1 Star 0 Fork 0

lzpong/Putty-cn

加入 Gitee
与超过 1200万 开发者一起发现、参与优秀开源项目,私有仓库也完全免费 :)
免费加入
文件
克隆/下载
sshdssg.c 4.75 KB
一键复制 编辑 原始数据 按行查看 历史
lzpong 提交于 2017-06-21 18:29 . No commit message
/*
* DSS key generation.
*/
#include "misc.h"
#include "ssh.h"
int dsa_generate(struct dss_key *key, int bits, progfn_t pfn,
void *pfnparam)
{
Bignum qm1, power, g, h, tmp;
unsigned pfirst, qfirst;
int progress;
/*
* Set up the phase limits for the progress report. We do this
* by passing minus the phase number.
*
* For prime generation: our initial filter finds things
* coprime to everything below 2^16. Computing the product of
* (p-1)/p for all prime p below 2^16 gives about 20.33; so
* among B-bit integers, one in every 20.33 will get through
* the initial filter to be a candidate prime.
*
* Meanwhile, we are searching for primes in the region of 2^B;
* since pi(x) ~ x/log(x), when x is in the region of 2^B, the
* prime density will be d/dx pi(x) ~ 1/log(B), i.e. about
* 1/0.6931B. So the chance of any given candidate being prime
* is 20.33/0.6931B, which is roughly 29.34 divided by B.
*
* So now we have this probability P, we're looking at an
* exponential distribution with parameter P: we will manage in
* one attempt with probability P, in two with probability
* P(1-P), in three with probability P(1-P)^2, etc. The
* probability that we have still not managed to find a prime
* after N attempts is (1-P)^N.
*
* We therefore inform the progress indicator of the number B
* (29.34/B), so that it knows how much to increment by each
* time. We do this in 16-bit fixed point, so 29.34 becomes
* 0x1D.57C4.
*/
pfn(pfnparam, PROGFN_PHASE_EXTENT, 1, 0x2800);
pfn(pfnparam, PROGFN_EXP_PHASE, 1, -0x1D57C4 / 160);
pfn(pfnparam, PROGFN_PHASE_EXTENT, 2, 0x40 * bits);
pfn(pfnparam, PROGFN_EXP_PHASE, 2, -0x1D57C4 / bits);
/*
* In phase three we are finding an order-q element of the
* multiplicative group of p, by finding an element whose order
* is _divisible_ by q and raising it to the power of (p-1)/q.
* _Most_ elements will have order divisible by q, since for a
* start phi(p) of them will be primitive roots. So
* realistically we don't need to set this much below 1 (64K).
* Still, we'll set it to 1/2 (32K) to be on the safe side.
*/
pfn(pfnparam, PROGFN_PHASE_EXTENT, 3, 0x2000);
pfn(pfnparam, PROGFN_EXP_PHASE, 3, -32768);
/*
* In phase four we are finding an element x between 1 and q-1
* (exclusive), by inventing 160 random bits and hoping they
* come out to a plausible number; so assuming q is uniformly
* distributed between 2^159 and 2^160, the chance of any given
* attempt succeeding is somewhere between 0.5 and 1. Lacking
* the energy to arrange to be able to specify this probability
* _after_ generating q, we'll just set it to 0.75.
*/
pfn(pfnparam, PROGFN_PHASE_EXTENT, 4, 0x2000);
pfn(pfnparam, PROGFN_EXP_PHASE, 4, -49152);
pfn(pfnparam, PROGFN_READY, 0, 0);
invent_firstbits(&pfirst, &qfirst);
/*
* Generate q: a prime of length 160.
*/
key->q = primegen(160, 2, 2, NULL, 1, pfn, pfnparam, qfirst);
/*
* Now generate p: a prime of length `bits', such that p-1 is
* divisible by q.
*/
key->p = primegen(bits-160, 2, 2, key->q, 2, pfn, pfnparam, pfirst);
/*
* Next we need g. Raise 2 to the power (p-1)/q modulo p, and
* if that comes out to one then try 3, then 4 and so on. As
* soon as we hit a non-unit (and non-zero!) one, that'll do
* for g.
*/
power = bigdiv(key->p, key->q); /* this is floor(p/q) == (p-1)/q */
h = bignum_from_long(1);
progress = 0;
while (1) {
pfn(pfnparam, PROGFN_PROGRESS, 3, ++progress);
g = modpow(h, power, key->p);
if (bignum_cmp(g, One) > 0)
break; /* got one */
tmp = h;
h = bignum_add_long(h, 1);
freebn(tmp);
}
key->g = g;
freebn(h);
/*
* Now we're nearly done. All we need now is our private key x,
* which should be a number between 1 and q-1 exclusive, and
* our public key y = g^x mod p.
*/
qm1 = copybn(key->q);
decbn(qm1);
progress = 0;
while (1) {
int i, v, byte, bitsleft;
Bignum x;
pfn(pfnparam, PROGFN_PROGRESS, 4, ++progress);
x = bn_power_2(159);
byte = 0;
bitsleft = 0;
for (i = 0; i < 160; i++) {
if (bitsleft <= 0)
bitsleft = 8, byte = random_byte();
v = byte & 1;
byte >>= 1;
bitsleft--;
bignum_set_bit(x, i, v);
}
if (bignum_cmp(x, One) <= 0 || bignum_cmp(x, qm1) >= 0) {
freebn(x);
continue;
} else {
key->x = x;
break;
}
}
freebn(qm1);
key->y = modpow(key->g, key->x, key->p);
return 1;
}
Loading...
马建仓 AI 助手
尝试更多
代码解读
代码找茬
代码优化
C
1
https://gitee.com/lzpong/Putty-cn.git
git@gitee.com:lzpong/Putty-cn.git
lzpong
Putty-cn
Putty-cn
master

搜索帮助

0d507c66 1850385 C8b1a773 1850385