我们可以写一个函数进行暴力破解凯撒密码。具体方法是从1到25枚举凯撒密码的密钥，每个密钥用来解密密文，并判断解密后的明文是否符合语句中的一些常见单词和短语，例如“the”、“and”、“of”等等。如果符合，就输出解密后的明文和密钥。以下是Python代码示例：

def brute_force(cipher_text):
    """ Bruteforcing the Caesar cipher """
    
    # English common words and phrases for checking plaintext
    common_words = ["the", "and", "of", "to", "in", "that", "with", "is", "for", "it", "on", "at", "by", "this", "we", "you", "not", "be", "are"]
    
    # Brute force the cipher key by enumerating from 1 to 25
    for key in range(1, 26):
        plain_text = ""
        
        # Decrypt the cipher text with current key
        for c in cipher_text:
            if c.isalpha():
                # Shift the character back by key positions
                shifted_c = ord(c)-key
                if shifted_c < 65:
                    shifted_c += 26
                plain_text += chr(shifted_c)
            else:
                plain_text += c
        
        # Check if the decrypted plaintext contains common words and phrases
        for word in common_words:
            if word in plain_text.lower().split():
                print("Key: ", key, " Plain text: ", plain_text)

该函数接收一个密文作为输入，从1到25枚举凯撒密码的密钥，并对密文进行解密。对于每个解密后的明文，函数检查其中是否包含了常见单词和短语，如果有，就输出解密出的明文和对应的密钥。