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Show that if P=NP then the RSA cryptosystem (Section 1.4.2) can be broken in polynomial time.

Short Answer

Expert verified

It is proved that RSA can be broken in polynomial time when P=NP.

Step by step solution

01

Explain RSA

One of the earliest public-key cryptography schemes was RSA. The basic concept is to encrypt a message using a public key and decrypt it with a private key. Factoring a semi prime that is a product of two primes appeared to be more difficult than multiplying two primes to figure out what semi prime they make up.

The general concept is to utilize a huge semi prime as your public key and the pair of primes that make up that semi prime as your private key.

Here, Fermat鈥檚 little theorem can be used. Fermat鈥檚 little theorem is given as, 鈥淚f N is prime, then cN=modN=cmodN.

Therefore, it requires a method consist of two steps so that the first step can be used for encryption and the second step can be used for decryption.

Consider the following rule to carry out encryption and decryption. 鈥淚f Nis a semi-prime, then N=p.q, where pand q are prime, and cp-1q-1modN=cmodN.

02

Encryption of Message

Encryption:

For a given pair of primesp and q ,p-1q-1+1 is a number that is not prime. Therefore, it can be represented using product of two whole numbers r and s.To encrypt the message, use the following steps;

Select two large prime numbersp and q

Calculate the product N.

Select r and s .

Now public key will be rand N. Message can be encrypted using public key.

03

Decryption of Message

Decryption:

Assume is a private key. To decrypt the encrypted message, increase the sthpower modN. It will generate the plaintext. If Nis known to attacker then pand qcan be computed easily. Thus, attacker can also find out the value of and decrypt the message.

In present, there is no method to factor large semi primes. If P=NP , then prime factorization can be solved easily.

So, anyone can calculate private keys quickly.

In this way, it is proved that RSA can be broken in polynomial time when P=NP.

Hence, the given statement is proved.

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Most popular questions from this chapter

Give a simple reduction from 3D MATCHING to SAT, and another from RUDRATA CYCLE to SAT.

(Hint: In the latter case you may use variables xijwhose intuitive meaning is 鈥渧ertex i is the j th vertex of the Hamilton cycle鈥; you then need to write clauses that express the constraints of the problem.)

Optimization versus search.Recall the traveling salesman problem:

TSP

Input: A matrix of distances; a budget b

Output: A tour which passes through all the cities and has lengthb, if such a tour exists.

The optimization version of this problem asks directly for the shortest tour.

TSP-OPT

Input:A matrix of distances

Output:The shortest tour which passes through all the cities.

Show that if TSP can be solved in polynomial time, then so can TSP-OPT.

In the NODE-DISJOINT PATHS problem, the input is an undirected graph in which some vertices have been specially marked: a certain number of 鈥渟ources鈥 s1,s2,,sk and an equal number of 鈥渄estinations鈥 t1,t2,,tk. The goal is to find k node-disjoint paths (that is, paths which have no nodes in common) where the ith path goes from si to ti. Show that this problem is NP-complete.Here is a sequence of progressively stronger hints.

  1. Reduce from 3SAT .
  2. For a 3SAT formula with m clauses and n variables, use k=m+n sources and destinations. Introduce one source/destination pair (sx,tx)for each variable x , and one source/destination pair (sc,tc) for each clause c .
  3. For each 3SAT clause, introducenew intermediate vertices, one for each literal occurring in that clause and one for its complement.

Notice that if the path from sc to tc goes through some intermediate vertex representing, say, an occurrence of variable x, then no other path can go through that vertex. What vertex would you like the other path to be forced to go through instead?

STINGY SAT is the following problem: given a set of clauses (each a disjunction of literals) and an integer K , find a satisfying assignment in which at most K variables are true, if such an assignment exists. Prove that isNP -complete.

Determine which of the following problems are NP-complete and which are solvable in polynomial time. In each problem you are given an undirected graph G=(V,E), along with:

(a)A set of nodesLV , and you must find a spanning tree such that its set of leaves includes the set L.

(b)A set of nodes LV, and you must find a spanning tree such that its set of leaves is precisely the set L.

(c)A set of nodesLV , and you must find a spanning tree such that its set of leaves is included in the set L.

(d)An integer k, and you must find a spanning tree withk or fewer leaves.

(e)An integer k, and you must find a spanning tree withk or more leaves.

(f)An integer k, and you must find a spanning tree with exactlyk leaves.

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