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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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Size: 1024 |
Author: 胡康康 |
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Description: Verilog编写的M序列发生器，希望能对大家带来帮助。
-Verilog prepared by the M-sequence generator, we hope to bring help.
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Size: 5120 |
Author: 张林 |
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Description: HMM(Hidden Markov Model)，狀態數目N=3，觀察符號數目M=2，時間長度T=3。
(a) Probability Evaluation: 給定狀態轉換機率A、狀態符號觀察機率B、和起始機率 ，求觀察序列 出現的機率。
(b) Optimal State Sequence: 給定狀態轉換機率A、狀態符號觀察機率B、起始機率 、和觀察序列 ，求一個狀態序列 使得O出現的機率最大。
(c) Parameter Estimation: 給定狀態轉換機率A、狀態符號觀察機率B、起始機率 、和觀察序列 ，求新的A、B、 ，使得O出現的機率最大。
-HMM (Hidden Markov Model), state the number of N = 3, Observation number of symbols M = 2, T = length of three. (A) Probability Evaluation : given state transition probability A, Observer status symbol probability of B, and initial probability for observation sequence in the octave. (B) Optimal State Sequence : given state transition probability A, Observer status symbol probability of B, the initial probability, and observation sequence, for a state sequence of O makes the greatest risk. (C) Parameter Estimation : given state transition probability A, Observer status symbol probability of B, the initial probability, and observation sequence, the way the A, B, and O makes the greatest risk.
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Size: 146432 |
Author: 章勝鈞 |
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Description: CHMMBOX, version 1.2, Iead Rezek, Oxford University, Feb 2001
Matlab toolbox for max. aposteriori estimation of two chain
Coupled Hidden Markov Models.
-CHMMBOX, version 1.2, Iead Rezek, Oxford University, Feb 2001 Matlab toolbox for max. aposteriori es timation of two chain Coupled Hidden Markov Mod els.
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Size: 823296 |
Author: Joyce |
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Description: USB Abstract Control Model driver for USB modems and ISDN adapters(for Linux)-USB Abstract Control Model driver for USB m odems and ISDN adapters (for Linux)
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Size: 9216 |
Author: 陶昱良 |
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Description: 在ns2.28中实现的多路径路由协议M-AODV-ns2.28 achieve in the multi-path routing protocol M-AODV
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Size: 452608 |
Author: |
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Description: 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)
= p1^(r1-1)*p2^(r2-1)* …… * pn^(rn-1)*phi(p1*p2*……*pn)
定理：若(a , m) = 1 则有 a^phi(m) = 1 (mod m) 即a^phi(m) - 1 整出m
在实际代码中可以用类似素数筛法求出
for (i = 1 i < MAXN i++)
phi[i] = i
for (i = 2 i < MAXN i++)
if (phi[i] == i)
{
for (j = i j < MAXN j += i)
{
phi[j] /= i
phi[j] *= i - 1
}
}
容斥原理：定义phi(p) 为比p小的与p互素的数的个数
设n的素因子有p1, p2, p3, … pk
包含p1, p2…的个数为n/p1, n/p2…
包含p1*p2, p2*p3…的个数为n/(p1*p2)…
phi(n) = n - sigm_[i = 1](n/pi) + sigm_[i!=j](n/(pi*pj)) - …… +- n/(p1*p2……pk)
= n*(1 - 1/p1)*(1 - 1/p2)*……*(1 - 1/pk)
-err
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Size: 1024 |
Author: Campbell |
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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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Size: 6144 |
Author: 阿达悟 |
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Description: 实现修改网卡地址，简单过程明细。直接修改，快速上网。-Modified to achieve LAN address, details of a simple process. Directly modify, fast Internet access.
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Size: 390144 |
Author: wlk |
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Description: embest 可以动态加载的led驱动 编译成.o文件后直接加载就行
-embest can be dynamically loaded into the led driver to compile. o files directly loaded on the line
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Size: 3072 |
Author: zhang |
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Description: Mod. 130 power supply
Schematic diagram
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Size: 159744 |
Author: Mario |
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Description: Mod. V-ULA 50 V-UHF linear amplifier
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Size: 758784 |
Author: Mario |
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Description: Mod. RT 30 SW tension reducer
Schematic diagram
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Size: 177152 |
Author: Mario |
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Description: Mod. 145 linear amplifier
Schematic diagram
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Size: 120832 |
Author: Mario |
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Description: This simulink model simulates the damped driven pendulum, showing it s chaotic motion.
theta = angle of pendulum
omega = (d/dt)theta = angular velocity
Gamma(t) = gcos(phi) = Force
omega_d = (d/dt) phi
Gamma(t) = (d/dt)omega + omega/Q + sin(theta)
Play with the initial conditions (omega_0, theta_0, phi_0 = omega(t=0), theta(t=0), phi(t=0)) and the system parameters (g, Q, omega_d) and the solver parameters/method.
Chaos can be seen for Q=2, omega_d=w/3.
The program outputs to Matlab time, theta(time) & omega(time).
Plot the phase space via:
plot(mod(theta+pi, 2*pi)-pi, omega, . )
Plot the Poincare sections using:
t_P = (0:2*pi/omega_d:max(time))
plot(mod(spline(time, theta+pi, t_P), 2*pi)-pi, spline(time, omega, t_P), . )
System is described in:
"Fractal basin boundaries and intermittency in the driven damped pendulum"
E. G. Gwinn and R. M. Westervelt
PRA 33(6):4143 (1986)
-This simulink model simulates the damped driven pendulum, showing it s chaotic motion.
theta = angle of pendulum
omega = (d/dt)theta = angular velocity
Gamma(t) = gcos(phi) = Force
omega_d = (d/dt) phi
Gamma(t) = (d/dt)omega+ omega/Q+ sin(theta)
Play with the initial conditions (omega_0, theta_0, phi_0 = omega(t=0), theta(t=0), phi(t=0)) and the system parameters (g, Q, omega_d) and the solver parameters/method.
Chaos can be seen for Q=2, omega_d=w/3.
The program outputs to Matlab time, theta(time) & omega(time).
Plot the phase space via:
plot(mod(theta+pi, 2*pi)-pi, omega, . )
Plot the Poincare sections using:
t_P = (0:2*pi/omega_d:max(time))
plot(mod(spline(time, theta+pi, t_P), 2*pi)-pi, spline(time, omega, t_P), . )
System is described in:
"Fractal basin boundaries and intermittency in the driven damped pendulum"
E. G. Gwinn and R. M. Westervelt
PRA 33(6):4143 (1986)
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Size: 8192 |
Author: Mike Gao |
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Description: 固定窗口的模重复平方算法，计算大数a的n次方模m余下的值，可以有效出理计算机计算过程中的大数溢出问题。-figure out the value of a^n(mod m) when a is large.
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Size: 1024 |
Author: xiaohe |
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Description: 求出一个数对于另一个数的指数，例如求a对模m的指数e,使 a^e(mod m)=1-Obtained a number to another number for the index, such as seeking a pair of mode indices m e, so that a ^ e (mod m) = 1
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Size: 1024 |
Author: xiaohe |
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Description: 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 theory, first of all define a Hash Function (hash (r) = r mod q), then the pattern string P and text string T of length m substring using Hash function to convert values . Clearly only those with the pattern string comparison Hash function values ??with the same substring. Of course, the existence of the conflict because of Hash, but also further string comparison, but as long as selecting the appropriate prime q, Hash is very small the probability of conflict
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Size: 218112 |
Author: chenb |
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Description: 5、设计一个程序计算 (mod m)。当a=31,n=48413,m=113时，计算其值。-5, to design a program to calculate (mod m). When a = 31, n = 48413, m = 113, the calculated value.
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Size: 11264 |
Author: 刘思 |
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Description: 设m不整除a，计算一次同余式ax=b(mod m)。当a=987，b=564，m=2005时，求出x。-Let m not divisible by a, calculate a congruence ax = b (mod m). When a = 987, b = 564, m = 2005, the calculated x.
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Size: 40960 |
Author: liuyingqiong |
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