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71 lines
2.6 KiB
C
71 lines
2.6 KiB
C
#include <stdint.h>
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/*-
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* Copyright (c) 2019 Christophe Meessen
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*
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* Permission is hereby granted, free of charge, to any person obtaining a copy
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* of this software and associated documentation files (the "Software"), to deal
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* in the Software without restriction, including without limitation the rights
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* to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
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* copies of the Software, and to permit persons to whom the Software is
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* furnished to do so, subject to the following conditions:
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*
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* The above copyright notice and this permission notice shall be included in all
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* copies or substantial portions of the Software.
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*
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* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
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* IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
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* FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
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* AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
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* LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
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* OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
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* SOFTWARE. */
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/*-
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* Computing the square root of an integer or a fixed point integer into a fixed
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* point integer. A fixed point is a 32 bit value with the comma between the
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* bits 15 and 16, where bit 0 is the less significant bit of the value.
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*
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* The algorithm uses the property that computing x² is trivial compared to the
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* sqrt. It will thus search the biggest x so that x² <= v, assuming we compute
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* sqrt(v).
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*
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* The algorithm tests each bit of x starting with the most significant toward
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* the less significant. It tests if the bit must be set or not.
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*
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* The algorithm uses the relation (x + a)² = x² + 2ax + a² to add the bit
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* efficiently. Instead of computing x² it keeps track of (x + a)² - x².
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*
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* When computing sqrt(v), r = v - x², q = 2ax, b = a² and t = 2ax + a2.
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*
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* Note that the input integers are signed and that the sign bit is used in the
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* computation. To accept unsigned integer as input, unfolding the initial loop
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* is required to handle this particular case. See the usenet discussion for the
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* proposed solution.
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*
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* Algorithm and code Author Christophe Meessen 1993. */
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static int32_t sqrt_fix16(int32_t value)
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{
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uint32_t t;
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uint32_t r = value;
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uint32_t b = 0x40000000;
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uint32_t q = 0;
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while (b > 0x40) {
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t = q + b;
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if (r >= t) {
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r -= t;
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q = t + b; /* Equivalent to q += 2 * b */
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}
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r <<= 1;
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b >>= 1;
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}
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q >>= 8;
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return q;
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}
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