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Mohamed G. Gouda						Fall 2006 
CS 395T: Network Protocol Security				Midterm 2
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1. 

Algorithm 

Input:  A certificate chain set CS where each certificate 
        appears in at most three chains

Output: An optimal dispersal D of CS

Steps: 
1. For each user u, that appears as a source or destination in CS, D.u := { }

2. For each certificate (u,v) in CS,
	if there is a user x where the source or destination
	   of each chain that has (u,v) is x    
	then D.x := D.x union {(u,v)}


	else if there are two users x and y where the source or destination
	     of each chain that has (u,v) is x or y
	     then D.x := D.x union {(u,v)} and 
		  D.y := D.y union {(u,v)}

	     else let x, y, and z be the sources of all chains that has (u,v)
		  D.x := D.x union {(u,v)}
		  D.y := D.y union {(u,v)}
		  D.z := D.z union {(u,v)}	
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2. (7 points)

Consider the following scenario where p ends up authenticating q even
though q does not receive any message and does not send any message:

1. p sends n to q
2. r intercepts n and sends n to p pretending to be q attempting to 
   authenticate p using the same protocol. 
3. p follows the protocol and sends the message I.p <n, q, SK> to ac.
4. r intercepts this last message and sends it to p pretending to be
   ac completing the authentication of q by p   

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