如果你也在 怎样代写密码学cryptography这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。密码学cryptography是在存在对抗行为的情况下安全通信技术的实践和研究。 更广泛地说，密码学是关于构建和分析防止第三方或公众阅读私人信息的协议；[信息安全的各个方面，如数据保密性、数据完整性、认证和不可抵赖性，是现代密码学的核心。现代密码学存在于数学、计算机科学、电子工程、通信科学和物理学等学科的交叉点。密码学的应用包括电子商务、基于芯片的支付卡、数字货币、计算机密码和军事通信。
密码学cryptography实际上是加密的同义词，将信息从可读状态转换为不可理解的废话。加密信息的发送者只与预期的接收者分享解码技术，以排除对手的访问。密码学文献通常用Alice（”A”）代表发送者，Bob（”B”）代表预定接收者，Eve（”窃听者”）代表对手。 自从第一次世界大战中转子密码机的发展和第二次世界大战中计算机的出现，密码学方法变得越来越复杂，其应用也越来越多。
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数学代写|密码学作业代写cryptography代考|Finding the Group Exponent
As explained in Section 7.1.2, we recall that the exponent of a group is the smallest nonnegative integer $\lambda$ such that $x^{\lambda}=1$ for all $x$ in the group (assuming that the group is multiplicatively denoted). For cyclic groups, the exponent is obviously the order of the group. For instance in $\mathbf{Z}{n}$ (which is additively denoted), the exponent is $n$ since $n$ is the smallest $x$ such that $x .1 \equiv 0(\bmod n)$ and we have $n . x \equiv 0(\bmod n)$ for any $x \in \mathbf{Z}{n}$.
In the case of $\mathbf{Z}{n}^{}, \lambda$ is denoted $\lambda(n)$ as the Carmichael function. We can easily prove that if $n=p{1}^{\alpha_{1}} \cdots p_{r}^{\alpha_{r}}$ is the factorization of $n$, then
$$
\lambda(n)=\operatorname{lcm}\left(\left(p_{1}-1\right) p_{1}^{\alpha_{1}-1}, \ldots,\left(p_{r}-1\right) p_{r}^{\alpha_{r}-1}\right)
$$
which should be compared to
$$
\varphi(n)=\left(p_{1}-1\right) p_{1}^{\alpha_{1}-1} \cdots\left(p_{r}-1\right) p_{r}^{\alpha_{r}-1} .
$$
Finding the exponent of $\mathbf{Z}_{n}^{}$ is not easy. It is actually as hard as factoring $n$ : obviously, the factorization of $n$ allows to compute $\lambda(n)$ by the above formula. The opposite is a little more subtle. Let us assume that we can compute $\lambda=\lambda(n)$ and let us factorize $n$.
数学代写|密码学作业代写CRYPTOGRAPHY代考|Computing Element Orders in Groups
As we saw in the previous section, computing element orders is at least as hard as computing the group exponent.
In some particular groups, computation is easy. For instance, in $\mathbf{Z}_{n}$, we can compute the order of $x$ even though we do not know how to factorize $n$. The order is the smallest nonnegative $k$ such that $k x \equiv 0(\bmod n)$, namely such that $n$ divides $k x$. The order is simply given by the formula $k=\frac{\operatorname{lcm}(n, x)}{x}$.
When the complete factorization of the exponent of a group $G$ is known, it is also easy to compute the order of any group element $x$ : let $\lambda$ be the exponent of $G$ and let $p_{1}^{\alpha_{1}} \cdots p_{r}^{\alpha_{r}}$ be the complete factorization of $\lambda$ (all $p_{i}$ are pairwise different primes, and all $\alpha_{i}$ are nonnegative integers). We want to compute the order of $x \in G$. We know that it is a factor of $\lambda$. We set $k=\lambda$ as a first approximation for the order of $x$. For any $i$ from 1 to $n$, we replace $k$ by $k / p_{i}$ as long as $x^{\frac{k}{p}}=1$ in $G$. (We assume the group to be multiplicatively denoted.) At the end we are ensured that $k$ is the smallest nonnegative power of $x$ which is such that $x^{k}=1$.
密码学代写
数学代写|密码学作业代写CRYPTOGRAPHY代考|FINDING THE GROUP EXPONENT
如第 7.1.2 节所述，我们记得一个群的指数是最小的非负整数λ这样$\lambda$ such that $x^{\lambda}=1$ for all $x$ in the group (assuming that the group is multiplicatively denoted). For cyclic groups, the exponent is obviously the order of the group. For instance in $\mathbf{Z}{n}$ (which is additively denoted), the exponent is $n$ since $n$ is the smallest $x$ such that $x .1 \equiv 0(\bmod n)$ and we have $n . x \equiv 0(\bmod n)$ for any $x \in \mathbf{Z}{n}$.
In the case of $\mathbf{Z}{n}^{}, \lambda$ is denoted $\lambda(n)$ as the Carmichael function. We can easily prove that if $n=p{1}^{\alpha_{1}} \cdots p_{r}^{\alpha_{r}}$ is the factorization of $n$, then
$$
\lambda(n)=\operatorname{lcm}\left(\left(p_{1}-1\right) p_{1}^{\alpha_{1}-1}, \ldots,\left(p_{r}-1\right) p_{r}^{\alpha_{r}-1}\right)
$$
which should be compared to
$$
\varphi(n)=\left(p_{1}-1\right) p_{1}^{\alpha_{1}-1} \cdots\left(p_{r}-1\right) p_{r}^{\alpha_{r}-1} .
$$
Finding the exponent of $\mathbf{Z}_{n}^{}$ is not easy. It is actually as hard as factoring $n$ : obviously, the factorization of $n$ allows to compute $\lambda(n)$ by the above formula. The opposite is a little more subtle. Let us assume that we can compute $\lambda=\lambda(n)$ and let us factorize $n$.
数学代写|密码学作业代写CRYPTOGRAPHY代考|COMPUTING ELEMENT ORDERS IN GROUPS
正如我们在上一节中看到的，计算元素阶至少与计算群指数一样难。
在某些特定的群体中，计算很容易。例如，在从n，我们可以计算X即使我们不知道如何分解n. 订单是最小的非负数ķ这样ķX≡0(反对n)，即使得n划分ķX. 顺序由公式简单给出ķ=厘米(n,X)X.
当群的指数完全分解时G已知，计算任何群元素的阶也很容易X： 让λ成为指数G然后让p1一种1⋯pr一种r是完全因式分解λ 一种ll$p一世$一种r和p一种一世r在一世s和d一世FF和r和n吨pr一世米和s,一种nd一种ll$一种一世$一种r和n这nn和G一种吨一世在和一世n吨和G和rs. 我们要计算的顺序X∈G. 我们知道这是一个因素λ. 我们设置ķ=λ作为顺序的第一个近似值X. 对于任何一世从 1 到n, 我们替换ķ经过ķ/p一世只要Xķp=1在G. 在和一种ss在米和吨H和Gr这在p吨这b和米在l吨一世pl一世C一种吨一世在和l是d和n这吨和d.最后，我们确保ķ是的最小非负幂X这是这样的Xķ=1.
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