practical cryptography exercise 1

7/27/2019 Practical Cryptography exercise 1

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Exercise 1

Explain why this form of internal feedback is much worse in practice in a non-ideal world, whenthe ciphertext may be corrupted by error.

Solution

In this form of internal feedback, the decryption will be such:

1i i i p Decrypt c p

And since each decryption depends on the previous decryption ( 1i p ), it will cause all the next

decryptions to fail. (And every single error will force all future blocks to be resent)


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However, in the other internal feedback structu re, wed get the decryption method:

1i i i p c Decrypt c

And therefore, each corrupted ciphertext block will prevent the decryption of at most two

blocks. (The corresponding plaintext block i p , and the one after it, 1i p )


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Exercise 2

When using multiple encryption, which internal feedback structure will guarantee no apparent

output structure, and no complete loss of data when 1 ciphertext block is corrupted?

Solution

An identical feedback structure to the one we saw in question 1:


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Observing the multiple encryption as a black box encryption module, will show that this indeeddoes not cause a complete loss of decryption when a single ciphertext block is corrupted. (As wesaw in question 1)

In addition, it does not generate any (apparent) output structure since each ciphertext block is

XORed with the new plaintext.


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Exercise 3

Calculate the probability, in the slidex attack, of:

1 1 2 2 1 2k x x c k w w e

(Where e is the common value 1 1 2 2w F x c w F x c )

Solution

For every 1 2 x x , and 1 1 2 1,c k x x k , the four terms:

1 1 2 1 1 2, , , x k x k x c x c Are different in pairs.

Therefore, choosing the permutation F uniformly at random, each of them is mapped u.a.r aswell. As a result, the probability that:

1 1 2 2 1 1 2 1

2 1 1 1 2 2 1 2 1 1 2 1

1 1 2 1 1 2 1 1 2 1

Pr | ,

Pr | ,

1Pr | ,

2

F

F

n F

w F x c w F x c c k x x k

k F x k F x c k F x k F x c c k x x k

F x k F x k F x c F x c c k x x k

(Since whatever the value of 1 1 2 1 1 F x k F x k F x c may be, the probability that

2 F x c will be the same is at most1

2 n.)

Therefore:

1 2

1 2

1 1 2 1 1 1 2 2

1 1 2 1 1 1 2 2

1Pr , |

2

1Pr , | 1

2

x x n F

x x n F

c k x x k w F x c w F x c

c k x x k w F x c w F x c


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And since both 1 1 2 1,c k x x k are equally likely when c is chosen u.a.r:

1 1 1 2 2 1,

1 2 1 1 1 2 2 1,

1 1Pr |

2 21 1

Pr | 2 2

nc F

nc F

c k w F x c w F x c

c x x k w F x c w F x c

Now, if 1 2 1c x x k , then:

1 2

2 1 2 1 2 1

2 1 2 1 2 1

2 1 1 1 1 2 1

Pr |

Pr |

Pr | 1

F

x x F

F

k w w e c x x k

k w F x c c x x k

k w F x k c x x k

However, if 1c k , then:

1 2

1

2 1 2 1 1 1 2 2 1

2 1 2 1 1 1 1 2 2 1

1 1 2 1 1 1 1 2 2 1

Pr |

Pr |

Pr | 0

F

x x F

x

F

k w w e w F x k w F x k

k w F x k w F x k w F x k

F x k F x k w F x k w F x k

Since 1 2 1 1 2 1 1 1 2 1 x x x k x k F x k F x k .

Therefore, the require probability is:

2 1 2 1 2 1 1 1 2 2

1 2 1 1 1 2 2 1,

Pr |

1 1Pr |

2 2

F

nc F

k w w e c x x k w F x c w F x c

c x x k w F x c w F x c

practical cryptography exercise 1

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