/*-
 * Copyright (c) 2015-2017 Taylor R. Campbell
 * All rights reserved.
 *
 * 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.
 *
 * THIS SOFTWARE IS PROVIDED BY THE AUTHOR 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 AUTHOR 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.
 */

#ifndef	MOD311_H
#define	MOD311_H

#include <stdint.h>

/*
 * Arithmetic modulo 2^31 - 1
 *
 * We represent a residue class of Z/(2^31 - 1)Z by a 32-bit
 * representative, not necessarily fully reduced.  Callers requiring
 * the least representative must use reduce311.
 */

#define	M31	UINT32_C(0x7fffffff)

/*
 * reduce311(x)
 *
 *	Given 0 <= x <= 2^32 - 1, yield a residue of x modulo 2^31 - 1.
 *	For 0 < x < 2^32 - 1, this is the least positive residue;
 *	otherwise, reduce311(0) = 0 and reduce311(2^32 - 1) = 2^31.
 *	Note that reduce311(2^31 - 1) = 2^31 - 1.
 */
static inline uint32_t
reduce311(uint32_t x)
{

	/* 2^31 = 1 (mod 2^31 - 1), so 2^31 h + l = h + l (mod 2^31 - 1).  */
	return (x >> 31) + (x & M31);
}

/*
 * add311(a, b)
 *
 *	Given a, b <= 2^31 - 1, compute an integer x <= 2^31 - 1
 *	congruent to a + b modulo 2^31 - 1.
 */
static inline uint32_t
add311(uint32_t a, uint32_t b)
{

	/*
	 * a, b <= 2^31 - 1, so a + b <= 2^32 - 2 < 2^32.  Further,
	 * reduce311(2^32 - 2) = 1 + (2^31 - 2) = 2^31 - 1 is the
	 * greatest possible image.
	 */
	return reduce311(a + b);
}

/*
 * neg311(x)
 *
 *	Given x <= 2^31 - 1, compute an integer y <= 2^31 - 1 congruent
 *	to 0 - x modulo 2^31 - 1.
 */
static inline uint32_t
neg311(uint32_t x)
{

	/* x <= 2^32 - 1, so 2^32 - 1 - x <= 2^32 - 1.  */
	return M31 - x;
}

/*
 * mul311(a, b)
 *
 *	Given a, b <= 2^31 - 1, compute an integer x <= 2^31 - 1
 *	congruent to a*b modulo 2^31 - 1.
 */
static inline uint32_t
mul311(uint32_t a, uint32_t b)
{
	const uint64_t p = ((uint64_t)a * (uint64_t)b);

	/*
	 * a, b <= 2^31 - 1, so that p = a * b <= 2^62 - 2^32 + 1.
	 * Hence (p >> 31) <= 2^31 - 2 and (p & M31) <= 2^31 - 1, so
	 * we can safely use add311 to do the reduction.
	 */
	return add311(p >> 31, p & M31);
}

#endif	/* MOD311_H */