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SHA-1将于2015年底失效,暴力破解成本仅为75k刀

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发表于 2017-11-24 21:05:10 | 只看该作者 回帖奖励 |倒序浏览 |阅读模式
根据这个报道,SHA-1今年底将失效,攻破成本仅为75k刀。

一个名叫Shappening的项目,仅用10天就攻破了SHA-1算法


测试器代码<div class="blockcode"><div id="code_Ea7"><ol>#include
#include
#include
#include
#include

#define KNRM "x1B[0m"
#define KRED "x1B[31m"
#define KGRN "x1B[32m"
#define KYEL "x1B[33m"
#define KBLU "x1B[34m"
#define KMAG "x1B[35m"
#define KCYN "x1B[36m"
#define KWHT "x1B[37m"
#define BK "x1B[40m" // black
#define BR "x1B[41m" // red
#define BG "x1B[42m" // green
#define BY "x1B[43m" // yellow (brown)
#define BB "x1B[44m" // blue
#define BM "x1B[45m" // magenta
#define BC "x1B[46m" // cyan
#define BW "x1B[47m" // white (L.gray)
#define CLEAR "x1B[0m"

#define K0 0x5a827999
#define K1 0x6ed9eba1
#define K2 0x8f1bbcdc
#define K3 0xca62c1d6


inline uint32_t rotate_left(uint32_t x, uint32_t r)
{
return ((x > (32 - r)));
}

inline uint32_t sha1_f1(uint32_t x, uint32_t y, uint32_t z)
{
return ((x & y) | (~x & z));
}

inline uint32_t sha1_f2(uint32_t x, uint32_t y, uint32_t z)
{
return (x ^ y ^ z);
}

inline uint32_t sha1_f3(uint32_t x, uint32_t y, uint32_t z)
{
return ((x & y) | (x & z) | (y & z));
}

inline uint32_t sha1_f4(uint32_t x, uint32_t y, uint32_t z)
{
return (x ^ y ^ z);
}

inline void sha1_step_round1(int t, uint32_t Q[85], uint32_t m[80])
{
Q[4+t+1] = rotate_left(Q[4+t],5) + rotate_left(Q[4+t-4],30) + sha1_f1(Q[4+t-1],rotate_left(Q[4+t-2],30),rotate_left(Q[4+t-3],30)) + K0 + m[t];
}
inline void sha1_step_round2(int t, uint32_t Q[85], uint32_t m[80])
{
Q[4+t+1] = rotate_left(Q[4+t],5) + rotate_left(Q[4+t-4],30) + sha1_f2(Q[4+t-1],rotate_left(Q[4+t-2],30),rotate_left(Q[4+t-3],30)) + K1 + m[t];
}
inline void sha1_step_round3(int t, uint32_t Q[85], uint32_t m[80])
{
Q[4+t+1] = rotate_left(Q[4+t],5) + rotate_left(Q[4+t-4],30) + sha1_f3(Q[4+t-1],rotate_left(Q[4+t-2],30),rotate_left(Q[4+t-3],30)) + K2 + m[t];
}
inline void sha1_step_round4(int t, uint32_t Q[85], uint32_t m[80])
{
Q[4+t+1] = rotate_left(Q[4+t],5) + rotate_left(Q[4+t-4],30) + sha1_f4(Q[4+t-1],rotate_left(Q[4+t-2],30),rotate_left(Q[4+t-3],30)) + K3 + m[t];
}

void sha1_me(uint32_t M[80])
{
for (int i = 16; i < 80; ++i)
{
M[i] = rotate_left(M[i - 3] ^ M[i - 8] ^ M[i - 14] ^ M[i - 16], 1);
}
}

void sha1_80(uint32_t out_cv[5], uint32_t Q[85], const uint32_t in_cv[5], const uint32_t m[16])
{
uint32_t W[80];
for (unsigned i = 0; i < 16; ++i)
W[i] = m[i];
sha1_me(W);

Q[4] = in_cv[0];
Q[3] = in_cv[1];
Q[2] = rotate_left(in_cv[2],2);
Q[1] = rotate_left(in_cv[3],2);
Q[0] = rotate_left(in_cv[4],2);

for (int t = 0; t < 20; ++t)
sha1_step_round1(t, Q, W);
for (int t = 20; t < 40; ++t)
sha1_step_round2(t, Q, W);
for (int t = 40; t < 60; ++t)
sha1_step_round3(t, Q, W);
for (int t = 60; t < 80; ++t)
sha1_step_round4(t, Q, W);

out_cv[0] = in_cv[0] + Q[4+80];
out_cv[1] = in_cv[1] + Q[4+79];
out_cv[2] = in_cv[2] + rotate_left(Q[4+78],30);
out_cv[3] = in_cv[3] + rotate_left(Q[4+77],30);
out_cv[4] = in_cv[4] + rotate_left(Q[4+76],30);
}

void test()
{
uint32_t iv1[5] = { 0x506b0178, 0xff6d1890, 0x202291fd, 0x3ade3871, 0xb2c665ea };
uint32_t iv2[5] = { 0x506b0178, 0xff6d1891, 0xa02291fd, 0x3ade3871, 0xb2c665ea };

uint32_t m1[16] = { 0x9d443828, 0xa5ea3df0, 0x86eaa0fa, 0x7783a736, 0x3324484d, 0xaf702aaa, 0xa3dab679, 0xd8a69e2d, 0x543820ed, 0xa7fffb52, 0xd3ff493f, 0xc3ff551e, 0xfbffd97f, 0x55feeef2, 0x085af312, 0x088688a9 };
uint32_t m2[16] = { 0x3f443838, 0x81ea3dec, 0xa0eaa0ee, 0x5183a72c, 0x3324485d, 0xab702ab6, 0x6fdab66d, 0xd4a69e2f, 0x943820fd, 0x13fffb4e, 0xefff493b, 0x7fff5504, 0xdbffd96f, 0x71feeeee, 0xe45af306, 0x048688ab };

uint32_t cv1[5], cv2[5];
uint32_t Q1[85], Q2[85];

sha1_80(cv1, Q1, iv1, m1);
sha1_80(cv2, Q2, iv2, m2);

for (int t = -4; t >b)&1;
if (q1b == q2b) {
std::cout
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