# 数学代写|抽象代数作业代写abstract algebra代考|MATH355

## 数学代写|抽象代数作业代写abstract algebra代考|Fast Exponentiation

Let $G$ be a group, let $g$ be an element in $G$, and let $n$ be a positive integer. To calculate the power $g^n$, one normally must calculate
$$g^n=\overbrace{g \cdot g \cdots g \text { times }}^n,$$
which involves $n-1$ operations. (If fact, when we implement this into a computer algorithm, since we must take into account the operation of incrementing a counter, the above direct calculation takes a minimum of $2 n-1$ computer operations.) If the order $|g|$ and the power $n$ are large, one may not notice any patterns in the powers of $g$ that would give us any shortcuts to determining $g^n$ with fewer than $n-1$ group operations.

The Fast Exponentiation Algorithm allows one to calculate $g^n$ with many fewer group operations than $n$, thus significantly reducing the calculation time.

The reason that $x$ has the value of $g^n$ at the end of the for loop is because when the algorithm terminates,
$$x=g^{b_k 2^k+b_{k-1} 2^{k-1}+\cdots+b_1 2+b_0},$$
which is precisely $g^n$. Note that in the binary expansion $n=\left(b_k b_{k-1} \cdots b_1 b_0\right)_2$, there is an assumption that $b_k=1$.

## 数学代写|抽象代数作业代写abstract algebra代考|Diffie-Hellman Algorithm

In order to describe how Diffie-Hellman works, we will introduce three players: Alice, Bob, and Eve. Alice and Bob want to talk secretly while Eve wants to eavesdrop on their conversation. We assume that Eve can hear ev erything that Alice and Bob say to each other. In the following diagram, that which is boxed stays secret to the individual and that which is not boxed is heard by everyone, including Eve. The Diffie-Hellman public key protocol works as follows (Figure 1.19).
(1) Alice (on the left) and Bob (on the right) settle on a group $G$ and a group element $g$, called the base. Ideally, the order of $g$ should be very large. (If you are doing calculations by hand, the order of $g$ should be in the hundreds. If we are using computers, the order of $g$ is ideally larger than what a typical for loop runs through, say $10^{15}$ or much more.)
(2) Alice chooses a relatively large integer $a$ and sends to Bob the group element $g^a$, calculated with Fast Exponentiation.
(3) Bob chooses a relatively large integer $b$ and sends to Alice the group element $g^b$, calculated with Fast Exponentiation.
(4) Alice calculates $\left(g^b\right)^a=g^{a b}$ using Fast Exponentiation.
(5) Bob calculates $\left(g^a\right)^b-g^{a b}$ using Fast Exponentiation.
(6) Alice and Bob will now use the group element $g^{a b}$ as the key.
The reason why Eve cannot easily figure out $g^{a b}$ is simply a matter of how long it would take her to do so. We should assume that Eve intercepts the group $G$, the base $g$, the element $g^a$, and the element $g^b$. However, Eve does not know the integers $a$ or $b$. In fact, Alice does not know $b$ and Bob does not know $a$. The reason why this is safe is that Fast Exponentiation makes it possible to calculate powers of group elements quickly while, on the other hand, the problem of determining $n$ from knowing $g$ and $g^n$ could take as long as $n$. If the power $n$ is very large, it simply takes far too long. Not knowing $a$ or $b$, Eve cannot quickly determine the key $g^{a b}$ if she only knows $g, g^a$, and $g^b$.

# 抽象代数代考

## 数学代写|抽象代数作业代写abstract algebra代考|Fast Exponentiation

$$g^n=\overbrace{g \cdot g \cdots g \text { times }}^n,$$

$$x=g^{b_k 2^k+b_{k-1} 2^{k-1}+\cdots+b_1 2+b_0},$$

## 数学代写|抽象代数作业代写abstract algebra代考|Diffie-Hellman Algorithm

(1) Alice (左边) 和 Bob (右边) 在一组 $G$ 和一个组元嗉 $g$ ，称为基数。理想情况下，顺序为 $g$ 应该很大。 (如果你是手工计算，顺序 $g$ 应该是几百。如果我们使用计算机，则顺序 $g$ 理想情况下，比典型的 for 循环 运行的要大，比如 $10^{15}$ 或更多。)
（2）爱丽丝选择一个相对较大的整数 $a$ 并将组元嗉发送给 Bob $g^a$ ，用快速指数计算。
(3) Bob选择了一个比较大的整数 $b$ 并将组元素发送给 Alice $g^b$ ，用快速指数计算。
(4) Alice 计算 $\left(g^b\right)^a=g^{a b}$ 使用快速指数。
(5) Bob 计算 $\left(g^a\right)^b-g^{a b}$ 使用快速指数。
(6) Alice 和 Bob 现在将使用 group 元素 $g^{a b}$ 作为关键。

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