Algorima Triangle Chaim Cipher dan contohnya
Algorima Triangle Chaim Cipher.
Algoritma kriptografi triangle chain atau umumnya dikenal dengan sebutan rantai segitiga merupakan cipher yang ide awalnya dari algoritma kriptografi One Time Pad, yaitu kunci yang dibangkitkan secara random dan panjang kunci sepanjang plainteks yang akan dienkripsi. Tetapi pada algoritma kriptogarfi rantai segitiga pembangkitan kunci-kunci tersebut secara otomatis dengan teknikberantai.
Teknik/Konsep :
- Melakukan perluasan dari teknik kriptografi algoritma caesar.
- Melakukan pengeseran karakter berdasarkan kunci dan faktor pengali.
- Melakukan enkripsu dan deskripsi secara berantai dan ganda artinya ada dua kali proses enkripsi maupun deskripsi.
- Formula Enkripsi :
a. Segitiga Pertama
Untuk baris 1 :
Mij = ( P [ j ] + ( K * R [ 1 ] ) ) mod 256
Untuk baris 2 :
Nilai J > 1 < N
Mij = (M[i-1] + (K*R[i])) mod 256
Cipher hasil segitiga pertama diambil dengan fomula :
Mij = [ i , ( n+1 ) – n ]
b. Segitiga Kedua
Plainteks untuk segitiga kedua adalah cipher hasil segitiga pertama.
a. untuk baris 1 :
Mij = [P(j) + (K*R(1))] mod 256
b. untuk baris 2 :
Nilai J > ( n+1 ) – i
Mij = [M(i,j) + (K*R(i))] mod 256
Cipherteks diambil dgn formula :
Formula Dekripsi :
Proses dekripsi dilakukan sebanyak 2 kali.
a. Deskripsi Segitiga pertama.
- Untuk Baris pertama
Mij = [C(j) –(K*R(i))] mod 256
- Untuk baris kedua dan seterusnya
J < (n+i) – i
Mij = [C(i-1),j – (K*R(i))] mod 256
Plaintext hasil segitiga pertama dapatkan dengan foemula
Mij = [ i, (n+1) – n]
b. Dekripsi Segitiga Kedua
Chiper yang diproses pada dekripsi segitiga kedua adalah hasil dekripsi segitiga pertama :
- Untuk segitiga pertama
Mij = [C(j) –(K*R(1))] mod 256
- Untuk Baris kedua ,dst :
Mij = [C(i-j),j – (K*R(i)] mod 256
Plainteks diambil dengan formula :
Mij = [(n+1)-i,1)
Keterangan :
P = Plainteks I = Indeks baris
C = Chipertext J = Indeks kolom
K = Kunci R = Faktor pengali
N = Jumlah Karakter plainteks
cipherteks
CONTOH :
Plainteks = CHRIS
K = 5
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C |
H |
R |
I |
S |
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67 |
72 |
82 |
73 |
83 |
R = R1 R2 R3 R4 R5
1 2 3 4 5
A. Enkripsi ∆ 1 :
Untuk baris 1 = P ( i-1)
i = 1
Mij = [ P (i) + ( K * R (i) ))] mod 256
M11 = ( C + (5*1) ) mod 256
=( 67 + 5 ) mod 256
= 72 è H
M12 = (H+ (5*1) ) mod 256
=( 72 + 5 ) mod 256
= 77 è M
M13 = ( R + (5*1) ) mod 256
=( 82 + 5 ) mod 256
= 87 è W
M14 = ( I + (5*1) ) mod 256
=( 73 + 5 ) mod 256
= 78 è N
M15 = ( S + (5*1) ) mod 256
=( 83 + 5 ) mod 256
= 88 è X
J = 1 2 3 4 5
R1 R2 R3 R4 R5
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C/67 |
H/72 |
R/82 |
I/73 |
S/83 |
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H/72 |
M/77 |
W/87 |
N/78 |
X/88 |
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W/87 |
a/97 |
X /88 |
b/98 |
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p/112 |
g/103 |
q/113 |
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{ / 123 |
…/133
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ž/158 |
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i/0
1
2
3
4
5
Baris ke 2 ( i = 2 ) :
Mij = M ( i -1 ), j + ( K * R ( i ) ) mod 256
M22 = M (2-1), 2 + (5 * 2 ) mod 256
= M (1,2 ) + ( 5 * 2 ) mod 256
= ( M + 10 ) mod 256
= ( 77 + 10 ) m od 256
= 87 è W
M23 = M (2-1), 3 + (5 * 2 ) mod 256
= M (1,3 ) + ( 5 * 2 ) mod 256
= ( W + 10 ) mod 256
= ( 87 + 10 ) m od 256
= 97 è a
M24 = M (2-1), 4 + (5 * 2 ) mod 256
= M (1,4 ) + ( 5 * 2 ) mod 256
= ( N + 10 ) mod 256
= ( 78 + 10 ) m od 256
= 88 è X
M25 = M (2-1), 5 + (5 * 2 ) mod 256
= M (1,5 ) + ( 5 * 2 ) mod 256
= ( X + 10 ) mod 256
= ( 88 + 10 ) m od 256
= 98 è b
Baris ke 3 ( i = 3 ) :
Mij = M ( i -1 ), j + ( K * R ( i ) ) mod 256
M33 = M (3-1), 3 + (5 * 3 ) mod 256
= M (2,3 ) + ( 5 * 3 ) mod 256
= ( a + 15 ) mod 256
= ( 97 + 15 ) mod 256
= 112 è p
M34 = M (3-1), 4 + (5 * 3 ) mod 256
= M (2,4 ) + ( 5 * 3 ) mod 256
= ( X + 15 ) mod 256
= ( 88 + 15 ) mod 256
= 103 è g
M35 = M (3-1), 5 + (5 * 3 ) mod 256
= M (2,5 ) + ( 5 * 3 ) mod 256
= ( b + 15 ) mod 256
= ( 98 + 15 ) mod 256
= 113 è q
Baris ke 4 ( i = 4 ) :
Mij = M ( i -1 ), j + ( K * R ( i ) ) mod 256
M44 = M (4-1), 4 + (5 * 4 ) mod 256
= M (3,4 ) + ( 5 * 4 ) mod 256
= ( g + 20 ) mod 256
= ( 103 + 20 ) mod 256
= 123 è {
M45 = M (4-1), 5 + (5 * 4 ) mod 256
= M (3,5 ) + ( 5 * 4 ) mod 256
= ( q + 20 ) mod 256
= ( 113 + 20 ) mod 256
= 133 è …
Baris ke 5 ( i = 5 ) :
Mij = M ( i -1 ), j + ( K * R ( i ) ) mod 256
M44 = M (5-1), 5 + (5 * 5 ) mod 256
= M (4,5 ) + ( 5 * 5 ) mod 256
= ( … + 25 ) mod 256
= ( 113 + 25 ) mod 256
= 158 è ž
Plainteks = C H R I S
Cipher ∆ 1 = H W p { ž
B. Segitiga kedua
P = H W p { ž
K = 5
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H/72 |
W/87 |
p/112 |
{ / 123 |
ž/158 |
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M/77 |
\ /92 |
u/117 |
€/128 |
£/163 |
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W/87 |
f/102 |
⌂/127 |
ž /138 |
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f/102 |
u/117 |
Ž/142 |
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z/122 |
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‰ /137 |
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“ /147 |
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i/0
1
2
3
4
5
Baris 1
Mij = ( P [i – 1 ] , j + ( K * R [i] )) mod 256
M11 = ( P [1-1] , 1 + ( 5 * 1 ) ) mod 256
= ( P (0,1) + 5 ) mod 256
= ( H + 5 ) mod 256
= ( 72 + 5 ) mod 256
= 77 è M
M12 = ( P [1-1] , 2 + ( 5 * 1 ) ) mod 256
= ( P (0,2) + 5 ) mod 256
= ( W + 5 ) mod 256
= ( 87 + 5 ) mod 256
= 92 è \
M13 = ( P [1-1] , 3 + ( 5 * 1 ) ) mod 256
= ( P (0,3) + 5 ) mod 256
= ( p + 5 ) mod 256
= ( 112 + 5 ) mod 256
= 117 è u
M14 = ( P [1-1] , 4 + ( 5 * 1 ) ) mod 256
= ( P (0,4) + 5 ) mod 256
= ( { + 5 ) mod 256
= ( 123 + 5 ) mod 256
= 128 è €
M15 = ( P [1-1] , 5 + ( 5 * 1 ) ) mod 256
= ( P (0,5) + 5 ) mod 256
= ( ž + 5 ) mod 256
= ( 158 + 5 ) mod 256
= 163 è £
Baris 2 i = 2
Mij = ( P [i – 1 ] , j + ( K * R [i] )) mod 256
M21 = ( P [2-1] , 1 + ( 5 * 2 ) ) mod 256
= ( P (1,1) + 10 ) mod 256
= ( M + 10 ) mod 256
= ( 77 + 10 ) mod 256
= 87 è W
M22 = ( P [2-1] , 2 + ( 5 * 2 ) ) mod 256
= ( P (1,2) + 10 ) mod 256
= ( \ + 10 ) mod 256
= ( 92 + 10 ) mod 256
= 102 è f
M23 = ( P [2-1] , 3 + ( 5 * 2 ) ) mod 256
= ( P (1,3) + 10 ) mod 256
= ( u + 10 ) mod 256
= ( 117 + 10 ) mod 256
= 127 è ⌂
M24 = ( P [2-1] , 4 + ( 5 * 2 ) ) mod 256
= ( P (1,4) + 10 ) mod 256
= ( € + 10 ) mod 256
= ( 128 + 10 ) mod 256
= 138 è Š
Baris 3 i = 3
Mij = ( P [i – 1 ] , j + ( K * R [i] )) mod 256
M31 = ( P [3-1] , 1 + ( 5 * 3 ) ) mod 256
= ( P (2,1) + 15 ) mod 256
= ( W + 15 ) mod 256
= ( 87 + 15 ) mod 256
= 102 è f
M32 = ( P [3-1] , 2 + ( 5 * 3 ) ) mod 256
= ( P (2,2) + 15 ) mod 256
= (f + 15 ) mod 256
= ( 102 + 15 ) mod 256
= 117 è u
M33 = ( P [3-1] , 3 + ( 5 * 3 ) ) mod 256
= ( P (2,3) + 15 ) mod 256
= ( ⌂ + 15 ) mod 256
= ( 127 + 15 ) mod 256
= 142 è Ž
Baris 4 i = 4
Mij = ( P [i – 1 ] , j + ( K * R [i] )) mod 256
M41 = ( P [4-1] , 1 + ( 5 * 4 ) ) mod 256
= ( P (3,1) + 20 ) mod 256
= ( f + 20 ) mod 256
= ( 102 + 20 ) mod 256
= 122 è z
M42 = ( P [4-1] , 2 + ( 5 * 4 ) ) mod 256
= ( P (3,2) + 20 ) mod 256
= ( u + 20 ) mod 256
= ( 117 + 20 ) mod 256
= 137 è ‰
Baris 5 i = 5
Mij = ( P [i – 1 ] , j + ( K * R [i] )) mod 256
M51 = ( P [5-1] , 1 + ( 5 * 5 ) ) mod 256
= ( P (4,1) + 25 ) mod 256
= ( z + 25 ) mod 256
= ( 122 + 25 ) mod 256
= 147 è “
Cipher ∆ II = “ ‰ Ž Š £
Dekripsi :
C = “ ‰ Ž Š £
K = 5
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“ |
‰ |
Ž |
Š |
£ |
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147 |
137 |
142 |
138 |
163 |
R1 R2 R3 R4 R5
R = 1 2 3 4 5
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“/147 |
‰/137 |
Ž/142 |
Š/138 |
£/163 |
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Ž/142 |
„/132 |
‰/137 |
…/133 |
ž/158 |
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z/122 |
⌂/127 |
{ /123 |
“/148 |
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p/112 |
l/108 |
…/133 |
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X / 88 |
q/113
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X/88 |
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i/0
1
2
3
4
5
Baris 1 i = 1
Mij = ( C [i – 1 ] , j - ( K * R [i] )) mod 256
M11 = ( C [1-1] , 1 - ( 5 * 1 ) ) mod 256
= ( C (0,1) - 5 ) mod 256
= ( “ - 5 ) mod 256
= ( 147 - 5 ) mod 256
= 142 è Ž
M12 = ( C [1-1] , 2 - ( 5 * 1 ) ) mod 256
= ( C (0,2) - 5 ) mod 256
= ( ‰ - 5 ) mod 256
= ( 137 - 5 ) mod 256
= 132 è „
M13 = ( C [1-1] , 3 - ( 5 * 1 ) ) mod 256
= ( C (0,3) - 5 ) mod 256
= ( Ž - 5 ) mod 256
= ( 142 - 5 ) mod 256
= 137 è ‰
M14 = ( C [1-1] , 4- ( 5 * 1 ) ) mod 256
= ( C (0,4) - 5 ) mod 256
= ( Š - 5 ) mod 256
= ( 138 - 5 ) mod 256
= 133 è …
M15 = ( C [1-1] , 5 - ( 5 * 1 ) ) mod 256
= ( C (0,5) - 5 ) mod 256
= ( £ - 5 ) mod 256
= ( 163 - 5 ) mod 256
= 158 è ž
Baris 2 i = 2
Mij = ( C [i – 1 ] , j - ( K * R [i] )) mod 256
M22 = ( C [2-1] , 2 - ( 5 * 2 ) ) mod 256
= ( C (1,2) - 10 ) mod 256
= ( „ - 10 ) mod 256
= ( 132 - 10 ) mod 256
= 122 è z
M23 = ( C [2-1] , 3 - ( 5 * 2 ) ) mod 256
= ( C (1,3) - 10 ) mod 256
= ( ‰ - 10 ) mod 256
= ( 137 - 10 ) mod 256
= 127 è ⌂
M24 = ( C [2-1] , 4 - ( 5 * 2 ) ) mod 256
= ( C (1,4) - 10 ) mod 256
= ( … - 10 ) mod 256
= ( 133 - 10 ) mod 256
= 123 è {
M25 = ( C [2-1] , 5 - ( 5 * 2 ) ) mod 256
= ( C (1,5) - 10 ) mod 256
= ( ž - 10 ) mod 256
= ( 158 - 10 ) mod 256
= 148 è ”
Baris 3 i = 3
Mij = ( C [i – 1 ] , j - ( K * R [i] )) mod 256
M33 = ( C [3-1] , 3 - ( 5 * 3 ) ) mod 256
= ( C (2,3) - 15 ) mod 256
= ( ⌂ - 15 ) mod 256
= ( 127 - 15 ) mod 256
= 112 è P
M34 = ( C [3-1] , 4 - ( 5 * 3 ) ) mod 256
= ( C (2,4) - 15 ) mod 256
= ( { - 15 ) mod 256
= ( 123 - 15 ) mod 256
= 108 è l
M35 = ( C [3-1] , 5 - ( 5 * 3 ) ) mod 256
= ( C (2,5) - 15 ) mod 256
= ( ” - 15 ) mod 256
= (148- 15 ) mod 256
= 133 è …
Baris 4 i = 4
Mij = ( C [i – 1 ] , j - ( K * R [i] )) mod 256
M44 = ( C [4-1] , 4 - ( 5 * 4 ) ) mod 256
= ( C (3,4) - 20 ) mod 256
= ( l - 20 ) mod 256
= ( 108 - 20 ) mod 256
= 88 è X
M45 = ( C [4-1] , 5 - ( 5 * 4 ) ) mod 256
= ( C (3,5) - 20 ) mod 256
= ( … - 20 ) mod 256
= ( 133 - 20 ) mod 256
= 113 è q
Baris 5 i = 5
Mij = ( C [i – 1 ] , j - ( K * R [i] )) mod 256
M55 = ( C [5-1] , 5 - ( 5 * 5 ) ) mod 256
= ( C (4,5) - 25 ) mod 256
= ( q - 25 ) mod 256
= ( 113 - 25 ) mod 256
= 88 è X
∆ 1 = Ž z p X X
C = Ž z p X X
142 122 112 88 88
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Ž/142 |
z/122 |
p/112 |
X / 88 |
X/88 |
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‰/137 |
u /117 |
k/107 |
S/83 |
S/83 |
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⌂/127 |
k/107 |
a/97 |
I /73 |
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P/112 |
\ /92 |
R/82 |
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\ /92 |
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H /72 |
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“C/67 |
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i/0
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2
3
4
5
Baris ke 1
Mij = ( C [i – 1 ] , j - ( K * R [i] )) mod 256
M11 = ( C [1-1] , 1 - ( 5 * 1 ) ) mod 256
= ( C (0,1) - 5 ) mod 256
= ( Ž - 5 ) mod 256
= ( 142 - 5 ) mod 256
= 137 è ‰
M12 = ( C [1-1] , 2 - ( 5 * 1 ) ) mod 256
= ( C (0,2) - 5 ) mod 256
= ( z - 5 ) mod 256
= ( 122 - 5 ) mod 256
= 117 è u
M13 = ( C [1-1] , 3 - ( 5 * 1 ) ) mod 256
= ( C (0,3) - 5 ) mod 256
= ( p - 5 ) mod 256
= ( 112 - 5 ) mod 256
= 107 è k
M14 = ( C [1-1] , 4 - ( 5 * 1 ) ) mod 256
= ( C (0,4) - 5 ) mod 256
= ( X - 5 ) mod 256
= ( 88 - 5 ) mod 256
= 83 è S
M15 = ( C [1-1] , 5 - ( 5 * 1 ) ) mod 256
= ( C (0,5) - 5 ) mod 256
= ( X - 5 ) mod 256
= ( 88 - 5 ) mod 256
= 83 è S
Baris ke 2
Mij = ( C [i – 1 ] , j - ( K * R [i] )) mod 256
M21 = ( C [2-1] , 1 - ( 5 * 2 ) ) mod 256
= ( C (1,1) - 10 ) mod 256
= ( ‰ - 10 ) mod 256
= ( 137 - 10 ) mod 256
= 127 è ⌂
M22 = ( C [2-1] , 2 - ( 5 * 2 ) ) mod 256
= ( C (1,2) - 10 ) mod 256
= ( u - 10 ) mod 256
= ( 117 - 10 ) mod 256
= 107 è k
M23 = ( C [2-1] , 3 - ( 5 * 2 ) ) mod 256
= ( C (1,3) - 10 ) mod 256
= ( k - 10 ) mod 256
= ( 107 - 10 ) mod 256
= 97 è a
M24 = ( C [2-1] , 4 - ( 5 * 2 ) ) mod 256
= ( C (1,4) - 10 ) mod 256
= ( S - 10 ) mod 256
= ( 83 - 10 ) mod 256
= 73 è I
Baris ke 3
Mij = ( C [i – 1 ] , j - ( K * R [i] )) mod 256
M31 = ( C [3-1] , 1 - ( 5 * 3 ) ) mod 256
= ( C (2,1) - 15 ) mod 256
= ( ⌂ - 15 ) mod 256
= ( 127 - 15 ) mod 256
= 112 è p
M32 = ( C [3-1] , 2 - ( 5 * 3 ) ) mod 256
= ( C (2,2) - 15 ) mod 256
= ( k - 15 ) mod 256
= ( 107 - 15 ) mod 256
= 92 è \
M33 = ( C [3-1] , 3 - ( 5 * 3 ) ) mod 256
= ( C (2,3) - 15 ) mod 256
= ( a - 15 ) mod 256
= ( 97 - 15 ) mod 256
= 82 è R
Baris ke 4
Mij = ( C [i – 1 ] , j - ( K * R [i] )) mod 256
M41 = ( C [4-1] , 1 - ( 5 * 4 ) ) mod 256
= ( C (3,1) - 20 ) mod 256
= ( p - 20 ) mod 256
= ( 112 - 20 ) mod 256
= 92 è \
M42 = ( C [4-1] , 2 - ( 5 * 4 ) ) mod 256
= ( C (3,2) - 20 ) mod 256
= ( \ - 20 ) mod 256
= ( 92 - 20 ) mod 256
= 72 è H
Baris ke 5
Mij = ( C [i – 1 ] , j - ( K * R [i] )) mod 256
M51 = ( C [5-1] , 1 - ( 5 * 5 ) ) mod 256
= ( C (4,1) - 25 ) mod 256
= ( \ - 25 ) mod 256
= ( 92 - 25 ) mod 256
= 67 è C
∆ ke II = C H R I S
10. Kriptografi Modern
Kriptografi modern berkerja dalam model operasi linier plainteks/cipherteks/kunci harus dikelompokan dalam jalan bit tertentu sesuai dengan algoritma-algoritma yang digunakan kegiatan dalam proses pekerjaan dibutuhkan karena setiap-setiap algoritma kriptografi modern berkerja secara berantai pada setiap aliran-aliran biner yang diterima.
Mode Operasi exit dalam kriptografi modern
- Mode electronic Code Book (ECB)
- Cipher Block Clinning (CBC)
- Cipher Ferd Book (CFB)
- Output Ferd Bock (OFB)
Keempat mode operasi ini melakukan operasinya berdasarkan operasi XOR, Operasi shift/wrapping dan bit padding.
Catatan : Shift/Wrapping pengeseran sejumlah operasinya berdasarkan kiri/kanan (left/right).
bit Padding = Penambah Sejumlah.
Hukum XOR ( Å )
1 Å 1 = 0
0 Å 0 = 0
1 Å 0 = 1
0 Å 1 = 1
Deskripsi => Pi = Ci Å K
- Konversi semua kelompok/nilai hexa cipher terbiner.
- Kelompokan menjadi 16 bit kelompok.
- Kembalikan sejumlah bit yang dipasar pada posisi semua ( shift 1 bit to right ).
- Hasil wrapping/shift di XOR dengan kunci.
- Kelompok menjadi 8 bit perkelompok.
Mode chiper book climming (Bc)
- Melakukan operasi manipulasi bit secara block.
- Membuuhkan block misiat (infial vektor/IV Co) sebagai block awal rantai proses.
- Block kunci dan IV/Co harus sama sesuai dengan jumlah block yang ditentukan.
- Block plainteks/cipher di XOR dengan setiap nilai Ci-1 kemudian hasilnya di XOR dengan kunci dengan melakukan shift/wrapping.
CONTOH :
Plainteks = C H R I S X } Jlh bit perkelompok = 16 bit
Kunci = DS
Plainteks :
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C |
H |
R |
I |
S |
X |
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67 |
72 |
82 |
73 |
83 |
88 |
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01000011 |
01001000 |
01010010 |
01001001 |
01010011 |
01011000 |
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Blok 1 ( P1 ) |
Blok 2 (P2 ) |
Blok 3 (P3 ) |
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D |
S |
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68 |
83 |
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01000100 |
01010011 |
Kunci :
Enkripsi dgn metode Mode XOR ( ECB )
Ci = Pi Å K
C1 = 0100001101001000 (P1)
= 0100010001010011 (K)
(Å)
C1` = 0000011100011011
C1 = 0000111000110110
C2 = 0101001001001001 (P2)
= 0100010001010011 (K)
(Å)
C2` = 0001011000011010
C2 = 0010110000110100
C3 = 0101001101011000 (P3)
= 0100010001010011 (K)
(Å)
C3` = 0001011100001011
C3 = 0010111000010110
Hasil Akhir :
C1 = 00001110 00110110
0E 36
C2 = 00101100 00110100
2C 34
C3 = 00101110 00010110
2E 16
Hasil / Cipher : 0E = 14
36 = 54
2C = 44
34 = 52
2E = 46
16 = 22
Cipher : Hexa = 0E 36 2C 34 2E 16
Symbol = ♫ 6 , 4 . ‒
Dekripsi :
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0E |
36 |
2C |
34 |
2E |
16
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14 |
54 |
44 |
52 |
46 |
22 |
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00001110 |
00110110 |
00101100 |
00110100 |
00101110 |
00010110 |
C1 = 000011100011011
= 
000011100011011
(Hasil Wrapping)
= 0100010001010011 ( K)
(Å)
= 01000011 01001000
= 01000011 = 67 è C
= 01001000 = 72 è H
C2 = 001011000011010
= 001011000011010
(Hasil Wrapping)
= 0100010001010011 ( K)
(Å)
= 01010010 01001001
= 01010010 = 82 è R
= 01001001 = 73 è I
C3 = 001011100001011
= 001011100001011
(Hasil Wrapping)
= 0100010001010011 ( K)
(Å)
= 01010011 01011000
= 01010011 = 83 è S
= 01011000 = 88 è X
Pengertian Kriptografi dan Contohnya
Algoritma Affine Chiper dan Contohnya
Kriptografi Model Transposition
Algoritma Coulumnar Transposition dan contohnya
Algoritma Vegeneere Cipher dab contohnya
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