Assignment #3 - University of Windsor



Assignment #3

My full name written in lower case is “yufei xu”. The ASSCII codes corresponding to each alphabet are in the following table:

|Alphabet |ASSCII Code (Ai) Decimal |

|y |121 |

|u |117 |

|f |102 |

|e |101 |

|i |105 |

|x |120 |

|u |117 |

V = [(121 + 117 + 102 + 101 + 105 + 120 + 117) mod 23] + 1 = 2

Therefore, I pick the second question in chapter 9.

Q_9.2 Perform encryption and decryption using the RSA algorithm, as in Figure 9.6 for the following:

1. p=3; q=11; e=7; M=5

Answer:

n = p * q = 3 * 11 = 33

f(n) = (p-1) * (q-1) = 2 * 10 = 20

Now, we need to compute d = e-1 mod f(n) by using backward substitution of GCD algorithm:

According to GCD:

20 = 7 * 2 + 6

7 = 6 * 1 + 1

6 = 1 * 6 + 0

Therefore, we have:

1 = 7 – 6

= 7 – (20 – 7 * 2)

= 7 – 20 + 7 * 2

= -20 + 7 * 3

Hence, we get d = e-1 mod f(n) = e-1 mod 20 = 3 mod 30 = 3

So, the public key is {7, 33} and the private key is {3, 33}, RSA encryption and decryption is following:

[pic]

2. p=5; q=11; e=3; M=9

Answer:

n = p * q = 5 * 11 = 55

f(n) = (p-1) * (q-1) = 4 * 10 = 40

Now, we need to compute d = e-1 mod f(n) by using backward substitution of GCD algorithm:

According to GCD:

40 = 3 * 13 + 1

13 = 1 * 13 + 0

Therefore, we have:

1 = 40 – 3 * 13

Hence, we get d = e-1 mod f(n) = e-1 mod 40 = -13 mod 40 = (27 – 40) mod 40 = 27

So, the public key is {3, 55} and the private key is {27, 55}, RSA encryption and decryption is following:

[pic]

3. p=7; q=11; e=17; M=8

Answer:

n = p * q = 7 * 11 = 77

f(n) = (p-1) * (q-1) = 6 * 10 = 60

Now, we need to compute d = e-1 mod f(n) by using backward substitution of GCD algorithm:

According to GCD:

60 = 17 * 3 + 9

17 = 9 * 1 + 8

9 = 8 * 1 + 1

8 = 1 * 8 + 0

Therefore, we have:

1 = 9 – 8

= 9 – (17 – 9)

= 9 – (17 – (60 – 17 * 3))

= 60 – 17*3 – (17 – 60 + 17*3)

= 60 – 17 *3 + 60 – 17*4

= 60*2 – 17*7

Hence, we get d = e-1 mod f(n) = e-1 mod 60 = -7 mod 60 = (53-60) mod 60 = 53

So, the public key is {17, 77} and the private key is {53, 77}, RSA encryption and decryption is following:

[pic]

4. p=11; q=13; e=11; M=7

Answer:

n = p * q = 11 * 13 = 143

f(n) = (p-1) * (q-1) = 10 * 12 = 120

Now, we need to compute d = e-1 mod f(n) by using backward substitution of GCD algorithm:

According to GCD:

120 = 11 * 10 + 10

11 = 10 * 1 + 1

10 = 1 * 10 + 0

Therefore, we have:

1 = 11 – 10

= 11 – (120 – 11 * 10)

= 11 – 120 + 11 * 10

= -120 + 11 * 11

Hence, we get d = e-1 mod f(n) = e-1 mod 120 = 11 mod 120 = 11

So, the public key is {11, 143} and the private key is {11, 143}, RSA encryption and decryption is following:

[pic]

5. p=17; q=31; e=7; M=2

n = p * q = 17 * 31 = 527

f(n) = (p-1) * (q-1) = 16 * 30 = 480

Now, we need to compute d = e-1 mod f(n) by using backward substitution of GCD algorithm:

According to GCD:

480 = 7 * 68 + 4

7 = 4 * 1 + 3

4 = 3 * 1 + 1

3 = 1 * 3 + 0

Therefore, we have:

1 = 4 – 3

= 4 – (7 – 4)

= 4 – (7 – (480 – 7*68))

= 4 – (7 – 480 + 7*68)

= 480 – 7*68 – 7 + 480 – 7*68

= 480*2 – 7*137

Hence, we get d = e-1 mod f(n) = e-1 mod 480 = -137 mod 480 = (343 – 480) mod 480 =343

So, the public key is {7, 527} and the private key is {343, 527}, RSA encryption and decryption is following:

[pic]

-----------------------

PR= (3, 33)

Plaintext

5

ciphertext

14

PU= (7, 33)

Plaintext

5

5

Decryption

143 Mod 33 = 5

Encryption

57 Mod 33= 14

93 Mod 55= 14

Encryption

1427 Mod 55 = 9

Decryption

Plaintext

9

5

PU= (3, 55)

ciphertext

14

Plaintext

9

PR= (27, 55)

817 Mod 77= 57

Encryption

5753 Mod 77 = 8

Decryption

Plaintext

8

5

PU= (17, 77)

ciphertext

57

Plaintext

8

PR= (53, 77)

711 Mod 143 = 106

Encryption

10611 Mod 143 = 8

Decryption

Plaintext

7

5

PU= (11, 143)

ciphertext

106

Plaintext

7

PR= (11, 143)

27 Mod 527 = 128

Encryption

128343 Mod 527 = 2

Decryption

Plaintext

2

5

PU= (7, 527)

ciphertext

128

Plaintext

2

PR= (343, 527)

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