《图解密码技术》小测验
第三章小测验：
小测试3 对称密码的密钥需要多长
A：需要521比特，B：1024比特；C：4096比特；D：至少需要1万比特；E：100万比特也不够。
答：The total number of 512-bit keys is 1.34
,we can calculate the total number of keys that can be tried is 366246060
=3.16
.So The total number of 512-bit keys is enough. When the key length reaches 512 bits, increasing the key length will only slow down the algorithm.
小测验4 对称密码的基础知识
1.在对称密码中，加密的密钥和解密的密钥是相等的。（right）
2.将来，当计算机的计算能力足够高时，就可以在现实的时间内破译一次性密码本的密文。（wrong）
答：One-time pad is not deciphered, there is no way to determine the deciphering is plaintext, and computing power has nothing to do with.
3.如果密钥长度为56比特，那么用暴力破解找到正确密钥平均尝试约
次。（wrong）
答：The average number of attempts is half the number of keys：
.
4.虽然AES是一致强度很高的对称密码算法，但在商用情况下需要向NIST支付授权费用。（wrong）
答：AES is free to use.
5.AES可以在现实的时间内被破译。（right）
6.AES标准所选定的密码算法叫做Rijndael. （right）
第四章小测验：
小测验1 比特与字节
答：The number of bits divided by 8 is the number of bytes. 128 bits is 16 bytes.
小测验2 对ECB模式的攻击
答: The ciphertext group with the name of the user name overrides the ciphertext group of the user’s password, and if the user name is known, we can log in。
小测验3 CBC模式的初始化向量
答：Every time the same initialization vector is used and the same key pair is used to encrypt the same plaintext, all ciphertext will be the same. When the cryptographer receives two identical ciphers at different times, he can judge that the decrypted plaintext is the same. We can avoid this problem by changing the value of the initialization vector every time.
小测验4 仿CBC模式
答：This model is virtually identical to the ECB model and it has the same character as the ECB.