序列密码之A5

一杯敬朝阳，一杯敬月光。

算法分析

1. A5算法已被应用于GSM通信系统中，用于加密从手机到基站的连接，以保护语音通信。一个GSM语言消息被转换成一系列的帧，每帧长228位，每帧用A5进行加密。
2. A5算法主要由三个长度不同的线性移位寄存器组成，即A, B, C。其中A有19位，B有22位，C有23位。
3. 移位由时钟控制的，且遵循“择多”的原则。即从每个寄存器中取出一个中间位，三个数中占多数的寄存器参加移位，其余的不移位。
   

可以参考：A5算法理解

算法实现

import re

def createkey(key,Fn):
    tem = strtobin(key)
    # 线性反馈移位寄存器LFSR A,B,C
    A = list(tem[:19])
    B = list(tem[19:41])
    C = list(tem[41:])
    res = []
    for i in range(114):
        resa = int(A[13]) ^ int(A[16]) ^ int(A[17]) ^ int(A[18])
        resb = int(B[12]) ^ int(A[16]) ^ int(B[20]) ^ int(B[21])
        resc = int(C[17]) ^ int(C[18]) ^ int(C[21]) ^ int(C[22])
        if int(A[9]) ^ int(B[11]) ^ int(C[11]) == 1:
            if  A[9] == '1' and B[11] == '1':
                A[1: 18] = A[0: 17]
                A[0] = str(resa)
                B[1: 21] = B[0: 20]
                B[0] = str(resb)
            elif A[9] == '1' and C[11] == '1':
                A[1: 18] = A[0: 17]
                A[0] = str(resa)
                C[1: 22] = C[0: 21]
                C[0] = str(resc)
            else:
                B[1: 21] = B[0: 20]
                B[0] = str(resb)
                C[1: 22] = C[0: 21]
                C[0] = str(resc)
        else:
            if A[9] == '0' and B[11] == '0':
                A[1: 18] = A[0: 17]
                A[0] = str(resa)
                B[1: 21] = B[0: 20]
                B[0] = str(resb)
            elif A[9] == '0' and C[11] == '0':
                A[1: 18] = A[0: 17]
                A[0] = str(resa)
                C[1: 22] = C[0: 21]
                C[0] = str(resc)
            else:
                B[1: 21] = B[0: 20]
                B[0] = str(resb)
                C[1: 22] = C[0: 21]
                C[0] = str(resc)
        res.append(int(A[18]) ^ int(B[21]) ^ int(C[22]))
    Fn = list(Fn)
    for i in range(22):
        B[i] = int(B[i]) ^ int(Fn[i])
    for i in range(114):
        resa = int(A[13]) ^ int(A[16]) ^ int(A[17]) ^ int(A[18])
        resb = int(B[12]) ^ int(A[16]) ^ int(B[20]) ^ int(B[21])
        resc = int(C[17]) ^ int(C[18]) ^ int(C[21]) ^ int(C[22])
        if int(A[9]) ^ int(B[11]) ^ int(C[11]) == 1:
            if A[9] == '1' and B[11] == '1':
                A[1: 18] = A[0: 17]
                A[0] = str(resa)
                B[1: 21] = B[0: 20]
                B[0] = str(resb)
            elif A[9] == '1' and C[11] == '1':
                A[1: 18] = A[0: 17]
                A[0] = str(resa)
                C[1: 22] = C[0: 21]
                C[0] = str(resc)
            else:
                B[1: 21] = B[0: 20]
                B[0] = str(resb)
                C[1: 22] = C[0: 21]
                C[0] = str(resc)
        else:
            if A[9] == '0' and B[11] == '0':
                A[1: 18] = A[0: 17]
                A[0] = str(resa)
                B[1: 21] = B[0: 20]
                B[0] = str(resb)
            elif A[9] == '0' and C[11] == '0':
                A[1: 18] = A[0: 17]
                A[0] = str(resa)
                C[1: 22] = C[0: 21]
                C[0] = str(resc)
            else:
                B[1: 21] = B[0: 20]
                B[0] = str(resb)
                C[1: 22] = C[0: 21]
                C[0] = str(resc)
        res.append(int(A[18]) ^ int(B[21]) ^ int(C[22]))
    return res

# 字符串转换为二进制字符串
def strtobin(s):
    res = []
    for c in s:
        tem = bin(ord(c)).replace('b', '')
        # 转为字符串时，后7位中，如果存在前面为0，会自动去掉，需要加回来，使之满足8位
        if len(tem) < 8:
            tem = "0" + tem
        res.append(tem)
    return ''.join(res)

# 二进制转字符串
def bintostr(s):
    tem = ""
    for i in s:
        tem += str(chr(int(i, base=2)))
    return tem

# 将明文字符串分割为指定长度字符串并存于列表中
def cut_text(text, lenth):
    tem = re.findall('.{' + str(lenth) + '}', text)
    tem.append(text[(len(tem) * lenth):])
    # 由于分割后，末尾出现一个空字符，故去掉
    result = [i for i in tem if i != '']
    return result

if __name__ == "__main__":
    # 明文
    plaintext = "ifnottothesunforsmilingwarmisstillinthesuntherebutwewilllaughmoreconfidentcalmifturnedtofoundhisownshadowappropriateescapethesunwillbethroughtheheartwarmeachplacebehindthecornerifanoutstretchedpalmcannotfallbutterflythenclenchedwavingarmsgivenpowerificanthavebrightsmileitwillfacetothesunshineandsunshinesmiletogetherinfullbloom"
    # 秘钥
    key = "asdfghjk"
    # 转换成二进制
    mtext = strtobin(plaintext)
    key = strtobin(key)
    # 帧号码Fn 22位
    Fn = "0101101000110101010101"
    # 生成密钥流
    keystream = createkey(key, Fn)

    # 加密
    mlist = cut_text(mtext, 228)
    ciphertext = ""
    for t in mlist:
         # 对每组明文分别加密
         for i in range(len(t)):
             ciphertext += str(int(keystream[i]) ^ int(t[i]))
    print("加密后得到的密文为： \n" + hex(int(ciphertext,2)).upper()[2:])

    # 解密
    clist = cut_text(ciphertext, 228)
    res = ""
    for t in clist:
        for i in range(len(t)):
            res += str(int(keystream[i]) ^ int(t[i]))
    result = cut_text(res, 8)
    end = bintostr(result)
    print("解密后得到的明文为： \n" + end)

加解密

标签：tem,22,int,18,A5,str,序列密码,21
来源： https://www.cnblogs.com/clwsec/p/15680055.html