-
Notifications
You must be signed in to change notification settings - Fork 0
/
Copy pathsshbn.c
2178 lines (1921 loc) · 59.6 KB
/
sshbn.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
/*
* Bignum routines for RSA and DH and stuff.
*/
#include <stdio.h>
#include <assert.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <ctype.h>
#include "misc.h"
#include "sshbn.h"
#define BIGNUM_INTERNAL
typedef BignumInt *Bignum;
#include "ssh.h"
#include "marshal.h"
BignumInt bnZero[1] = { 0 };
BignumInt bnOne[2] = { 1, 1 };
BignumInt bnTen[2] = { 1, 10 };
/*
* The Bignum format is an array of `BignumInt'. The first
* element of the array counts the remaining elements. The
* remaining elements express the actual number, base 2^BIGNUM_INT_BITS, _least_
* significant digit first. (So it's trivial to extract the bit
* with value 2^n for any n.)
*
* All Bignums in this module are positive. Negative numbers must
* be dealt with outside it.
*
* INVARIANT: the most significant word of any Bignum must be
* nonzero.
*/
Bignum Zero = bnZero, One = bnOne, Ten = bnTen;
static Bignum newbn(int length)
{
Bignum b;
assert(length >= 0 && length < INT_MAX / BIGNUM_INT_BITS);
b = snewn(length + 1, BignumInt);
memset(b, 0, (length + 1) * sizeof(*b));
b[0] = length;
return b;
}
void bn_restore_invariant(Bignum b)
{
while (b[0] > 1 && b[b[0]] == 0)
b[0]--;
}
Bignum copybn(Bignum orig)
{
Bignum b = snewn(orig[0] + 1, BignumInt);
if (!b)
abort(); /* FIXME */
memcpy(b, orig, (orig[0] + 1) * sizeof(*b));
return b;
}
void freebn(Bignum b)
{
/*
* Burn the evidence, just in case.
*/
smemclr(b, sizeof(b[0]) * (b[0] + 1));
sfree(b);
}
Bignum bn_power_2(int n)
{
Bignum ret;
assert(n >= 0);
ret = newbn(n / BIGNUM_INT_BITS + 1);
bignum_set_bit(ret, n, 1);
return ret;
}
/*
* Internal addition. Sets c = a - b, where 'a', 'b' and 'c' are all
* big-endian arrays of 'len' BignumInts. Returns the carry off the
* top.
*/
static BignumCarry internal_add(const BignumInt *a, const BignumInt *b,
BignumInt *c, int len)
{
int i;
BignumCarry carry = 0;
for (i = len-1; i >= 0; i--)
BignumADC(c[i], carry, a[i], b[i], carry);
return (BignumInt)carry;
}
/*
* Internal subtraction. Sets c = a - b, where 'a', 'b' and 'c' are
* all big-endian arrays of 'len' BignumInts. Any borrow from the top
* is ignored.
*/
static void internal_sub(const BignumInt *a, const BignumInt *b,
BignumInt *c, int len)
{
int i;
BignumCarry carry = 1;
for (i = len-1; i >= 0; i--)
BignumADC(c[i], carry, a[i], ~b[i], carry);
}
/*
* Compute c = a * b.
* Input is in the first len words of a and b.
* Result is returned in the first 2*len words of c.
*
* 'scratch' must point to an array of BignumInt of size at least
* mul_compute_scratch(len). (This covers the needs of internal_mul
* and all its recursive calls to itself.)
*/
#define KARATSUBA_THRESHOLD 50
static int mul_compute_scratch(int len)
{
int ret = 0;
while (len > KARATSUBA_THRESHOLD) {
int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
int midlen = botlen + 1;
ret += 4*midlen;
len = midlen;
}
return ret;
}
static void internal_mul(const BignumInt *a, const BignumInt *b,
BignumInt *c, int len, BignumInt *scratch)
{
if (len > KARATSUBA_THRESHOLD) {
int i;
/*
* Karatsuba divide-and-conquer algorithm. Cut each input in
* half, so that it's expressed as two big 'digits' in a giant
* base D:
*
* a = a_1 D + a_0
* b = b_1 D + b_0
*
* Then the product is of course
*
* ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
*
* and we compute the three coefficients by recursively
* calling ourself to do half-length multiplications.
*
* The clever bit that makes this worth doing is that we only
* need _one_ half-length multiplication for the central
* coefficient rather than the two that it obviouly looks
* like, because we can use a single multiplication to compute
*
* (a_1 + a_0) (b_1 + b_0) = a_1 b_1 + a_1 b_0 + a_0 b_1 + a_0 b_0
*
* and then we subtract the other two coefficients (a_1 b_1
* and a_0 b_0) which we were computing anyway.
*
* Hence we get to multiply two numbers of length N in about
* three times as much work as it takes to multiply numbers of
* length N/2, which is obviously better than the four times
* as much work it would take if we just did a long
* conventional multiply.
*/
int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
int midlen = botlen + 1;
BignumCarry carry;
#ifdef KARA_DEBUG
int i;
#endif
/*
* The coefficients a_1 b_1 and a_0 b_0 just avoid overlapping
* in the output array, so we can compute them immediately in
* place.
*/
#ifdef KARA_DEBUG
printf("a1,a0 = 0x");
for (i = 0; i < len; i++) {
if (i == toplen) printf(", 0x");
printf("%0*x", BIGNUM_INT_BITS/4, a[i]);
}
printf("\n");
printf("b1,b0 = 0x");
for (i = 0; i < len; i++) {
if (i == toplen) printf(", 0x");
printf("%0*x", BIGNUM_INT_BITS/4, b[i]);
}
printf("\n");
#endif
/* a_1 b_1 */
internal_mul(a, b, c, toplen, scratch);
#ifdef KARA_DEBUG
printf("a1b1 = 0x");
for (i = 0; i < 2*toplen; i++) {
printf("%0*x", BIGNUM_INT_BITS/4, c[i]);
}
printf("\n");
#endif
/* a_0 b_0 */
internal_mul(a + toplen, b + toplen, c + 2*toplen, botlen, scratch);
#ifdef KARA_DEBUG
printf("a0b0 = 0x");
for (i = 0; i < 2*botlen; i++) {
printf("%0*x", BIGNUM_INT_BITS/4, c[2*toplen+i]);
}
printf("\n");
#endif
/* Zero padding. midlen exceeds toplen by at most 2, so just
* zero the first two words of each input and the rest will be
* copied over. */
scratch[0] = scratch[1] = scratch[midlen] = scratch[midlen+1] = 0;
for (i = 0; i < toplen; i++) {
scratch[midlen - toplen + i] = a[i]; /* a_1 */
scratch[2*midlen - toplen + i] = b[i]; /* b_1 */
}
/* compute a_1 + a_0 */
scratch[0] = internal_add(scratch+1, a+toplen, scratch+1, botlen);
#ifdef KARA_DEBUG
printf("a1plusa0 = 0x");
for (i = 0; i < midlen; i++) {
printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);
}
printf("\n");
#endif
/* compute b_1 + b_0 */
scratch[midlen] = internal_add(scratch+midlen+1, b+toplen,
scratch+midlen+1, botlen);
#ifdef KARA_DEBUG
printf("b1plusb0 = 0x");
for (i = 0; i < midlen; i++) {
printf("%0*x", BIGNUM_INT_BITS/4, scratch[midlen+i]);
}
printf("\n");
#endif
/*
* Now we can do the third multiplication.
*/
internal_mul(scratch, scratch + midlen, scratch + 2*midlen, midlen,
scratch + 4*midlen);
#ifdef KARA_DEBUG
printf("a1plusa0timesb1plusb0 = 0x");
for (i = 0; i < 2*midlen; i++) {
printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);
}
printf("\n");
#endif
/*
* Now we can reuse the first half of 'scratch' to compute the
* sum of the outer two coefficients, to subtract from that
* product to obtain the middle one.
*/
scratch[0] = scratch[1] = scratch[2] = scratch[3] = 0;
for (i = 0; i < 2*toplen; i++)
scratch[2*midlen - 2*toplen + i] = c[i];
scratch[1] = internal_add(scratch+2, c + 2*toplen,
scratch+2, 2*botlen);
#ifdef KARA_DEBUG
printf("a1b1plusa0b0 = 0x");
for (i = 0; i < 2*midlen; i++) {
printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);
}
printf("\n");
#endif
internal_sub(scratch + 2*midlen, scratch,
scratch + 2*midlen, 2*midlen);
#ifdef KARA_DEBUG
printf("a1b0plusa0b1 = 0x");
for (i = 0; i < 2*midlen; i++) {
printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);
}
printf("\n");
#endif
/*
* And now all we need to do is to add that middle coefficient
* back into the output. We may have to propagate a carry
* further up the output, but we can be sure it won't
* propagate right the way off the top.
*/
carry = internal_add(c + 2*len - botlen - 2*midlen,
scratch + 2*midlen,
c + 2*len - botlen - 2*midlen, 2*midlen);
i = 2*len - botlen - 2*midlen - 1;
while (carry) {
assert(i >= 0);
BignumADC(c[i], carry, c[i], 0, carry);
i--;
}
#ifdef KARA_DEBUG
printf("ab = 0x");
for (i = 0; i < 2*len; i++) {
printf("%0*x", BIGNUM_INT_BITS/4, c[i]);
}
printf("\n");
#endif
} else {
int i;
BignumInt carry;
const BignumInt *ap, *bp;
BignumInt *cp, *cps;
/*
* Multiply in the ordinary O(N^2) way.
*/
for (i = 0; i < 2 * len; i++)
c[i] = 0;
for (cps = c + 2*len, ap = a + len; ap-- > a; cps--) {
carry = 0;
for (cp = cps, bp = b + len; cp--, bp-- > b ;)
BignumMULADD2(carry, *cp, *ap, *bp, *cp, carry);
*cp = carry;
}
}
}
/*
* Variant form of internal_mul used for the initial step of
* Montgomery reduction. Only bothers outputting 'len' words
* (everything above that is thrown away).
*/
static void internal_mul_low(const BignumInt *a, const BignumInt *b,
BignumInt *c, int len, BignumInt *scratch)
{
if (len > KARATSUBA_THRESHOLD) {
int i;
/*
* Karatsuba-aware version of internal_mul_low. As before, we
* express each input value as a shifted combination of two
* halves:
*
* a = a_1 D + a_0
* b = b_1 D + b_0
*
* Then the full product is, as before,
*
* ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
*
* Provided we choose D on the large side (so that a_0 and b_0
* are _at least_ as long as a_1 and b_1), we don't need the
* topmost term at all, and we only need half of the middle
* term. So there's no point in doing the proper Karatsuba
* optimisation which computes the middle term using the top
* one, because we'd take as long computing the top one as
* just computing the middle one directly.
*
* So instead, we do a much more obvious thing: we call the
* fully optimised internal_mul to compute a_0 b_0, and we
* recursively call ourself to compute the _bottom halves_ of
* a_1 b_0 and a_0 b_1, each of which we add into the result
* in the obvious way.
*
* In other words, there's no actual Karatsuba _optimisation_
* in this function; the only benefit in doing it this way is
* that we call internal_mul proper for a large part of the
* work, and _that_ can optimise its operation.
*/
int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
/*
* Scratch space for the various bits and pieces we're going
* to be adding together: we need botlen*2 words for a_0 b_0
* (though we may end up throwing away its topmost word), and
* toplen words for each of a_1 b_0 and a_0 b_1. That adds up
* to exactly 2*len.
*/
/* a_0 b_0 */
internal_mul(a + toplen, b + toplen, scratch + 2*toplen, botlen,
scratch + 2*len);
/* a_1 b_0 */
internal_mul_low(a, b + len - toplen, scratch + toplen, toplen,
scratch + 2*len);
/* a_0 b_1 */
internal_mul_low(a + len - toplen, b, scratch, toplen,
scratch + 2*len);
/* Copy the bottom half of the big coefficient into place */
for (i = 0; i < botlen; i++)
c[toplen + i] = scratch[2*toplen + botlen + i];
/* Add the two small coefficients, throwing away the returned carry */
internal_add(scratch, scratch + toplen, scratch, toplen);
/* And add that to the large coefficient, leaving the result in c. */
internal_add(scratch, scratch + 2*toplen + botlen - toplen,
c, toplen);
} else {
int i;
BignumInt carry;
const BignumInt *ap, *bp;
BignumInt *cp, *cps;
/*
* Multiply in the ordinary O(N^2) way.
*/
for (i = 0; i < len; i++)
c[i] = 0;
for (cps = c + len, ap = a + len; ap-- > a; cps--) {
carry = 0;
for (cp = cps, bp = b + len; bp--, cp-- > c ;)
BignumMULADD2(carry, *cp, *ap, *bp, *cp, carry);
}
}
}
/*
* Montgomery reduction. Expects x to be a big-endian array of 2*len
* BignumInts whose value satisfies 0 <= x < rn (where r = 2^(len *
* BIGNUM_INT_BITS) is the Montgomery base). Returns in the same array
* a value x' which is congruent to xr^{-1} mod n, and satisfies 0 <=
* x' < n.
*
* 'n' and 'mninv' should be big-endian arrays of 'len' BignumInts
* each, containing respectively n and the multiplicative inverse of
* -n mod r.
*
* 'tmp' is an array of BignumInt used as scratch space, of length at
* least 3*len + mul_compute_scratch(len).
*/
static void monty_reduce(BignumInt *x, const BignumInt *n,
const BignumInt *mninv, BignumInt *tmp, int len)
{
int i;
BignumInt carry;
/*
* Multiply x by (-n)^{-1} mod r. This gives us a value m such
* that mn is congruent to -x mod r. Hence, mn+x is an exact
* multiple of r, and is also (obviously) congruent to x mod n.
*/
internal_mul_low(x + len, mninv, tmp, len, tmp + 3*len);
/*
* Compute t = (mn+x)/r in ordinary, non-modular, integer
* arithmetic. By construction this is exact, and is congruent mod
* n to x * r^{-1}, i.e. the answer we want.
*
* The following multiply leaves that answer in the _most_
* significant half of the 'x' array, so then we must shift it
* down.
*/
internal_mul(tmp, n, tmp+len, len, tmp + 3*len);
carry = internal_add(x, tmp+len, x, 2*len);
for (i = 0; i < len; i++)
x[len + i] = x[i], x[i] = 0;
/*
* Reduce t mod n. This doesn't require a full-on division by n,
* but merely a test and single optional subtraction, since we can
* show that 0 <= t < 2n.
*
* Proof:
* + we computed m mod r, so 0 <= m < r.
* + so 0 <= mn < rn, obviously
* + hence we only need 0 <= x < rn to guarantee that 0 <= mn+x < 2rn
* + yielding 0 <= (mn+x)/r < 2n as required.
*/
if (!carry) {
for (i = 0; i < len; i++)
if (x[len + i] != n[i])
break;
}
if (carry || i >= len || x[len + i] > n[i])
internal_sub(x+len, n, x+len, len);
}
static void internal_add_shifted(BignumInt *number,
BignumInt n, int shift)
{
int word = 1 + (shift / BIGNUM_INT_BITS);
int bshift = shift % BIGNUM_INT_BITS;
BignumInt addendh, addendl;
BignumCarry carry;
addendl = n << bshift;
addendh = (bshift == 0 ? 0 : n >> (BIGNUM_INT_BITS - bshift));
assert(word <= number[0]);
BignumADC(number[word], carry, number[word], addendl, 0);
word++;
if (!addendh && !carry)
return;
assert(word <= number[0]);
BignumADC(number[word], carry, number[word], addendh, carry);
word++;
while (carry) {
assert(word <= number[0]);
BignumADC(number[word], carry, number[word], 0, carry);
word++;
}
}
static int bn_clz(BignumInt x)
{
/*
* Count the leading zero bits in x. Equivalently, how far left
* would we need to shift x to make its top bit set?
*
* Precondition: x != 0.
*/
/* FIXME: would be nice to put in some compiler intrinsics under
* ifdef here */
int i, ret = 0;
for (i = BIGNUM_INT_BITS / 2; i != 0; i >>= 1) {
if ((x >> (BIGNUM_INT_BITS-i)) == 0) {
x <<= i;
ret += i;
}
}
return ret;
}
static BignumInt reciprocal_word(BignumInt d)
{
BignumInt dshort, recip, prodh, prodl;
int corrections;
/*
* Input: a BignumInt value d, with its top bit set.
*/
assert(d >> (BIGNUM_INT_BITS-1) == 1);
/*
* Output: a value, shifted to fill a BignumInt, which is strictly
* less than 1/(d+1), i.e. is an *under*-estimate (but by as
* little as possible within the constraints) of the reciprocal of
* any number whose first BIGNUM_INT_BITS bits match d.
*
* Ideally we'd like to _totally_ fill BignumInt, i.e. always
* return a value with the top bit set. Unfortunately we can't
* quite guarantee that for all inputs and also return a fixed
* exponent. So instead we take our reciprocal to be
* 2^(BIGNUM_INT_BITS*2-1) / d, so that it has the top bit clear
* only in the exceptional case where d takes exactly the maximum
* value BIGNUM_INT_MASK; in that case, the top bit is clear and
* the next bit down is set.
*/
/*
* Start by computing a half-length version of the answer, by
* straightforward division within a BignumInt.
*/
dshort = (d >> (BIGNUM_INT_BITS/2)) + 1;
recip = (BIGNUM_TOP_BIT + dshort - 1) / dshort;
recip <<= BIGNUM_INT_BITS - BIGNUM_INT_BITS/2;
/*
* Newton-Raphson iteration to improve that starting reciprocal
* estimate: take f(x) = d - 1/x, and then the N-R formula gives
* x_new = x - f(x)/f'(x) = x - (d-1/x)/(1/x^2) = x(2-d*x). Or,
* taking our fixed-point representation into account, take f(x)
* to be d - K/x (where K = 2^(BIGNUM_INT_BITS*2-1) as discussed
* above) and then we get (2K - d*x) * x/K.
*
* Newton-Raphson doubles the number of correct bits at every
* iteration, and the initial division above already gave us half
* the output word, so it's only worth doing one iteration.
*/
BignumMULADD(prodh, prodl, recip, d, recip);
prodl = ~prodl;
prodh = ~prodh;
{
BignumCarry c;
BignumADC(prodl, c, prodl, 1, 0);
prodh += c;
}
BignumMUL(prodh, prodl, prodh, recip);
recip = (prodh << 1) | (prodl >> (BIGNUM_INT_BITS-1));
/*
* Now make sure we have the best possible reciprocal estimate,
* before we return it. We might have been off by a handful either
* way - not enough to bother with any better-thought-out kind of
* correction loop.
*/
BignumMULADD(prodh, prodl, recip, d, recip);
corrections = 0;
if (prodh >= BIGNUM_TOP_BIT) {
do {
BignumCarry c = 1;
BignumADC(prodl, c, prodl, ~d, c); prodh += BIGNUM_INT_MASK + c;
recip--;
corrections++;
} while (prodh >= ((BignumInt)1 << (BIGNUM_INT_BITS-1)));
} else {
while (1) {
BignumInt newprodh, newprodl;
BignumCarry c = 0;
BignumADC(newprodl, c, prodl, d, c); newprodh = prodh + c;
if (newprodh >= BIGNUM_TOP_BIT)
break;
prodh = newprodh;
prodl = newprodl;
recip++;
corrections++;
}
}
return recip;
}
/*
* Compute a = a % m.
* Input in first alen words of a and first mlen words of m.
* Output in first alen words of a
* (of which first alen-mlen words will be zero).
* Quotient is accumulated in the `quotient' array, which is a Bignum
* rather than the internal bigendian format.
*
* 'recip' must be the result of calling reciprocal_word() on the top
* BIGNUM_INT_BITS of the modulus (denoted m0 in comments below), with
* the topmost set bit normalised to the MSB of the input to
* reciprocal_word. 'rshift' is how far left the top nonzero word of
* the modulus had to be shifted to set that top bit.
*/
static void internal_mod(BignumInt *a, int alen,
BignumInt *m, int mlen,
BignumInt *quot, BignumInt recip, int rshift)
{
int i, k;
#ifdef DIVISION_DEBUG
{
int d;
printf("start division, m=0x");
for (d = 0; d < mlen; d++)
printf("%0*llx", BIGNUM_INT_BITS/4, (unsigned long long)m[d]);
printf(", recip=%#0*llx, rshift=%d\n",
BIGNUM_INT_BITS/4, (unsigned long long)recip, rshift);
}
#endif
/*
* Repeatedly use that reciprocal estimate to get a decent number
* of quotient bits, and subtract off the resulting multiple of m.
*
* Normally we expect to terminate this loop by means of finding
* out q=0 part way through, but one way in which we might not get
* that far in the first place is if the input a is actually zero,
* in which case we'll discard zero words from the front of a
* until we reach the termination condition in the for statement
* here.
*/
for (i = 0; i <= alen - mlen ;) {
BignumInt product;
BignumInt aword, q;
int shift, full_bitoffset, bitoffset, wordoffset;
#ifdef DIVISION_DEBUG
{
int d;
printf("main loop, a=0x");
for (d = 0; d < alen; d++)
printf("%0*llx", BIGNUM_INT_BITS/4, (unsigned long long)a[d]);
printf("\n");
}
#endif
if (a[i] == 0) {
#ifdef DIVISION_DEBUG
printf("zero word at i=%d\n", i);
#endif
i++;
continue;
}
aword = a[i];
shift = bn_clz(aword);
aword <<= shift;
if (shift > 0 && i+1 < alen)
aword |= a[i+1] >> (BIGNUM_INT_BITS - shift);
{
BignumInt unused;
BignumMUL(q, unused, recip, aword);
(void)unused;
}
#ifdef DIVISION_DEBUG
printf("i=%d, aword=%#0*llx, shift=%d, q=%#0*llx\n",
i, BIGNUM_INT_BITS/4, (unsigned long long)aword,
shift, BIGNUM_INT_BITS/4, (unsigned long long)q);
#endif
/*
* Work out the right bit and word offsets to use when
* subtracting q*m from a.
*
* aword was taken from a[i], which means its LSB was at bit
* position (alen-1-i) * BIGNUM_INT_BITS. But then we shifted
* it left by 'shift', so now the low bit of aword corresponds
* to bit position (alen-1-i) * BIGNUM_INT_BITS - shift, i.e.
* aword is approximately equal to a / 2^(that).
*
* m0 comes from the top word of mod, so its LSB is at bit
* position (mlen-1) * BIGNUM_INT_BITS - rshift, i.e. it can
* be considered to be m / 2^(that power). 'recip' is the
* reciprocal of m0, times 2^(BIGNUM_INT_BITS*2-1), i.e. it's
* about 2^((mlen+1) * BIGNUM_INT_BITS - rshift - 1) / m.
*
* Hence, recip * aword is approximately equal to the product
* of those, which simplifies to
*
* a/m * 2^((mlen+2+i-alen)*BIGNUM_INT_BITS + shift - rshift - 1)
*
* But we've also shifted recip*aword down by BIGNUM_INT_BITS
* to form q, so we have
*
* q ~= a/m * 2^((mlen+1+i-alen)*BIGNUM_INT_BITS + shift - rshift - 1)
*
* and hence, when we now compute q*m, it will be about
* a*2^(all that lot), i.e. the negation of that expression is
* how far left we have to shift the product q*m to make it
* approximately equal to a.
*/
full_bitoffset = -((mlen+1+i-alen)*BIGNUM_INT_BITS + shift-rshift-1);
#ifdef DIVISION_DEBUG
printf("full_bitoffset=%d\n", full_bitoffset);
#endif
if (full_bitoffset < 0) {
/*
* If we find ourselves needing to shift q*m _right_, that
* means we've reached the bottom of the quotient. Clip q
* so that its right shift becomes zero, and if that means
* q becomes _actually_ zero, this loop is done.
*/
if (full_bitoffset <= -BIGNUM_INT_BITS)
break;
q >>= -full_bitoffset;
full_bitoffset = 0;
if (!q)
break;
#ifdef DIVISION_DEBUG
printf("now full_bitoffset=%d, q=%#0*llx\n",
full_bitoffset, BIGNUM_INT_BITS/4, (unsigned long long)q);
#endif
}
wordoffset = full_bitoffset / BIGNUM_INT_BITS;
bitoffset = full_bitoffset % BIGNUM_INT_BITS;
#ifdef DIVISION_DEBUG
printf("wordoffset=%d, bitoffset=%d\n", wordoffset, bitoffset);
#endif
/* wordoffset as computed above is the offset between the LSWs
* of m and a. But in fact m and a are stored MSW-first, so we
* need to adjust it to be the offset between the actual array
* indices, and flip the sign too. */
wordoffset = alen - mlen - wordoffset;
if (bitoffset == 0) {
BignumCarry c = 1;
BignumInt prev_hi_word = 0;
for (k = mlen - 1; wordoffset+k >= i; k--) {
BignumInt mword = k<0 ? 0 : m[k];
BignumMULADD(prev_hi_word, product, q, mword, prev_hi_word);
#ifdef DIVISION_DEBUG
printf(" aligned sub: product word for m[%d] = %#0*llx\n",
k, BIGNUM_INT_BITS/4,
(unsigned long long)product);
#endif
#ifdef DIVISION_DEBUG
printf(" aligned sub: subtrahend for a[%d] = %#0*llx\n",
wordoffset+k, BIGNUM_INT_BITS/4,
(unsigned long long)product);
#endif
BignumADC(a[wordoffset+k], c, a[wordoffset+k], ~product, c);
}
} else {
BignumInt add_word = 0;
BignumInt c = 1;
BignumInt prev_hi_word = 0;
for (k = mlen - 1; wordoffset+k >= i; k--) {
BignumInt mword = k<0 ? 0 : m[k];
BignumMULADD(prev_hi_word, product, q, mword, prev_hi_word);
#ifdef DIVISION_DEBUG
printf(" unaligned sub: product word for m[%d] = %#0*llx\n",
k, BIGNUM_INT_BITS/4,
(unsigned long long)product);
#endif
add_word |= product << bitoffset;
#ifdef DIVISION_DEBUG
printf(" unaligned sub: subtrahend for a[%d] = %#0*llx\n",
wordoffset+k,
BIGNUM_INT_BITS/4, (unsigned long long)add_word);
#endif
BignumADC(a[wordoffset+k], c, a[wordoffset+k], ~add_word, c);
add_word = product >> (BIGNUM_INT_BITS - bitoffset);
}
}
if (quot) {
#ifdef DIVISION_DEBUG
printf("adding quotient word %#0*llx << %d\n",
BIGNUM_INT_BITS/4, (unsigned long long)q, full_bitoffset);
#endif
internal_add_shifted(quot, q, full_bitoffset);
#ifdef DIVISION_DEBUG
{
int d;
printf("now quot=0x");
for (d = quot[0]; d > 0; d--)
printf("%0*llx", BIGNUM_INT_BITS/4,
(unsigned long long)quot[d]);
printf("\n");
}
#endif
}
}
#ifdef DIVISION_DEBUG
{
int d;
printf("end main loop, a=0x");
for (d = 0; d < alen; d++)
printf("%0*llx", BIGNUM_INT_BITS/4, (unsigned long long)a[d]);
if (quot) {
printf(", quot=0x");
for (d = quot[0]; d > 0; d--)
printf("%0*llx", BIGNUM_INT_BITS/4,
(unsigned long long)quot[d]);
}
printf("\n");
}
#endif
/*
* The above loop should terminate with the remaining value in a
* being strictly less than 2*m (if a >= 2*m then we should always
* have managed to get a nonzero q word), but we can't guarantee
* that it will be strictly less than m: consider a case where the
* remainder is 1, and another where the remainder is m-1. By the
* time a contains a value that's _about m_, you clearly can't
* distinguish those cases by looking at only the top word of a -
* you have to go all the way down to the bottom before you find
* out whether it's just less or just more than m.
*
* Hence, we now do a final fixup in which we subtract one last
* copy of m, or don't, accordingly. We should never have to
* subtract more than one copy of m here.
*/
for (i = 0; i < alen; i++) {
/* Compare a with m, word by word, from the MSW down. As soon
* as we encounter a difference, we know whether we need the
* fixup. */
int mindex = mlen-alen+i;
BignumInt mword = mindex < 0 ? 0 : m[mindex];
if (a[i] < mword) {
#ifdef DIVISION_DEBUG
printf("final fixup not needed, a < m\n");
#endif
return;
} else if (a[i] > mword) {
#ifdef DIVISION_DEBUG
printf("final fixup is needed, a > m\n");
#endif
break;
}
/* If neither of those cases happened, the words are the same,
* so keep going and look at the next one. */
}
#ifdef DIVISION_DEBUG
if (i == mlen) /* if we printed neither of the above diagnostics */
printf("final fixup is needed, a == m\n");
#endif
/*
* If we got here without returning, then a >= m, so we must
* subtract m, and increment the quotient.
*/
{
BignumCarry c = 1;
for (i = alen - 1; i >= 0; i--) {
int mindex = mlen-alen+i;
BignumInt mword = mindex < 0 ? 0 : m[mindex];
BignumADC(a[i], c, a[i], ~mword, c);
}
}
if (quot)
internal_add_shifted(quot, 1, 0);
#ifdef DIVISION_DEBUG
{
int d;
printf("after final fixup, a=0x");
for (d = 0; d < alen; d++)
printf("%0*llx", BIGNUM_INT_BITS/4, (unsigned long long)a[d]);
if (quot) {
printf(", quot=0x");
for (d = quot[0]; d > 0; d--)
printf("%0*llx", BIGNUM_INT_BITS/4,
(unsigned long long)quot[d]);
}
printf("\n");
}
#endif
}
/*
* Compute (base ^ exp) % mod, the pedestrian way.
*/
Bignum modpow_simple(Bignum base_in, Bignum exp, Bignum mod)
{
BignumInt *a, *b, *n, *m, *scratch;
BignumInt recip;
int rshift;
int mlen, scratchlen, i, j;
Bignum base, result;
/*
* The most significant word of mod needs to be non-zero. It
* should already be, but let's make sure.
*/
assert(mod[mod[0]] != 0);
/*
* Make sure the base is smaller than the modulus, by reducing
* it modulo the modulus if not.
*/
base = bigmod(base_in, mod);
/* Allocate m of size mlen, copy mod to m */
/* We use big endian internally */
mlen = mod[0];
m = snewn(mlen, BignumInt);
for (j = 0; j < mlen; j++)
m[j] = mod[mod[0] - j];
/* Allocate n of size mlen, copy base to n */
n = snewn(mlen, BignumInt);
i = mlen - base[0];
for (j = 0; j < i; j++)
n[j] = 0;
for (j = 0; j < (int)base[0]; j++)
n[i + j] = base[base[0] - j];
/* Allocate a and b of size 2*mlen. Set a = 1 */
a = snewn(2 * mlen, BignumInt);
b = snewn(2 * mlen, BignumInt);
for (i = 0; i < 2 * mlen; i++)
a[i] = 0;
a[2 * mlen - 1] = 1;
/* Scratch space for multiplies */
scratchlen = mul_compute_scratch(mlen);
scratch = snewn(scratchlen, BignumInt);
/* Skip leading zero bits of exp. */
i = 0;
j = BIGNUM_INT_BITS-1;
while (i < (int)exp[0] && (exp[exp[0] - i] & ((BignumInt)1 << j)) == 0) {
j--;
if (j < 0) {
i++;
j = BIGNUM_INT_BITS-1;
}
}
/* Compute reciprocal of the top full word of the modulus */
{