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Chapter 1: Q43E (page 52)

In the RSA cryptosystem, Alice鈥檚 public key (N,e)is available to everyone. Suppose that her private key d is compromised and becomes known to Eve. Show that e=3if (a common choice) then Eve can efficiently factor N.

Short Answer

Expert verified

It can be shown thate=3 allows Eve to calculate the factor N efficiently.

Step by step solution

01

Explain RSA cryptosystem

RSA cryptosystem is cryptographic algorithm that is asymmetric, that works on both public and private keys. The RSA is based on the larger integer that is difficult to factorize.

02

. Show that if e =3 (a common choice) then Eve can efficiently factor N.

Consider that N=pq,C=p-1q-10, then ed=lm0dC. Substitute the value ofe by 3.

3d=kC+1k=3d-1C

For,0<d<C, the values of k will be 1,2. Using the values of k heuristically C calculates, N.

Therefore, it is shown thate=3 allows Eve to calculate the factor N efficiently.

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Most popular questions from this chapter

Let[m]denote the set{0,1,鈥�,m鈭�1}. For each of the following families of hash functions, say whether or not it is universal, and determine how many random bits are needed to choose a function from the family.

(a) H={ha1,a2:a1,a2鈭�[m]}, wheremis a fixed prime and

ha1路ha1,a2(x1,x2)=a1x1+a2x2modm

Notice that each of these functions has signatureha1,a2:[m]2鈫�[m]that is, it maps a pair of integers in[m]to a single integer in[m].

(b) His as before, except that nowm=2kis some fixed power of.2

(c) His the set of all functionsf:[m]鈫�[m鈭�1].

A d-ary tree is a rooted tree in which each node has at most dchildren. Show that any d-ary tree with n nodes must have a depth of惟(lognlogd) .Can you give a precise formula for the minimum depth it could possibly have?


Show that in any base b鈮�2, the sum of any three single-digit numbers is at most two digits long.

Show that if xis a nontrivial square root of 1 modulo N , that is if x2鈮�1modNbut x鈮�卤1modN, thenN must be composite. (For instance,42鈮�1mod15but4鈮�卤1mod15; thus 4 is a nontrivial square root of 1 modulo 15.)

The grade-school algorithm for multiplying two n-bit binary numbers x and y consist of addingtogethern copies of r, each appropriately left-shifted. Each copy, when shifted, is at most 2n bits long.
In this problem, we will examine a scheme for adding n binary numbers, each m bits long, using a circuit or a parallel architecture. The main parameter of interest in this question is therefore the depth of the circuit or the longest path from the input to the output of the circuit. This determines the total time taken for computing the function.
To add two m-bit binary numbers naively, we must wait for the carry bit from position i-1before we can figure out the ith bit of the answer. This leads to a circuit of depth螣(m). However, carry-lookahead circuits (seewikipedia.comif you want to know more about this) can add in螣(logn)depth.

  1. Assuming you have carry-lookahead circuits for addition, show how to add n numbers eachm bits long using a circuit of depth 螣(lognlogm).
  2. When adding three m-bit binary numbers x+y+z, there is a trick we can use to parallelize the process. Instead of carrying out the addition completely, we can re-express the result as the sum of just two binary numbersr+s, such that the ith bits of r and s can be computedindependently of the other bits. Show how this can be done. (Hint: One of the numbers represents carry bits.)
  3. Show how to use the trick from the previous part to design a circuit of depth螣(logn)for multiplying two n-bit numbers.
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