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My first initial commit with all of mine experimental changes.

Dmitry Kabak (userdima committed 13 years ago

Commit d37b88d1 parent 3f5b46618352 files · +2598093 −-8690Browse files

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1/*
2 * Copyright (c) 1999, 2007, Oracle and/or its affiliates. All rights reserved.
3 * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
4 *
5 * This code is free software; you can redistribute it and/or modify it
6 * under the terms of the GNU General Public License version 2 only, as
7 * published by the Free Software Foundation. Oracle designates this
8 * particular file as subject to the "Classpath" exception as provided
9 * by Oracle in the LICENSE file that accompanied this code.
10 *
11 * This code is distributed in the hope that it will be useful, but WITHOUT
12 * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
13 * FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
14 * version 2 for more details (a copy is included in the LICENSE file that
15 * accompanied this code).
16 *
17 * You should have received a copy of the GNU General Public License version
18 * 2 along with this work; if not, write to the Free Software Foundation,
19 * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
20 *
21 * Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA
22 * or visit www.oracle.com if you need additional information or have any
23 * questions.
24 */
25
26package java.math;
27
28/**
29 * A class used to represent multiprecision integers that makes efficient
30 * use of allocated space by allowing a number to occupy only part of
31 * an array so that the arrays do not have to be reallocated as often.
32 * When performing an operation with many iterations the array used to
33 * hold a number is only reallocated when necessary and does not have to
34 * be the same size as the number it represents. A mutable number allows
35 * calculations to occur on the same number without having to create
36 * a new number for every step of the calculation as occurs with
37 * BigIntegers.
38 *
39 * @see BigInteger
40 * @author Michael McCloskey
41 * @since 1.3
42 */
43
44import java.util.Arrays;
45
46import static java.math.BigInteger.LONG_MASK;
47import static java.math.BigDecimal.INFLATED;
48
49class MutableBigInteger {
50 /**
51 * Holds the magnitude of this MutableBigInteger in big endian order.
52 * The magnitude may start at an offset into the value array, and it may
53 * end before the length of the value array.
54 */
55 int[] value;
56
57 /**
58 * The number of ints of the value array that are currently used
59 * to hold the magnitude of this MutableBigInteger. The magnitude starts
60 * at an offset and offset + intLen may be less than value.length.
61 */
62 int intLen;
63
64 /**
65 * The offset into the value array where the magnitude of this
66 * MutableBigInteger begins.
67 */
68 int offset = 0;
69
70 // Constants
71 /**
72 * MutableBigInteger with one element value array with the value 1. Used by
73 * BigDecimal divideAndRound to increment the quotient. Use this constant
74 * only when the method is not going to modify this object.
75 */
76 static final MutableBigInteger ONE = new MutableBigInteger(1);
77
78 // Constructors
79
80 /**
81 * The default constructor. An empty MutableBigInteger is created with
82 * a one word capacity.
83 */
84 MutableBigInteger() {
85 value = new int[1];
86 intLen = 0;
87 }
88
89 /**
90 * Construct a new MutableBigInteger with a magnitude specified by
91 * the int val.
92 */
93 MutableBigInteger(int val) {
94 value = new int[1];
95 intLen = 1;
96 value[0] = val;
97 }
98
99 /**
100 * Construct a new MutableBigInteger with the specified value array
101 * up to the length of the array supplied.
102 */
103 MutableBigInteger(int[] val) {
104 value = val;
105 intLen = val.length;
106 }
107
108 /**
109 * Construct a new MutableBigInteger with a magnitude equal to the
110 * specified BigInteger.
111 */
112 MutableBigInteger(BigInteger b) {
113 intLen = b.mag.length;
114 value = Arrays.copyOf(b.mag, intLen);
115 }
116
117 /**
118 * Construct a new MutableBigInteger with a magnitude equal to the
119 * specified MutableBigInteger.
120 */
121 MutableBigInteger(MutableBigInteger val) {
122 intLen = val.intLen;
123 value = Arrays.copyOfRange(val.value, val.offset, val.offset + intLen);
124 }
125
126 /**
127 * Internal helper method to return the magnitude array. The caller is not
128 * supposed to modify the returned array.
129 */
130 private int[] getMagnitudeArray() {
131 if (offset > 0 || value.length != intLen)
132 return Arrays.copyOfRange(value, offset, offset + intLen);
133 return value;
134 }
135
136 /**
137 * Convert this MutableBigInteger to a long value. The caller has to make
138 * sure this MutableBigInteger can be fit into long.
139 */
140 private long toLong() {
141 assert (intLen <= 2) : "this MutableBigInteger exceeds the range of long";
142 if (intLen == 0)
143 return 0;
144 long d = value[offset] & LONG_MASK;
145 return (intLen == 2) ? d << 32 | (value[offset + 1] & LONG_MASK) : d;
146 }
147
148 /**
149 * Convert this MutableBigInteger to a BigInteger object.
150 */
151 BigInteger toBigInteger(int sign) {
152 if (intLen == 0 || sign == 0)
153 return BigInteger.ZERO;
154 return new BigInteger(getMagnitudeArray(), sign);
155 }
156
157 /**
158 * Convert this MutableBigInteger to BigDecimal object with the specified sign
159 * and scale.
160 */
161 BigDecimal toBigDecimal(int sign, int scale) {
162 if (intLen == 0 || sign == 0)
163 return BigDecimal.valueOf(0, scale);
164 int[] mag = getMagnitudeArray();
165 int len = mag.length;
166 int d = mag[0];
167 // If this MutableBigInteger can't be fit into long, we need to
168 // make a BigInteger object for the resultant BigDecimal object.
169 if (len > 2 || (d < 0 && len == 2))
170 return new BigDecimal(new BigInteger(mag, sign), INFLATED, scale, 0);
171 long v = (len == 2) ?
172 ((mag[1] & LONG_MASK) | (d & LONG_MASK) << 32) :
173 d & LONG_MASK;
174 return new BigDecimal(null, sign == -1 ? -v : v, scale, 0);
175 }
176
177 /**
178 * Clear out a MutableBigInteger for reuse.
179 */
180 void clear() {
181 offset = intLen = 0;
182 for (int index=0, n=value.length; index < n; index++)
183 value[index] = 0;
184 }
185
186 /**
187 * Set a MutableBigInteger to zero, removing its offset.
188 */
189 void reset() {
190 offset = intLen = 0;
191 }
192
193 /**
194 * Compare the magnitude of two MutableBigIntegers. Returns -1, 0 or 1
195 * as this MutableBigInteger is numerically less than, equal to, or
196 * greater than <tt>b</tt>.
197 */
198 final int compare(MutableBigInteger b) {
199 int blen = b.intLen;
200 if (intLen < blen)
201 return -1;
202 if (intLen > blen)
203 return 1;
204
205 // Add Integer.MIN_VALUE to make the comparison act as unsigned integer
206 // comparison.
207 int[] bval = b.value;
208 for (int i = offset, j = b.offset; i < intLen + offset; i++, j++) {
209 int b1 = value[i] + 0x80000000;
210 int b2 = bval[j] + 0x80000000;
211 if (b1 < b2)
212 return -1;
213 if (b1 > b2)
214 return 1;
215 }
216 return 0;
217 }
218
219 /**
220 * Compare this against half of a MutableBigInteger object (Needed for
221 * remainder tests).
222 * Assumes no leading unnecessary zeros, which holds for results
223 * from divide().
224 */
225 final int compareHalf(MutableBigInteger b) {
226 int blen = b.intLen;
227 int len = intLen;
228 if (len <= 0)
229 return blen <=0 ? 0 : -1;
230 if (len > blen)
231 return 1;
232 if (len < blen - 1)
233 return -1;
234 int[] bval = b.value;
235 int bstart = 0;
236 int carry = 0;
237 // Only 2 cases left:len == blen or len == blen - 1
238 if (len != blen) { // len == blen - 1
239 if (bval[bstart] == 1) {
240 ++bstart;
241 carry = 0x80000000;
242 } else
243 return -1;
244 }
245 // compare values with right-shifted values of b,
246 // carrying shifted-out bits across words
247 int[] val = value;
248 for (int i = offset, j = bstart; i < len + offset;) {
249 int bv = bval[j++];
250 long hb = ((bv >>> 1) + carry) & LONG_MASK;
251 long v = val[i++] & LONG_MASK;
252 if (v != hb)
253 return v < hb ? -1 : 1;
254 carry = (bv & 1) << 31; // carray will be either 0x80000000 or 0
255 }
256 return carry == 0? 0 : -1;
257 }
258
259 /**
260 * Return the index of the lowest set bit in this MutableBigInteger. If the
261 * magnitude of this MutableBigInteger is zero, -1 is returned.
262 */
263 private final int getLowestSetBit() {
264 if (intLen == 0)
265 return -1;
266 int j, b;
267 for (j=intLen-1; (j>0) && (value[j+offset]==0); j--)
268 ;
269 b = value[j+offset];
270 if (b==0)
271 return -1;
272 return ((intLen-1-j)<<5) + Integer.numberOfTrailingZeros(b);
273 }
274
275 /**
276 * Return the int in use in this MutableBigInteger at the specified
277 * index. This method is not used because it is not inlined on all
278 * platforms.
279 */
280 private final int getInt(int index) {
281 return value[offset+index];
282 }
283
284 /**
285 * Return a long which is equal to the unsigned value of the int in
286 * use in this MutableBigInteger at the specified index. This method is
287 * not used because it is not inlined on all platforms.
288 */
289 private final long getLong(int index) {
290 return value[offset+index] & LONG_MASK;
291 }
292
293 /**
294 * Ensure that the MutableBigInteger is in normal form, specifically
295 * making sure that there are no leading zeros, and that if the
296 * magnitude is zero, then intLen is zero.
297 */
298 final void normalize() {
299 if (intLen == 0) {
300 offset = 0;
301 return;
302 }
303
304 int index = offset;
305 if (value[index] != 0)
306 return;
307
308 int indexBound = index+intLen;
309 do {
310 index++;
311 } while(index < indexBound && value[index]==0);
312
313 int numZeros = index - offset;
314 intLen -= numZeros;
315 offset = (intLen==0 ? 0 : offset+numZeros);
316 }
317
318 /**
319 * If this MutableBigInteger cannot hold len words, increase the size
320 * of the value array to len words.
321 */
322 private final void ensureCapacity(int len) {
323 if (value.length < len) {
324 value = new int[len];
325 offset = 0;
326 intLen = len;
327 }
328 }
329
330 /**
331 * Convert this MutableBigInteger into an int array with no leading
332 * zeros, of a length that is equal to this MutableBigInteger's intLen.
333 */
334 int[] toIntArray() {
335 int[] result = new int[intLen];
336 for(int i=0; i<intLen; i++)
337 result[i] = value[offset+i];
338 return result;
339 }
340
341 /**
342 * Sets the int at index+offset in this MutableBigInteger to val.
343 * This does not get inlined on all platforms so it is not used
344 * as often as originally intended.
345 */
346 void setInt(int index, int val) {
347 value[offset + index] = val;
348 }
349
350 /**
351 * Sets this MutableBigInteger's value array to the specified array.
352 * The intLen is set to the specified length.
353 */
354 void setValue(int[] val, int length) {
355 value = val;
356 intLen = length;
357 offset = 0;
358 }
359
360 /**
361 * Sets this MutableBigInteger's value array to a copy of the specified
362 * array. The intLen is set to the length of the new array.
363 */
364 void copyValue(MutableBigInteger src) {
365 int len = src.intLen;
366 if (value.length < len)
367 value = new int[len];
368 System.arraycopy(src.value, src.offset, value, 0, len);
369 intLen = len;
370 offset = 0;
371 }
372
373 /**
374 * Sets this MutableBigInteger's value array to a copy of the specified
375 * array. The intLen is set to the length of the specified array.
376 */
377 void copyValue(int[] val) {
378 int len = val.length;
379 if (value.length < len)
380 value = new int[len];
381 System.arraycopy(val, 0, value, 0, len);
382 intLen = len;
383 offset = 0;
384 }
385
386 /**
387 * Returns true iff this MutableBigInteger has a value of one.
388 */
389 boolean isOne() {
390 return (intLen == 1) && (value[offset] == 1);
391 }
392
393 /**
394 * Returns true iff this MutableBigInteger has a value of zero.
395 */
396 boolean isZero() {
397 return (intLen == 0);
398 }
399
400 /**
401 * Returns true iff this MutableBigInteger is even.
402 */
403 boolean isEven() {
404 return (intLen == 0) || ((value[offset + intLen - 1] & 1) == 0);
405 }
406
407 /**
408 * Returns true iff this MutableBigInteger is odd.
409 */
410 boolean isOdd() {
411 return isZero() ? false : ((value[offset + intLen - 1] & 1) == 1);
412 }
413
414 /**
415 * Returns true iff this MutableBigInteger is in normal form. A
416 * MutableBigInteger is in normal form if it has no leading zeros
417 * after the offset, and intLen + offset <= value.length.
418 */
419 boolean isNormal() {
420 if (intLen + offset > value.length)
421 return false;
422 if (intLen ==0)
423 return true;
424 return (value[offset] != 0);
425 }
426
427 /**
428 * Returns a String representation of this MutableBigInteger in radix 10.
429 */
430 public String toString() {
431 BigInteger b = toBigInteger(1);
432 return b.toString();
433 }
434
435 /**
436 * Right shift this MutableBigInteger n bits. The MutableBigInteger is left
437 * in normal form.
438 */
439 void rightShift(int n) {
440 if (intLen == 0)
441 return;
442 int nInts = n >>> 5;
443 int nBits = n & 0x1F;
444 this.intLen -= nInts;
445 if (nBits == 0)
446 return;
447 int bitsInHighWord = BigInteger.bitLengthForInt(value[offset]);
448 if (nBits >= bitsInHighWord) {
449 this.primitiveLeftShift(32 - nBits);
450 this.intLen--;
451 } else {
452 primitiveRightShift(nBits);
453 }
454 }
455
456 /**
457 * Left shift this MutableBigInteger n bits.
458 */
459 void leftShift(int n) {
460 /*
461 * If there is enough storage space in this MutableBigInteger already
462 * the available space will be used. Space to the right of the used
463 * ints in the value array is faster to utilize, so the extra space
464 * will be taken from the right if possible.
465 */
466 if (intLen == 0)
467 return;
468 int nInts = n >>> 5;
469 int nBits = n&0x1F;
470 int bitsInHighWord = BigInteger.bitLengthForInt(value[offset]);
471
472 // If shift can be done without moving words, do so
473 if (n <= (32-bitsInHighWord)) {
474 primitiveLeftShift(nBits);
475 return;
476 }
477
478 int newLen = intLen + nInts +1;
479 if (nBits <= (32-bitsInHighWord))
480 newLen--;
481 if (value.length < newLen) {
482 // The array must grow
483 int[] result = new int[newLen];
484 for (int i=0; i<intLen; i++)
485 result[i] = value[offset+i];
486 setValue(result, newLen);
487 } else if (value.length - offset >= newLen) {
488 // Use space on right
489 for(int i=0; i<newLen - intLen; i++)
490 value[offset+intLen+i] = 0;
491 } else {
492 // Must use space on left
493 for (int i=0; i<intLen; i++)
494 value[i] = value[offset+i];
495 for (int i=intLen; i<newLen; i++)
496 value[i] = 0;
497 offset = 0;
498 }
499 intLen = newLen;
500 if (nBits == 0)
501 return;
502 if (nBits <= (32-bitsInHighWord))
503 primitiveLeftShift(nBits);
504 else
505 primitiveRightShift(32 -nBits);
506 }
507
508 /**
509 * A primitive used for division. This method adds in one multiple of the
510 * divisor a back to the dividend result at a specified offset. It is used
511 * when qhat was estimated too large, and must be adjusted.
512 */
513 private int divadd(int[] a, int[] result, int offset) {
514 long carry = 0;
515
516 for (int j=a.length-1; j >= 0; j--) {
517 long sum = (a[j] & LONG_MASK) +
518 (result[j+offset] & LONG_MASK) + carry;
519 result[j+offset] = (int)sum;
520 carry = sum >>> 32;
521 }
522 return (int)carry;
523 }
524
525 /**
526 * This method is used for division. It multiplies an n word input a by one
527 * word input x, and subtracts the n word product from q. This is needed
528 * when subtracting qhat*divisor from dividend.
529 */
530 private int mulsub(int[] q, int[] a, int x, int len, int offset) {
531 long xLong = x & LONG_MASK;
532 long carry = 0;
533 offset += len;
534
535 for (int j=len-1; j >= 0; j--) {
536 long product = (a[j] & LONG_MASK) * xLong + carry;
537 long difference = q[offset] - product;
538 q[offset--] = (int)difference;
539 carry = (product >>> 32)
540 + (((difference & LONG_MASK) >
541 (((~(int)product) & LONG_MASK))) ? 1:0);
542 }
543 return (int)carry;
544 }
545
546 /**
547 * Right shift this MutableBigInteger n bits, where n is
548 * less than 32.
549 * Assumes that intLen > 0, n > 0 for speed
550 */
551 private final void primitiveRightShift(int n) {
552 int[] val = value;
553 int n2 = 32 - n;
554 for (int i=offset+intLen-1, c=val[i]; i>offset; i--) {
555 int b = c;
556 c = val[i-1];
557 val[i] = (c << n2) | (b >>> n);
558 }
559 val[offset] >>>= n;
560 }
561
562 /**
563 * Left shift this MutableBigInteger n bits, where n is
564 * less than 32.
565 * Assumes that intLen > 0, n > 0 for speed
566 */
567 private final void primitiveLeftShift(int n) {
568 int[] val = value;
569 int n2 = 32 - n;
570 for (int i=offset, c=val[i], m=i+intLen-1; i<m; i++) {
571 int b = c;
572 c = val[i+1];
573 val[i] = (b << n) | (c >>> n2);
574 }
575 val[offset+intLen-1] <<= n;
576 }
577
578 /**
579 * Adds the contents of two MutableBigInteger objects.The result
580 * is placed within this MutableBigInteger.
581 * The contents of the addend are not changed.
582 */
583 void add(MutableBigInteger addend) {
584 int x = intLen;
585 int y = addend.intLen;
586 int resultLen = (intLen > addend.intLen ? intLen : addend.intLen);
587 int[] result = (value.length < resultLen ? new int[resultLen] : value);
588
589 int rstart = result.length-1;
590 long sum;
591 long carry = 0;
592
593 // Add common parts of both numbers
594 while(x>0 && y>0) {
595 x--; y--;
596 sum = (value[x+offset] & LONG_MASK) +
597 (addend.value[y+addend.offset] & LONG_MASK) + carry;
598 result[rstart--] = (int)sum;
599 carry = sum >>> 32;
600 }
601
602 // Add remainder of the longer number
603 while(x>0) {
604 x--;
605 if (carry == 0 && result == value && rstart == (x + offset))
606 return;
607 sum = (value[x+offset] & LONG_MASK) + carry;
608 result[rstart--] = (int)sum;
609 carry = sum >>> 32;
610 }
611 while(y>0) {
612 y--;
613 sum = (addend.value[y+addend.offset] & LONG_MASK) + carry;
614 result[rstart--] = (int)sum;
615 carry = sum >>> 32;
616 }
617
618 if (carry > 0) { // Result must grow in length
619 resultLen++;
620 if (result.length < resultLen) {
621 int temp[] = new int[resultLen];
622 // Result one word longer from carry-out; copy low-order
623 // bits into new result.
624 System.arraycopy(result, 0, temp, 1, result.length);
625 temp[0] = 1;
626 result = temp;
627 } else {
628 result[rstart--] = 1;
629 }
630 }
631
632 value = result;
633 intLen = resultLen;
634 offset = result.length - resultLen;
635 }
636
637
638 /**
639 * Subtracts the smaller of this and b from the larger and places the
640 * result into this MutableBigInteger.
641 */
642 int subtract(MutableBigInteger b) {
643 MutableBigInteger a = this;
644
645 int[] result = value;
646 int sign = a.compare(b);
647
648 if (sign == 0) {
649 reset();
650 return 0;
651 }
652 if (sign < 0) {
653 MutableBigInteger tmp = a;
654 a = b;
655 b = tmp;
656 }
657
658 int resultLen = a.intLen;
659 if (result.length < resultLen)
660 result = new int[resultLen];
661
662 long diff = 0;
663 int x = a.intLen;
664 int y = b.intLen;
665 int rstart = result.length - 1;
666
667 // Subtract common parts of both numbers
668 while (y>0) {
669 x--; y--;
670
671 diff = (a.value[x+a.offset] & LONG_MASK) -
672 (b.value[y+b.offset] & LONG_MASK) - ((int)-(diff>>32));
673 result[rstart--] = (int)diff;
674 }
675 // Subtract remainder of longer number
676 while (x>0) {
677 x--;
678 diff = (a.value[x+a.offset] & LONG_MASK) - ((int)-(diff>>32));
679 result[rstart--] = (int)diff;
680 }
681
682 value = result;
683 intLen = resultLen;
684 offset = value.length - resultLen;
685 normalize();
686 return sign;
687 }
688
689 /**
690 * Subtracts the smaller of a and b from the larger and places the result
691 * into the larger. Returns 1 if the answer is in a, -1 if in b, 0 if no
692 * operation was performed.
693 */
694 private int difference(MutableBigInteger b) {
695 MutableBigInteger a = this;
696 int sign = a.compare(b);
697 if (sign ==0)
698 return 0;
699 if (sign < 0) {
700 MutableBigInteger tmp = a;
701 a = b;
702 b = tmp;
703 }
704
705 long diff = 0;
706 int x = a.intLen;
707 int y = b.intLen;
708
709 // Subtract common parts of both numbers
710 while (y>0) {
711 x--; y--;
712 diff = (a.value[a.offset+ x] & LONG_MASK) -
713 (b.value[b.offset+ y] & LONG_MASK) - ((int)-(diff>>32));
714 a.value[a.offset+x] = (int)diff;
715 }
716 // Subtract remainder of longer number
717 while (x>0) {
718 x--;
719 diff = (a.value[a.offset+ x] & LONG_MASK) - ((int)-(diff>>32));
720 a.value[a.offset+x] = (int)diff;
721 }
722
723 a.normalize();
724 return sign;
725 }
726
727 /**
728 * Multiply the contents of two MutableBigInteger objects. The result is
729 * placed into MutableBigInteger z. The contents of y are not changed.
730 */
731 void multiply(MutableBigInteger y, MutableBigInteger z) {
732 int xLen = intLen;
733 int yLen = y.intLen;
734 int newLen = xLen + yLen;
735
736 // Put z into an appropriate state to receive product
737 if (z.value.length < newLen)
738 z.value = new int[newLen];
739 z.offset = 0;
740 z.intLen = newLen;
741
742 // The first iteration is hoisted out of the loop to avoid extra add
743 long carry = 0;
744 for (int j=yLen-1, k=yLen+xLen-1; j >= 0; j--, k--) {
745 long product = (y.value[j+y.offset] & LONG_MASK) *
746 (value[xLen-1+offset] & LONG_MASK) + carry;
747 z.value[k] = (int)product;
748 carry = product >>> 32;
749 }
750 z.value[xLen-1] = (int)carry;
751
752 // Perform the multiplication word by word
753 for (int i = xLen-2; i >= 0; i--) {
754 carry = 0;
755 for (int j=yLen-1, k=yLen+i; j >= 0; j--, k--) {
756 long product = (y.value[j+y.offset] & LONG_MASK) *
757 (value[i+offset] & LONG_MASK) +
758 (z.value[k] & LONG_MASK) + carry;
759 z.value[k] = (int)product;
760 carry = product >>> 32;
761 }
762 z.value[i] = (int)carry;
763 }
764
765 // Remove leading zeros from product
766 z.normalize();
767 }
768
769 /**
770 * Multiply the contents of this MutableBigInteger by the word y. The
771 * result is placed into z.
772 */
773 void mul(int y, MutableBigInteger z) {
774 if (y == 1) {
775 z.copyValue(this);
776 return;
777 }
778
779 if (y == 0) {
780 z.clear();
781 return;
782 }
783
784 // Perform the multiplication word by word
785 long ylong = y & LONG_MASK;
786 int[] zval = (z.value.length<intLen+1 ? new int[intLen + 1]
787 : z.value);
788 long carry = 0;
789 for (int i = intLen-1; i >= 0; i--) {
790 long product = ylong * (value[i+offset] & LONG_MASK) + carry;
791 zval[i+1] = (int)product;
792 carry = product >>> 32;
793 }
794
795 if (carry == 0) {
796 z.offset = 1;
797 z.intLen = intLen;
798 } else {
799 z.offset = 0;
800 z.intLen = intLen + 1;
801 zval[0] = (int)carry;
802 }
803 z.value = zval;
804 }
805
806 /**
807 * This method is used for division of an n word dividend by a one word
808 * divisor. The quotient is placed into quotient. The one word divisor is
809 * specified by divisor.
810 *
811 * @return the remainder of the division is returned.
812 *
813 */
814 int divideOneWord(int divisor, MutableBigInteger quotient) {
815 long divisorLong = divisor & LONG_MASK;
816
817 // Special case of one word dividend
818 if (intLen == 1) {
819 long dividendValue = value[offset] & LONG_MASK;
820 int q = (int) (dividendValue / divisorLong);
821 int r = (int) (dividendValue - q * divisorLong);
822 quotient.value[0] = q;
823 quotient.intLen = (q == 0) ? 0 : 1;
824 quotient.offset = 0;
825 return r;
826 }
827
828 if (quotient.value.length < intLen)
829 quotient.value = new int[intLen];
830 quotient.offset = 0;
831 quotient.intLen = intLen;
832
833 // Normalize the divisor
834 int shift = Integer.numberOfLeadingZeros(divisor);
835
836 int rem = value[offset];
837 long remLong = rem & LONG_MASK;
838 if (remLong < divisorLong) {
839 quotient.value[0] = 0;
840 } else {
841 quotient.value[0] = (int)(remLong / divisorLong);
842 rem = (int) (remLong - (quotient.value[0] * divisorLong));
843 remLong = rem & LONG_MASK;
844 }
845
846 int xlen = intLen;
847 int[] qWord = new int[2];
848 while (--xlen > 0) {
849 long dividendEstimate = (remLong<<32) |
850 (value[offset + intLen - xlen] & LONG_MASK);
851 if (dividendEstimate >= 0) {
852 qWord[0] = (int) (dividendEstimate / divisorLong);
853 qWord[1] = (int) (dividendEstimate - qWord[0] * divisorLong);
854 } else {
855 divWord(qWord, dividendEstimate, divisor);
856 }
857 quotient.value[intLen - xlen] = qWord[0];
858 rem = qWord[1];
859 remLong = rem & LONG_MASK;
860 }
861
862 quotient.normalize();
863 // Unnormalize
864 if (shift > 0)
865 return rem % divisor;
866 else
867 return rem;
868 }
869
870 /**
871 * Calculates the quotient of this div b and places the quotient in the
872 * provided MutableBigInteger objects and the remainder object is returned.
873 *
874 * Uses Algorithm D in Knuth section 4.3.1.
875 * Many optimizations to that algorithm have been adapted from the Colin
876 * Plumb C library.
877 * It special cases one word divisors for speed. The content of b is not
878 * changed.
879 *
880 */
881 MutableBigInteger divide(MutableBigInteger b, MutableBigInteger quotient) {
882 if (b.intLen == 0)
883 throw new ArithmeticException("BigInteger divide by zero");
884
885 // Dividend is zero
886 if (intLen == 0) {
887 quotient.intLen = quotient.offset;
888 return new MutableBigInteger();
889 }
890
891 int cmp = compare(b);
892 // Dividend less than divisor
893 if (cmp < 0) {
894 quotient.intLen = quotient.offset = 0;
895 return new MutableBigInteger(this);
896 }
897 // Dividend equal to divisor
898 if (cmp == 0) {
899 quotient.value[0] = quotient.intLen = 1;
900 quotient.offset = 0;
901 return new MutableBigInteger();
902 }
903
904 quotient.clear();
905 // Special case one word divisor
906 if (b.intLen == 1) {
907 int r = divideOneWord(b.value[b.offset], quotient);
908 if (r == 0)
909 return new MutableBigInteger();
910 return new MutableBigInteger(r);
911 }
912
913 // Copy divisor value to protect divisor
914 int[] div = Arrays.copyOfRange(b.value, b.offset, b.offset + b.intLen);
915 return divideMagnitude(div, quotient);
916 }
917
918 /**
919 * Internally used to calculate the quotient of this div v and places the
920 * quotient in the provided MutableBigInteger object and the remainder is
921 * returned.
922 *
923 * @return the remainder of the division will be returned.
924 */
925 long divide(long v, MutableBigInteger quotient) {
926 if (v == 0)
927 throw new ArithmeticException("BigInteger divide by zero");
928
929 // Dividend is zero
930 if (intLen == 0) {
931 quotient.intLen = quotient.offset = 0;
932 return 0;
933 }
934 if (v < 0)
935 v = -v;
936
937 int d = (int)(v >>> 32);
938 quotient.clear();
939 // Special case on word divisor
940 if (d == 0)
941 return divideOneWord((int)v, quotient) & LONG_MASK;
942 else {
943 int[] div = new int[]{ d, (int)(v & LONG_MASK) };
944 return divideMagnitude(div, quotient).toLong();
945 }
946 }
947
948 /**
949 * Divide this MutableBigInteger by the divisor represented by its magnitude
950 * array. The quotient will be placed into the provided quotient object &
951 * the remainder object is returned.
952 */
953 private MutableBigInteger divideMagnitude(int[] divisor,
954 MutableBigInteger quotient) {
955
956 // Remainder starts as dividend with space for a leading zero
957 MutableBigInteger rem = new MutableBigInteger(new int[intLen + 1]);
958 System.arraycopy(value, offset, rem.value, 1, intLen);
959 rem.intLen = intLen;
960 rem.offset = 1;
961
962 int nlen = rem.intLen;
963
964 // Set the quotient size
965 int dlen = divisor.length;
966 int limit = nlen - dlen + 1;
967 if (quotient.value.length < limit) {
968 quotient.value = new int[limit];
969 quotient.offset = 0;
970 }
971 quotient.intLen = limit;
972 int[] q = quotient.value;
973
974 // D1 normalize the divisor
975 int shift = Integer.numberOfLeadingZeros(divisor[0]);
976 if (shift > 0) {
977 // First shift will not grow array
978 BigInteger.primitiveLeftShift(divisor, dlen, shift);
979 // But this one might
980 rem.leftShift(shift);
981 }
982
983 // Must insert leading 0 in rem if its length did not change
984 if (rem.intLen == nlen) {
985 rem.offset = 0;
986 rem.value[0] = 0;
987 rem.intLen++;
988 }
989
990 int dh = divisor[0];
991 long dhLong = dh & LONG_MASK;
992 int dl = divisor[1];
993 int[] qWord = new int[2];
994
995 // D2 Initialize j
996 for(int j=0; j<limit; j++) {
997 // D3 Calculate qhat
998 // estimate qhat
999 int qhat = 0;
1000 int qrem = 0;
1001 boolean skipCorrection = false;
1002 int nh = rem.value[j+rem.offset];
1003 int nh2 = nh + 0x80000000;
1004 int nm = rem.value[j+1+rem.offset];
1005
1006 if (nh == dh) {
1007 qhat = ~0;
1008 qrem = nh + nm;
1009 skipCorrection = qrem + 0x80000000 < nh2;
1010 } else {
1011 long nChunk = (((long)nh) << 32) | (nm & LONG_MASK);
1012 if (nChunk >= 0) {
1013 qhat = (int) (nChunk / dhLong);
1014 qrem = (int) (nChunk - (qhat * dhLong));
1015 } else {
1016 divWord(qWord, nChunk, dh);
1017 qhat = qWord[0];
1018 qrem = qWord[1];
1019 }
1020 }
1021
1022 if (qhat == 0)
1023 continue;
1024
1025 if (!skipCorrection) { // Correct qhat
1026 long nl = rem.value[j+2+rem.offset] & LONG_MASK;
1027 long rs = ((qrem & LONG_MASK) << 32) | nl;
1028 long estProduct = (dl & LONG_MASK) * (qhat & LONG_MASK);
1029
1030 if (unsignedLongCompare(estProduct, rs)) {
1031 qhat--;
1032 qrem = (int)((qrem & LONG_MASK) + dhLong);
1033 if ((qrem & LONG_MASK) >= dhLong) {
1034 estProduct -= (dl & LONG_MASK);
1035 rs = ((qrem & LONG_MASK) << 32) | nl;
1036 if (unsignedLongCompare(estProduct, rs))
1037 qhat--;
1038 }
1039 }
1040 }
1041
1042 // D4 Multiply and subtract
1043 rem.value[j+rem.offset] = 0;
1044 int borrow = mulsub(rem.value, divisor, qhat, dlen, j+rem.offset);
1045
1046 // D5 Test remainder
1047 if (borrow + 0x80000000 > nh2) {
1048 // D6 Add back
1049 divadd(divisor, rem.value, j+1+rem.offset);
1050 qhat--;
1051 }
1052
1053 // Store the quotient digit
1054 q[j] = qhat;
1055 } // D7 loop on j
1056
1057 // D8 Unnormalize
1058 if (shift > 0)
1059 rem.rightShift(shift);
1060
1061 quotient.normalize();
1062 rem.normalize();
1063 return rem;
1064 }
1065
1066 /**
1067 * Compare two longs as if they were unsigned.
1068 * Returns true iff one is bigger than two.
1069 */
1070 private boolean unsignedLongCompare(long one, long two) {
1071 return (one+Long.MIN_VALUE) > (two+Long.MIN_VALUE);
1072 }
1073
1074 /**
1075 * This method divides a long quantity by an int to estimate
1076 * qhat for two multi precision numbers. It is used when
1077 * the signed value of n is less than zero.
1078 */
1079 private void divWord(int[] result, long n, int d) {
1080 long dLong = d & LONG_MASK;
1081
1082 if (dLong == 1) {
1083 result[0] = (int)n;
1084 result[1] = 0;
1085 return;
1086 }
1087
1088 // Approximate the quotient and remainder
1089 long q = (n >>> 1) / (dLong >>> 1);
1090 long r = n - q*dLong;
1091
1092 // Correct the approximation
1093 while (r < 0) {
1094 r += dLong;
1095 q--;
1096 }
1097 while (r >= dLong) {
1098 r -= dLong;
1099 q++;
1100 }
1101
1102 // n - q*dlong == r && 0 <= r <dLong, hence we're done.
1103 result[0] = (int)q;
1104 result[1] = (int)r;
1105 }
1106
1107 /**
1108 * Calculate GCD of this and b. This and b are changed by the computation.
1109 */
1110 MutableBigInteger hybridGCD(MutableBigInteger b) {
1111 // Use Euclid's algorithm until the numbers are approximately the
1112 // same length, then use the binary GCD algorithm to find the GCD.
1113 MutableBigInteger a = this;
1114 MutableBigInteger q = new MutableBigInteger();
1115
1116 while (b.intLen != 0) {
1117 if (Math.abs(a.intLen - b.intLen) < 2)
1118 return a.binaryGCD(b);
1119
1120 MutableBigInteger r = a.divide(b, q);
1121 a = b;
1122 b = r;
1123 }
1124 return a;
1125 }
1126
1127 /**
1128 * Calculate GCD of this and v.
1129 * Assumes that this and v are not zero.
1130 */
1131 private MutableBigInteger binaryGCD(MutableBigInteger v) {
1132 // Algorithm B from Knuth section 4.5.2
1133 MutableBigInteger u = this;
1134 MutableBigInteger r = new MutableBigInteger();
1135
1136 // step B1
1137 int s1 = u.getLowestSetBit();
1138 int s2 = v.getLowestSetBit();
1139 int k = (s1 < s2) ? s1 : s2;
1140 if (k != 0) {
1141 u.rightShift(k);
1142 v.rightShift(k);
1143 }
1144
1145 // step B2
1146 boolean uOdd = (k==s1);
1147 MutableBigInteger t = uOdd ? v: u;
1148 int tsign = uOdd ? -1 : 1;
1149
1150 int lb;
1151 while ((lb = t.getLowestSetBit()) >= 0) {
1152 // steps B3 and B4
1153 t.rightShift(lb);
1154 // step B5
1155 if (tsign > 0)
1156 u = t;
1157 else
1158 v = t;
1159
1160 // Special case one word numbers
1161 if (u.intLen < 2 && v.intLen < 2) {
1162 int x = u.value[u.offset];
1163 int y = v.value[v.offset];
1164 x = binaryGcd(x, y);
1165 r.value[0] = x;
1166 r.intLen = 1;
1167 r.offset = 0;
1168 if (k > 0)
1169 r.leftShift(k);
1170 return r;
1171 }
1172
1173 // step B6
1174 if ((tsign = u.difference(v)) == 0)
1175 break;
1176 t = (tsign >= 0) ? u : v;
1177 }
1178
1179 if (k > 0)
1180 u.leftShift(k);
1181 return u;
1182 }
1183
1184 /**
1185 * Calculate GCD of a and b interpreted as unsigned integers.
1186 */
1187 static int binaryGcd(int a, int b) {
1188 if (b==0)
1189 return a;
1190 if (a==0)
1191 return b;
1192
1193 // Right shift a & b till their last bits equal to 1.
1194 int aZeros = Integer.numberOfTrailingZeros(a);
1195 int bZeros = Integer.numberOfTrailingZeros(b);
1196 a >>>= aZeros;
1197 b >>>= bZeros;
1198
1199 int t = (aZeros < bZeros ? aZeros : bZeros);
1200
1201 while (a != b) {
1202 if ((a+0x80000000) > (b+0x80000000)) { // a > b as unsigned
1203 a -= b;
1204 a >>>= Integer.numberOfTrailingZeros(a);
1205 } else {
1206 b -= a;
1207 b >>>= Integer.numberOfTrailingZeros(b);
1208 }
1209 }
1210 return a<<t;
1211 }
1212
1213 /**
1214 * Returns the modInverse of this mod p. This and p are not affected by
1215 * the operation.
1216 */
1217 MutableBigInteger mutableModInverse(MutableBigInteger p) {
1218 // Modulus is odd, use Schroeppel's algorithm
1219 if (p.isOdd())
1220 return modInverse(p);
1221
1222 // Base and modulus are even, throw exception
1223 if (isEven())
1224 throw new ArithmeticException("BigInteger not invertible.");
1225
1226 // Get even part of modulus expressed as a power of 2
1227 int powersOf2 = p.getLowestSetBit();
1228
1229 // Construct odd part of modulus
1230 MutableBigInteger oddMod = new MutableBigInteger(p);
1231 oddMod.rightShift(powersOf2);
1232
1233 if (oddMod.isOne())
1234 return modInverseMP2(powersOf2);
1235
1236 // Calculate 1/a mod oddMod
1237 MutableBigInteger oddPart = modInverse(oddMod);
1238
1239 // Calculate 1/a mod evenMod
1240 MutableBigInteger evenPart = modInverseMP2(powersOf2);
1241
1242 // Combine the results using Chinese Remainder Theorem
1243 MutableBigInteger y1 = modInverseBP2(oddMod, powersOf2);
1244 MutableBigInteger y2 = oddMod.modInverseMP2(powersOf2);
1245
1246 MutableBigInteger temp1 = new MutableBigInteger();
1247 MutableBigInteger temp2 = new MutableBigInteger();
1248 MutableBigInteger result = new MutableBigInteger();
1249
1250 oddPart.leftShift(powersOf2);
1251 oddPart.multiply(y1, result);
1252
1253 evenPart.multiply(oddMod, temp1);
1254 temp1.multiply(y2, temp2);
1255
1256 result.add(temp2);
1257 return result.divide(p, temp1);
1258 }
1259
1260 /*
1261 * Calculate the multiplicative inverse of this mod 2^k.
1262 */
1263 MutableBigInteger modInverseMP2(int k) {
1264 if (isEven())
1265 throw new ArithmeticException("Non-invertible. (GCD != 1)");
1266
1267 if (k > 64)
1268 return euclidModInverse(k);
1269
1270 int t = inverseMod32(value[offset+intLen-1]);
1271
1272 if (k < 33) {
1273 t = (k == 32 ? t : t & ((1 << k) - 1));
1274 return new MutableBigInteger(t);
1275 }
1276
1277 long pLong = (value[offset+intLen-1] & LONG_MASK);
1278 if (intLen > 1)
1279 pLong |= ((long)value[offset+intLen-2] << 32);
1280 long tLong = t & LONG_MASK;
1281 tLong = tLong * (2 - pLong * tLong); // 1 more Newton iter step
1282 tLong = (k == 64 ? tLong : tLong & ((1L << k) - 1));
1283
1284 MutableBigInteger result = new MutableBigInteger(new int[2]);
1285 result.value[0] = (int)(tLong >>> 32);
1286 result.value[1] = (int)tLong;
1287 result.intLen = 2;
1288 result.normalize();
1289 return result;
1290 }
1291
1292 /*
1293 * Returns the multiplicative inverse of val mod 2^32. Assumes val is odd.
1294 */
1295 static int inverseMod32(int val) {
1296 // Newton's iteration!
1297 int t = val;
1298 t *= 2 - val*t;
1299 t *= 2 - val*t;
1300 t *= 2 - val*t;
1301 t *= 2 - val*t;
1302 return t;
1303 }
1304
1305 /*
1306 * Calculate the multiplicative inverse of 2^k mod mod, where mod is odd.
1307 */
1308 static MutableBigInteger modInverseBP2(MutableBigInteger mod, int k) {
1309 // Copy the mod to protect original
1310 return fixup(new MutableBigInteger(1), new MutableBigInteger(mod), k);
1311 }
1312
1313 /**
1314 * Calculate the multiplicative inverse of this mod mod, where mod is odd.
1315 * This and mod are not changed by the calculation.
1316 *
1317 * This method implements an algorithm due to Richard Schroeppel, that uses
1318 * the same intermediate representation as Montgomery Reduction
1319 * ("Montgomery Form"). The algorithm is described in an unpublished
1320 * manuscript entitled "Fast Modular Reciprocals."
1321 */
1322 private MutableBigInteger modInverse(MutableBigInteger mod) {
1323 MutableBigInteger p = new MutableBigInteger(mod);
1324 MutableBigInteger f = new MutableBigInteger(this);
1325 MutableBigInteger g = new MutableBigInteger(p);
1326 SignedMutableBigInteger c = new SignedMutableBigInteger(1);
1327 SignedMutableBigInteger d = new SignedMutableBigInteger();
1328 MutableBigInteger temp = null;
1329 SignedMutableBigInteger sTemp = null;
1330
1331 int k = 0;
1332 // Right shift f k times until odd, left shift d k times
1333 if (f.isEven()) {
1334 int trailingZeros = f.getLowestSetBit();
1335 f.rightShift(trailingZeros);
1336 d.leftShift(trailingZeros);
1337 k = trailingZeros;
1338 }
1339
1340 // The Almost Inverse Algorithm
1341 while(!f.isOne()) {
1342 // If gcd(f, g) != 1, number is not invertible modulo mod
1343 if (f.isZero())
1344 throw new ArithmeticException("BigInteger not invertible.");
1345
1346 // If f < g exchange f, g and c, d
1347 if (f.compare(g) < 0) {
1348 temp = f; f = g; g = temp;
1349 sTemp = d; d = c; c = sTemp;
1350 }
1351
1352 // If f == g (mod 4)
1353 if (((f.value[f.offset + f.intLen - 1] ^
1354 g.value[g.offset + g.intLen - 1]) & 3) == 0) {
1355 f.subtract(g);
1356 c.signedSubtract(d);
1357 } else { // If f != g (mod 4)
1358 f.add(g);
1359 c.signedAdd(d);
1360 }
1361
1362 // Right shift f k times until odd, left shift d k times
1363 int trailingZeros = f.getLowestSetBit();
1364 f.rightShift(trailingZeros);
1365 d.leftShift(trailingZeros);
1366 k += trailingZeros;
1367 }
1368
1369 while (c.sign < 0)
1370 c.signedAdd(p);
1371
1372 return fixup(c, p, k);
1373 }
1374
1375 /*
1376 * The Fixup Algorithm
1377 * Calculates X such that X = C * 2^(-k) (mod P)
1378 * Assumes C<P and P is odd.
1379 */
1380 static MutableBigInteger fixup(MutableBigInteger c, MutableBigInteger p,
1381 int k) {
1382 MutableBigInteger temp = new MutableBigInteger();
1383 // Set r to the multiplicative inverse of p mod 2^32
1384 int r = -inverseMod32(p.value[p.offset+p.intLen-1]);
1385
1386 for(int i=0, numWords = k >> 5; i<numWords; i++) {
1387 // V = R * c (mod 2^j)
1388 int v = r * c.value[c.offset + c.intLen-1];
1389 // c = c + (v * p)
1390 p.mul(v, temp);
1391 c.add(temp);
1392 // c = c / 2^j
1393 c.intLen--;
1394 }
1395 int numBits = k & 0x1f;
1396 if (numBits != 0) {
1397 // V = R * c (mod 2^j)
1398 int v = r * c.value[c.offset + c.intLen-1];
1399 v &= ((1<<numBits) - 1);
1400 // c = c + (v * p)
1401 p.mul(v, temp);
1402 c.add(temp);
1403 // c = c / 2^j
1404 c.rightShift(numBits);
1405 }
1406
1407 // In theory, c may be greater than p at this point (Very rare!)
1408 while (c.compare(p) >= 0)
1409 c.subtract(p);
1410
1411 return c;
1412 }
1413
1414 /**
1415 * Uses the extended Euclidean algorithm to compute the modInverse of base
1416 * mod a modulus that is a power of 2. The modulus is 2^k.
1417 */
1418 MutableBigInteger euclidModInverse(int k) {
1419 MutableBigInteger b = new MutableBigInteger(1);
1420 b.leftShift(k);
1421 MutableBigInteger mod = new MutableBigInteger(b);
1422
1423 MutableBigInteger a = new MutableBigInteger(this);
1424 MutableBigInteger q = new MutableBigInteger();
1425 MutableBigInteger r = b.divide(a, q);
1426
1427 MutableBigInteger swapper = b;
1428 // swap b & r
1429 b = r;
1430 r = swapper;
1431
1432 MutableBigInteger t1 = new MutableBigInteger(q);
1433 MutableBigInteger t0 = new MutableBigInteger(1);
1434 MutableBigInteger temp = new MutableBigInteger();
1435
1436 while (!b.isOne()) {
1437 r = a.divide(b, q);
1438
1439 if (r.intLen == 0)
1440 throw new ArithmeticException("BigInteger not invertible.");
1441
1442 swapper = r;
1443 a = swapper;
1444
1445 if (q.intLen == 1)
1446 t1.mul(q.value[q.offset], temp);
1447 else
1448 q.multiply(t1, temp);
1449 swapper = q;
1450 q = temp;
1451 temp = swapper;
1452 t0.add(q);
1453
1454 if (a.isOne())
1455 return t0;
1456
1457 r = b.divide(a, q);
1458
1459 if (r.intLen == 0)
1460 throw new ArithmeticException("BigInteger not invertible.");
1461
1462 swapper = b;
1463 b = r;
1464
1465 if (q.intLen == 1)
1466 t0.mul(q.value[q.offset], temp);
1467 else
1468 q.multiply(t0, temp);
1469 swapper = q; q = temp; temp = swapper;
1470
1471 t1.add(q);
1472 }
1473 mod.subtract(t1);
1474 return mod;
1475 }
1476
1477}