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1.1 root 1: /* $NetBSD: fpu_sqrt.c,v 1.8 2013/03/26 11:30:21 isaki Exp $ */
2:
3: /*
4: * Copyright (c) 1992, 1993
5: * The Regents of the University of California. All rights reserved.
6: *
7: * This software was developed by the Computer Systems Engineering group
8: * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
9: * contributed to Berkeley.
10: *
11: * All advertising materials mentioning features or use of this software
12: * must display the following acknowledgement:
13: * This product includes software developed by the University of
14: * California, Lawrence Berkeley Laboratory.
15: *
16: * Redistribution and use in source and binary forms, with or without
17: * modification, are permitted provided that the following conditions
18: * are met:
19: * 1. Redistributions of source code must retain the above copyright
20: * notice, this list of conditions and the following disclaimer.
21: * 2. Redistributions in binary form must reproduce the above copyright
22: * notice, this list of conditions and the following disclaimer in the
23: * documentation and/or other materials provided with the distribution.
24: * 3. Neither the name of the University nor the names of its contributors
25: * may be used to endorse or promote products derived from this software
26: * without specific prior written permission.
27: *
28: * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
29: * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
30: * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
31: * ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
32: * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
33: * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
34: * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
35: * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
36: * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
37: * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
38: * SUCH DAMAGE.
39: *
40: * @(#)fpu_sqrt.c 8.1 (Berkeley) 6/11/93
41: */
42:
43: /*
44: * Perform an FPU square root (return sqrt(x)).
45: */
46:
47: #include "fpu_arith.h"
48: #include "fpu_emulate.h"
49:
50: /*
51: * Our task is to calculate the square root of a floating point number x0.
52: * This number x normally has the form:
53: *
54: * exp
55: * x = mant * 2 (where 1 <= mant < 2 and exp is an integer)
56: *
57: * This can be left as it stands, or the mantissa can be doubled and the
58: * exponent decremented:
59: *
60: * exp-1
61: * x = (2 * mant) * 2 (where 2 <= 2 * mant < 4)
62: *
63: * If the exponent `exp' is even, the square root of the number is best
64: * handled using the first form, and is by definition equal to:
65: *
66: * exp/2
67: * sqrt(x) = sqrt(mant) * 2
68: *
69: * If exp is odd, on the other hand, it is convenient to use the second
70: * form, giving:
71: *
72: * (exp-1)/2
73: * sqrt(x) = sqrt(2 * mant) * 2
74: *
75: * In the first case, we have
76: *
77: * 1 <= mant < 2
78: *
79: * and therefore
80: *
81: * sqrt(1) <= sqrt(mant) < sqrt(2)
82: *
83: * while in the second case we have
84: *
85: * 2 <= 2*mant < 4
86: *
87: * and therefore
88: *
89: * sqrt(2) <= sqrt(2*mant) < sqrt(4)
90: *
91: * so that in any case, we are sure that
92: *
93: * sqrt(1) <= sqrt(n * mant) < sqrt(4), n = 1 or 2
94: *
95: * or
96: *
97: * 1 <= sqrt(n * mant) < 2, n = 1 or 2.
98: *
99: * This root is therefore a properly formed mantissa for a floating
100: * point number. The exponent of sqrt(x) is either exp/2 or (exp-1)/2
101: * as above. This leaves us with the problem of finding the square root
102: * of a fixed-point number in the range [1..4).
103: *
104: * Though it may not be instantly obvious, the following square root
105: * algorithm works for any integer x of an even number of bits, provided
106: * that no overflows occur:
107: *
108: * let q = 0
109: * for k = NBITS-1 to 0 step -1 do -- for each digit in the answer...
110: * x *= 2 -- multiply by radix, for next digit
111: * if x >= 2q + 2^k then -- if adding 2^k does not
112: * x -= 2q + 2^k -- exceed the correct root,
113: * q += 2^k -- add 2^k and adjust x
114: * fi
115: * done
116: * sqrt = q / 2^(NBITS/2) -- (and any remainder is in x)
117: *
118: * If NBITS is odd (so that k is initially even), we can just add another
119: * zero bit at the top of x. Doing so means that q is not going to acquire
120: * a 1 bit in the first trip around the loop (since x0 < 2^NBITS). If the
121: * final value in x is not needed, or can be off by a factor of 2, this is
122: * equivalant to moving the `x *= 2' step to the bottom of the loop:
123: *
124: * for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done
125: *
126: * and the result q will then be sqrt(x0) * 2^floor(NBITS / 2).
127: * (Since the algorithm is destructive on x, we will call x's initial
128: * value, for which q is some power of two times its square root, x0.)
129: *
130: * If we insert a loop invariant y = 2q, we can then rewrite this using
131: * C notation as:
132: *
133: * q = y = 0; x = x0;
134: * for (k = NBITS; --k >= 0;) {
135: * #if (NBITS is even)
136: * x *= 2;
137: * #endif
138: * t = y + (1 << k);
139: * if (x >= t) {
140: * x -= t;
141: * q += 1 << k;
142: * y += 1 << (k + 1);
143: * }
144: * #if (NBITS is odd)
145: * x *= 2;
146: * #endif
147: * }
148: *
149: * If x0 is fixed point, rather than an integer, we can simply alter the
150: * scale factor between q and sqrt(x0). As it happens, we can easily arrange
151: * for the scale factor to be 2**0 or 1, so that sqrt(x0) == q.
152: *
153: * In our case, however, x0 (and therefore x, y, q, and t) are multiword
154: * integers, which adds some complication. But note that q is built one
155: * bit at a time, from the top down, and is not used itself in the loop
156: * (we use 2q as held in y instead). This means we can build our answer
157: * in an integer, one word at a time, which saves a bit of work. Also,
158: * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are
159: * `new' bits in y and we can set them with an `or' operation rather than
160: * a full-blown multiword add.
161: *
162: * We are almost done, except for one snag. We must prove that none of our
163: * intermediate calculations can overflow. We know that x0 is in [1..4)
164: * and therefore the square root in q will be in [1..2), but what about x,
165: * y, and t?
166: *
167: * We know that y = 2q at the beginning of each loop. (The relation only
168: * fails temporarily while y and q are being updated.) Since q < 2, y < 4.
169: * The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and.
170: * Furthermore, we can prove with a bit of work that x never exceeds y by
171: * more than 2, so that even after doubling, 0 <= x < 8. (This is left as
172: * an exercise to the reader, mostly because I have become tired of working
173: * on this comment.)
174: *
175: * If our floating point mantissas (which are of the form 1.frac) occupy
176: * B+1 bits, our largest intermediary needs at most B+3 bits, or two extra.
177: * In fact, we want even one more bit (for a carry, to avoid compares), or
178: * three extra. There is a comment in fpu_emu.h reminding maintainers of
179: * this, so we have some justification in assuming it.
180: */
181: struct fpn *
182: fpu_sqrt(struct fpemu *fe)
183: {
184: struct fpn *x = &fe->fe_f2;
185: uint32_t bit, q, tt;
186: uint32_t x0, x1, x2;
187: uint32_t y0, y1, y2;
188: uint32_t d0, d1, d2;
189: int e;
190: FPU_DECL_CARRY
191:
192: /*
193: * Take care of special cases first. In order:
194: *
195: * sqrt(NaN) = NaN
196: * sqrt(+0) = +0
197: * sqrt(-0) = -0
198: * sqrt(x < 0) = NaN (including sqrt(-Inf))
199: * sqrt(+Inf) = +Inf
200: *
201: * Then all that remains are numbers with mantissas in [1..2).
202: */
203: if (ISNAN(x) || ISZERO(x))
204: return (x);
205: if (x->fp_sign)
206: return (fpu_newnan(fe));
207: if (ISINF(x))
208: return (x);
209:
210: /*
211: * Calculate result exponent. As noted above, this may involve
212: * doubling the mantissa. We will also need to double x each
213: * time around the loop, so we define a macro for this here, and
214: * we break out the multiword mantissa.
215: */
216: #ifdef FPU_SHL1_BY_ADD
217: #define DOUBLE_X { \
218: FPU_ADDS(x2, x2, x2); \
219: FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \
220: }
221: #else
222: #define DOUBLE_X { \
223: x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \
224: x2 <<= 1; \
225: }
226: #endif
227: #if (FP_NMANT & 1) != 0
228: # define ODD_DOUBLE DOUBLE_X
229: # define EVEN_DOUBLE /* nothing */
230: #else
231: # define ODD_DOUBLE /* nothing */
232: # define EVEN_DOUBLE DOUBLE_X
233: #endif
234: x0 = x->fp_mant[0];
235: x1 = x->fp_mant[1];
236: x2 = x->fp_mant[2];
237: e = x->fp_exp;
238: if (e & 1) /* exponent is odd; use sqrt(2mant) */
239: DOUBLE_X;
240: /* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */
241: x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */
242:
243: /*
244: * Now calculate the mantissa root. Since x is now in [1..4),
245: * we know that the first trip around the loop will definitely
246: * set the top bit in q, so we can do that manually and start
247: * the loop at the next bit down instead. We must be sure to
248: * double x correctly while doing the `known q=1.0'.
249: *
250: * We do this one mantissa-word at a time, as noted above, to
251: * save work. To avoid `(1 << 31) << 1', we also do the top bit
252: * outside of each per-word loop.
253: *
254: * The calculation `t = y + bit' breaks down into `t0 = y0, ...,
255: * t2 = y2, t? |= bit' for the appropriate word. Since the bit
256: * is always a `new' one, this means that three of the `t?'s are
257: * just the corresponding `y?'; we use `#define's here for this.
258: * The variable `tt' holds the actual `t?' variable.
259: */
260:
261: /* calculate q0 */
262: #define t0 tt
263: bit = FP_1;
264: EVEN_DOUBLE;
265: /* if (x >= (t0 = y0 | bit)) { */ /* always true */
266: q = bit;
267: x0 -= bit;
268: y0 = bit << 1;
269: /* } */
270: ODD_DOUBLE;
271: while ((bit >>= 1) != 0) { /* for remaining bits in q0 */
272: EVEN_DOUBLE;
273: t0 = y0 | bit; /* t = y + bit */
274: if (x0 >= t0) { /* if x >= t then */
275: x0 -= t0; /* x -= t */
276: q |= bit; /* q += bit */
277: y0 |= bit << 1; /* y += bit << 1 */
278: }
279: ODD_DOUBLE;
280: }
281: x->fp_mant[0] = q;
282: #undef t0
283:
284: /* calculate q1. note (y0&1)==0. */
285: #define t0 y0
286: #define t1 tt
287: q = 0;
288: y1 = 0;
289: bit = 1 << 31;
290: EVEN_DOUBLE;
291: t1 = bit;
292: FPU_SUBS(d1, x1, t1);
293: FPU_SUBC(d0, x0, t0); /* d = x - t */
294: if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */
295: x0 = d0, x1 = d1; /* x -= t */
296: q = bit; /* q += bit */
297: y0 |= 1; /* y += bit << 1 */
298: }
299: ODD_DOUBLE;
300: while ((bit >>= 1) != 0) { /* for remaining bits in q1 */
301: EVEN_DOUBLE; /* as before */
302: t1 = y1 | bit;
303: FPU_SUBS(d1, x1, t1);
304: FPU_SUBC(d0, x0, t0);
305: if ((int)d0 >= 0) {
306: x0 = d0, x1 = d1;
307: q |= bit;
308: y1 |= bit << 1;
309: }
310: ODD_DOUBLE;
311: }
312: x->fp_mant[1] = q;
313: #undef t1
314:
315: /* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */
316: #define t1 y1
317: #define t2 tt
318: q = 0;
319: y2 = 0;
320: bit = 1 << 31;
321: EVEN_DOUBLE;
322: t2 = bit;
323: FPU_SUBS(d2, x2, t2);
324: FPU_SUBCS(d1, x1, t1);
325: FPU_SUBC(d0, x0, t0);
326: if ((int)d0 >= 0) {
327: x0 = d0, x1 = d1, x2 = d2;
328: q |= bit;
329: y1 |= 1; /* now t1, y1 are set in concrete */
330: }
331: ODD_DOUBLE;
332: while ((bit >>= 1) != 0) {
333: EVEN_DOUBLE;
334: t2 = y2 | bit;
335: FPU_SUBS(d2, x2, t2);
336: FPU_SUBCS(d1, x1, t1);
337: FPU_SUBC(d0, x0, t0);
338: if ((int)d0 >= 0) {
339: x0 = d0, x1 = d1, x2 = d2;
340: q |= bit;
341: y2 |= bit << 1;
342: }
343: ODD_DOUBLE;
344: }
345: x->fp_mant[2] = q;
346: #undef t2
347:
348: /*
349: * The result, which includes guard and round bits, is exact iff
350: * x is now zero; any nonzero bits in x represent sticky bits.
351: */
352: x->fp_sticky = x0 | x1 | x2;
353: return (x);
354: }
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