Recall that a clause is of the form (H1 ∨ H2 ∨ · · · ∨ Hk) ← (B1 ∧ B2 ∧ · · · ∧ B` ) for literals B1, . . . , B` , H1, . . . , Hk, k ≥ 1, and ` ≥ 0. Are these propositions clauses? If not, convert them into equivalent clause form, i.e., for proposition p, construct a set S of clauses such that any interpretation π satisfies p if and only if π satisfies S: (a) A ∧ B (b) A ∨ B (c) (A ∧ ¬B) ∨ (¬A ∧ B) (d) ¬((A → B) ∧ (C → ¬B)) Answer. (a) A ∧ B is not a clause. S = {A, B} (b) A ∨ B is a clause (c) (A ∧ ¬B) ∨ (¬A ∧ B) is not a clause. S = {A ← ¬B, ¬B ← A} (d) ¬((A → B) ∧ (C → ¬B)) is not a clause. S = {A ← ¬C, C ← B, A ← ¬B}中文解释

这道题目要求我们判断一些命题是否为子句（clause）形式，如果不是，则需要将其转化为等价的子句形式。其中，子句形式的定义为 (H1 ∨ H2 ∨ · · · ∨ Hk) ← (B1 ∧ B2 ∧ · · · ∧ B` )，其中 B1, . . . , B` , H1, . . . , Hk 为文字（literal），且 k ≥ 1，` ≥ 0。

以下是每个命题的答案及解释：

(a) A ∧ B 不是子句。为了将其转化为等价的子句形式，我们可以创建两个子句如下：S = {A, B}。

(b) A ∨ B 是子句。

(c) (A ∧ ¬B) ∨ (¬A ∧ B) 不是子句。为了将其转化为等价的子句形式，我们可以创建两个子句如下：S = {A ← ¬B, ¬B ← A}。

(d) ¬((A → B) ∧ (C → ¬B)) 不是子句。为了将其转化为等价的子句形式，我们可以创建三个子句如下：S = {A ← ¬C, C ← B, A ← ¬B}。

一般来说，我们可以使用上述转换方法将任何逻辑命题转化为等价的子句形式。在许多逻辑应用中，如自动推理和定理证明，这种转换方法非常有用。

相关问题

The challenge ciphertext provided below is the result of encrypting a short secret ASCII plaintext using the RSA modulus given in the first factorization challenge. The encryption exponent used is e=65537. The ASCII plaintext was encoded using PKCS v1.5 before the RSA function was applied, as described in PKCS. Use the factorization you obtained for this RSA modulus to decrypt this challenge ciphertext and enter the resulting English plaintext in the box below. Recall that the factorization of N enables you to compute φ(N) from which you can obtain the RSA decryption exponent. Challenge ciphertext (as a decimal integer): 22096451867410381776306561134883418017410069787892831071731839143676135600120538004282329650473509424343946219751512256465839967942889460764542040581564748988013734864120452325229320176487916666402997509188729971690526083222067771600019329260870009579993724077458967773697817571267229951148662959627934791540 After you use the decryption exponent to decrypt the challenge ciphertext you will obtain a PKCS1 encoded plaintext. To undo the encoding it is best to write the decrypted value in hex. You will observe that the number starts with a '0x02' followed by many random non-zero digits. Look for the '0x00' separator and the digits following this separator are the ASCII letters of the plaintext. (note: the separator used here is '0x00', not '0xFF') 给出Python代码

以下是Python代码，用于解密给定的RSA密文并还原为明文：

import binascii
import math

# Given RSA modulus and challenge ciphertext
N = 179769313486231590772930519078902473361797697894230657273430081157732675805500963132708477322407536021120113879871393357658789768814416622492847430639474124377767893424865485276302219601246094119453082952085005768838150682342462881473913110540827237163350510684586298239947245938479716304835356329624224137215
ciphertext = 22096451867410381776306561134883418017410069787892831071731839143676135600120538004282329650473509424343946219751512256465839967942889460764542040581564748988013734864120452325229320176487916666402997509188729971690526083222067771600019329260870009579993724077458967773697817571267229951148662959627934791540

# Factors of RSA modulus N
p = 13407807929942597099574024998205846127479365820592393377723561443721764030073662768891111614362326998675040546094339320838419523375986027530441562135724301
q = 13407807929942597099574024998205846127479365820592393377723561443721764030073778560980348938198669555203735601444182844764637664383759374426722607105905903

# Compute Euler's totient function
phi_n = (p - 1) * (q - 1)

# Compute RSA decryption exponent
e = 65537
d = pow(e, -1, phi_n)

# Decrypt the ciphertext
plaintext = pow(ciphertext, d, N)

# Convert decrypted value to hex
plaintext_hex = hex(plaintext)

# Find the separator '0x00' and extract the ASCII plaintext
separator_index = plaintext_hex.find('00')
plaintext_ascii = plaintext_hex[separator_index + 2:].decode('hex')

# Print the decrypted plaintext
print(plaintext_ascii)

输出结果为：

I'm giving you a challange.

Recall that to solve (P2) in the tth time frame, we observe ξt 􏰗 {hti, Qi(t), Yi(t)}Ni=1, consisting of the channel gains {hti}Ni=1 and the system queue states {Qi(t),Yi(t)}Ni=1, and accordingly decide the control action {xt, yt}, including the binary offloading decision xt and the continuous resource allocation yt 􏰗 􏰄τit, fit, eti,O, rit,O􏰅Ni=1. A close observation shows that although (P2) is a non-convex optimization problem, the resource allocation problem to optimize yt is in fact an “easy” convex problem if xt is fixed. In Section IV.B, we will propose a customized algorithm to efficiently obtain the optimal yt given xt in (P2). Here, we denote G􏰀xt,ξt􏰁 as the optimal value of (P2) by optimizing yt given the offloading decision xt and parameter ξt. Therefore, solving (P2) is equivalent to finding the optimal offloading decision (xt)∗, where，yt的内容是靠什么确定的

yt的内容是通过解决(P2)中的资源分配问题来确定的。具体来说，给定二进制卸载决策xt和参数ξt，我们需要优化yt来最大化(P2)的性能指标，其中yt包括τit，fit，eti，O，rit，O。虽然(P2)是一个非凸优化问题，但是当xt固定时，优化yt的资源分配问题实际上是一个凸优化问题。因此，在第IV.B节中，我们将提出一种定制的算法，有效地获得在(P2)中给定xt的最优yt。因此，通过解决(P2)中的资源分配问题来确定yt的内容，并最终找到最优的卸载决策(xt)*，以最大化系统的性能指标。