Sabtu, 30 Oktober 2010

Kriptografi


  1. Dengan menggunakan ketiga cipher diatas maka enkripsi kemudian dekripsi plaintext berikut :

“ TERJAWAB SUDAH TEKA TEKI YANG BEREDAR DI MASYARAKAT TENTANG SIAPA KAPOLRI BARU UNTUK MENGGANTIKAN BAMBANG KOMA TERNYATA TIMUR PRADOPA “

Ketentuan :

· Kunci tentukan sendiri

· Banyaknya kunci minimal 3 untuk Polyalphabetic

· Buat program untuk ketiga cipher tersebut diatas. Terserah menggunakan bahasa pemograman.

· Dikumpul waktu MID Text

2. “ TERJAWAB SUDAH TEKA TEKI YANG BEREDAR DI MASYARAKAT TENTANG SIAPA KAPOLRI BARU UNTUK MENGGANTIKAN BAMBANG KOMA TERNYATA TIMUR PRADOPA “

Mengunakan :

· Affine Cipher

· Playfair

· Metode Kolom

Penyelesaian Tugas No. 1.


· Caser :

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

D E F G H I J K L M N O P Q R S T U V W X Y Z A B C

T E R J A W A B S U D A H T E K A T E K I Y A N G

W H U M D Z D E V X G D K W H N D W H N L B D Q J

B E R E D A R D I M A S Y A R A K A T T E N T A N G

E H U H G D U G L P D V B D U D N D W W H Q W D Q J

S I A P A K A P O L R I B A R U U N T U K M E N G G

V L D S D N D S R O U L E D U X X Q W X N P H Q J J

A N T I K A N B A M B A N G K O M A T E R N Y A T

D Q W L N D Q E D P E D Q J N R P D W H U Q B D W

A T I M U R P R A D O P A

D W L P X U S U D G R S D

· Polyalphaberic :

Kunci 1: R O Y

Kunci 2: S A H A L A

Kunci 3: A R U A N

F1: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

R O Y A B C D E F G H I J K L M N P Q S T U V W X Z

F2: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

S A H L B C D E F G I J K M N O P Q R T U V W X Y Z

F3: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

A R U N B C D E F G H I J K L M O P Q S T V W X Y Z

Penyelesaian :

T E R J A W A B S U D A H T E K A T E K I Y A N G B E R E D A R D I

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1

M A S Y A R A K A T T E N T A N G S I A P A K A P O L R I B A R U U N

2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

T U K M E N G G A N T I K A N B A M B A N G K O M A T E R N Y A T

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

A T I M U R P R A D O P A

1 2 3 1 2 3 1 2 3 1 2 3 1

Ciphertex s:

SBPGSWRAQTLAETBHSSBIFXSKDABPBNRQNFKAQYAPSHRTSBMSR MDQFAMAHROLIQFOSPTUKSUHJBKDDAKTFHSKOSJOSKDILJSSBQKXSSRTFJUPMQAANMR

CAESAR CIPHER

program caesar;

uses wincrt;

label 10;

var judul,kalimat:string;

baru:array[1..100] of char;

pil,i,geser,b:integer;

skalimat:char;

begin

repeat

clrscr;

gotoxy(52-length(judul),2);

write(judul);

gotoxy(30,4);

for i := 30 to 49 do

begin

write('-');

end;

gotoxy(30,5);

writeln('1. Enkripsi Tulisan.');

gotoxy(30,6);

writeln('2. Dekripsi Tulisan.');

gotoxy(30,7);

Writeln('3. Exit');

gotoxy(30,8);

for i := 30 to 49 do

begin

write('-');

end;

gotoxy(1,10);

write('Masukkan pilihan anda : ');readln(pil);

if pil = 1 then

begin

write('Masukkan kalimat yang ingin di geser : ');readln(kalimat);

write('Masukkan banyaknya pergeseran (dalam angka) : ');readln(geser);

for i := 1 to length(kalimat) do

begin

while geser > 26 do

begin

geser:=geser - 26;

end;

skalimat := upcase(kalimat[i]);

if ord(skalimat) = 32 then

begin

b := 32;

end

else

b := ord(skalimat) + geser;

if b > 90 then

begin

b:=64+(b-90);

end;

baru[i] := chr(b);

end;

write('Hasilnya adalah : ');

for i := 1 to length(kalimat) do

begin

write(baru[i]);

end;

readkey;

end;

if pil = 2 then

begin

write('Masukkan kalimat yang ingin di geser : ');readln(kalimat);

write('Masukkan banyaknya pergeseran (dalam angka) : ');readln(geser);

for i := 1 to length(kalimat) do

begin

while geser > 26 do

begin

geser:=geser - 26;

end;

skalimat := upcase(kalimat[i]);

if ord(skalimat) = 32 then

begin

b := 32;

goto 10;

end

else

b := ord(skalimat) - geser;

if b < 65 then

begin

b:=90-(64-b);

end;

10:

baru[i] := chr(b);

end;

write('Hasilnya adalah : ');

for i := 1 to length(kalimat) do

begin

write(baru[i]);

end;

readkey;

end;

until pil = 3;

end.

Hasil Program :



2. “ TERJAWAB SUDAH TEKA TEKI YANG BEREDAR DI MASYARAKAT TENTANG SIAPA KAPOLRI BARU UNTUK MENGGANTIKAN BAMBANG KOMA TERNYATA TIMUR PRADOPA “

* Menggunakan Affine Cipher :

C ≡ 3 P + 10 (Mod 26)

P1 = T = 19 = C1 = 3.19 + 10 = 67 = 15 (Mod 26) = “ P“

P2 = E = 4 = C2 = 3.4 + 10 = 22 = 0 (Mod 26) = “ A“

P3 = R = 17 = C3 = 3.17 + 10 = 61 = 9 (Mod 26) = “ J “

P4 = J = 9 = C4 = 3.9 + 10 = 37 = 11 (Mod 26) = “ L “

P5 = A = 0 = C5 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A “

P6 = W = 22 = C6 = 3.22 + 10 = 76 = 24 (Mod 26) = “ Y “

P7 = A = 0 = C7 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A “

P8 = B = 1 = C8 = 3.1 + 10 = 13 = 0 (Mod 26) = “ A“

P9 = S = 18 = C9 = 3.18 + 10 = 64 = 12 (Mod 26) = “ M “

P10 = U = 20 = C10 = 3.20 + 10 = 70 = 12 (Mod 26) = “ M “

P11 = D = 3 = C11 = 3.3 + 10 = 19 = 18 (Mod 26) = “ S “

P12 = A = 0 = C12 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

P13 = H = 7 = C13 = 3.7 + 10 = 31 = 0 (Mod 26) = “ A“

P14 = T = 19 = C14 = 3.19 + 10 = 67 = 5 (Mod 26) = “ F “

P15 = E = 4 = C15 = 3.4 + 10 = 22 = 15 (Mod 26) = “ P “

P16 = K = 10 = C16 = 3.10 + 10 = 40 = 0 (Mod 26) = “ A“

P17 = A = 0 = C17 = 3.0 + 10 = 10 = 14 (Mod 26) = “O “

P18 = T = 19 = C18 = 3.19 + 10 = 67 = 0 (Mod 26) = “ A“

P19 = E = 4 = C19 = 3.4 + 10 = 22 = 0 (Mod 26) = “A “

P20 = K = 10 = C20 = 3.10 + 10 = 40 = 0 (Mod 26) = “ A“

P21 = I = 8 = C21 = 3.8 + 10 = 34 = 14 (Mod 26) = “ O“

P22 = Y = 24 = C22 = 3.24 + 10 = 82 = 4 (Mod 26) = “ E “

P23 = A = 0 = C23 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

P24 = N = 13 = C24 = 3.13 + 10 = 49 = 23 (Mod 26) = “ X“

P25 = G = 6 = C25 = 3.6 + 10 = 28 = 2 (Mod 26) = “ C“

P26 = B = 1 = C26 = 3.1 + 10 = 13 = 0 (Mod 26) = “ A“

P27 = E = 4 = C27 = 3.4 + 10 = 22 = 0 (Mod 26) = “ A“

P28 = R = 17 = C28 = 3.17 + 10 = 61 = 9 (Mod 26) = “ J “

P29 = E = 4 = C29 = 3.4 + 10 = 22 = 0 (Mod 26) = “ A“

P30 = D = 3 = C30 = 3.3 + 10 = 19 = 18 (Mod 26) = “ S“

P31 = A = 0 = C31 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

P32 = R = 17 = C32 = 3.17 + 10 = 61 = 9 (Mod 26) = “ J “

P33 = D = 3 = C33 = 3.3 + 10 = 19 = 18 (Mod 26) = “ S“

P34 = I = 8 = C34 = 3.8 + 10 = 34 = 14 (Mod 26) = “ O“

P35 = M = 12 = C35 = 3.12 + 10 = 46 = 20 (Mod 26) = “ U“

P36 = A = 0 = C36 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

P37 = S = 18 = C37 = 3.18 + 10 = 64 = 12 (Mod 26) = “ M“

P38 = Y = 24 = C38 = 3.24 + 10 = 82 = 4 (Mod 26) = “ E “

P39 = A = 0 = C39 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

P40 = R = 17 = C40 = 3.17 + 10 = 61 = 9 (Mod 26) = “ J “

P41 = A = 0 = C41 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

P42 = K = 10 = C42 = 3.10 + 10 = 40 = 14 (Mod 26) = “ O“

P43 = A = 0 = C43 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

P44 = T = 19 = C44 = 3.19 + 10 = 67 = 15 (Mod 26) = “ P“

P45 = T = 19 = C45 = 3.19 + 10 = 67 = 15 (Mod 26) = “ P“

P46 = E = 4 = C46 = 3.4 + 10 = 22 = 0 (Mod 26) = “ A“

P47 = N = 13 = C47 = 3.13 + 10 = 49 = 23 (Mod 26) = “ X“

P48 = T = 19 = C48 = 3.19 + 10 = 67 = 15 (Mod 26) = “ P“

P49 = A = 0 = C49 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

P50 = N = 13 = C50 = 3.13 + 10 = 49 = 23 (Mod 26) = “ X “

P51 = G = 6 = C51 = 3.6 + 10 = 28 = 2 (Mod 26) = “ C“

P52 = S = 18 = C52 = 3.18 + 10 = 64 = 12 (Mod 26) = “ M“

P53 = I = 8 = C53 = 3.8 + 10 = 34 = 14 (Mod 26) = “ O“

P54 = A = 0 = C54 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

P55 = P = 15 = C55 = 3.15 + 10 = 55 = 3 (Mod 26) = “ D “

P56 = A = 0 = C56 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

P57 = K = 10 = C57 = 3.10 + 10 = 40 = 14 (Mod 26) = “ O“

P58 = A = 0 = C58 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

P59 = P = 15 = C59 = 3.15 + 10 = 55 = 3 (Mod 26) = “ D “

P60 = O = 14 = C60 = 3.14 + 10 = 52 = 0 (Mod 26) = “ A “

P61 = L = 11 = C61 = 3.11 + 10 = 43 = 17 (Mod 26) = “ R “

P62 = R = 17 = C62 = 3.17 + 10 = 61 = 9 (Mod 26) = “J “

P63 = I = 8 = C63 = 3.8 + 10 = 34 = 14 (Mod 26) = “ O“

P64 = B = 1 = C64 = 3.1 + 10 = 13 = 0 (Mod 26) = “ A“

P65 = A = 0 = C65 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

P66 = R = 17 = C66 = 3.17 + 10 = 61 = 9 (Mod 26) = “ J “

P67 = U = 20 = C67 = 3.20 + 10 = 70 = 18 (Mod 26) = “ S “

P68 = U = 20 = C68 = 3.20 + 10 = 70 = 18 (Mod 26) = “ S “

P69 = N = 13 = C69 = 3.13 + 10 = 49 = 23 (Mod 26) = “ X“

P70 = T = 19 = C70 = 3.19 + 10 = 67 = 15 (Mod 26) = “ P“

P71 = U = 20 = C71 = 3.20 + 10 = 70 = 18 (Mod 26) = “ S “

P72 = K = 10 = C72 = 3.10 + 10 = 40 = 14 (Mod 26) = “ O“

P73 = M = 12 = C73 = 3.12 + 10 = 46 = 20 (Mod 26) = “ U“

P74 = E = 4 = C74 = 3.4 + 10 = 22 = 0 (Mod 26) = “ A“

P75 = N = 13 = C75 = 3.13 + 10 = 49 = 23 (Mod 26) = “ X“

P76 = G = 6 = C76 = 3.6 + 10 = 28 = 2 (Mod 26) = “ C“

P77 = G = 6 = C77 = 3.6 + 10 = 28 = 2 (Mod 26) = “ C“

P78 = A = 0 = C78 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

P79 = N = 13 = C79 = 3.13 + 10 = 49 = 23 (Mod 26) = “ X“

P80 = T = 19 = C80 = 3.19 + 10 = 67 = 15 (Mod 26) = “ P“

P81 = I = 8 = C81 = 3.8 + 10 = 34 = 14 (Mod 26) = “ O“

P82 = K = 10 = C82 = 3.10 + 10 = 40 = 14 (Mod 26) = “ O“

P83 = A = 0 = C83 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

P84 = N = 13 = C84 = 3.13 + 10 = 49 = 23 (Mod 26) = “ X“

P85 = B = 1 = C85 = 3.1 + 10 = 13 = 0 (Mod 26) = “ A“

P86 = A = 0 = C86 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

P87 = M = 12 = C87 = 3.12 + 10 = 46 = 20 (Mod 26) = “ U“

P88 = B = 1 = C88 = 3.1 + 10 = 13 = 0 (Mod 26) = “ A“

P89 = A = 0 = C89 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

P90 = N = 13 = C90 = 3.13 + 10 = 49 = 23 (Mod 26) = “ X“

P91 = G = 6 = C91 = 3.6 + 10 = 28 = 2 (Mod 26) = “ C“

P92 = K = 10 = C92 = 3.10 + 10 = 40 = 14 (Mod 26) = “ O“

P93 = O = 14 = C93 = 3.14 + 10 = 52 = 0 (Mod 26) = “ A “

P94 = M = 12 = C94 = 3.12 + 10 = 46 = 20 (Mod 26) = “ U“

P95 = A = 0 = C95 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

P96 = T = 19 = C96 = 3.19 + 10 = 67 = 15 (Mod 26) = “ P“

P97 = E = 4 = C97 = 3.4 + 10 = 22 = 0 (Mod 26) = “ A“

P98 = R = 17 = C98 = 3.17 + 10 = 61 = 9 (Mod 26) = “J “

P99 = N = 13 = C99 = 3.13 + 10 = 49 = 23 (Mod 26) = “ X“

P100 = Y = 24 = C100 = 3.24 + 10 = 82 = 4 (Mod 26) = “ E “

P101 = A = 0 = C101 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

P102 = T = 19 = C102 = 3.19 + 10 = 67 = 15 (Mod 26) = “ P“

P103 = A = 0 = C103 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

P104 = T = 19 = C104 = 3.19 + 10 = 67 = 15 (Mod 26) = “ P“

P105 = I = 8 = C105 = 3.8 + 10 = 34 = 14 (Mod 26) = “ O“

P106 = M = 12 = C106 = 3.12 + 10 = 46 = 20 (Mod 26) = “ U“

P107 = U = 20 = C107 = 3.20 + 10 = 70 = 18 (Mod 26) = “ S “

P108 = R = 17 = C108 = 3.17 + 10 = 61 = 9 (Mod 26) = “ J “

P109 = P = 15 = C109 = 3.15 + 10 = 55 = 3 (Mod 26) = “ D “

P110 = R = 17 = C110 = 3.17 + 10 = 61 = 9 (Mod 26) = “ J “

P111 = A = 0 = C111 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

P112 = D = 3 = C112 = 3.3 + 10 = 19 = 18 (Mod 26) = “ S “

P113 = O = 14 = C113 = 3.14 + 10 = 52 = 0 (Mod 26) = “ A “

P114 = P = 15 = C114 = 3.15 + 10 = 55 = 3 (Mod 26) = “ D “

P115 = A = 0 = C115 = 3.0 + 10 = 10 = 0 (Mod 26) = “ A“

* Menggunakan Playfair :

Pesan : TE RJ AW AB SU DA HT EK AT EK IY AN GB ER ED AR DI MA SY AR AK AT TE NT AN GS IA PA KA PO LR IB AR UU NT UK ME NG GA NT IK AN BA MB AN GK OM AT ER NY AT AT IM UR PR AD OP AN



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