RiveSt Shamir Adleman (RSA) Encryption Introduction and Sample Code
Seraph chutium - http://com.6to23.com/
First, this site's message is a point about RSA.
61 Date: 2002-04-29 06:41:04 ChutiumRSA is a ciphertext that obtains encrypted ciphertext by mathematical transformation of the entire plain text with the interactivity of the two general prime numbers (a, b). N = ab, f (n) = (a-1) (b-1), encryp * decryp = 1 mod f (n)}. Why not necessarily display? This information should be very good online, but there are very few words and its mathematical changes, you are best to see the BBS of foreign universities. These things in the university textbook should have more detailed introduction, but who is there really in the year ...
67 Date: 2002-04-30 06:52:38 LIYXRSA's public key and private key are two functions greater than 100 decided positions. It is speculated that the difficulty of clearing the plain text from a key and ciphertext is equivalent to decomposing two largest accumulation. Select two large numbers, p and q when the key is generated. Calculate: n = p * q then randomly select the encryption key E, require E and (P-1) * (Q-1) mutual. Finally, the decryption key D is calculated using the Euclid algorithm to satisfy E * D = 1 (MOD (P-1) * (Q-1)) where n and d are also mutual. Number E and N are public key, D is the private key (ie you said Decrypt). Two prime numbers P and Q are no longer needed, they should be discarded. When encrypting information m (binary representation), the m is first divided into a equal length data block M1, M2, ..., Mi, block length S, where 2 ^ s <= n, and s is as large as possible. The corresponding ciphertext is: CI = Mi ^ e (MOD N) (a) decrypts as follows: Mi = Ci ^ D (MOD N) (b) RSA digital signature is used (A) signature, (b) Verification. Considering the factors such as safety and M information when considering the factors of safety and M information, they are generally ash as a HASH operation.
If you are an encrypted algorithm, you can see two stickers. Below is the more detailed encryption process of the two stickers and some information (see the end), there is a better mathematical friend, if you are interested Please continue.
It is recommended that the reader basically understand the number of knowledge of "factor, rigmer, the same, the same, the fare, the Euler theorem, Wilson, LUCAS theorem".
This encryption method is the above three scientists published in 1978, which steps are as follows
1. Reporters take two different substances
p,
Q and another
P-1)
Q-1) number of mutual
A,
a
w let
W = ("
P-1)
Q-1),
m =
PQ
and
p,
The smaller number of q (ten into the position) is
k.
2. (Open) tell the reporter
K,
M
a.
3. The reporter divides his signal into many segments, including each section
K-1 digits (ten into position) (if
K = 3 (ie
p,
Q is not less than two digits), then signal
331414320001
It should be divided into
33, 14, 14, 32, 100, 01
One a consideration is issued), one is one of the signals of the reporter
X (
The K-1 digits, that is, 33, or 14, or 32, ...) in the upper example, then he makes it
issue.
4. Revenue received
After C, you can put the original
X find out, because
a with
W Mutual, by theorem 2.2 and well, we can find two integers
D,
E;
D> 0 makes
AD WE = 1
make
Then
Y is the reporter
x. Let's prove first
y =
x.
Because
W,
a,
D is an integer greater than 1, so
e must be a negative number, ie -
E is a positive number, but
X is smaller than the number
P,
Q, so
x
M mutual, is based on theorem 2.3
Because
Y and
X is less than
The number of M, so
y =
x. Therefore, the correctness of this program is certified.
I am relatively weak, so the calculation of the crack year is very unhappy. Here is the information on my hand "mathematics communication", which has addition.
The key is the key to the two large numbers of P, Q, according to this paper, decomposition M becomes P, Q is an extremely time, if the decomposition does not open M, then I can't find W and D, therefore Unable to solve x from C, soon, to break down a number of factors still stay at almost harmful stages, that is, from 2, 3, 5, 7, ..., always try N ^ (1/2) Stop it. If N is 50 bits, the decomposing M is divided by approximately 1025 times, if an electronic computer is 106 high-speed operations per second, this is still a 1011 job, currently due to everyone At this area, the time to break down a 50-digit number can be shortened to 1010 computing. The current (1980) is listed in the table below to break down a large number of probably required time.
M of the number of digits of M, the fastest number of computers (1980) Computer Time 501.4 x 10103.9 hours 701.0 x 1012104 Day 801.0 x 1013150 Day 1002.3 x 10 1774 2001.2 x 10233.8 x 109
(As for how, I don't know how to use the combination and statistical knowledge, I am in this respect)
Cryptographic information
1. Rivest, R. L; SHEMIR, A .; Adleman, L. "a Method for Obtaining Digital Signature and public-key cryptasystem" Communication Of ACM, 1978, PP. 120-126.
2. Simmons, G. "CRYPTOLOGY: The Mathematics of SEWRE Communication" The Mathematical Intelligencer, 1978, PP. 233-246.
3. Hellman, M.E. "The Mathematics of Public-Key Cryptography" Scientific American, August, 1979, PP. 146-157.
4. Pomerat, C. "The Search for Prime Numbers" Scientific American, December, 1982, PP. 136-147.
5. Paul Fahn, "About Today's Cryptography V2.0 Draft 2F" RSA Laboratories, September 20, 1993 (http://www.surfsites.net/chutium/rsa/rsafaq.txt)
6. The Mathematical Basis of RSA Key-Pair Generation (http://www.surfsites.net/chutium/rsa/rsake.htm)
7. The Latest RSA FAQ (http://www.rsarated.com/rsalabs/faq/3-1-1.html)
8. Yang Zhaokun (Taiwan) "Mathematics Communication"
Documentation
RSA Sample Code (C ) rsa.cpp rsa.hpp vlong.cpp vlong.hpp; RSA Sample Code (VB) RSAVB.TXT
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