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lecture05 - CS 70 Fall 2004 Discrete Mathematics for CS Rao...

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CS 70 Discrete Mathematics for CS Fall 2004 Rao Lecture 5 The next sequence of lectures is on the topic of Arithmetic Algorithms . We shall build up to an understanding of the RSA public-key cryptosystem. Primality and Factoring You are given a natural number — say, 307131961967 — and you are asked: Is it a prime number? You must have faced this familiar kind of question in the past. How does one decide if a given number is prime? There is, of course, an algorithm for deciding this: algorithm prime(x) y := 2 repeat if x mod y = 0 then return(false); y := y + 1 until y = x return(true) Here by x mod y we mean the remainder of the division of x by y negationslash = 0 ( x % y in the C family). This algorithm correctly determines if a natural number x > 2 is a prime. It implements the definition of a prime number by checking exhaustively all possible divisors, from 2 up to x - 1. But it is not a useful algorithm: it would be impractical to apply it even to the relatively modest specimen 307131961967 (and, as we shall see, modern cryptography requires that numbers with several hundreds of digits be tested for primality). It takes a number of steps that is proportional to its argument x —and, as we shall see, this is bad. algorithm fasterprime(x) y := 2 repeat if x mod y = 0 then return(false); y := y + 1 until y * y x return(true) Now, this is a little better. This algorithm checks all possible divisors up to the square root of x . And this suffices, because, if x had any divisors besides 1 and itself, then consider the smallest among these, and call it y . Thus, x = y · z for some integer z which is also a divisor of x other than one and itself. And since y is the smallest divisor, z y . It follows that y · y y · z = x , and hence y is no larger than the square root of x . And the second algorithm does indeed look for such a y . Still, this algorithm is not satisfactory: in a certain well-defined sense it is “as exponential” as the exhaustive CS 70, Fall 2004, Lecture 5 1
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algorithm for satisfiability. To see why, we must understand how we evaluate the running time of algorithms with arguments that are natural numbers. And you know such algorithms: e.g., the methods you learned in elementary school for adding, multiplying, and dividing whole numbers. To add two numbers, you have to carry out several elementary operations (adding two digits, remembering the carry, etc.), and the number of these operations is proportional to the number of digits n in the input: we express this by saying that the number of such operations is O ( n ) (pro- nounced “big-Oh of n ”). To multiply two numbers, you need a number of elementary operations (looking up the multiplication table, remembering the carry, etc.) that is proportional to the square of the number of digits, i.e., O ( n 2 ) . (Make sure you understand why it is n 2 ).
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