pqc/external/flint-2.4.3/fq_poly_templates/divrem_divconquer_recursive.c

150 lines
4.9 KiB
C
Raw Normal View History

2014-05-18 22:03:37 +00:00
/*=============================================================================
This file is part of FLINT.
FLINT is free software; you can redistribute it and/or modify
it under the terms of the GNU General Public License as published by
the Free Software Foundation; either version 2 of the License, or
(at your option) any later version.
FLINT is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
GNU General Public License for more details.
You should have received a copy of the GNU General Public License
along with FLINT; if not, write to the Free Software
Foundation, Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA
=============================================================================*/
/******************************************************************************
Copyright (C) 2008, 2009 William Hart
Copyright (C) 2010, 2011 Sebastian Pancratz
Copyright (C) 2013 Mike Hansen
******************************************************************************/
#ifdef T
#include "templates.h"
void
_TEMPLATE(T, poly_divrem_divconquer_recursive) (
TEMPLATE(T, struct) * Q,
TEMPLATE(T, struct) * BQ,
TEMPLATE(T, struct) * W,
const TEMPLATE(T, struct) * A,
const TEMPLATE(T, struct) * B, slong lenB,
const TEMPLATE(T, t) invB,
const TEMPLATE(T, ctx_t) ctx)
{
if (lenB <= TEMPLATE(CAP_T, POLY_DIVREM_DIVCONQUER_CUTOFF))
{
_TEMPLATE(T, vec_zero) (BQ, lenB - 1, ctx);
_TEMPLATE(T, vec_set) (BQ + (lenB - 1), A + (lenB - 1), lenB, ctx);
_TEMPLATE(T, poly_divrem_basecase) (Q, BQ, BQ, 2 * lenB - 1, B, lenB,
invB, ctx);
_TEMPLATE(T, poly_neg) (BQ, BQ, lenB - 1, ctx);
_TEMPLATE(T, vec_set) (BQ + (lenB - 1), A + (lenB - 1), lenB, ctx);
}
else
{
const slong n2 = lenB / 2;
const slong n1 = lenB - n2;
TEMPLATE(T, struct) * W1 = W;
TEMPLATE(T, struct) * W2 = W + lenB;
const TEMPLATE(T, struct) * p1 = A + 2 * n2;
const TEMPLATE(T, struct) * p2;
const TEMPLATE(T, struct) * d1 = B + n2;
const TEMPLATE(T, struct) * d2 = B;
const TEMPLATE(T, struct) * d3 = B + n1;
const TEMPLATE(T, struct) * d4 = B;
TEMPLATE(T, struct) * q1 = Q + n2;
TEMPLATE(T, struct) * q2 = Q;
TEMPLATE(T, struct) * dq1 = BQ + n2;
TEMPLATE(T, struct) * d1q1 = BQ + 2 * n2;
TEMPLATE(T, struct) * d2q1, *d3q2, *d4q2, *t;
/*
Set q1 to p1 div d1, a 2 n1 - 1 by n1 division so q1 ends up
being of length n1; d1q1 = d1 q1 is of length 2 n1 - 1
*/
_TEMPLATE(T, poly_divrem_divconquer_recursive) (q1, d1q1, W1,
p1, d1, n1, invB, ctx);
/*
Compute d2q1 = d2 q1, of length lenB - 1
*/
d2q1 = W1;
_TEMPLATE(T, poly_mul) (d2q1, q1, n1, d2, n2, ctx);
/*
Compute dq1 = d1 q1 x^n2 + d2 q1, of length 2 n1 + n2 - 1
*/
_TEMPLATE(T, vec_swap) (dq1, d2q1, n2, ctx);
_TEMPLATE(T, poly_add) (dq1 + n2, dq1 + n2, n1 - 1, d2q1 + n2, n1 - 1,
ctx);
/*
Compute t = A/x^n2 - dq1, which has length 2 n1 + n2 - 1, but we
are not interested in the top n1 coeffs as they will be zero, so
this has effective length n1 + n2 - 1
For the following division, we want to set {p2, 2 n2 - 1} to the
top 2 n2 - 1 coeffs of this
Since the bottom n2 - 1 coeffs of p2 are irrelevant for the
division, we in fact set {t, n2} to the relevant coeffs
*/
t = BQ;
_TEMPLATE(T, poly_sub) (t, A + n2 + (n1 - 1), n2, dq1 + (n1 - 1), n2,
ctx);
p2 = t - (n2 - 1);
/*
Compute q2 = t div d3, a 2 n2 - 1 by n2 division, so q2 will have
length n2; let d3q2 = d3 q2, of length 2 n2 - 1
*/
d3q2 = W1;
_TEMPLATE(T, poly_divrem_divconquer_recursive) (q2, d3q2, W2,
p2, d3, n2, invB, ctx);
/*
Compute d4q2 = d4 q2, of length n1 + n2 - 1 = lenB - 1
*/
d4q2 = W2;
_TEMPLATE(T, poly_mul) (d4q2, d4, n1, q2, n2, ctx);
/*
Compute dq2 = d3q2 x^n1 + d4q2, of length n1 + 2 n2 - 1
*/
_TEMPLATE(T, vec_swap) (BQ, d4q2, n2, ctx);
_TEMPLATE(T, poly_add) (BQ + n2, BQ + n2, n1 - 1, d4q2 + n2, n1 - 1,
ctx);
_TEMPLATE(T, poly_add) (BQ + n1, BQ + n1, 2 * n2 - 1, d3q2, 2 * n2 - 1,
ctx);
/*
Note Q = q1 x^n2 + q2, and BQ = dq1 x^n2 + dq2
*/
}
}
#endif