/* $NetBSD: fpu_sqrt.c,v 1.7 2022/08/28 22:09:26 rin Exp $ */
/*
* Copyright (c) 1992, 1993
* The Regents of the University of California. All rights reserved.
*
* This software was developed by the Computer Systems Engineering group
* at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
* contributed to Berkeley.
*
* All advertising materials mentioning features or use of this software
* must display the following acknowledgement:
* This product includes software developed by the University of
* California, Lawrence Berkeley Laboratory.
*
* Redistribution and use in source and binary forms, with or without
* modification, are permitted provided that the following conditions
* are met:
* 1. Redistributions of source code must retain the above copyright
* notice, this list of conditions and the following disclaimer.
* 2. Redistributions in binary form must reproduce the above copyright
* notice, this list of conditions and the following disclaimer in the
* documentation and/or other materials provided with the distribution.
* 3. Neither the name of the University nor the names of its contributors
* may be used to endorse or promote products derived from this software
* without specific prior written permission.
*
* THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
* ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
* IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
* ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
* FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
* DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
* OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
* HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
* LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
* OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
* SUCH DAMAGE.
*
* @(#)fpu_sqrt.c 8.1 (Berkeley) 6/11/93
*/
/*
* Perform an FPU square root (return sqrt(x)).
*/
/*
* Our task is to calculate the square root of a floating point number x0.
* This number x normally has the form:
*
* exp
* x = mant * 2 (where 1 <= mant < 2 and exp is an integer)
*
* This can be left as it stands, or the mantissa can be doubled and the
* exponent decremented:
*
* exp-1
* x = (2 * mant) * 2 (where 2 <= 2 * mant < 4)
*
* If the exponent `exp' is even, the square root of the number is best
* handled using the first form, and is by definition equal to:
*
* exp/2
* sqrt(x) = sqrt(mant) * 2
*
* If exp is odd, on the other hand, it is convenient to use the second
* form, giving:
*
* (exp-1)/2
* sqrt(x) = sqrt(2 * mant) * 2
*
* In the first case, we have
*
* 1 <= mant < 2
*
* and therefore
*
* sqrt(1) <= sqrt(mant) < sqrt(2)
*
* while in the second case we have
*
* 2 <= 2*mant < 4
*
* and therefore
*
* sqrt(2) <= sqrt(2*mant) < sqrt(4)
*
* so that in any case, we are sure that
*
* sqrt(1) <= sqrt(n * mant) < sqrt(4), n = 1 or 2
*
* or
*
* 1 <= sqrt(n * mant) < 2, n = 1 or 2.
*
* This root is therefore a properly formed mantissa for a floating
* point number. The exponent of sqrt(x) is either exp/2 or (exp-1)/2
* as above. This leaves us with the problem of finding the square root
* of a fixed-point number in the range [1..4).
*
* Though it may not be instantly obvious, the following square root
* algorithm works for any integer x of an even number of bits, provided
* that no overflows occur:
*
* let q = 0
* for k = NBITS-1 to 0 step -1 do -- for each digit in the answer...
* x *= 2 -- multiply by radix, for next digit
* if x >= 2q + 2^k then -- if adding 2^k does not
* x -= 2q + 2^k -- exceed the correct root,
* q += 2^k -- add 2^k and adjust x
* fi
* done
* sqrt = q / 2^(NBITS/2) -- (and any remainder is in x)
*
* If NBITS is odd (so that k is initially even), we can just add another
* zero bit at the top of x. Doing so means that q is not going to acquire
* a 1 bit in the first trip around the loop (since x0 < 2^NBITS). If the
* final value in x is not needed, or can be off by a factor of 2, this is
* equivalent to moving the `x *= 2' step to the bottom of the loop:
*
* for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done
*
* and the result q will then be sqrt(x0) * 2^floor(NBITS / 2).
* (Since the algorithm is destructive on x, we will call x's initial
* value, for which q is some power of two times its square root, x0.)
*
* If we insert a loop invariant y = 2q, we can then rewrite this using
* C notation as:
*
* q = y = 0; x = x0;
* for (k = NBITS; --k >= 0;) {
* #if (NBITS is even)
* x *= 2;
* #endif
* t = y + (1 << k);
* if (x >= t) {
* x -= t;
* q += 1 << k;
* y += 1 << (k + 1);
* }
* #if (NBITS is odd)
* x *= 2;
* #endif
* }
*
* If x0 is fixed point, rather than an integer, we can simply alter the
* scale factor between q and sqrt(x0). As it happens, we can easily arrange
* for the scale factor to be 2**0 or 1, so that sqrt(x0) == q.
*
* In our case, however, x0 (and therefore x, y, q, and t) are multiword
* integers, which adds some complication. But note that q is built one
* bit at a time, from the top down, and is not used itself in the loop
* (we use 2q as held in y instead). This means we can build our answer
* in an integer, one word at a time, which saves a bit of work. Also,
* since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are
* `new' bits in y and we can set them with an `or' operation rather than
* a full-blown multiword add.
*
* We are almost done, except for one snag. We must prove that none of our
* intermediate calculations can overflow. We know that x0 is in [1..4)
* and therefore the square root in q will be in [1..2), but what about x,
* y, and t?
*
* We know that y = 2q at the beginning of each loop. (The relation only
* fails temporarily while y and q are being updated.) Since q < 2, y < 4.
* The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and.
* Furthermore, we can prove with a bit of work that x never exceeds y by
* more than 2, so that even after doubling, 0 <= x < 8. (This is left as
* an exercise to the reader, mostly because I have become tired of working
* on this comment.)
*
* If our floating point mantissas (which are of the form 1.frac) occupy
* B+1 bits, our largest intermediary needs at most B+3 bits, or two extra.
* In fact, we want even one more bit (for a carry, to avoid compares), or
* three extra. There is a comment in fpu_emu.h reminding maintainers of
* this, so we have some justification in assuming it.
*/
struct fpn *
fpu_sqrt(struct fpemu *fe)
{
struct fpn *x = &fe->fe_f1;
u_int bit, q, tt;
u_int x0, x1, x2, x3;
u_int y0, y1, y2, y3;
u_int d0, d1, d2, d3;
int e;
/*
* Take care of special cases first. In order:
*
* sqrt(NaN) = NaN
* sqrt(+0) = +0
* sqrt(-0) = -0
* sqrt(x < 0) = NaN (including sqrt(-Inf))
* sqrt(+Inf) = +Inf
*
* Then all that remains are numbers with mantissas in [1..2).
*/
if (ISNAN(x) || ISZERO(x))
return (x);
if (x->fp_sign)
return (fpu_newnan(fe));
if (ISINF(x))
return (x);
/*
* Calculate result exponent. As noted above, this may involve
* doubling the mantissa. We will also need to double x each
* time around the loop, so we define a macro for this here, and
* we break out the multiword mantissa.
*/
#ifdef FPU_SHL1_BY_ADD
#define DOUBLE_X { \
FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \
FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \
}
#else
#define DOUBLE_X { \
x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \
x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \
}
#endif
#if (FP_NMANT & 1) != 0
# define ODD_DOUBLE DOUBLE_X
# define EVEN_DOUBLE /* nothing */
#else
# define ODD_DOUBLE /* nothing */
# define EVEN_DOUBLE DOUBLE_X
#endif
x0 = x->fp_mant[0];
x1 = x->fp_mant[1];
x2 = x->fp_mant[2];
x3 = x->fp_mant[3];
e = x->fp_exp;
if (e & 1) /* exponent is odd; use sqrt(2mant) */
DOUBLE_X;
/* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */
x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */
/*
* Now calculate the mantissa root. Since x is now in [1..4),
* we know that the first trip around the loop will definitely
* set the top bit in q, so we can do that manually and start
* the loop at the next bit down instead. We must be sure to
* double x correctly while doing the `known q=1.0'.
*
* We do this one mantissa-word at a time, as noted above, to
* save work. To avoid `(1 << 31) << 1', we also do the top bit
* outside of each per-word loop.
*
* The calculation `t = y + bit' breaks down into `t0 = y0, ...,
* t3 = y3, t? |= bit' for the appropriate word. Since the bit
* is always a `new' one, this means that three of the `t?'s are
* just the corresponding `y?'; we use `#define's here for this.
* The variable `tt' holds the actual `t?' variable.
*/
/* calculate q0 */
#define t0 tt
bit = FP_1;
EVEN_DOUBLE;
/* if (x >= (t0 = y0 | bit)) { */ /* always true */
q = bit;
x0 -= bit;
y0 = bit << 1;
/* } */
ODD_DOUBLE;
while ((bit >>= 1) != 0) { /* for remaining bits in q0 */
EVEN_DOUBLE;
t0 = y0 | bit; /* t = y + bit */
if (x0 >= t0) { /* if x >= t then */
x0 -= t0; /* x -= t */
q |= bit; /* q += bit */
y0 |= bit << 1; /* y += bit << 1 */
}
ODD_DOUBLE;
}
x->fp_mant[0] = q;
#undef t0
/* calculate q1. note (y0&1)==0. */
#define t0 y0
#define t1 tt
q = 0;
y1 = 0;
bit = 1 << 31;
EVEN_DOUBLE;
t1 = bit;
FPU_SUBS(d1, x1, t1);
FPU_SUBC(d0, x0, t0); /* d = x - t */
if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */
x0 = d0, x1 = d1; /* x -= t */
q = bit; /* q += bit */
y0 |= 1; /* y += bit << 1 */
}
ODD_DOUBLE;
while ((bit >>= 1) != 0) { /* for remaining bits in q1 */
EVEN_DOUBLE; /* as before */
t1 = y1 | bit;
FPU_SUBS(d1, x1, t1);
FPU_SUBC(d0, x0, t0);
if ((int)d0 >= 0) {
x0 = d0, x1 = d1;
q |= bit;
y1 |= bit << 1;
}
ODD_DOUBLE;
}
x->fp_mant[1] = q;
#undef t1
/*
* The result, which includes guard and round bits, is exact iff
* x is now zero; any nonzero bits in x represent sticky bits.
*/
x->fp_sticky = x0 | x1 | x2 | x3;
return (x);
}
