24 24 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 $2N$ $N(N-1)$ $\dfrac{N(N-1)}{2}$ $(N-1)^{2}$ Computer Networks gatecse-2015-set1 computer-networks network-security normal out-of-gatecse-syllabus + – Misbah Ghaya 10.5k views answer comment Share Follow Print See all 2 Comments 2 2 Comments reply Akash Papnai commented Dec 13, 2019 reply Follow flag Same Question as this one: https://gateoverflow.in/3384/gate2008-it-70 0 0 replyShare Siddharth_Perkar commented Aug 2 reply Follow flag Removed from current syllabus 0 0 replyShare Please log in or register to add a comment.
Best answer 41 41 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 Arjun answered Feb 15, 2015 • edited Jun 29, 2018 by Milicevic3306 Arjun comment Share Follow See all 3 Comments 3 3 Comments reply Santanu Naskar commented Aug 20, 2015 reply Follow flag I don't understand this. My point would be like this: As they have asked the number of keys as a whole and not summed up on individual knowledge of no of keys for each person, hence what I could think of is: As per symmetric cryptography we can communicate with a person secretly if we have his public key, and in turn that person will use his own private key to decrypt the message and no one else. So each person posses a Public Key & a Private Key each, and so for N persons with 2 keys each. the total number of keys in the system would be 2*N = 2N to suffice a confidential communication. To think of the whole system I would think like: I have a huge whiteboard on which N public keys are written and each N person possesses his own private key which is secret. Please correct me if I am wrong somewhere. 3 3 replyShare Arjun commented Aug 20, 2015 reply Follow flag You are correct but that is for Public Key cryptography. https://en.wikipedia.org/wiki/Symmetric-key_algorithm 4 4 replyShare zeeshanmohnavi commented Dec 22, 2018 reply Follow flag @Arjun Since both parties participating in a communication share one secret key, I assume that each party is in possession of a copy of the key. Also, the inclusion of the phrase "as a whole" in the problem, and $N(N-1)$ as one of the option do require your comments on why the number of the secret keys used per communication not be counted as $2$? 0 0 replyShare Please log in or register to add a comment.
10 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 Shubham Pandey 2 answered Oct 13, 2016 Shubham Pandey 2 comment Share Follow See 1 comment 1 1 comment reply Ankit pipaliya commented Jan 22, 2020 reply Follow flag correct. 0 0 replyShare Please log in or register to add a comment.
4 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 Regina Phalange answered Apr 30, 2017 Regina Phalange comment Share Follow 0 reply Please log in or register to add a comment.