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// Copyright 2021 the V8 project authors. All rights reserved.
// Use of this source code is governed by a BSD-style license that can be
// found in the LICENSE file.
// Burnikel-Ziegler division.
// Reference: "Fast Recursive Division" by Christoph Burnikel and Joachim
// Ziegler, found at http://cr.yp.to/bib/1998/burnikel.ps
#include <string.h>
#include "src/bigint/bigint-internal.h"
#include "src/bigint/digit-arithmetic.h"
#include "src/bigint/div-helpers.h"
#include "src/bigint/util.h"
#include "src/bigint/vector-arithmetic.h"
namespace v8 {
namespace bigint {
namespace {
// Compares [a_high, A] with B.
// Returns:
// - a value < 0 if [a_high, A] < B
// - 0 if [a_high, A] == B
// - a value > 0 if [a_high, A] > B.
int SpecialCompare(digit_t a_high, Digits A, Digits B) {
B.Normalize();
int a_len;
if (a_high == 0) {
A.Normalize();
a_len = A.len();
} else {
a_len = A.len() + 1;
}
int diff = a_len - B.len();
if (diff != 0) return diff;
int i = a_len - 1;
if (a_high != 0) {
if (a_high > B[i]) return 1;
if (a_high < B[i]) return -1;
i--;
}
while (i >= 0 && A[i] == B[i]) i--;
if (i < 0) return 0;
return A[i] > B[i] ? 1 : -1;
}
void SetOnes(RWDigits X) {
memset(X.digits(), 0xFF, X.len() * sizeof(digit_t));
}
// Since the Burnikel-Ziegler method is inherently recursive, we put
// non-changing data into a container object.
class BZ {
public:
BZ(ProcessorImpl* proc, int scratch_space)
: proc_(proc),
scratch_mem_(scratch_space >= kBurnikelThreshold ? scratch_space : 0) {}
void DivideBasecase(RWDigits Q, RWDigits R, Digits A, Digits B);
void D3n2n(RWDigits Q, RWDigits R, Digits A1A2, Digits A3, Digits B);
void D2n1n(RWDigits Q, RWDigits R, Digits A, Digits B);
private:
ProcessorImpl* proc_;
Storage scratch_mem_;
};
void BZ::DivideBasecase(RWDigits Q, RWDigits R, Digits A, Digits B) {
A.Normalize();
B.Normalize();
DCHECK(B.len() > 0);
int cmp = Compare(A, B);
if (cmp <= 0) {
Q.Clear();
if (cmp == 0) {
// If A == B, then Q=1, R=0.
R.Clear();
Q[0] = 1;
} else {
// If A < B, then Q=0, R=A.
PutAt(R, A, R.len());
}
return;
}
if (B.len() == 1) {
return proc_->DivideSingle(Q, R.digits(), A, B[0]);
}
return proc_->DivideSchoolbook(Q, R, A, B);
}
// Algorithm 2 from the paper. Variable names same as there.
// Returns Q(uotient) and R(emainder) for A/B, with B having two thirds
// the size of A = [A1, A2, A3].
void BZ::D3n2n(RWDigits Q, RWDigits R, Digits A1A2, Digits A3, Digits B) {
DCHECK((B.len() & 1) == 0);
int n = B.len() / 2;
DCHECK(A1A2.len() == 2 * n);
// Actual condition is stricter than length: A < B * 2^(kDigitBits * n)
DCHECK(Compare(A1A2, B) < 0);
DCHECK(A3.len() == n);
DCHECK(Q.len() == n);
DCHECK(R.len() == 2 * n);
// 1. Split A into three parts A = [A1, A2, A3] with Ai < 2^(kDigitBits * n).
Digits A1(A1A2, n, n);
// 2. Split B into two parts B = [B1, B2] with Bi < 2^(kDigitBits * n).
Digits B1(B, n, n);
Digits B2(B, 0, n);
// 3. Distinguish the cases A1 < B1 or A1 >= B1.
RWDigits Qhat = Q;
RWDigits R1(R, n, n);
digit_t r1_high = 0;
if (Compare(A1, B1) < 0) {
// 3a. If A1 < B1, compute Qhat = floor([A1, A2] / B1) with remainder R1
// using algorithm D2n1n.
D2n1n(Qhat, R1, A1A2, B1);
if (proc_->should_terminate()) return;
} else {
// 3b. If A1 >= B1, set Qhat = 2^(kDigitBits * n) - 1 and set
// R1 = [A1, A2] - [B1, 0] + [0, B1]
SetOnes(Qhat);
// Step 1: compute A1 - B1, which can't underflow because of the comparison
// guarding this else-branch, and always has a one-digit result because
// of this function's preconditions.
RWDigits temp = R1;
Subtract(temp, A1, B1);
temp.Normalize();
DCHECK(temp.len() <= 1);
if (temp.len() > 0) r1_high = temp[0];
// Step 2: compute A2 + B1.
Digits A2(A1A2, 0, n);
r1_high += AddAndReturnCarry(R1, A2, B1);
}
// 4. Compute D = Qhat * B2 using (Karatsuba) multiplication.
RWDigits D(scratch_mem_.get(), 2 * n);
proc_->Multiply(D, Qhat, B2);
if (proc_->should_terminate()) return;
// 5. Compute Rhat = R1*2^(kDigitBits * n) + A3 - D = [R1, A3] - D.
PutAt(R, A3, n);
// 6. As long as Rhat < 0, repeat:
while (SpecialCompare(r1_high, R, D) < 0) {
// 6a. Rhat = Rhat + B
r1_high += AddAndReturnCarry(R, R, B);
// 6b. Qhat = Qhat - 1
Subtract(Qhat, 1);
}
// 5. Compute Rhat = R1*2^(kDigitBits * n) + A3 - D = [R1, A3] - D.
digit_t borrow = SubtractAndReturnBorrow(R, R, D);
DCHECK(borrow == r1_high);
DCHECK(Compare(R, B) < 0);
(void)borrow;
// 7. Return R = Rhat, Q = Qhat.
}
// Algorithm 1 from the paper. Variable names same as there.
// Returns Q(uotient) and (R)emainder for A/B, with A twice the size of B.
void BZ::D2n1n(RWDigits Q, RWDigits R, Digits A, Digits B) {
int n = B.len();
DCHECK(A.len() <= 2 * n);
// A < B * 2^(kDigitsBits * n)
DCHECK(Compare(Digits(A, n, n), B) < 0);
DCHECK(Q.len() == n);
DCHECK(R.len() == n);
// 1. If n is odd or smaller than some convenient constant, compute Q and R
// by school division and return.
if ((n & 1) == 1 || n < kBurnikelThreshold) {
return DivideBasecase(Q, R, A, B);
}
// 2. Split A into four parts A = [A1, ..., A4] with
// Ai < 2^(kDigitBits * n / 2). Split B into two parts [B2, B1] with
// Bi < 2^(kDigitBits * n / 2).
Digits A1A2(A, n, n);
Digits A3(A, n / 2, n / 2);
Digits A4(A, 0, n / 2);
// 3. Compute the high part Q1 of floor(A/B) as
// Q1 = floor([A1, A2, A3] / [B1, B2]) with remainder R1 = [R11, R12],
// using algorithm D3n2n.
RWDigits Q1(Q, n / 2, n / 2);
ScratchDigits R1(n);
D3n2n(Q1, R1, A1A2, A3, B);
if (proc_->should_terminate()) return;
// 4. Compute the low part Q2 of floor(A/B) as
// Q2 = floor([R11, R12, A4] / [B1, B2]) with remainder R, using
// algorithm D3n2n.
RWDigits Q2(Q, 0, n / 2);
D3n2n(Q2, R, R1, A4, B);
// 5. Return Q = [Q1, Q2] and R.
}
} // namespace
// Algorithm 3 from the paper. Variables names same as there.
// Returns Q(uotient) and R(emainder) for A/B (no size restrictions).
// R is optional, Q is not.
void ProcessorImpl::DivideBurnikelZiegler(RWDigits Q, RWDigits R, Digits A,
Digits B) {
DCHECK(A.len() >= B.len());
DCHECK(R.len() == 0 || R.len() >= B.len());
DCHECK(Q.len() > A.len() - B.len());
int r = A.len();
int s = B.len();
// The requirements are:
// - n >= s, n as small as possible.
// - m must be a power of two.
// 1. Set m = min {2^k | 2^k * kBurnikelThreshold > s}.
int m = 1 << BitLength(s / kBurnikelThreshold);
// 2. Set j = roundup(s/m) and n = j * m.
int j = DIV_CEIL(s, m);
int n = j * m;
// 3. Set sigma = max{tao | 2^tao * B < 2^(kDigitBits * n)}.
int sigma = CountLeadingZeros(B[s - 1]);
int digit_shift = n - s;
// 4. Set B = B * 2^sigma to normalize B. Shift A by the same amount.
ScratchDigits B_shifted(n);
LeftShift(B_shifted + digit_shift, B, sigma);
for (int i = 0; i < digit_shift; i++) B_shifted[i] = 0;
B = B_shifted;
// We need an extra digit if A's top digit does not have enough space for
// the left-shift by {sigma}. Additionally, the top bit of A must be 0
// (see "-1" in step 5 below), which combined with B being normalized (i.e.
// B's top bit is 1) ensures the preconditions of the helper functions.
int extra_digit = CountLeadingZeros(A[r - 1]) < (sigma + 1) ? 1 : 0;
r = A.len() + digit_shift + extra_digit;
ScratchDigits A_shifted(r);
LeftShift(A_shifted + digit_shift, A, sigma);
for (int i = 0; i < digit_shift; i++) A_shifted[i] = 0;
A = A_shifted;
// 5. Set t = min{t >= 2 | A < 2^(kDigitBits * t * n - 1)}.
int t = std::max(DIV_CEIL(r, n), 2);
// 6. Split A conceptually into t blocks.
// 7. Set Z_(t-2) = [A_(t-1), A_(t-2)].
int z_len = n * 2;
ScratchDigits Z(z_len);
PutAt(Z, A + n * (t - 2), z_len);
// 8. For i from t-2 downto 0 do:
BZ bz(this, n);
ScratchDigits Ri(n);
{
// First iteration unrolled and specialized.
// We might not have n digits at the top of Q, so use temporary storage
// for Qi...
ScratchDigits Qi(n);
bz.D2n1n(Qi, Ri, Z, B);
if (should_terminate()) return;
// ...but there *will* be enough space for any non-zero result digits!
Qi.Normalize();
RWDigits target = Q + n * (t - 2);
DCHECK(Qi.len() <= target.len());
PutAt(target, Qi, target.len());
}
// Now loop over any remaining iterations.
for (int i = t - 3; i >= 0; i--) {
// 8b. If i > 0, set Z_(i-1) = [Ri, A_(i-1)].
// (De-duped with unrolled first iteration, hence reading A_(i).)
PutAt(Z + n, Ri, n);
PutAt(Z, A + n * i, n);
// 8a. Using algorithm D2n1n compute Qi, Ri such that Zi = B*Qi + Ri.
RWDigits Qi(Q, i * n, n);
bz.D2n1n(Qi, Ri, Z, B);
if (should_terminate()) return;
}
// 9. Return Q = [Q_(t-2), ..., Q_0] and R = R_0 * 2^(-sigma).
#if DEBUG
for (int i = 0; i < digit_shift; i++) {
DCHECK(Ri[i] == 0);
}
#endif
if (R.len() != 0) {
Digits Ri_part(Ri, digit_shift, Ri.len());
Ri_part.Normalize();
DCHECK(Ri_part.len() <= R.len());
RightShift(R, Ri_part, sigma);
}
}
} // namespace bigint
} // namespace v8
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