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gen_key.h
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gen_key.h
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/************************************************************************************
This program 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.
作者：李恒林 henglinli@gmail.com
************************************************************************************/
#ifndef H__GEN_KEY
#define H__GEN_KEY
#include "gen_primer.h"
//#include "gmpxx.h"
//该函数生成公钥(n,e)和私钥(n.d)。
//公钥为(RSAn,tmp_encryption),私钥为(RSAn,decryption)
int gen_key(mpz_t RSAn,mpz_t decryption)
{
int random_primer_p=0,
random_primer_q=0,
encryption;//e
mpz_t primer_p,//p
primer_q,//q
coprime_encryption,//z
primer_p_sub_1,//p-1
primer_q_sub_1;//q-1
mpz_init(primer_p); //大数t使用前要进行初始化，以便动态分配空间
mpz_init(primer_q);
mpz_init(primer_p_sub_1);
mpz_init(primer_q_sub_1);
mpz_init(coprime_encryption);
random_primer_p=random_primer();
//printf("random_primer_p=%d\n",random_primer_p);
random_primer_q=random_primer_next();
//printf("random_primer_q=%d\n",random_primer_q);
//产生大素数
gen_primer(primer_p,random_primer_p);
//gmp_printf("primer_p=%Zd\n",primer_p);
gen_primer(primer_q,random_primer_q);
//gmp_printf("primer_q=%Zd\n",primer_q);
mpz_mul(RSAn,primer_p,primer_q);//RSAn=primer_p*primer_q,得到RSAn
mpz_sub_ui(primer_p_sub_1,primer_p,1);//primer_p-1
mpz_sub_ui(primer_q_sub_1,primer_q,1);//primer_1-1
mpz_mul(coprime_encryption,primer_p_sub_1,primer_q_sub_1);//coprime_encryption=(primer_p-1)*(primer_q-1)
mpz_clear(primer_p_sub_1);//释放内存
mpz_clear(primer_q_sub_1);
mpz_clear(primer_p);
mpz_clear(primer_q);
/*************************************************
//在98个素数中随机选一个作为encryption公钥。
int tmp_encryption=random_primer();
//判断encryption和coprime_encryption是否互素。
while (1)
{
mpz_init_set_ui(encryption,tmp_encryption);
//
//int mpz_invert (mpz t rop,mpz t op1,mpz t op2)
//Computetheinverseof op1 modulo op2 andputtheresultin rop.Iftheinverseexists,the
//returnvalueisnon-zeroand rop willsatisfy0 ≤ rop < op2.Ifaninversedoesn’texistthe
//returnvalueiszeroand rop isundefned
if (mpz_invert (decryption, encryption,coprime_encryption) == 0)
//abort ();
tmp_encryption=random_primer_next();
else
break;
}
//计算decryption。
//gmp_printf("公钥为(n,e)=(%Zd,%Zd)\n",RSAn,encryption);
//gmp_printf("私钥为(n,d)=(%Zd,%Zd)\n",RSAn,decryption);
*************************************************************/
//判断tmp_encryption和coprime_encryption是否互素。
encryption=random_primer();
while (mpz_sgn(primer_p))
{
mpz_mod_ui(primer_p,coprime_encryption,encryption);//取模prmer_p=coprime_encryption MOD encryption
if (mpz_sgn(primer_p))//如果模为不为0，则退出循环，得到encyption.因为encryption本身就
//就是素数，只需判断模 prmer_p=coprime_encryption MOD encryption是否为0
//就可知道coprime_encryption和encryption是否互素。
break;
else mpz_set_ui(primer_p,1);//如果模为0，则设置primer_p为1，使循环继续，以便重新得到encryption。
//恰巧98个素数都能被coprime_encryption整除的概率几乎为0或者为0。本人未有严格证明。实际应用足矣。
}
//寻找decryption私钥。
//encryption*decryption-1=n*coprime_encryption --> decryption = (n*coprime_encryption+1)/encryption且余数为0；
int n=1;
mpz_t tmp_nz;
mpz_t tmp_nz_add_1;
mpz_t tmp_mod;
mpz_init(tmp_nz);
mpz_init(tmp_nz_add_1);
mpz_init(tmp_mod);
while (n)
{
mpz_mul_ui(tmp_nz,coprime_encryption,n);//n*coprime_encryption
mpz_add_ui(tmp_nz_add_1,tmp_nz,1);//n*coprime_encryption+1
mpz_mod_ui(tmp_mod,tmp_nz_add_1,encryption);//(n*coprime_encryption+1)/encryption
if mpz_sgn(tmp_mod)//如果tmp_mod不为0，则继续循环
{
n++;
continue;
}
else //如果tmp_mod为0，即n*coprime_encryption+1整除encryption,的到decryption.
{
mpz_cdiv_q_ui(decryption,tmp_nz_add_1,encryption);
break;
}
}
mpz_clear(tmp_nz);
mpz_clear(tmp_nz_add_1);
mpz_clear(tmp_mod);
mpz_clear(coprime_encryption);
return encryption;
}
#endif