ele*_*ent 10 python cryptography des brute-force
我正在修读一门关于密码学的课程,我被困在一项任务上.说明如下:
明文plain6.txt已使用DES加密加密6.dat,使用64位密钥作为8个字符的字符串给出(64位,每8位被忽略),所有字符都是字母(小写或更高 - case)和数字(0到9).
要完成分配,请在23.59年2月12日之前将加密密钥发送给我.
注意:我希望得到一个8字节(64位)的密钥.每个字节应该与我的密钥中的相应字节一致,除了在DES中没有使用的最低有效位,因此可以是任意的.
这是我在Python中第一次尝试的代码:
import time
from Crypto.Cipher import DES
class BreakDES(object):
def __init__(self, file, passwordLength = 8, testLength = 8):
self.file = file
self.passwordLength = passwordLength
self.testLength = testLength
self.EncryptedFile = open(file + '.des')
self.DecryptedFile = open(file + '.txt')
self.encryptedChunk = self.EncryptedFile.read(self.testLength)
self.decryptedChunk = self.DecryptedFile.read(self.testLength)
self.start = time.time()
self.counter = 0
self.chars = range(48, 58) + range(65, 91) + range(97, 123)
self.key = False
self.broken = False
self.testPasswords(passwordLength, 0, '')
if not self.broken:
print "Password not found."
duration = time.time() - self.start
print "Brute force took %.2f" % duration
print "Tested %.2f per second" % (self.counter / duration)
def decrypt(self):
des = DES.new(self.key.decode('hex'))
if des.decrypt(self.encryptedChunk) == self.decryptedChunk:
self.broken = True
print "Password found: 0x%s" % self.key
self.counter += 1
def testPasswords(self, width, position, baseString):
for char in self.chars:
if(not self.broken):
if position < width:
self.testPasswords(width, position + 1, baseString + "%c" % char)
self.key = (baseString + "%c" % char).encode('hex').zfill(16)
self.decrypt()
# run it for password length 4
BreakDES("test3", 4)
我的速度达到60,000次/秒.密码为8个字节,超过62个字符,可提供13万亿个可能性,这意味着在这个速度下,我需要130年才能解决.我知道这不是一个有效的实现,我可以用更快的语言(如C或它的味道)获得更好的速度,但我从来没有编程.即使我加速10,我们仍然可以从每秒10,000,000,000的巨大飞跃中获得小时范围.
我错过了什么？这应该是一个弱键:).好吧,弱于完整的256个字符集.
编辑
由于对分配的一些模糊性,这里有完整的描述和一些校准的测试文件:http://users.abo.fi/ipetre/crypto/assignment6.html
编辑2
这是一个粗略的C实现,在i7 2600K上每个核心获得大约2.000.000个密码/秒.您必须指定密码的第一个字符,并且可以在不同的核心/计算机上手动运行多个实例.我设法在四台计算机上用几个小时就解决了这个问题.
#include <stdio.h> /* fprintf */
#include <stdlib.h> /* malloc, free, exit */
#include <unistd.h>
#include <string.h> /* strerror */
#include <signal.h>
#include <openssl/des.h>
static long long unsigned nrkeys = 0; // performance counter
char *
Encrypt( char *Key, char *Msg, int size)
{
static char* Res;
free(Res);
int n=0;
DES_cblock Key2;
DES_key_schedule schedule;
Res = ( char * ) malloc( size );
/* Prepare the key for use with DES_ecb_encrypt */
memcpy( Key2, Key,8);
DES_set_odd_parity( &Key2 );
DES_set_key_checked( &Key2, &schedule );
/* Encryption occurs here */
DES_ecb_encrypt( ( unsigned char (*) [8] ) Msg, ( unsigned char (*) [8] ) Res,
&schedule, DES_ENCRYPT );
return (Res);
}
char *
Decrypt( char *Key, char *Msg, int size)
{
static char* Res;
free(Res);
int n=0;
DES_cblock Key2;
DES_key_schedule schedule;
Res = ( char * ) malloc( size );
/* Prepare the key for use with DES_ecb_encrypt */
memcpy( Key2, Key,8);
DES_set_odd_parity( &Key2 );
DES_set_key_checked( &Key2, &schedule );
/* Decryption occurs here */
DES_ecb_encrypt( ( unsigned char (*) [8]) Msg, ( unsigned char (*) [8]) Res,
&schedule, DES_DECRYPT );
return (Res);
}
void ex_program(int sig);
int main(int argc, char *argv[])
{
(void) signal(SIGINT, ex_program);
if ( argc != 4 ) /* argc should be 2 for correct execution */
{
printf( "Usage: %s ciphertext plaintext keyspace \n", argv[0] );
exit(1);
}
FILE *f, *g;
int counter, i, prime = 0, len = 8;
char cbuff[8], mbuff[8];
char letters[] = "02468ACEGIKMOQSUWYacegikmoqsuwy";
int nbletters = sizeof(letters)-1;
int entry[len];
char *password, *decrypted, *plain;
if(atoi(argv[3]) > nbletters-2) {
printf("The range must be between 0-%d\n", nbletters-2);
exit(1);
}
prime = atoi(argv[1])
// read first 8 bytes of the encrypted file
f = fopen(argv[1], "rb");
if(!f) {
printf("Unable to open the file\n");
return 1;
}
for (counter = 0; counter < 8; counter ++) cbuff[counter] = fgetc(f);
fclose(f);
// read first 8 bytes of the plaintext file
g = fopen(argv[2], "r");
if(!f) {
printf("Unable to open the file\n");
return 1;
}
for (counter = 0; counter < 8; counter ++) mbuff[counter] = fgetc(g);
fclose(g);
plain = malloc(8);
memcpy(plain, mbuff, 8);
// fill the keys
for(i=0 ; i<len ; i++) entry[i] = 0;
entry[len-1] = prime;
// loop until the length is reached
do {
password = malloc(8);
decrypted = malloc(8);
// build the pasword
for(i=0 ; i<len ; i++) password[i] = letters[entry[i]];
nrkeys++;
// end of range and notices
if(nrkeys % 10000000 == 0) {
printf("Current key: %s\n", password);
printf("End of range ");
for(i=0; i<len; i++) putchar(letters[lastKey[i]]);
putchar('\n');
}
// decrypt
memcpy(decrypted,Decrypt(password,cbuff,8), 8);
// compare the decrypted with the mbuff
// if they are equal, exit the loop, we have the password
if (strcmp(mbuff, decrypted) == 0)
{
printf("We've got it! The key is: %s\n", password);
printf("%lld keys searched\n", nrkeys);
exit(0);
}
free(password);
free(decrypted);
// spin up key until it overflows
for(i=0 ; i<len && ++entry[i] == nbletters; i++) entry[i] = 0;
} while(i<len);
return 0;
}
void ex_program(int sig) {
printf("\n\nProgram terminated %lld keys searched.\n", nrkeys);
(void) signal(SIGINT, SIG_DFL);
exit(0);
}
我认为所需的解决方案是实际实现算法.然后,由于你自己解密,你可以提前保释,假设纯文本也是A-Za-z0-9,你在解密单个字节后有98%的机会能够停止,99.97%解密2个字节后有机会停止,3个字节后有99.9995%的机会停止.
此外,使用C或Ocaml或类似的东西.你可能花费大量时间进行字符串操作而不是进行加密.或者,至少使用多处理并启动所有核心......
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