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Cryptology
CAESAR CIPHER
How to encrypt a message?
 By using a formula .
 What is the formula?
 F(p) = (p+3) mod 26 (Encryption formula)
 F−1
(p) = (p-3) mod 26 (Decryption formula)
 The values of p is 0 to 25
 These are the numeric values of the characters (A-Z)
 Like A = 0 , B = 1, C = 2 , D = 3, E = 4 etc.
Write down the following chart before every
problem of cryptology in exam hall.
Characters Values
A 0
B 1
C 2
D 3
E 4
F 5
G 6
H 7
I 8
J 9
Characters Values
K 10
L 11
M 12
N 13
O 14
P 15
Q 16
R 17
S 18
T 19
Characters Values
U 20
V 21
W 22
X 23
Y 24
Z 25
Steps of Encryption/Decryption
1. First find out the letters(given) numeric values.
2. Then use the appropriate formula.(Encryption/ Decryption)
3. After using the appropriate formula you will get values from 0-25
for each character.
4. Then find the values from the chart.
5. Use new character according to the assigned values in the chart
for every character.
6. Its done.
Lets Encrypt a message.
The unique letters we found from the following message are T, E, S, O, D, A, Y
F(p) = (p+3) mod 26 (Encryption formula)
F(19) = (19+3) mod 26 = 22 mod 26 = 22 = W
F(4) = (4+3) mod 26 = 7 mod 26 = 7 = H
F(18) = (18+3) mod 26 = 21 mod 26 = 21 = V
Others
F(14) = (14+3) mod 26 = 17 mod 26 = 17 = R
F(3) = (3+3) mod 26 = 6 mod 26 = 6 = G
F(0) = (0+3) mod 26 = 3 mod 26 = 3 = D
F(24) = (24+3) mod 26 = 27 mod 26 = 1 = B
Now replace the characters with new
characters.
TEST TODAY = WHVW WRGDB
So our encrypted message is WHVW WRGDB
Lets Decrypt a message.
The unique letters we found from the following message are O, R, Y, H
F−1
(p) = (p-3) mod 26 (Decryption formula)
𝐹−1
(14) = (14-3) mod 26 = 11 mod 26 = 11 = L
𝐹−1
(17) = (17-3) mod 26 = 14 mod 26 = 14 = O
𝐹−1
(24) = (24-3) mod 26 = 21 mod 26 = 21 = V
𝐹−1(7) = (7-3) mod 26 = 4 mod 26 = 4 = E
Now replace the characters with new
characters.
ORYH = LOVE
So our Decrypted message is LOVE
Exercises
1.ENCRYPT THE MESSAGE
“KOS KI MAMA ”
2.DECRYPT THE MESSAGE “L
ORYH EDQJODGHVK”

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Cryptology Caesar Cipher

  • 2. How to encrypt a message?  By using a formula .  What is the formula?  F(p) = (p+3) mod 26 (Encryption formula)  F−1 (p) = (p-3) mod 26 (Decryption formula)  The values of p is 0 to 25  These are the numeric values of the characters (A-Z)  Like A = 0 , B = 1, C = 2 , D = 3, E = 4 etc.
  • 3. Write down the following chart before every problem of cryptology in exam hall. Characters Values A 0 B 1 C 2 D 3 E 4 F 5 G 6 H 7 I 8 J 9 Characters Values K 10 L 11 M 12 N 13 O 14 P 15 Q 16 R 17 S 18 T 19 Characters Values U 20 V 21 W 22 X 23 Y 24 Z 25
  • 4. Steps of Encryption/Decryption 1. First find out the letters(given) numeric values. 2. Then use the appropriate formula.(Encryption/ Decryption) 3. After using the appropriate formula you will get values from 0-25 for each character. 4. Then find the values from the chart. 5. Use new character according to the assigned values in the chart for every character. 6. Its done.
  • 5. Lets Encrypt a message. The unique letters we found from the following message are T, E, S, O, D, A, Y F(p) = (p+3) mod 26 (Encryption formula) F(19) = (19+3) mod 26 = 22 mod 26 = 22 = W F(4) = (4+3) mod 26 = 7 mod 26 = 7 = H F(18) = (18+3) mod 26 = 21 mod 26 = 21 = V
  • 6. Others F(14) = (14+3) mod 26 = 17 mod 26 = 17 = R F(3) = (3+3) mod 26 = 6 mod 26 = 6 = G F(0) = (0+3) mod 26 = 3 mod 26 = 3 = D F(24) = (24+3) mod 26 = 27 mod 26 = 1 = B
  • 7. Now replace the characters with new characters. TEST TODAY = WHVW WRGDB So our encrypted message is WHVW WRGDB
  • 8. Lets Decrypt a message. The unique letters we found from the following message are O, R, Y, H F−1 (p) = (p-3) mod 26 (Decryption formula) 𝐹−1 (14) = (14-3) mod 26 = 11 mod 26 = 11 = L 𝐹−1 (17) = (17-3) mod 26 = 14 mod 26 = 14 = O 𝐹−1 (24) = (24-3) mod 26 = 21 mod 26 = 21 = V 𝐹−1(7) = (7-3) mod 26 = 4 mod 26 = 4 = E
  • 9. Now replace the characters with new characters. ORYH = LOVE So our Decrypted message is LOVE
  • 10. Exercises 1.ENCRYPT THE MESSAGE “KOS KI MAMA ” 2.DECRYPT THE MESSAGE “L ORYH EDQJODGHVK”