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21 votes
21 votes

Suppose that everyone in a group on $N$ people wants to communicate secretly with the $(\text{N - 1})$ others using symmetric Key cryptographic system. The communication between any two person should not be decodable by the others in the group. The numbers of keys required in the system as a whole to satisfy the confidentiality requirement is

  1. $2N$
  2. $N(N-1)$
  3. $\dfrac{N(N-1)}{2}$
  4. $(N-1)^{2}$
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3 Answers

Best answer
38 votes
38 votes

In symmetric key cryptographic system, both parties have access to key.
So, the first person has $\text{N-1 keys}$ with other $\text{N-1}$ people,
second one has another $\text{N-2}$ with $\text{N-2}$ people ( $1$ we already
considered ) and so on till $1.$

So, Total number of keys required $= N-1 + N-2 +\ldots + 1$

$=\dfrac{N(N-1)}{2}$

C choice.

Had we been using Public key cryptography we needed just $2N$ keys in the system. 

Reference: https://en.wikipedia.org/wiki/Symmetric-key_algorithm

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10 votes
10 votes

IT is complete Graph of N vertex 

where each edge B/W two Vertex need Distinct Key

Total number of Edges in Complete Graph is      NC2   = N(N-1)/2 

4 votes
4 votes

In Symmetric Key Cryptography, access of key is with both the parties. It implies every person needs to communicate N-1 other users using different keys i.e 1+2+3...N-2+N-1 This is like number of edges needed in a complete graph with N vertices is N(N-1)/2. Answer is therefore C

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