Transcript Kunci 32 bit p A B GE NG ε  Plaintext : 3946183951 : 537680305 : 1059676324 : [1152222263 3133703258] : 3946206427 : 100 : [3539395206 1802765602] : Mat lpesan  Panjang pesan = 3 (banyaknya karakter.

Kunci 32 bit
p
A
B
GE
NG
ε

Plaintext
: 3946183951
: 537680305
: 1059676324
: [1152222263 3133703258]
: 3946206427
: 100
: [3539395206 1802765602]
: Mat
lpesan  Panjang pesan = 3 (banyaknya karakter plaintext)
bpesan  ceil ( (32/8) – 1 ) = 3 (dipotong setiap 3 karakter)
ipesan  ceil (lpesan / bpesan) = 3/3 = 1 (pemotongan pesan)
pesan: Mat
M
ASCII
a
ASCII
t
ASCII
77
Biner
97
Biner
116
Biner
1001101
Biner 8 bit
01001101
1100001
Biner 8 bit
01100001
1110101
Biner 8 bit
01110100
Desimal
m
5071220
pesan: Mat
m = 5071220
xj = m. ε + j , j [0, ε -1]  Sj = xj3 + A.xj + B (mod p)
yj = akar dari Sj
PM = (xj , yj)
xDipilih
5071220(100)+0
j [0, ε -1] = =[0,507122003
507122000
99]  j = 
95S30 = 2413459687
211020104
30 = 5071220(100)+3
(p-1)/2
=
akar
S
=
=Tdk
3301895794
memiliki
dan
akar.
644288157
S
= -1(mod p)
xy95
=5071220(100)+82=507122003
30
30
0 S95 =375659790
yx95
akar S
1075291432
dan 2870892519
==5071220(100)+1
= 507122001
, 3301895794)
 S1 =3164013225
dan
95 =(507122003
1 Titiknya
y1 Titiknya
= akar S1
=Tdk
(507122003
memiliki,, 1075291432)
644288157)
akar. S1(p-1)/2 dan
= -1(mod p)
(507122095
(507122095
, 2870892519)
x2 = 5071220(100)+2
…………
sampai
j= =
ε -1=99.
507122002
 S2 =1267370450
y2 = akar
S2 =Tdk
akar. S,2(p-1)/2
= -1(mod p)
Dipilih
1 titik
( PM memiliki
)  ( 507122095
1075291432)
pesan: Mat
m = 5071220
PM = ( 507122082 , 307629979 )
ENKRIPSI ElGamal ECC
k [1, NG – 1] = [1,3946206426]  912120288
P1 = k.GE = (3227285189 , 2965666882)

P2 = PM + k.  = PM + (736261778 , 2068514041)
= (3828826938 , 1504649631)
PC = (P1,P2)
= ((3227285189, 2965666882) , (3828826398,1504649631))
PC = ( 3227285189 2965666882 3828826398 1504649631 )