the tree proper
* will be used.
*/
void *findrelpos234(tree234 * t, void *e, cmpfn234 cmp,
int relation, int *index)
{
search234_state ss;
int reldir = (relation == REL234_LT || relation == REL234_LE ? -1 :
relation == REL234_GT || relation == REL234_GE ? +1 : 0);
bool equal_permitted = (relation != REL234_LT && relation != REL234_GT);
void *toret;
/* Only LT / GT relations are permitted with a null query element. */
assert(!(equal_permitted && !e));
if (cmp == NULL)
cmp = t->cmp;
search234_start(&ss, t);
while (ss.element) {
int cmpret;
if (e) {
cmpret = cmp(e, ss.element);
} else {
cmpret = -reldir; /* invent a fixed compare result */
}
if (cmpret == 0) {
/*
* We've found an element that compares exactly equal to
* the query element.
*/
if (equal_permitted) {
/* If our search relation permits equality, we've
* finished already. */
if (index)
*index = ss.index;
return ss.element;
} else {
/* Otherwise, pretend this element was slightly too
* big/small, according to the direction of search. */
cmpret = reldir;
}
}
search234_step(&ss, cmpret);
}
/*
* No element compares equal to the one we were after, but
* ss.index indicates the index that element would have if it were
* inserted.
*
* So if our search relation is EQ, we must simply return failure.
*/
if (relation == REL234_EQ)
return NULL;
/*
* Otherwise, we must do an index lookup for the previous index
* (if we're going left - LE or LT) or this index (if we're going
* right - GE or GT).
*/
if (relation == REL234_LT || relation == REL234_LE) {
ss.index--;
}
/*
* We know the index of the element we want; just call index234
* to do the rest. This will return NULL if the index is out of
* bounds, which is exactly what we want.
*/
toret = index234(t, ss.index);
if (toret && index)
*index = ss.index;
return toret;
}
void *find234(tree234 * t, void *e, cmpfn234 cmp)
{
return findrelpos234(t, e, cmp, REL234_EQ, NULL);
}
void *findrel234(tree234 * t, void *e, cmpfn234 cmp, int relation)
{
return findrelpos234(t, e, cmp, relation, NULL);
}
void *findpos234(tree234 * t, void *e, cmpfn234 cmp, int *index)
{
return findrelpos234(t, e, cmp, REL234_EQ, index);
}
void search234_start(search234_state *state, tree234 *t)
{
state->_node = t->root;
state->_base = 0; /* index of first element in this node's subtree */
state->_last = -1; /* indicate that this node is not previously visted */
search234_step(state, 0);
}
void search234_step(search234_state *state, int direction)
{
node234 *node = state->_node;
int i;
if (!node) {
state->element = NULL;
state->index = 0;
return;
}
if (state->_last != -1) {
/*
* We're already pointing at some element of a node, so we
* should restrict to the elements left or right of it,
* depending on the requested search direction.
*/
assert(direction);
assert(node);
if (direction > 0)
state->_lo = state->_last + 1;
else
state->_hi = state->_last - 1;
if (state->_lo > state->_hi) {
/*
* We've run out of elements in this node, i.e. we've
* narrowed to nothing but a child pointer. Descend to
* that child, and update _base to the leftmost index of
* its subtree.
*/
for (i = 0; i < state->_lo; i++)
state->_base += 1 + node->counts[i];
state->_node = node = node->kids[state->_lo];
state->_last = -1;
}
}
if (state->_last == -1) {
/*
* We've just entered a new node - either because of the above
* code, or because we were called from search234_start - and
* anything in that node is a viable answer.
*/
state->_lo = 0;
state->_hi = node ? elements234(node)-1 : 0;
}
/*
* Now we've got something we can return.
*/
if (!node) {
state->element = NULL;
state->index = state->_base;
} else {
state->_last = (state->_lo + state->_hi) / 2;
state->element = node->elems[state->_last];
state->index = state->_base + state->_last;
for (i = 0; i <= state->_last; i++)
state->index += node->counts[i];
}
}
/*
* Delete an element e in a 2-3-4 tree. Does not free the element,
* merely removes all links to it from the tree nodes.
*/
static void *delpos234_internal(tree234 * t, int index)
{
node234 *n;
void *retval;
int ei = -1;
retval = 0;
n = t->root;
LOG(("deleting item %d from tree %p\n", index, t));
while (1) {
while (n) {
int ki;
node234 *sub;
LOG(
(" node %p: %p/%d [%p] %p/%d [%p] %p/%d [%p] %p/%d index=%d\n",
n, n->kids[0], n->counts[0], n->elems[0], n->kids[1],
n->counts[1], n->elems[1], n->kids[2], n->counts[2],
n->elems[2], n->kids[3], n->counts[3], index));
if (index < n->counts[0]) {
ki = 0;
} else if (index -= n->counts[0] + 1, index < 0) {
ei = 0;
break;
} else if (index < n->counts[1]) {
ki = 1;
} else if (index -= n->counts[1] + 1, index < 0) {
ei = 1;
break;
} else if (index < n->counts[2]) {
ki = 2;
} else if (index -= n->counts[2] + 1, index < 0) {
ei = 2;
break;
} else {
ki = 3;
}
/*
* Recurse down to subtree ki. If it has only one element,
* we have to do some transformation to start with.
*/
LOG((" moving to subtree %d\n", ki));
sub = n->kids[ki];
if (!sub->elems[1]) {
LOG((" subtree has only one element!\n"));
if (ki > 0 && n->kids[ki - 1]->elems[1]) {
/*
* Case 3a, left-handed variant. Child ki has
* only one element, but child ki-1 has two or
* more. So we need to move a subtree from ki-1
* to ki.
*
* . C . . B .
* / \ -> / \
* [more] a A b B c d D e [more] a A b c C d D e
*/
node234 *sib = n->kids[ki - 1];
int lastelem = (sib->elems[2] ? 2 :
sib->elems[1] ? 1 : 0);
sub->kids[2] = sub->kids[1];
sub->counts[2] = sub->counts[1];
sub->elems[1] = sub->elems[0];
sub->kids[1] = sub->kids[0];
sub->counts[1] = sub->counts[0];
sub->elems[0] = n->elems[ki - 1];
sub->kids[0] = sib->kids[lastelem + 1];
sub->counts[0] = sib->counts[lastelem + 1];
if (sub->kids[0])
sub->kids[0]->parent = sub;
n->elems[ki - 1] = sib->elems[lastelem];
sib->kids[lastelem + 1] = NULL;
sib->counts[lastelem + 1] = 0;
sib->elems[lastelem] = NULL;
n->counts[ki] = countnode234(sub);
LOG((" case 3a left\n"));
LOG(
(" index and left subtree count before adjustment: %d, %d\n",
index, n->counts[ki - 1]));
index += n->counts[ki - 1];
n->counts[ki - 1] = countnode234(sib);
index -= n->counts[ki - 1];
LOG(
(" index and left subtree count after adjustment: %d, %d\n",
index, n->counts[ki - 1]));
} else if (ki < 3 && n->kids[ki + 1]
&& n->kids[ki + 1]->elems[1]) {
/*
* Case 3a, right-handed variant. ki has only
* one element but ki+1 has two or more. Move a
* subtree from ki+1 to ki.
*
* . B . . C .
* / \ -> / \
* a A b c C d D e [more] a A b B c d D e [more]
*/
node234 *sib = n->kids[ki + 1];
int j;
sub->elems[1] = n->elems[ki];
sub->kids[2] = sib->kids[0];
sub->counts[2] = sib->counts[0];
if (sub->kids[2])
sub->kids[2]->parent = sub;
n->elems[ki] = sib->elems[0];
sib->kids[0] = sib->kids[1];
sib->counts[0] = sib->counts[1];
for (j = 0; j < 2 && sib->elems[j + 1]; j++) {
sib->kids[j + 1] = sib->kids[j + 2];
sib->counts[j + 1] = sib->counts[j + 2];
sib->elems[j] = sib->elems[j + 1];
}
sib->kids[j + 1] = NULL;
sib->counts[j + 1] = 0;
sib->elems[j] = NULL;
n->counts[ki] = countnode234(sub);
n->counts[ki + 1] = countnode234(sib);
LOG((" case 3a right\n"));
} else {
/*
* Case 3b. ki has only one element, and has no
* neighbour with more than one. So pick a
* neighbour and merge it with ki, taking an
* element down from n to go in the middle.
*
* . B . .
* / \ -> |
* a A b c C d a A b B c C d
*
* (Since at all points we have avoided
* descending to a node with only one element,
* we can be sure that n is not reduced to
* nothingness by this move, _unless_ it was
* the very first node, ie the root of the
* tree. In that case we remove the now-empty
* root and replace it with its single large
* child as shown.)
*/
node234 *sib;
int j;
if (ki > 0) {
ki--;
index += n->counts[ki] + 1;
}
sib = n->kids[ki];
sub = n->kids[ki + 1];
sub->kids[3] = sub->kids[1];
sub->counts[3] = sub->counts[1];
sub->elems[2] = sub->elems[0];
sub->kids[2] = sub->kids[0];
sub->counts[2] = sub->counts[0];
sub->elems[1] = n->elems[ki];
sub->kids[1] = sib->kids[1];
sub->counts[1] = sib->counts[1];
if (sub->kids[1])
sub->kids[1]->parent = sub;
sub->elems[0] = sib->elems[0];
sub->kids[0] = sib->kids[0];
sub->counts[0] = sib->counts[0];
if (sub->kids[0])
sub->kids[0]->parent = sub;
n->counts[ki + 1] = countnode234(sub);
sfree(sib);
/*
* That's built the big node in sub. Now we
* need to remove the reference to sib in n.
*/
for (j = ki; j < 3 && n->kids[j + 1]; j++) {
n->kids[j] = n->kids[j + 1];
n->counts[j] = n->counts[j + 1];
n->elems[j] = j < 2 ? n->elems[j + 1] : NULL;
}
n->kids[j] = NULL;
n->counts[j] = 0;
if (j < 3)
n->elems[j] = NULL;
LOG((" case 3b ki=%d\n", ki));
if (!n->elems[0]) {
/*
* The root is empty and needs to be
* removed.
*/
LOG((" shifting root!\n"));
t->root = sub;
sub->parent = NULL;
sfree(n);
}
}
}
n = sub;
}
if (!retval)
retval = n->elems[ei];
if (ei == -1)
return NULL; /* although this shouldn't happen */
/*
* Treat special case: this is the one remaining item in
* the tree. n is the tree root (no parent), has one
* element (no elems[1]), and has no kids (no kids[0]).
*/
if (!n->parent && !n->elems[1] && !n->kids[0]) {
LOG((" removed last element in tree\n"));
sfree(n);
t->root = NULL;
return retval;
}
/*
* Now we have the element we want, as n->elems[ei], and we
* have also arranged for that element not to be the only
* one in its node. So...
*/
if (!n->kids[0] && n->elems[1]) {
/*
* Case 1. n is a leaf node with more than one element,
* so it's _really easy_. Just delete the thing and
* we're done.
*/
int i;
LOG((" case 1\n"));
for (i = ei; i < 2 && n->elems[i + 1]; i++)
n->elems[i] = n->elems[i + 1];
n->elems[i] = NULL;
/*
* Having done that to the leaf node, we now go back up
* the tree fixing the counts.
*/
while (n->parent) {
int childnum;
childnum = (n->parent->kids[0] == n ? 0 :
n->parent->kids[1] == n ? 1 :
n->parent->kids[2] == n ? 2 : 3);
n->parent->counts[childnum]--;
n = n->parent;
}
return retval; /* finished! */
} else if (n->kids[ei]->elems[1]) {
/*
* Case 2a. n is an internal node, and the root of the
* subtree to the left of e has more than one element.
* So find the predecessor p to e (ie the largest node
* in that subtree), place it where e currently is, and
* then start the deletion process over again on the
* subtree with p as target.
*/
node234 *m = n->kids[ei];
void *target;
LOG((" case 2a\n"));
while (m->kids[0]) {
m = (m->kids[3] ? m->kids[3] :
m->kids[2] ? m->kids[2] :
m->kids[1] ? m->kids[1] : m->kids[0]);
}
target = (m->elems[2] ? m->elems[2] :
m->elems[1] ? m->elems[1] : m->elems[0]);
n->elems[ei] = target;
index = n->counts[ei] - 1;
n = n->kids[ei];
} else if (n->kids[ei + 1]->elems[1]) {
/*
* Case 2b, symmetric to 2a but s/left/right/ and
* s/predecessor/successor/. (And s/largest/smallest/).
*/
node234 *m = n->kids[ei + 1];
void *target;
LOG((" case 2b\n"));
while (m->kids[0]) {
m = m->kids[0];
}
target = m->elems[0];
n->elems[ei] = target;
n = n->kids[ei + 1];
index = 0;
} else {
/*
* Case 2c. n is an internal node, and the subtrees to
* the left and right of e both have only one element.
* So combine the two subnodes into a single big node
* with their own elements on the left and right and e
* in the middle, then restart the deletion process on
* that subtree, with e still as target.
*/
node234 *a = n->kids[ei], *b = n->kids[ei + 1];
int j;
LOG((" case 2c\n"));
a->elems[1] = n->elems[ei];
a->kids[2] = b->kids[0];
a->counts[2] = b->counts[0];
if (a->kids[2])
a->kids[2]->parent = a;
a->elems[2] = b->elems[0];
a->kids[3] = b->kids[1];
a->counts[3] = b->counts[1];
if (a->kids[3])
a->kids[3]->parent = a;
sfree(b);
n->counts[ei] = countnode234(a);
/*
* That's built the big node in a, and destroyed b. Now
* remove the reference to b (and e) in n.
*/
for (j = ei; j < 2 && n->elems[j + 1]; j++) {
n->elems[j] = n->elems[j + 1];