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HILL CHIPHER

HILL CHIPHER. Langkah-Langkah Enkripsi: Tentukan Plain Text Tentukan Key Dalam Bentuk Matriks 3 x 3 K Ubah Plain Text Ke Bentuk Angka  P n Kalikan Matriks K dengan Plain Text P n Tp n Moduluskan Tp n dengan 26 Cn Ubah Cn ke bentuk huruf Chipher Text. Contoh. 16 -17 -7

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HILL CHIPHER

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  1. HILL CHIPHER Langkah-Langkah Enkripsi: Tentukan Plain Text Tentukan Key Dalam Bentuk Matriks 3 x 3 K Ubah Plain Text Ke Bentuk Angka Pn Kalikan Matriks K dengan Plain Text Pn Tpn Moduluskan Tpn dengan 26 Cn Ubah Cn ke bentuk huruf Chipher Text

  2. Contoh 16-17-7 -1115-12 19-151 (2) Key = 4 13 19 Tp4 Tp5 = Tp6 16-17-7 -1115-12 19-151 18 8 11 Tp1 Tp2 = Tp3 (4) Dik : Plain Text = S I L E N T (1) Dit : Lakukan Enkripsi Jwb: Plain Text = S I L E N T Misal Index Huruf A = 0 Pn =18 8 11 4 13 19 (3)

  3. Tp1 = (16 * 18) + (-17*8)+(-7*11) = (288) + (-136) +(-77) = 75 Tp2 = (-11 * 18) + (15*8)+(-12*11) = (-198) + (120) +(-132) = -210 Tp3 = (19 * 18) + (-15*8)+(1*11) = (342) + (-120) +(11) = 233 Tp4 = (16 * 4) + (-17*13)+(-7*19) (4) = (64) + (-221) +(-133) = -290 Tp5 = (-11 * 4) + (15*13)+(-12*19) = (-44) + (195) +(-228) = -77 Tp6 = (19 * 4) + (-15*13)+(1*19) = (76) + (-195) +(19) = -100

  4. C4 = Tp4 mod 26 = (-290) mod 26  290 mod 26 = 4 = 26 – 4 =22 C5 = Tp5 mod 26 = (-77) mod 26  77 mod 26 = 25 = 26 – 25 = 1 C6 = Tp6 mod 26 = (-100) mod 26  100 mod 26 = 22 = 26 – 22 =4 Cn= Tpn Mod 26 (5) C1 = Tp1 mod 26 = 75 mod 26 = 23 C2 = Tp2 mod 26 = (-210) mod 26  210 mod 26 =2 = 26 – 2 = 24 C3 = Tp3mod 26 = 233 mod 26 = 25

  5. Cipher Text Yang Dihasilkan Dari Kata SILENT Cn = 23 24 25 22 1 4 (6) Ciphertext = X Y Z W B E

  6. HILL CHIPHER Langkah-Langkah Dekripsi: Tentukan Cipher Text Ubah Cipher Text Ke Bentuk Angka Cn Tentukan Key Dalam Bentuk Matriks 3 x 3 K Hitung Determinan dari Matriks K D(K) Moduluskan D(K)dengan 26 Z Cari Inverse dari Z dimodulus 26 Z-1 Cari Adjoin dari Matriks K Adj(K) Kalikan Z-1* Adj(K) mod 26 K-1 Kalikan Matriks K-1dengan Cn Pn Ubah Pn ke dalam bentuk huruf  Plain Text

  7. 16 -17 -7 • K = -11 15 -12 • 19 -15 1 (3) • 16 -17 -7 • Det(K) = -11 15 -12 • 19 -15 1 • 16 -17 • -11 15 • 19 -15 Det(K)= ((16*15*1)+(-17*-12*19)+(-7*-11*-15)) – ((19*15*-7)+(-15*-12*16)+(11*-11*-17)) Contoh : Carilah Plain Text dari Ciphertext Dibawah Ini. Ciphertext = X Y Z W B E (1) Cn = 23 24 25 22 1 4 (2) (4) Det(K) = 2961 – 1072 = 1889

  8. Langkah Ke Lima Mencari Nilai det(K) Modulus 26 (5) Z = det(K) mod 26 = 1889 mod 26 = 17 Langkah Ke Enam Mencari Invers Z Modulus 26 (6) Tabel Mencari Invers Mod 26 Dari tabel dapat diketahui bahwa Z-1(17)=23 Z-1 = 23 Note: Jika nilai det(K) mod 26 tidak terdapat pada tabel Z, maka dapat dipastikan bahwa permasalahan tidak akan terpecahkan(terdapat kesalahan pada matriks key).Pada Kasus Hill Cipher, nilai determinan Matriks Key di Mod 26, harus terdapat pada tabel Z.

  9. Langkah Ke Tujuh: Mencari Adjoin Matrik K dengan Kofaktor Matriks (+)a11(-)a21(+)a31 (-)a12(+)a22(-)a32 (+)a13(-)a23(+)a33 Adj(K) =

  10. a11 = +((15 * 1) – (-15 * -12)) a11 = +((15) - (180)) a11 = +(-165) = 165 a12 = - (( -11 * 1) – ( 19 * -12 )) a12 = - (( -11 ) - ( -228)) a12 = - (217) = -217 a13 = +( ( -11 * -15 ) – ( 19 * 15 )) a13 = +(( 165 ) - ( 285)) a13 = +(-120) = -120 16 -17 -7 a11= -11 15 -12 19 -15 1 16 -17 -7 a12= -11 15 -12 19 -15 1 16 -17 -7 a13= -11 15 -12 19 -15 1

  11. a21 = - ((-17 * 1) – (-15 * -7)) a21 = - ((-17) - (105)) a21 = - (-122) = 122 a22 = + (( 16 * 1) – ( 19 * -7 )) a22 = + (( 16 ) - ( -133)) a22 = + (149) = 149 a23 = - (( 16 * -15) – ( 19 * -17 )) a23 = - (( -240 ) - ( -323)) a23 = - (83) = -83 16 -17 -7 a21= -11 15 -12 19 -15 1 16 -17 -7 a22= -11 15 -12 19 -15 1 16 -17 -7 a23= -11 15 -12 19 -15 1

  12. a31 = + ((-17 * -12) – (15 * -7)) a31 = + ((204) - (-105)) a31 = + (309) = 309 a32 = - (( 16 * -12) – ( -11 * -7 )) a32 = - (( -192 ) - (77)) a32 = - (-269) = 269 a33 = + (( 16 * 15) – ( -11 * -17 )) a33 = + ((240 ) - ( 187)) a33 = + (53) = 53 16 -17 -7 a31= -11 15 -12 19 -15 1 16 -17 -7 a32= -11 15 -12 19 -15 1 16 -17 -7 a33= -11 15 -12 19 -15 1

  13. Adjoint dari Matriks Key(K) ( 8) Dari Operasi Kofaktor didapat Adjoint dari Matrik Key(K) Yaitu -165122309 -217149269 -120-8353 Adj(K) =

  14. Langkah Ke-8 Mencari Nilai Matrik K-1 -165122309 -217149269 -120-8353 (8) -165122309 -217149269 -120-8353 -3795 28067107 -4991 34276187 -2760 -19091219 K-1 = (Z-1 * Adj(K)) mod 26 Z-1 = 23 (6) Adj(K)= K-1 = 23* mod 26 K-1 = mod 26

  15. Langkah Ke-8 Mencari Nilai Matrik K-1(Cont) 1249 12125 221523 Langkah Ke-9 Mencari Nilai Pn (9) P1 P2 = P3 P4 P5 = P6 1249 12125 221523 C1 C2 C3 C4 C5 C6 K-1 = Pn = (K-1* Cn) mod 26 mod 26

  16. Mencari Nilai Pn (Cont) P4 P5 = P6 P1 P2 = P3 22 1 4 1249 12125 221523 1249 12125 221523 23 24 25 P1 P2 = P3 824 1152 1441 P1 P2 = P3 18 8 11 P4 P5 = P6 P4 P5 = P6 4 13 19 82 143 591 mod 26  Pn = 18 8 11 4 13 19 Plaintext = S I L E N T mod 26 mod 26 

  17. Selamat Belajar I LOVE KRIPTOGRAFI

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