/* $Id$ */ /* * Copyright (C) 1998-2002 RSA Security Inc. All rights reserved. * * This work contains proprietary information of RSA Security. * Distribution is limited to authorized licensees of RSA * Security. Any unauthorized reproduction, distribution or * modification of this work is strictly prohibited. * */ #include "bn_lcl.h" void BN_RECP_CTX_init(recp) BN_RECP_CTX *recp; { BN_init(&(recp->N)); BN_init(&(recp->Nr)); recp->num_bits=0; recp->flags=0; } BN_RECP_CTX *BN_RECP_CTX_new() { BN_RECP_CTX *ret; if ((ret=(BN_RECP_CTX *)Malloc(sizeof(BN_RECP_CTX))) == NULL) return(NULL); BN_RECP_CTX_init(ret); ret->flags=BN_FLG_MALLOCED; return(ret); } void BN_RECP_CTX_free(recp) BN_RECP_CTX *recp; { BN_free(&(recp->N)); BN_free(&(recp->Nr)); if (recp->flags & BN_FLG_MALLOCED) Free(recp); } int BN_RECP_CTX_set(recp,d,ctx) BN_RECP_CTX *recp; BIGNUM *d; BN_CTX *ctx; { (void)BN_copy(&(recp->N),d); (void)BN_zero(&(recp->Nr)); recp->num_bits=BN_num_bits(d); recp->shift=0; ctx=ctx; /* Stop warning */ return(1); } int BN_mod_mul_reciprocal(r, x, y, recp, ctx) BIGNUM *r; BIGNUM *x; BIGNUM *y; BN_RECP_CTX *recp; BN_CTX *ctx; { int ret=0; BIGNUM *a; bn_check_top(x); bn_check_top(y); a= &(ctx->bn[ctx->tos++]); if (y != NULL) { if (x == y) { if (!BN_sqr(a,x,ctx)) goto err; } else { if (!BN_mul(a,x,y,ctx)) goto err; } } else a=x; /* Just do the mod */ (void)BN_div_recp(NULL,r,a,recp,ctx); ret=1; err: ctx->tos--; return(ret); } int BN_div_recp(dv,rem,m,recp,ctx) BIGNUM *dv; BIGNUM *rem; BIGNUM *m; BN_RECP_CTX *recp; BN_CTX *ctx; { int i,j,tos,ret=0,ex; BIGNUM *a,*b,*d,*r; bn_check_top(m); tos=ctx->tos; a= &(ctx->bn[ctx->tos++]); b= &(ctx->bn[ctx->tos++]); if (dv != NULL) d=dv; else d= &(ctx->bn[ctx->tos++]); if (rem != NULL) r=rem; else r= &(ctx->bn[ctx->tos++]); if (BN_ucmp(m,&(recp->N)) < 0) { (void)BN_zero(d); (void)BN_copy(r,m); ctx->tos=tos; return(1); } /* We want the remainder * Given input of ABCDEF / ab * we need multiply ABCDEF by 3 digests of the reciprocal of ab * */ i=BN_num_bits(m); j=recp->num_bits*2; if (j > i) { i=j; ex=0; } else { ex=(i-j)/2; } j=i/2; if (i != recp->shift) recp->shift=BN_reciprocal(&(recp->Nr),&(recp->N), i,ctx); if (!BN_rshift(a,m,j-ex)) goto err; if (!BN_mul(b,a,&(recp->Nr),ctx)) goto err; if (!BN_rshift(d,b,j+ex)) goto err; d->neg=0; if (!BN_mul(b,&(recp->N),d,ctx)) goto err; if (!BN_usub(r,m,b)) goto err; r->neg=0; j=0; #if 1 while (BN_ucmp(r,&(recp->N)) >= 0) { if (j++ > 2) { #ifndef NO_ERR BNerr(BN_F_BN_DIV_RECP,BN_R_BAD_RECIPROCAL); #endif goto err; } if (!BN_usub(r,r,&(recp->N))) goto err; if (!BN_add_word(d,1)) goto err; } #endif r->neg=BN_is_zero(r)?0:m->neg; d->neg=m->neg^recp->N.neg; ret=1; err: ctx->tos=tos; return(ret); } /* len is the expected size of the result * We actually calculate with an extra word of precision, so * we can do faster division if the remainder is not required. */ int BN_reciprocal(r,m,len,ctx) BIGNUM *r; BIGNUM *m; int len; BN_CTX *ctx; { int ret= -1; BIGNUM t; bn_check_top(m); BN_init(&t); (void)BN_zero(&t); if (!BN_set_bit(&t,len)) goto err; if (!BN_div(r,NULL,&t,m,ctx)) goto err; ret=len; err: BN_free(&t); return(ret); }