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Description: RSA算法 :首先, 找出三个数, p, q, r, 其中 p, q 是两个相异的质数, r 是与 (p-1)(q-1) 互质的数...... p, q, r 这三个数便是 person_key，接著, 找出 m, 使得 r^m == 1 mod (p-1)(q-1)..... 这个 m 一定存在, 因为 r 与 (p-1)(q-1) 互质, 用辗转相除法就可以得到了..... 再来, 计算 n = pq....... m, n 这两个数便是 public_key ，编码过程是, 若资料为 a, 将其看成是一个大整数, 假设 a < n.... 如果 a >= n 的话, 就将 a 表成 s 进位 (s <= n, 通常取 s = 2^t), 则每一位数均小于 n, 然後分段编码...... 接下来, 计算 b == a^m mod n, (0 <= b < n), b 就是编码后的资料。解码的过程是, 计算 c == b^r mod pq (0 <= c < pq), 于是, 解码完毕-RSA algorithm : First, find a few 3, p, q, r, p, q is the two different prime number, with the r (p-1) (q-1) coprime several ... p, q, r it person_key number is three, and then find m, making r ^ m == a mod (p-1) (q-1 m ...................................... this must exist, because r (p-1) (q - 1) coprime, with the division algorithm can be a ..... Next, calculate n = pq ....... m, n is the number two public_key, coding process is that if a data and to treat it as a big Integer assuming a <n. ... If a> = n, a table will be rounding into s (s <= n, usually from 2 ^ s = t), each n is less than the median, and then sub-coding ..... . Next, calculate a == b ^ m mod n, (0 <= b <n), b is encoded information. Decoding is the process of calculating c == b ^ r mod pq (0 <= c &
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Author: 胡康康 |
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Description: RSA算法 :首先, 找出三个数, p, q, r, 其中 p, q 是两个相异的质数, r 是与 (p-1)(q-1) 互质的数...... p, q, r 这三个数便是 person_key，接著, 找出 m, 使得 r^m == 1 mod (p-1)(q-1)..... 这个 m 一定存在, 因为 r 与 (p-1)(q-1) 互质, 用辗转相除法就可以得到了..... 再来, 计算 n = pq....... m, n 这两个数便是 public_key ，编码过程是, 若资料为 a, 将其看成是一个大整数, 假设 a < n.... 如果 a >= n 的话, 就将 a 表成 s 进位 (s <= n, 通常取 s = 2^t), 则每一位数均小于 n, 然後分段编码...... 接下来, 计算 b == a^m mod n, (0 <= b < n), b 就是编码后的资料。解码的过程是, 计算 c == b^r mod pq (0 <= c < pq), 于是, 解码完毕-RSA algorithm : First, find a few 3, p, q, r, p, q is the two different prime number, with the r (p-1) (q-1) coprime several ... p, q, r it person_key number is three, and then find m, making r ^ m == a mod (p-1) (q-1 m ...................................... this must exist, because r (p-1) (q- 1) coprime, with the division algorithm can be a ..... Next, calculate n = pq ....... m, n is the number two public_key, coding process is that if a data and to treat it as a big Integer assuming a <n. ... If a> = n, a table will be rounding into s (s <= n, usually from 2 ^ s = t), each n is less than the median, and then sub-coding ..... . Next, calculate a == b ^ m mod n, (0 <= b <n), b is encoded information. Decoding is the process of calculating c == b ^ r mod pq (0 <= c &
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Author: 胡康康 |
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Description: 一、RSA基本原理
对明文分组M和密文分组C，加密与解密过程如下：
C = POW (M , e) mod n
M = POW(C , d) mod n = POW(POW( M ,e), d) mod n=POW( M,e*d)
其中POW是指数函数，mod是求余数函数。
其中收发双方均已知n，发送放已知e，只有接受方已知d，因此公钥加密算法的公钥为
KU={ e , n},私钥为KR={d , n}。该算法要能用做公钥加密，必须满足下列条件：
1. 可以找到e ，d和n，使得对所有M<n ,POW(M ,e*d)=M mod n .
2. 对所有 M<n,计算POW (M , e)和POW(C , d)是比较容易的。
3. 由e 和n确定d是不可行的
-one, the basic tenets of RSA expressly group M and cipher block C, encryption and decryption process is as follows : C = POW (M, e) mod n = M POW (C, d) mod n = POW (POW (M, e), d) mod n = POW (M, e* d), which is an exponential function POW, mod is the pursuit of the remaining functions. Transceivers which both known n, send e Fang known, the only known recipient d, therefore the public key encryption algorithm for public key e KU = (n), private key for KR = (d, n). The algorithm could be used to be a public key encryption, must meet the following conditions : 1. E can be found, and d n, making all the right M
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Description: 加密算法
Test Driver for Crypto++, a C++ Class Library of Cryptographic Primitives:
- To generate an RSA key
cryptest g
- To encrypt and decrypt a string using RSA
cryptest r
- To calculate MD5, SHS, and RIPEMD-160 message digests:
cryptest m file
- To encrypt and decrypt a string using DES-EDE in CBC mode:
cryptest t
- To encrypt or decrypt a file
cryptest e|d input output
- To share a file into shadows:
cryptest s <pieces> <pieces-needed> file
(make sure file has no extension, if you re running this under DOS)
- To reconstruct a file from shadows:
cryptest j output file1 file2 [....]
- To gzip a file:
cryptest z <compression-level> input output
- To gunzip a file:
cryptest u input output
- To run validation tests:
cryptest v
- To run benchmarks:
cryptest b [time for each benchmark in seconds]
-encryption algorithm Test Driver for Crypto. a Class C Library of spreadsheets Primitives :- To generate an RSA key cryptest g-To encrypt an d decrypt a string using RSA cryptest r-To calcu late MD5, SHS, and RIPEMD algorithms-160 message digests : cryptest m file-To encrypt and decrypt a string using DES-EDE in CBC mode : cryptest t-To encrypt or decrypt a file cryptes t e | d input output- To share a file into shadows : cryptest's
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Author: Nikii |
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Description: 包含6个*.m文件，分别是adline网络，bp网络，hopfiled网络，字符识别，学习速度自适应，和增强型lms算法的六个仿真算法程序，真是我的珍藏，这次全抖出来了。-contains 6 m*. documents were adline network bp network hopfiled network, character recognition, adaptive learning speed, lms and enhanced algorithm simulation algorithm six procedures, I really treasure this all shake out.
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Author: mahb |
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Description: RSA Encryption and Decryption using Matlab
by Thunyawat Rajatasereekul and Voranon Kiettrsalpipop
ECE575, Winter 2002, Oregon State University
Prof, Cetin Keya Koc
The program set contains thirteen files listed below.
errormeg.fig
errormsg.m
helpmsg.fig
helpmsg.m
inputmsg.fig
inputmsg.m
mesgcut.m
pro2.fig
pro2.m
rsacore.m
readme.txt
screenshot.gif
In order to run the program please call pro2.m under Matlab 6.0 Environment.
System Requirement
Matlab version 6.0 and if necessary, Maple version 6.0 on any platform
PentiumII 300 MHz or equivalent.-RSA Encryption and Decryption using Matlabby Thunyawat Rajatasereekul and Voranon KiettrsalpipopECE575, Winter 2002, Oregon State UniversityProf, Cetin Keya KocThe program set contains thirteen files listed below.errormeg.figerrormsg.mhelpmsg.fighelpmsg.minputmsg.figinputmsg.mmesgcut.mpro2.figpro2.mrsacore. mreadme.txtscreenshot.gifIn order to run the program please call pro2.m under Matlab 6.0 Environment.System RequirementMatlab version 6.0 and if necessary, Maple version 6.0 on any platformPentiumII 300 MHz or equivalent.
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Author: 张汉江 |
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Description: 实现DES和RSA的加密解密,MD5还有数字签名。-The realization of DES and RSA s encryption and decryption, MD5 digital signature also.
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Author: 天天 |
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Description: Enclosed in this distribution are four projects:
MD5DLLTest: Visual basic project which runs the MD5 test suite against a
MD5 DLL (see below).
VbMD5: A Visual Basic native MD5 message digest class based on the
RSA reference implementation.
md5DLL: A C project which generates a simple Win32 DLL with
the MD5 message digest routines using the RSA reference
implementation.
MD5Java: A Java implementation of the MD5 message digest algorithm based
on the RSA reference implmentation.
Let me know if this has any use!
Robert M. Hubley
hubley@u.washington.edu
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Author: 0940566827 |
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Description: Digital Signature Algorithm (DSA)是Schnorr和ElGamal签名算法的变种，被美国NIST作为DSS(DigitalSignature Standard)。算法中应用了下述参数：
p：L bits长的素数。L是64的倍数，范围是512到1024；
q：p - 1的160bits的素因子；
g：g = h^((p-1)/q) mod p，h满足h < p - 1, h^((p-1)/q) mod p > 1；
x：x < q，x为私钥 ；
y：y = g^x mod p ，( p, q, g, y )为公钥；
H( x )：One-Way Hash函数。DSS中选用SHA( Secure Hash Algorithm )。
p, q, g可由一组用户共享，但在实际应用中，使用公共模数可能会带来一定的威胁。签名及验证协议如下：
1. P产生随机数k，k < q；
2. P计算 r = ( g^k mod p ) mod q
s = ( k^(-1) (H(m) + xr)) mod q
签名结果是( m, r, s )。
3. 验证时计算 w = s^(-1)mod q
u1 = ( H( m ) * w ) mod q
u2 = ( r * w ) mod q
v = (( g^u1 * y^u2 ) mod p ) mod q
若v = r，则认为签名有效。
DSA是基于整数有限域离散对数难题的，其安全性与RSA相比差不多。DSA的一个重要特点是两个素数公开，这样，当使用别人的p和q时，即使不知道私钥，你也能确认它们是否是随机产生的，还是作了手脚。RSA算法却作不到。
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Author: wildkaede |
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Description: RSA核心运算使用的乘模算法就是 M(A*B)。虽然M(A*B)并不是乘模所需要的真正结果，但只要在幂模算法中进行相应的修改，就可以调用这个乘模算法进行计算了。本软件起初未使用Montgomery 乘模算法时，加密速度比使用Montgomery乘模算法慢，但速度相差不到一个数量级。
将上述乘模算法结合前面叙述的幂模算法，构成标准Montgomery幂模算法，即本软件所使用的流程
-RSA computation used by the core modulus algorithm is M (A* B). Although the M (A* B) is not required by the real-mode, but as long as the power modulus algorithm in the corresponding changes, you can call this by the calculated modulus algorithm. This software was initially not used by Montgomery modulus algorithm, the encryption faster than the use of Montgomery modulus algorithm by slow, but the speed was less than an order of magnitude. Modulus algorithm by the above described combination of the previous power-modulus algorithm, which constitute the standard Montgomery idempotent modulus algorithm, that is, the software used in flow
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Author: 拉萨 |
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Description: 1) 找出两个相异的大素数P和Q，令N＝P×Q，M＝（P－1）（Q－1）。
2) 找出与M互素的大数E，用欧氏算法计算出大数D，使D×E≡1 MOD M。
3) 丢弃P和Q，公开E，D和N。E和N即加密密钥，D和N即解密密钥。
-1) to identify two different large prime numbers P and Q, so N = P × Q, M = (P-1) (Q-1). 2) to identify and M large numbers coprime E, Euclidean algorithm used to calculate lump sum for the D, so that D × E ≡ 1 MOD M. 3) disposed of P and Q, the open E, D and N. E and N that encryption keys, D and N that the decryption key.
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Author: 阿达悟 |
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Description: 实现通信过程中的数据加密（AES, DES, SHA-256,SHA-384,SHA512, RSA）。
测试加解密效率等-To achieve communication in the process of data encryption (AES, DES, SHA-256, SHA-384, SHA512, RSA). Encryption and decryption efficiency test
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Author: BY |
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Description: RSA ( Rivest Shamir Adleman )is crypthograph system that used to give a secret information and digital signature . Its security based on Integer Factorization Problem (IFP). RSA uses an asymetric key. RSA was created by Rivest, Shamir, and Adleman in 1977. Every user have a pair of key, public key and private key. Public key (e) . You may choose any number for e with these requirements, 1< e <Æ (n), where Æ (n)= (p-1) (q-1) ( p and q are first-rate), gcd (e,Æ (n))=1 (gcd= greatest common divisor). Private key (d). d=(1/e) mod(Æ (n)) Encyption (C) . C=Mª mod(n), a = e (public key), n=pq Descryption (D) . D=C° mod(n), o = d (private key- RSA ( Rivest Shamir Adleman )is crypthograph system that used to give a secret information and digital signature . Its security based on Integer Factorization Problem (IFP). RSA uses an asymetric key. RSA was created by Rivest, Shamir, and Adleman in 1977. Every user have a pair of key, public key and private key. Public key (e) . You may choose any number for e with these requirements, 1< e <Æ (n), where Æ (n)= (p-1) (q-1) ( p and q are first-rate), gcd (e,Æ (n))=1 (gcd= greatest common divisor). Private key (d). d=(1/e) mod(Æ (n)) Encyption (C) . C=Mª mod(n), a = e (public key), n=pq Descryption (D) . D=C° mod(n), o = d (private key
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Author: nb |
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Description: RSA-MFC,基于MFC的RSA加密算法的实现，可以用于16进制文件的加解密-RSA-MFC, the MFC based on the realization of RSA encryption algorithm, can be used for 16-band file encryption and decryption
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Author: 高飞 |
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Description: RSA算法实验报告和代码
1.选取两个素数p,q（不可相差悬殊）
2.计算n=pq,f(n)=(p-1)(q-1)
3.选取e,满足1<e<f(n),则gcd(e,f(n))=1
4.计算d，满足de=1 mod f(n)。一般d>=[n的四分之一方],(e,n)为公钥，(p,q,d)为私钥，将明文0，1序列分组，使每组十进制小于n。c=[m的e次方] mod n,m=[c的d次方] mod n。-RSA algorithm and code an experimental report. Choose two primes p, q (non-significant differences between) 2. Calculate n = pq, f (n) = (p-1) (q-1) 3. Select the e, to satisfy a <e<f(n),则gcd(e,f(n))=1
4.计算d，满足de=1 mod f(n)。一般d> = [n the fourth side], (e, n) for the public key, (p, q, d) for the private key will be expressly 0,1 sequence packet, so that each of the decimal is less than n. c = [m of the e-th power] mod n, m = [c of the d-th power] mod n.
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Author: jhp627 |
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Description: rsa&md5 公钥和私钥就是从两个文件PublicKey.xml、PrivateKey.xml中读取出来，相当于是现实中的指定公钥和私钥；再验证时我为了方便直接验证的A计算出的HASH码，现实中应该是由B重新计算出文件M的HASH码-ras&md5
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Author: 123 |
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Description: MFC实现RSA的加密与解密，界面比较简洁清楚，对RSA初学者有较大帮助！-MFC to achieve RSA encryption and decryption, more concise and clear interface, great help for beginners on the RSA!
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Author: 郭子 |
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Description: CipherWall Client, code to work with sockets, sqlite database and CA realization:
- RSA 2048-4096 bit
- digit auth on RSA
- 3-step cert auth
- Blowfish in CFB mode (448 bit)
- SHA-256 и SHA-512
- HMAC on base SHA-256
- random-digit genereator ANSI X9.17
- CRC32
- DoD-5220.22-M
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Author: Andrey |
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Description: 编程实现RSA算法。包括：生成公钥(e, n)和私钥d，对明文m加密，对密文m解密。
注：实际应用中，512比特的n 已经不够安全，所以建议公司用1024比特的n，及其重要的场合用2048比特的 n。所以大家要选择大整数n。-Programming RSA algorithm. Include: Creation of a public key (e, n) and private key d, m the plaintext encryption, decryption of ciphertext m. Note: The actual application, the 512-bit n is not secure, it is recommended that companies with 1024-bit n, and the important occasion with a 2048-bit n. Therefore, we should choose a large integer n.
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Author: semmir |
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Description: 1．问题描述
RSA密码系统可具体描述为：取两个大素数p和q，令n=pq,N=(p-1)(q-1),随机选择整数d,满足gcd(d,N)=1,ed=1 modN。
公开密钥：k1=(n,e)
私有密钥：k2=(p,q,d)
加密算法：对于待加密消息m，其对应的密文为c=E(m)=me(modn)
解密算法：D(c)=cd(modn)
2．基本要求
p,q,d,e参数选取合理，程序要求界面友好，自动化程度高。
4． 实现提示
要实现一个真实的RSA密码系统，主要考虑对大整数的处理。P和q是1024位的，n取2048位。(1. problem description
The RSA cryptosystem can be specifically described as: take two large prime numbers P and Q, make n=pq, N= (p-1) (Q-1), select integer D randomly, and satisfy GCD (D, N) =1.
Public key: k1= (n, e)
Private key: k2= (P, Q, d)
Encryption algorithm: for the encrypted message M, its corresponding ciphertext is c=E (m) =me (MODN)
Decryption algorithm: D (c) =cd (MODN)
2. basic requirements
P, Q, D, e parameters are selected reasonably, the program requires friendly interface and high degree of automation.
4. realization hints
To implement a real RSA cryptosystem, the main consideration is to deal with large integers. P and Q are 1024 bits, and N takes 2048.)
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Author: Appoint |
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