4611413041_pipit riski s
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kriptografiTRANSCRIPT
Pipit Riski S4611413041
SOAL
A. ENKRIPSI dan DEKRIPSI HILL CIPHER secara MANUAL dan MATLAB
K=(2 73 11)
Plaintext: CRYPTOGRAPHY
B. ENKRIPSI dan DEKRIPSI HILL CIPHER dengan MATLAB
K = [29 12 23 9 5 721 14 22 8 3 415 11 10 1 0 010 10 21 8 3 43 2 12 7 3 413 1 13 8 5 7
]Plaintext = CRYPTOGRAPHY
JAWABAN A
1. ENKRIPSIPlaintext: CRYPTOGRAPHY(C,R) , (Y,P) , (T,O) (G,R) (A,P) (H,Y)(2,17) , (24,15) , (19,14) , (6,17) , (0,15) , (7,24)
a. (X1 , X2) = (C,R) = (2,17)
(Y1 , Y2) = (2 17 )(2 73 11)=(55 ,201 )=(3 ,19 )=(D ,T )
b. (X1 , X2) = (Y,P) = (24,15)
(Y1 , Y2) = (24 15 )(2 73 11)= (93 ,333 )=(15 ,21 )=(P ,V )
c. (X1 , X2) = (T,O) = (19,14)
(Y1 , Y2) = (19 14 )(2 73 11)=( 80287 )=(2 ,1 )=(C ,B)
d. (X1 , X2) = (G,R) = (6,17)
(Y1 , Y2) = (6 17 )(2 73 11)= (63 ,229 )=(11 ,21 )=(L,V )
e. (X1 , X2) = (A,P) = (0,15)
(Y1 , Y2) = (0 15 )(2 73 11)=( 45 ,165 )=(19 ,9 )=(T , J )
f. (X1 , X2) = (H,Y) = (7,24)
(Y1 , Y2) = (7 24 )(2 73 11)=(86 ,313 )=(8 ,1 )= (I ,B )
Pipit Riski S4611413041
CIPHERTEXT: DTOVCBLVTJIB
2. DEKRIPSICiphertext: DTOVCBLVTJIB(D,T) , (P,V) , (C,B) , (L,V) , (T,J) , (I,B)(3,19) , (15,21) , (2,1) , (11,21) , (19,9) , (8,1)
K−1= 1det K ( 11 −7
−3 2 )¿ 1
22−21 ( 11 −7−3 2 )
¿1( 11 −7−3 2 )
¿(11 1923 2 )
a. (Y1 , Y2) = (D,T) = (3,19)
(X1 , X2) = (3 19 )(11 1923 2 )=(470 ,95 )=(2 ,17 )=(C , R)
b. (Y1 , Y2) = (P,V) = (15,21)
(X1 , X2) = (15 21 )(11 1923 2 )=(648 ,327 )=(24 ,15 )=(Y ,P)
c. (Y1 , Y2) = (C,B) = (2,1)
(X1 , X2) = (2 1 )(11 1923 2 )=(45 ,40 )=(19 ,14 )=(T ,O)
d. (Y1 , Y2) = (L,V) = (11,21)
(X1 , X2) = (11 21 )(11 1923 2 )=(144 ,211 )=(6 ,17 )=(G ,R)
e. (Y1 , Y2) = (T,J) = (19,9)
(X1 , X2) = (19 9 )(11 1923 2 )=( 416 ,379 )=(0 ,15 )=(A ,P)
f. (Y1 , Y2) = (I,B) = (8,1)
(X1 , X2) = (8 1 )(11 1923 2 )=(111 ,154 )=(7 ,24 )=(H ,Y )
Pipit Riski S4611413041
Pipit Riski S4611413041
JAWABAN A dengan MATLAB
>> k=[2 7;3 11]
k =
2 7
3 11
>> X1=[2 17]
X1 =
2 17
>> X2=[24 15]
X2 =
24 15
>> X3=[19 14]
X3 =
19 14
Pipit Riski S4611413041
>> X4=[6 17]
X4 =
6 17
>> X5=[0 15]
X5 =
0 15
>> X6=[7 24]
X6 =
7 24
>> enkripsi1=X1*k;
>> enkripsiCR=mod(enkripsi1,26)
enkripsiCR =
3 19
>> enkripsi2=X2*k;
>> enkripsiIP=mod(enkripsi2,26)
Pipit Riski S4611413041
enkripsiIP =
15 21
>> enkripsi3=X3*k;
>> enkripsiTO=mod(enkripsi3,26)
enkripsiTO =
2 1
>> enkripsi4=X4*k;
>> enkripsiGR=mod(enkripsi4,26)
enkripsiGR =
11 21
>> enkripsi5=X5*k;
>> enkripsiAP=mod(enkripsi5,26)
enkripsiAP =
19 9
>> enkripsi6=X6*k;
Pipit Riski S4611413041
>> enkripsiHY=mod(enkripsi6,26)
enkripsiHY =
8 1
>> inversK=inv(k)
inversK =
11.0000 -7.0000
-3.0000 2.0000
>> invershasil=mod(inversK,26)
invershasil =
11.0000 19.0000
23.0000 2.0000
>> dekripsi1=(enkripsiCR*invershasil)
dekripsi1 =
470.0000 95.0000
>> decripsiCR=mod(dekripsi1,26)
Pipit Riski S4611413041
decripsiCR =
2.0000 17.0000
>> dekripsi2=(enkripsiIP*invershasil)
dekripsi2 =
648.0000 327.0000
>> decripsiIP=mod(dekripsi2,26)
decripsiIP =
24.0000 15.0000
>> dekripsi3=(enkripsiTO*invershasil)
dekripsi3 =
45.0000 40.0000
>> decripsiTO=mod(dekripsi3,26)
decripsiTO =
Pipit Riski S4611413041
19.0000 14.0000
>> dekripsi4=(enkripsiGR*invershasil)
dekripsi4 =
604.0000 251.0000
>> decripsiGR=mod(dekripsi4,26)
decripsiGR =
6.0000 17.0000
>> dekripsi5=(enkripsiAP*invershasil)
dekripsi5 =
416.0000 379.0000
>> decripsiAP=mod(dekripsi5,26)
decripsiAP =
0.0000 15.0000
>> dekripsi6=(enkripsiHY*invershasil)
Pipit Riski S4611413041
dekripsi6 =
111.0000 154.0000
>> decripsiHY=mod(dekripsi6,26)
decripsiHY =
7.0000 24.0000
Pipit Riski S4611413041
JAWABAN B
Pipit Riski S4611413041