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GassianXY
- 利用二元域的高斯消元法得到输入矩阵H对应的生成矩阵G，同时返回与G满足mod(G*P ,2)=0的矩阵P，其中P 表示P的转置 使用方法：[P，G]=Gaussian(H,x)，x=1 or 2,1表示G的左边为单位阵-binary domain PGE law input matrix corresponding to the formation of H matrix G, Meanwhile the return of mod meet with the G (G * P, 2) =
ImgRotate
- Dim DP() As Byte, c As Long, d As Long Dim i As Long, j As Long Dim Color As Long, SRad As Double, CRad As Double Dim p As Double, q As Double, m As Double, n As Double Dim m1 As Double, n1 As Double, m2 As Double, n2 As Double, t As RGB, t1
ercitongyufangchengdeqiujie
- 求解 形如 a*x^2+b*x+c=0 (mod p)的二次同余方程，其中p为任意素数，a,b,c为任意整数.
DSA
- 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
雅可比法
- 新手上路 用雅可比符号来计算勒让德符号的值 用于判断与素数p互素的正整数n是否是mod p 的二次剩余-Started Jacobian symbols used to calculate Legendre symbol for the value judgment and mutual-p-a positive integer n is whether the mod p Quadratic Residue
RSA解密和加密算法的实现和应用
- 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,
Mod.rar
- 包含12个MATLAB函数，分别长生2PSK 4psk 8psk信号 和16QAM信号以及星座图 噪声程序等,12 MATLAB includes a function of longevity separately 2PSK 4psk 8psk signal and 16QAM signal constellation diagram, as well as procedures such as noise
momi
- 大数的模幂算法(GUI)，用密码学课本中的算法，快速、高效。计算(x的r次方) mod p 的值-Large numbers of modular exponentiation algorithm (GUI), used cryptography algorithm textbooks, fast and efficient. Calculation of (x of the r-th power) mod p value
Euler_fuction
- Euler函数: m = p1^r1 * p2^r2 * …… * pn^rn ai >= 1 , 1 <= i <= n Euler函数： 定义：phi(m) 表示小于等于m并且与m互质的正整数的个数。 phi(m) = p1^(r1-1)*(p1-1) * p2^(r2-1)*(p2-1) * …… * pn^(rn-1)*(pn-1) = m*(1 - 1/p1)*(1 - 1/p2)*……*(1 - 1/pn)
rsa
- 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 cop
iml
- IML package provides efficient routines to solve nonsingular systems of linear equations, certified solve any shape systems of linear equations, and perform mod p matrix operations, such as computing row-echelon form, determinant, rank profi
p
- 基于马尔萨斯及Logistic人口模型，本文做了相应改进，根据已有数据确立了我们的人口模型。人口净增长率与人口密度比是简单常数值或者是简单的指数函数，所建立的模型来预测人口总数都是不完备的，所以我们假定人口净增长率是随总人口变化的函数，而这个函数是我们通过数据拟合可以近似得到的，由人口净增长率函数的确立，我们可以以微分方程的形式建立人口模型-Based on the Malthusian and Logistic population model, this article has done a
BasicRSA_latest.tar
- 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 i
Prime_Numbers
- Threw history,prime numbers has always be a great fascination for mathematicians. Even in modern times,primes numbers continues to fascinate many people.Probably the main reason why prime numbers continues to create such great interest would be beca
lisanduishu
- 设p是素数，a是p的本原根，即a1,a2,a3,…,ap-1在mod p下产生1到p-1的所有值，所以对任何b属于{1,…,p-1},有唯一的i属于{1,…,p-1}使得ai mod p 等于p。称i为模p下以a为底b的离散对数。-P is a prime number based, a is the primitive root p, that is, a1, a2, a3, ..., ap-1 in the mod p under 1 to p-1 for all values, so fo
cannon
- cannon 并行程序 Matrix multiplication with the Cannon Algorithm: The matrices A and B are stored in files. The file names have to be specified as parameter. The root process reads the matrices and distributes the respective values to all processes,
RK
- 实验RK算法，即利用Hash方法和素数理论，首先定义一个Hash函数（hash (r) = r mod q），然后将模式串P和文本串T中长度为m的子串利用Hash函数转换成数值。显然只需比较那些与模式串具有相同Hash函数值的子串。 当然因为Hash冲突的存在，还要进一步进行字符串比较，但只要选择适当的素数q, Hash冲突的概率就会很小 -Experimental RK algorithm, namely the use of Hash methods and prime number
powmod
- An effecient algorithm to calculate the power mod (b^p)(mod n) using Dynamic Programming
1-p^2-1mod-p
- 从1-p^2-1中找出与p互素的元素，计算1-p^2-1模p的值，并按模值分类。已验证p=5,7。-Find numbers 1-p^2-1 which is co-prime to p，compute values mod p, and classify them according to values mod p.
ModOfPower
- Write an efficient algorithm to calculate R=B^P mod M