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use crate::{
add::{add_in_place, add_word_in_place, sub_in_place, sub_one_in_place},
arch::word::{DoubleWord, Word},
div,
math::FastDivideNormalized2,
memory::{self, Memory},
mul::add_mul_word_in_place,
primitive::{double_word, extend_word, highest_dword, split_dword, WORD_BITS},
shift::shr_in_place_with_carry,
sqr,
};
use alloc::alloc::Layout;
use dashu_base::{DivRem, SquareRootRem};
// n is the size of the output, or half the size of the input
pub fn memory_requirement_sqrt_rem(n: usize) -> Layout {
if n == 2 {
memory::zero_layout()
} else {
// We need to perform a squaring with n words and an n by n/2 division
memory::max_layout(
sqr::memory_requirement_exact(n),
div::memory_requirement_exact(n, n - n / 2),
)
}
}
// Requires a is normalized to 2n words (length must be even)
// Returns the carry of the remainder
pub fn sqrt_rem(b: &mut [Word], a: &mut [Word], memory: &mut Memory) -> bool {
debug_assert!(a.len() % 2 == 0);
debug_assert!(a.len() >= 4, "use native sqrt when a has less than 2 words");
debug_assert!(a.len() == b.len() * 2);
// shortcut when a has exactly 4 words
if a.len() == 4 {
return sqrt_rem_42(b, a);
}
/*
* the "Karatsuba Square Root" algorithm:
* assume n = a*B^2 + b1*B + b0, B=2^k, a has 2k bits and
* is normalized (the top two bits of a are not all zeros)
* 1. calculate sqrt on high part:
* s1, r1 = sqrt_rem(a) (r1 <= 2*s1)
* 2. estimate the root with low part
* q, u = div_rem(r1*B + b1, 2*s1)
* s = s1*B + q
* r = u*B + b0 - q^2
* at this step, since a is normalized, we have s1 >= B/2,
* therefore q <= floor((r1*B + b1) / B) <= r1 <= 2*s1
* also notice b1 < B <= 2*s1, so q <= B
*
* 3. if a3 is normalized, then s is either correct or 1 too big.
* r is negative in the latter case, needs adjustment
* if r < 0 {
* r += 2*s - 1
* s -= 1
* }
*
* Reference: Zimmermann, P. (1999). Karatsuba square root (Doctoral dissertation, INRIA).
* https://hal.inria.fr/inria-00072854/en/
*/
let n = a.len() / 2; // the length of a
let split = n / 2; // the length of b0
// step1: sqrt on the higher half
// afterwards, s1 = b[split..], r1 = a[2*split..split + n]
let r1_top = sqrt_rem(&mut b[split..], &mut a[2 * split..], memory);
if r1_top {
// if the remainder `r1` has a carry, subtract `s1` from it so that the carry is removed
// so later when calculate 2*q = (r1*B + b1) / s1, the result is actually one less
let carry = sub_in_place(&mut a[2 * split..split + n], &b[split..]);
debug_assert!(carry);
}
// step2: estimate the result with lower half
let fast_div_top = FastDivideNormalized2::new(highest_dword(b));
let carry = div::div_rem_in_place(&mut a[split..split + n], &b[split..], fast_div_top, memory);
let (a_lo, a_hi) = a.split_at_mut(n);
b[..split].copy_from_slice(&a_hi[..split]);
// by now 2*q = b[..split], u = a[split..n], carry is true only if r1 >= s1.
// also notice that r1 <= 2 * s1, if r1 was subtracted by s1, then r1 <= s1.
// so r_top and carry are both true only if r1 == 2 * s1 at the beginning.
// the top bit of q is true if either r_top or carry is true, but not both
let _ =
shr_in_place_with_carry(&mut b[..split], 1, ((r1_top ^ carry) as Word) << (WORD_BITS - 1));
let q_top = r1_top && carry; // true only when q = B, and then b[..split] = 0
let mut c = 0i8; // stores final carry (top bit) of the remainder
if a_hi[0] & 1 != 0 {
// this step fixes the error in u caused by using s1 as divisor instead of 2*s1
c = add_in_place(&mut a_lo[split..], &b[split..]) as i8;
}
// store q^2 in high part of a, ignoring q_top.
// afterwards, the q_top flag will be considered in the subtraction,
a_hi.fill(0);
if !q_top {
// if q_top is True, then q^2 = B^2, so we don't need to do squaring
if split == 1 {
let (b2_lo, b2_hi) = split_dword(extend_word(b[0]) * extend_word(b[0]));
a_hi[0] = b2_lo;
a_hi[1] = b2_hi;
} else {
sqr::sqr(&mut a_hi[..2 * split], &b[..split], memory);
}
}
if 2 * split < n {
a_hi[2 * split] = q_top as Word;
} else {
c -= q_top as i8;
}
c -= sub_in_place(a_lo, a_hi) as i8;
// step3: fix the estimation error if necessary
if c < 0 {
// r += 2*s - 1; s -= 1;
// apply the q_top to s first, and then adjust s and r
let overflow = add_word_in_place(&mut b[split..], q_top as _);
c += add_mul_word_in_place(a_lo, 2, b) as i8 + 2 * overflow as i8;
c -= sub_one_in_place(a_lo) as i8;
let borrow = sub_one_in_place(b);
debug_assert!(!(overflow ^ borrow)); // borrow should happen if and only if when overflow is true
}
c > 0
}
// Special case when a has exactly 4 Words
fn sqrt_rem_42(b: &mut [Word], a: &mut [Word]) -> bool {
debug_assert!(a.len() == 4 && b.len() == 2);
// see sqrt_rem() for algorithm explanation
// step1: sqrt on the higher half
let (s1, r1) = highest_dword(a).sqrt_rem();
let s1 = s1 as Word;
// step2: estimate the result with lower half
// here r0 = (r1*B + b1) / 2
let (r1_lo, r1_hi) = split_dword(r1);
let r0_hi = r1_hi << (WORD_BITS - 1) | r1_lo >> 1;
let r0_lo = r1_lo << (WORD_BITS - 1) | a[1] >> 1;
let (mut q, mut u) = double_word(r0_lo, r0_hi).div_rem(s1 as DoubleWord);
if q >> WORD_BITS > 0 {
// if q >= B (then q = B), reduce the overestimate
q -= 1;
u += s1 as DoubleWord;
}
u = u << 1 | (a[1] & 1) as DoubleWord;
let q = q as Word; // now q must fit in a Word
let (u_lo, u_hi) = split_dword(u);
let mut s = double_word(q, s1);
let q2 = extend_word(q) * extend_word(q);
let (mut r, borrow) = double_word(a[0], u_lo).overflowing_sub(q2);
let mut c: i8 = u_hi as i8 - borrow as i8;
// step3: fix the estimation error if necessary
if c < 0 {
let (new_r, c1) = r.overflowing_add(s);
s -= 1;
let (new_r, c2) = new_r.overflowing_add(s);
c += c1 as i8 + c2 as i8;
r = new_r;
}
let (r_lo, r_hi) = split_dword(r);
let (s_lo, s_hi) = split_dword(s);
a[0] = r_lo;
a[1] = r_hi;
b[0] = s_lo;
b[1] = s_hi;
c > 0
}