des举例(gz).doc

假设m=computer=63,6F,6D,70,75,74,65,72k=network=6E,65,74,77,6F,72,6B请用DES加密法加密，写出加密过程：解：1、先求 密钥 K1因为：k=01101100，01100101，01110100，01110111， 01101111，01110010，0110101111 先拓展到64位：在每七位后加一位校验位 K(temp)=0110110*,0011001*,0101110*,1000111*,0111011*,0111101*,1100100*,1101011*排成8*8矩阵如下：0110110*0 011001*0101110*1000111*0111011*0111101*1100100*1101011*12 根据置换选择 pc-1的进行变换,其过程如下：1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 57494133251791585042342618102595143352719113605244366355473931231576254463830221466153453729211352820124PC-1=0110110*置换选择 pc-111001000011001*01111010101110*01001101000111*01110110111011*10111010111101*01001111100100*01011011101011*1010110故有：规律: 右表的前28位,是从左表的左下角开始,由下向上依次取,并按7位分组填入右表; 右表的后28位,是从左表的7右下角开始,由下向上依次取,并按7位分组填入右表.(注:最后一行是区剩下的).1.3 将右表分为两组:前28位一组C0, 后28位一组D0可得如下:C0=1100100,0111101,0100110,0111011D0=1011101,0100111,0101101,10101101. 4将C0, D0左循环移位,根据下表确定移多少位:轮数:12345678910111213141516位数:1122222212222221所以这里左移一位.C0=1100100,0111101,0100110,0111011C1=100100,0111101,0100110,01110111D0=1011101,0100111,0101101,1010110 D1=011101,0100111,0101101,101011011.5 在将C1,D1组合成8行*7列的矩阵,如下: C1=100100,0111101,0100110,01110111D1=011101,0100111,0101101,1010110114171124153281562110231912426816727201324152313747553040514533484449395634534642503629321.6 根据置换选择PC-2进行变换过程如下:1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556 10010001111010100110011101110111010100111010110110101101PC-2规律:将9,18,22,25,35,38,43,54去掉;其他的没有规律.PC-2故得到48位的K1,00111001100111011100101010100011110101111110010110010001111010100110011101110111010100111010110110101101K1=001110,011001,110111,001010,101000,111101,011111,100101算完K1后，取C1,D1从1.4开始迭代16次2、明文用密钥加密2.1、变换明文，分为两组m=computer=63,6F,6D,70,75,74,65,72m=01100011,01101111,01101101,01110000,01110101,01110100,01100101,01110010组成8*8矩阵如下0110001101101111011011010111000001110101011101000110010101110010进行初始变换IP，1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 58 50 42 34 26 18 10 2 60 52 44 36 28 20 12 4 62 54 46 38 30 22 14 6 64 56 48 40 32 24 16 8 57 49 41 33 25 17 9 1 59 51 43 35 27 19 11 3 61 53 45 37 29 21 13 5 63 55 47 39 31 23 15 7 =IP故有初始变换: IP 01100011011011110110110101110000011101010111010001100101011100101111111110111000011101100101011100000000111111110000011010000011分成两组:前32位一组L0, 后32位一组R0L0=11111111，10111000，01110110，01010111R0=00000000，11111111，00000110，10000011并且 将R0=00000000，11111111，00000110，10000011组合成8行*4列矩阵如下：000000001111111100000110100000112.2、将R0扩展为48位根据选择扩展运算E，来扩展 E 12345678910111213141516171819202122232425262728293031323212345456789891011121312131415161716171819202120212223242524252627282928293031321故有：00000000111111110000011010000011100000000001011111111110100000001101010000000110因此，R0(t48)= 100000,000001,011111,111110,100000,001101,010000,0001102.3、将R0(t48)与密钥K1进行 XORR0(t48)= 100000,000001,011111,111110,100000,001101,010000,000110K1=001110,011001,110111,001010,101000,111101,011111,100101 得出48位结果 R0(X48)= 101110,011000,101000,110100,001000,110000,001111,1000112.4 、进行S盒输出，其定义如下：S1:012345678910111213141501441312151183106125907101574142131106121195382411481362111512973105031512824917511314100613S2:012345678910111213141501518146113497213120510131347152814120110691152014711104131581269321531381013154211671205149S3:012345678910111213141501009146315511312711428113709346102851412111512136498153011121251014731101306987415143115212S4:012345678910111213141507131430691012851112415113811561503472121101492106901211713151314528433150610113894511127214S5:012345678910111213141502124171011685315130149114112124713150151039862421111013781591256301431181271142136150910453S6: 012345678910111213141501211015926801334147511110154271295611314011382914155281237041011311634321295151011141760813S7:012345678910111213141504112141508133129751061113011749110143512215862141113123714101568059236111381410795015142312S8:012345678910111213141501328461511110931450127111513810374125611014922711419121420610131535832114741081315129035611 故有：查表过程如下： 原始数据行，列盒子查得数据二进制表示1011102，7S11110110110000，12S21211001010002，4S3810001101002，10S4300110010000，4S5701111100002，8S6701110011111，7S71010101000113，1S810001R0(S32)= 1011,1100,1000,0011,0111,0111,1010,00012.4 、对R0(S32)进行置换运算P 置换运算 P12345678910111213141516171819202122232425262728293031321672021291228171152326518311028241432273919133062211425故有：1010000011101100001011111001101110111100100000110111011110100001置换运算P 所以，R0(P32)=1010,0000,1110,1100,0010,1111,1001,10112.5 、将R0(P32)与L0进行XOR运算得 R1R0(P32)=1010,0000,1110,1100,0010,1111,1001,1011L0=1111,1111,1011,1000,0111,0110,0101,0111运算得R1=01