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Consider the problem of computing x y for given integers x and y: we want the whole answer, not modulo a third integer. We know two algorithms for doing this: the iterative algorithm which performs y 鈭 1 multiplications by x; and the recursive algorithm based on the binary expansion of y. Compare the time requirements of these two algorithms, assuming that the time to multiply an n-bit number by an m-bit number is O(mn).

Short Answer

Expert verified

Time complexity of both the algorithms are compared.

Step by step solution

01

Explain Time Complexity

Time complexity is defined as the total amount of time taken to run and complete the function. It is being observed step by step at each statement of the function.Two integers x and y are considered and the exponential is found out by iterative and recursive approach and by comparing their time complexities.

02

Iterative algorithm

Iterative Algorithm is defined as:

defiterative(x,y):Input=x,yOutput=xyFinal=xForiinrange(1,y):Final=xReturnFinal

Total time complexity of the given algorithm is O(2n).

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Most popular questions from this chapter

In an RSA cryptosystem, p = 7and q = 11(as in Figure 1.9). Find appropriate exponents and .

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RSA and digital signatures. Recall that in the RSA public-key cryptosystem, each user has a public key P=(N,e) and a secret key d. In a digital signature scheme, there are two algorithms, sign and verify. The sign procedure takes a message and a secret key, then outputs a signature . The verify procedure takes a public key (N,e), a signature , and a message M, then returns 鈥渢rue鈥 if could have been created by sign (when called with message M and the secret key (N, e) corresponding to the public key ); 鈥渇alse鈥 otherwise.

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(c) Generate your own RSA modulus, N=pq public key e, and private key d (you don鈥檛 need to use a computer). Pick p and q so you have a 4-digit modulus and work by hand. Now sign your name using the private exponent of this RSA modulus. To do this you will need to specify some one-to-one mapping from strings to integers in [0,N-1]. Specify any mapping you like. Give the mapping from your name to numbers m1,m2,...mk,then sign the first number by giving the value md1(modN), and finally show that .

(md1)e=m1(modN)

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