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# 数学代写|密码学代写Cryptography Theory代考|CS355 Knapsack Encryption

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## 数学代写|密码学Cryptography Theory代考|Knapsack Encryption

The knapsack problem is known to be NP-complete, so it should make a very firm foundation to build a cryptosystem upon. After all, factoring, although believed to be hard, has eluded attempts to pin it down to being NP-complete, and look at how successful RSA has been. A system with a firmer foundation might be used with even greater confidence, but how could it be done? Merkle and Hellman worked out the details for their 1978 paper referred to above. Their solution follows. We’ll use the knapsack from our brief list of NP-complete problems:
$$S={4,8,13,17,21,33,95,104,243,311,400,620,698,805,818,912}$$
Given a selection of elements of $S$, their sum is easily computed, but the inverse problem appears intractable (not in $\mathbf{P}$ ), so we’ll use it for a one-way function.

## 数学代写|密码学Cryptography Theory代考|Elgamal Encryption

The difficulty of solving the discrete log problem (see Section 14.2) was used by Taher Elgamal ${ }^{41}$ (Figure 16.9), an Egyptian-born American cryptographer, to create the Elgamal public key cryptosystem in $1985 .{ }^{42}$ Alice and Bob will illustrate how Elgamal works.

Alice begins by picking a large prime $p$ and a generator $g$ of the multiplicative group of integers modulo p. Recall from Section 14.2 that taking consecutive powers of a generator $(\bmod p)$, until we reach 1 , will result in the entire group being generated. Alice then selects a private key $a$ and computes $A=g^a \bmod p$. She publishes $g, p$, and $A$, only keeping the value a secret.

Bob wishes to send a message $M$, which he must put in the form of a number between 2 and $p$. If the message is too long for this, it can be broken into pieces. Bob then selects a key $k(\bmod p)$ and computes $C_1=g^k(\bmod p)$ and $C_2=M A^k(\bmod p)$. He then sends Alice both $C_1$ and $C_2$. Thus, this system has the disadvantage that the ciphertext is twice as long as the message.

To recover the message $M$, Alice computes $x=C_1^a(\bmod p)$. From this, she is able to then calculate $x^{-1}$ by using the Euclidean algorithm, as detailed in Section 14.3.
Finally, Alice computes $x^{-1} C_2(\bmod p)$, which reveals the message, because
$$x^{-1} C_2=\left(C_1^a\right)^{-1} C_2=\left(g^{a k}\right)^{-1} M A^k=\left(g^{a k}\right)^{-1} M\left(g^a\right)^k=M\left(g^{a k}\right)^{-1} g^{a k}=M$$

## 数学代写|密码学Cryptography Theory代考|Knapsack Encryption

$$S=4,8,13,17,21,33,95,104,243,311,400,620,698,805,818,912$$

## 数学代写|密码学Cryptography Theory代考|Elgamal Encryption

Bob 希望发送诮息 $M$ ，他必须以 2 到p. 如果消息太长，可以将其分成几部分。Bob 然后选择一个键 $k(\bmod p)$ 并计算 $C_1=g^k(\bmod p)$ 和 $C_2=M A^k(\bmod p)$. 然后他把两个都发给爱丽丝 $C_1$ 和 $C_2$. 因此，该系统的缺点是密文的长度是消息的两倍。

$$x^{-1} C_2=\left(C_1^a\right)^{-1} C_2=\left(g^{a k}\right)^{-1} M A^k=\left(g^{a k}\right)^{-1} M\left(g^a\right)^k=M\left(g^{a k}\right)^{-1} g^{a k}=M$$

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