By Karp R.

**Read Online or Download Combinatorics, Complexity and Randomness (Turing Award lecture) PDF**

**Best combinatorics books**

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**Additional info for Combinatorics, Complexity and Randomness (Turing Award lecture)**

**Sample text**

6. What is a situation where it is more important that a message be authenticated (that is, digitally signed), than it be secret? 7. What are some of your day-to-day activities that will involve the use of a cryptosystem? Beginning Exercises 8. Suppose Alice is logging on to her computer from a distant terminal, using the public-key authentication protocol. In what ways can Mallory, who may tamper with any messages between the host computer and Alice, foil Alice’s attempts to log on? 9. How can the protocol be improved to foil Mallory?

By definition rk is a divisor of rk−1 , and hence by the previous line it is a divisor of rk−2 . Moving back up through the list of equations it is easy to see that rk must be a divisor of both m and n. This means that rk must be a divisor of gcd(m, n). But, suppose that some other value g also divides both m and n. By moving down through the equations, starting with m = xg and n = yg it can be seen that g must divide rk . Thus rk is equal to gcd(m, n). This algorithm can be presented as an iteration.

This is readily checked: 11 5 = 1 5 = 1. 11 7 = 4 7 = and 2 7 2 = 1. So given 11 = 1 (mod 5) 11 = 4 (mod 7) the above congruence can be reduced to the two simpler congruences x2 = 1 (mod 5), x2 = 4 (mod 7). 36 Cryptography with Open-Source Software By inspection, the solutions to these are 1, −1, and 2, −2. The Chinese remainder theorem can now be used to solve the following systems of congruences: x= 1 x = −1 x= 1 x = −1 (mod 5) (mod 5) (mod 5) (mod 5) and and and and x= 2 x= 2 x = −2 x = −2 (mod 7) (mod 7) (mod 7) (mod 7).