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Diffstat (limited to 'third_party/bigint/BigUnsigned.cc')
-rw-r--r-- | third_party/bigint/BigUnsigned.cc | 697 |
1 files changed, 697 insertions, 0 deletions
diff --git a/third_party/bigint/BigUnsigned.cc b/third_party/bigint/BigUnsigned.cc new file mode 100644 index 0000000000..d7f9889cc6 --- /dev/null +++ b/third_party/bigint/BigUnsigned.cc @@ -0,0 +1,697 @@ +#include "BigUnsigned.hh" + +// Memory management definitions have moved to the bottom of NumberlikeArray.hh. + +// The templates used by these constructors and converters are at the bottom of +// BigUnsigned.hh. + +BigUnsigned::BigUnsigned(unsigned long x) { initFromPrimitive (x); } +BigUnsigned::BigUnsigned(unsigned int x) { initFromPrimitive (x); } +BigUnsigned::BigUnsigned(unsigned short x) { initFromPrimitive (x); } +BigUnsigned::BigUnsigned( long x) { initFromSignedPrimitive(x); } +BigUnsigned::BigUnsigned( int x) { initFromSignedPrimitive(x); } +BigUnsigned::BigUnsigned( short x) { initFromSignedPrimitive(x); } + +unsigned long BigUnsigned::toUnsignedLong () const { return convertToPrimitive <unsigned long >(); } +unsigned int BigUnsigned::toUnsignedInt () const { return convertToPrimitive <unsigned int >(); } +unsigned short BigUnsigned::toUnsignedShort() const { return convertToPrimitive <unsigned short>(); } +long BigUnsigned::toLong () const { return convertToSignedPrimitive< long >(); } +int BigUnsigned::toInt () const { return convertToSignedPrimitive< int >(); } +short BigUnsigned::toShort () const { return convertToSignedPrimitive< short>(); } + +// BIT/BLOCK ACCESSORS + +void BigUnsigned::setBlock(Index i, Blk newBlock) { + if (newBlock == 0) { + if (i < len) { + blk[i] = 0; + zapLeadingZeros(); + } + // If i >= len, no effect. + } else { + if (i >= len) { + // The nonzero block extends the number. + allocateAndCopy(i+1); + // Zero any added blocks that we aren't setting. + for (Index j = len; j < i; j++) + blk[j] = 0; + len = i+1; + } + blk[i] = newBlock; + } +} + +/* Evidently the compiler wants BigUnsigned:: on the return type because, at + * that point, it hasn't yet parsed the BigUnsigned:: on the name to get the + * proper scope. */ +BigUnsigned::Index BigUnsigned::bitLength() const { + if (isZero()) + return 0; + else { + Blk leftmostBlock = getBlock(len - 1); + Index leftmostBlockLen = 0; + while (leftmostBlock != 0) { + leftmostBlock >>= 1; + leftmostBlockLen++; + } + return leftmostBlockLen + (len - 1) * N; + } +} + +void BigUnsigned::setBit(Index bi, bool newBit) { + Index blockI = bi / N; + Blk block = getBlock(blockI), mask = Blk(1) << (bi % N); + block = newBit ? (block | mask) : (block & ~mask); + setBlock(blockI, block); +} + +// COMPARISON +BigUnsigned::CmpRes BigUnsigned::compareTo(const BigUnsigned &x) const { + // A bigger length implies a bigger number. + if (len < x.len) + return less; + else if (len > x.len) + return greater; + else { + // Compare blocks one by one from left to right. + Index i = len; + while (i > 0) { + i--; + if (blk[i] == x.blk[i]) + continue; + else if (blk[i] > x.blk[i]) + return greater; + else + return less; + } + // If no blocks differed, the numbers are equal. + return equal; + } +} + +// COPY-LESS OPERATIONS + +/* + * On most calls to copy-less operations, it's safe to read the inputs little by + * little and write the outputs little by little. However, if one of the + * inputs is coming from the same variable into which the output is to be + * stored (an "aliased" call), we risk overwriting the input before we read it. + * In this case, we first compute the result into a temporary BigUnsigned + * variable and then copy it into the requested output variable *this. + * Each put-here operation uses the DTRT_ALIASED macro (Do The Right Thing on + * aliased calls) to generate code for this check. + * + * I adopted this approach on 2007.02.13 (see Assignment Operators in + * BigUnsigned.hh). Before then, put-here operations rejected aliased calls + * with an exception. I think doing the right thing is better. + * + * Some of the put-here operations can probably handle aliased calls safely + * without the extra copy because (for example) they process blocks strictly + * right-to-left. At some point I might determine which ones don't need the + * copy, but my reasoning would need to be verified very carefully. For now + * I'll leave in the copy. + */ +#define DTRT_ALIASED(cond, op) \ + if (cond) { \ + BigUnsigned tmpThis; \ + tmpThis.op; \ + *this = tmpThis; \ + return; \ + } + + + +void BigUnsigned::add(const BigUnsigned &a, const BigUnsigned &b) { + DTRT_ALIASED(this == &a || this == &b, add(a, b)); + // If one argument is zero, copy the other. + if (a.len == 0) { + operator =(b); + return; + } else if (b.len == 0) { + operator =(a); + return; + } + // Some variables... + // Carries in and out of an addition stage + bool carryIn, carryOut; + Blk temp; + Index i; + // a2 points to the longer input, b2 points to the shorter + const BigUnsigned *a2, *b2; + if (a.len >= b.len) { + a2 = &a; + b2 = &b; + } else { + a2 = &b; + b2 = &a; + } + // Set prelimiary length and make room in this BigUnsigned + len = a2->len + 1; + allocate(len); + // For each block index that is present in both inputs... + for (i = 0, carryIn = false; i < b2->len; i++) { + // Add input blocks + temp = a2->blk[i] + b2->blk[i]; + // If a rollover occurred, the result is less than either input. + // This test is used many times in the BigUnsigned code. + carryOut = (temp < a2->blk[i]); + // If a carry was input, handle it + if (carryIn) { + temp++; + carryOut |= (temp == 0); + } + blk[i] = temp; // Save the addition result + carryIn = carryOut; // Pass the carry along + } + // If there is a carry left over, increase blocks until + // one does not roll over. + for (; i < a2->len && carryIn; i++) { + temp = a2->blk[i] + 1; + carryIn = (temp == 0); + blk[i] = temp; + } + // If the carry was resolved but the larger number + // still has blocks, copy them over. + for (; i < a2->len; i++) + blk[i] = a2->blk[i]; + // Set the extra block if there's still a carry, decrease length otherwise + if (carryIn) + blk[i] = 1; + else + len--; +} + +void BigUnsigned::subtract(const BigUnsigned &a, const BigUnsigned &b) { + DTRT_ALIASED(this == &a || this == &b, subtract(a, b)); + if (b.len == 0) { + // If b is zero, copy a. + operator =(a); + return; + } else if (a.len < b.len) + // If a is shorter than b, the result is negative. + throw "BigUnsigned::subtract: " + "Negative result in unsigned calculation"; + // Some variables... + bool borrowIn, borrowOut; + Blk temp; + Index i; + // Set preliminary length and make room + len = a.len; + allocate(len); + // For each block index that is present in both inputs... + for (i = 0, borrowIn = false; i < b.len; i++) { + temp = a.blk[i] - b.blk[i]; + // If a reverse rollover occurred, + // the result is greater than the block from a. + borrowOut = (temp > a.blk[i]); + // Handle an incoming borrow + if (borrowIn) { + borrowOut |= (temp == 0); + temp--; + } + blk[i] = temp; // Save the subtraction result + borrowIn = borrowOut; // Pass the borrow along + } + // If there is a borrow left over, decrease blocks until + // one does not reverse rollover. + for (; i < a.len && borrowIn; i++) { + borrowIn = (a.blk[i] == 0); + blk[i] = a.blk[i] - 1; + } + /* If there's still a borrow, the result is negative. + * Throw an exception, but zero out this object so as to leave it in a + * predictable state. */ + if (borrowIn) { + len = 0; + throw "BigUnsigned::subtract: Negative result in unsigned calculation"; + } else + // Copy over the rest of the blocks + for (; i < a.len; i++) + blk[i] = a.blk[i]; + // Zap leading zeros + zapLeadingZeros(); +} + +/* + * About the multiplication and division algorithms: + * + * I searched unsucessfully for fast C++ built-in operations like the `b_0' + * and `c_0' Knuth describes in Section 4.3.1 of ``The Art of Computer + * Programming'' (replace `place' by `Blk'): + * + * ``b_0[:] multiplication of a one-place integer by another one-place + * integer, giving a two-place answer; + * + * ``c_0[:] division of a two-place integer by a one-place integer, + * provided that the quotient is a one-place integer, and yielding + * also a one-place remainder.'' + * + * I also missed his note that ``[b]y adjusting the word size, if + * necessary, nearly all computers will have these three operations + * available'', so I gave up on trying to use algorithms similar to his. + * A future version of the library might include such algorithms; I + * would welcome contributions from others for this. + * + * I eventually decided to use bit-shifting algorithms. To multiply `a' + * and `b', we zero out the result. Then, for each `1' bit in `a', we + * shift `b' left the appropriate amount and add it to the result. + * Similarly, to divide `a' by `b', we shift `b' left varying amounts, + * repeatedly trying to subtract it from `a'. When we succeed, we note + * the fact by setting a bit in the quotient. While these algorithms + * have the same O(n^2) time complexity as Knuth's, the ``constant factor'' + * is likely to be larger. + * + * Because I used these algorithms, which require single-block addition + * and subtraction rather than single-block multiplication and division, + * the innermost loops of all four routines are very similar. Study one + * of them and all will become clear. + */ + +/* + * This is a little inline function used by both the multiplication + * routine and the division routine. + * + * `getShiftedBlock' returns the `x'th block of `num << y'. + * `y' may be anything from 0 to N - 1, and `x' may be anything from + * 0 to `num.len'. + * + * Two things contribute to this block: + * + * (1) The `N - y' low bits of `num.blk[x]', shifted `y' bits left. + * + * (2) The `y' high bits of `num.blk[x-1]', shifted `N - y' bits right. + * + * But we must be careful if `x == 0' or `x == num.len', in + * which case we should use 0 instead of (2) or (1), respectively. + * + * If `y == 0', then (2) contributes 0, as it should. However, + * in some computer environments, for a reason I cannot understand, + * `a >> b' means `a >> (b % N)'. This means `num.blk[x-1] >> (N - y)' + * will return `num.blk[x-1]' instead of the desired 0 when `y == 0'; + * the test `y == 0' handles this case specially. + */ +inline BigUnsigned::Blk getShiftedBlock(const BigUnsigned &num, + BigUnsigned::Index x, unsigned int y) { + BigUnsigned::Blk part1 = (x == 0 || y == 0) ? 0 : (num.blk[x - 1] >> (BigUnsigned::N - y)); + BigUnsigned::Blk part2 = (x == num.len) ? 0 : (num.blk[x] << y); + return part1 | part2; +} + +void BigUnsigned::multiply(const BigUnsigned &a, const BigUnsigned &b) { + DTRT_ALIASED(this == &a || this == &b, multiply(a, b)); + // If either a or b is zero, set to zero. + if (a.len == 0 || b.len == 0) { + len = 0; + return; + } + /* + * Overall method: + * + * Set this = 0. + * For each 1-bit of `a' (say the `i2'th bit of block `i'): + * Add `b << (i blocks and i2 bits)' to *this. + */ + // Variables for the calculation + Index i, j, k; + unsigned int i2; + Blk temp; + bool carryIn, carryOut; + // Set preliminary length and make room + len = a.len + b.len; + allocate(len); + // Zero out this object + for (i = 0; i < len; i++) + blk[i] = 0; + // For each block of the first number... + for (i = 0; i < a.len; i++) { + // For each 1-bit of that block... + for (i2 = 0; i2 < N; i2++) { + if ((a.blk[i] & (Blk(1) << i2)) == 0) + continue; + /* + * Add b to this, shifted left i blocks and i2 bits. + * j is the index in b, and k = i + j is the index in this. + * + * `getShiftedBlock', a short inline function defined above, + * is now used for the bit handling. It replaces the more + * complex `bHigh' code, in which each run of the loop dealt + * immediately with the low bits and saved the high bits to + * be picked up next time. The last run of the loop used to + * leave leftover high bits, which were handled separately. + * Instead, this loop runs an additional time with j == b.len. + * These changes were made on 2005.01.11. + */ + for (j = 0, k = i, carryIn = false; j <= b.len; j++, k++) { + /* + * The body of this loop is very similar to the body of the first loop + * in `add', except that this loop does a `+=' instead of a `+'. + */ + temp = blk[k] + getShiftedBlock(b, j, i2); + carryOut = (temp < blk[k]); + if (carryIn) { + temp++; + carryOut |= (temp == 0); + } + blk[k] = temp; + carryIn = carryOut; + } + // No more extra iteration to deal with `bHigh'. + // Roll-over a carry as necessary. + for (; carryIn; k++) { + blk[k]++; + carryIn = (blk[k] == 0); + } + } + } + // Zap possible leading zero + if (blk[len - 1] == 0) + len--; +} + +/* + * DIVISION WITH REMAINDER + * This monstrous function mods *this by the given divisor b while storing the + * quotient in the given object q; at the end, *this contains the remainder. + * The seemingly bizarre pattern of inputs and outputs was chosen so that the + * function copies as little as possible (since it is implemented by repeated + * subtraction of multiples of b from *this). + * + * "modWithQuotient" might be a better name for this function, but I would + * rather not change the name now. + */ +void BigUnsigned::divideWithRemainder(const BigUnsigned &b, BigUnsigned &q) { + /* Defending against aliased calls is more complex than usual because we + * are writing to both *this and q. + * + * It would be silly to try to write quotient and remainder to the + * same variable. Rule that out right away. */ + if (this == &q) + throw "BigUnsigned::divideWithRemainder: Cannot write quotient and remainder into the same variable"; + /* Now *this and q are separate, so the only concern is that b might be + * aliased to one of them. If so, use a temporary copy of b. */ + if (this == &b || &q == &b) { + BigUnsigned tmpB(b); + divideWithRemainder(tmpB, q); + return; + } + + /* + * Knuth's definition of mod (which this function uses) is somewhat + * different from the C++ definition of % in case of division by 0. + * + * We let a / 0 == 0 (it doesn't matter much) and a % 0 == a, no + * exceptions thrown. This allows us to preserve both Knuth's demand + * that a mod 0 == a and the useful property that + * (a / b) * b + (a % b) == a. + */ + if (b.len == 0) { + q.len = 0; + return; + } + + /* + * If *this.len < b.len, then *this < b, and we can be sure that b doesn't go into + * *this at all. The quotient is 0 and *this is already the remainder (so leave it alone). + */ + if (len < b.len) { + q.len = 0; + return; + } + + // At this point we know (*this).len >= b.len > 0. (Whew!) + + /* + * Overall method: + * + * For each appropriate i and i2, decreasing: + * Subtract (b << (i blocks and i2 bits)) from *this, storing the + * result in subtractBuf. + * If the subtraction succeeds with a nonnegative result: + * Turn on bit i2 of block i of the quotient q. + * Copy subtractBuf back into *this. + * Otherwise bit i2 of block i remains off, and *this is unchanged. + * + * Eventually q will contain the entire quotient, and *this will + * be left with the remainder. + * + * subtractBuf[x] corresponds to blk[x], not blk[x+i], since 2005.01.11. + * But on a single iteration, we don't touch the i lowest blocks of blk + * (and don't use those of subtractBuf) because these blocks are + * unaffected by the subtraction: we are subtracting + * (b << (i blocks and i2 bits)), which ends in at least `i' zero + * blocks. */ + // Variables for the calculation + Index i, j, k; + unsigned int i2; + Blk temp; + bool borrowIn, borrowOut; + + /* + * Make sure we have an extra zero block just past the value. + * + * When we attempt a subtraction, we might shift `b' so + * its first block begins a few bits left of the dividend, + * and then we'll try to compare these extra bits with + * a nonexistent block to the left of the dividend. The + * extra zero block ensures sensible behavior; we need + * an extra block in `subtractBuf' for exactly the same reason. + */ + Index origLen = len; // Save real length. + /* To avoid an out-of-bounds access in case of reallocation, allocate + * first and then increment the logical length. */ + allocateAndCopy(len + 1); + len++; + blk[origLen] = 0; // Zero the added block. + + // subtractBuf holds part of the result of a subtraction; see above. + Blk *subtractBuf = new Blk[len]; + + // Set preliminary length for quotient and make room + q.len = origLen - b.len + 1; + q.allocate(q.len); + // Zero out the quotient + for (i = 0; i < q.len; i++) + q.blk[i] = 0; + + // For each possible left-shift of b in blocks... + i = q.len; + while (i > 0) { + i--; + // For each possible left-shift of b in bits... + // (Remember, N is the number of bits in a Blk.) + q.blk[i] = 0; + i2 = N; + while (i2 > 0) { + i2--; + /* + * Subtract b, shifted left i blocks and i2 bits, from *this, + * and store the answer in subtractBuf. In the for loop, `k == i + j'. + * + * Compare this to the middle section of `multiply'. They + * are in many ways analogous. See especially the discussion + * of `getShiftedBlock'. + */ + for (j = 0, k = i, borrowIn = false; j <= b.len; j++, k++) { + temp = blk[k] - getShiftedBlock(b, j, i2); + borrowOut = (temp > blk[k]); + if (borrowIn) { + borrowOut |= (temp == 0); + temp--; + } + // Since 2005.01.11, indices of `subtractBuf' directly match those of `blk', so use `k'. + subtractBuf[k] = temp; + borrowIn = borrowOut; + } + // No more extra iteration to deal with `bHigh'. + // Roll-over a borrow as necessary. + for (; k < origLen && borrowIn; k++) { + borrowIn = (blk[k] == 0); + subtractBuf[k] = blk[k] - 1; + } + /* + * If the subtraction was performed successfully (!borrowIn), + * set bit i2 in block i of the quotient. + * + * Then, copy the portion of subtractBuf filled by the subtraction + * back to *this. This portion starts with block i and ends-- + * where? Not necessarily at block `i + b.len'! Well, we + * increased k every time we saved a block into subtractBuf, so + * the region of subtractBuf we copy is just [i, k). + */ + if (!borrowIn) { + q.blk[i] |= (Blk(1) << i2); + while (k > i) { + k--; + blk[k] = subtractBuf[k]; + } + } + } + } + // Zap possible leading zero in quotient + if (q.blk[q.len - 1] == 0) + q.len--; + // Zap any/all leading zeros in remainder + zapLeadingZeros(); + // Deallocate subtractBuf. + // (Thanks to Brad Spencer for noticing my accidental omission of this!) + delete [] subtractBuf; +} + +/* BITWISE OPERATORS + * These are straightforward blockwise operations except that they differ in + * the output length and the necessity of zapLeadingZeros. */ + +void BigUnsigned::bitAnd(const BigUnsigned &a, const BigUnsigned &b) { + DTRT_ALIASED(this == &a || this == &b, bitAnd(a, b)); + // The bitwise & can't be longer than either operand. + len = (a.len >= b.len) ? b.len : a.len; + allocate(len); + Index i; + for (i = 0; i < len; i++) + blk[i] = a.blk[i] & b.blk[i]; + zapLeadingZeros(); +} + +void BigUnsigned::bitOr(const BigUnsigned &a, const BigUnsigned &b) { + DTRT_ALIASED(this == &a || this == &b, bitOr(a, b)); + Index i; + const BigUnsigned *a2, *b2; + if (a.len >= b.len) { + a2 = &a; + b2 = &b; + } else { + a2 = &b; + b2 = &a; + } + allocate(a2->len); + for (i = 0; i < b2->len; i++) + blk[i] = a2->blk[i] | b2->blk[i]; + for (; i < a2->len; i++) + blk[i] = a2->blk[i]; + len = a2->len; + // Doesn't need zapLeadingZeros. +} + +void BigUnsigned::bitXor(const BigUnsigned &a, const BigUnsigned &b) { + DTRT_ALIASED(this == &a || this == &b, bitXor(a, b)); + Index i; + const BigUnsigned *a2, *b2; + if (a.len >= b.len) { + a2 = &a; + b2 = &b; + } else { + a2 = &b; + b2 = &a; + } + allocate(a2->len); + for (i = 0; i < b2->len; i++) + blk[i] = a2->blk[i] ^ b2->blk[i]; + for (; i < a2->len; i++) + blk[i] = a2->blk[i]; + len = a2->len; + zapLeadingZeros(); +} + +void BigUnsigned::bitShiftLeft(const BigUnsigned &a, int b) { + DTRT_ALIASED(this == &a, bitShiftLeft(a, b)); + if (b < 0) { + if (b << 1 == 0) + throw "BigUnsigned::bitShiftLeft: " + "Pathological shift amount not implemented"; + else { + bitShiftRight(a, -b); + return; + } + } + Index shiftBlocks = b / N; + unsigned int shiftBits = b % N; + // + 1: room for high bits nudged left into another block + len = a.len + shiftBlocks + 1; + allocate(len); + Index i, j; + for (i = 0; i < shiftBlocks; i++) + blk[i] = 0; + for (j = 0, i = shiftBlocks; j <= a.len; j++, i++) + blk[i] = getShiftedBlock(a, j, shiftBits); + // Zap possible leading zero + if (blk[len - 1] == 0) + len--; +} + +void BigUnsigned::bitShiftRight(const BigUnsigned &a, int b) { + DTRT_ALIASED(this == &a, bitShiftRight(a, b)); + if (b < 0) { + if (b << 1 == 0) + throw "BigUnsigned::bitShiftRight: " + "Pathological shift amount not implemented"; + else { + bitShiftLeft(a, -b); + return; + } + } + // This calculation is wacky, but expressing the shift as a left bit shift + // within each block lets us use getShiftedBlock. + Index rightShiftBlocks = (b + N - 1) / N; + unsigned int leftShiftBits = N * rightShiftBlocks - b; + // Now (N * rightShiftBlocks - leftShiftBits) == b + // and 0 <= leftShiftBits < N. + if (rightShiftBlocks >= a.len + 1) { + // All of a is guaranteed to be shifted off, even considering the left + // bit shift. + len = 0; + return; + } + // Now we're allocating a positive amount. + // + 1: room for high bits nudged left into another block + len = a.len + 1 - rightShiftBlocks; + allocate(len); + Index i, j; + for (j = rightShiftBlocks, i = 0; j <= a.len; j++, i++) + blk[i] = getShiftedBlock(a, j, leftShiftBits); + // Zap possible leading zero + if (blk[len - 1] == 0) + len--; +} + +// INCREMENT/DECREMENT OPERATORS + +// Prefix increment +void BigUnsigned::operator ++() { + Index i; + bool carry = true; + for (i = 0; i < len && carry; i++) { + blk[i]++; + carry = (blk[i] == 0); + } + if (carry) { + // Allocate and then increase length, as in divideWithRemainder + allocateAndCopy(len + 1); + len++; + blk[i] = 1; + } +} + +// Postfix increment: same as prefix +void BigUnsigned::operator ++(int) { + operator ++(); +} + +// Prefix decrement +void BigUnsigned::operator --() { + if (len == 0) + throw "BigUnsigned::operator --(): Cannot decrement an unsigned zero"; + Index i; + bool borrow = true; + for (i = 0; borrow; i++) { + borrow = (blk[i] == 0); + blk[i]--; + } + // Zap possible leading zero (there can only be one) + if (blk[len - 1] == 0) + len--; +} + +// Postfix decrement: same as prefix +void BigUnsigned::operator --(int) { + operator --(); +} |