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Mohamed G. Gouda			            Network Protocol Security
CS 395T					            HW1 Solution
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1. 
process p

const  T     : integer,     /round-trip delay between p and q
       r     : integer,     /max. value for the shared key
       sk    : 0..r-1,      /shared key between p and q

var    t     : integer,     /text of current data msg
       nc    : integer,     /nonce of current data msg, init. nc.p=nc.q 
       nc'   : integer,     /nonce of next data msg 
       c     : integer,     /received ack
       ready : boolean      /initially true

begin
   ready     -->           ready := false;
                           t     := any;
                           nc'   := NNC;
                           send data(NCR(sk, t|nc|nc')) to q

[] rcv ack(c) from q  -->  if nc' = c  --> ready := true; nc := nc'
                           [] nc' <> c --> skip
                           fi

[] timeout
   ~ready ^ 
    T time units have passed since first action is executed last
             --> send data(NCR(sk, t|nc|nc')) to q
end


process q

const r    : integer,       /max. value for the shared key
      sk   : 0..r-1         /shared key between p and q

var   t    : integer,       /text of received data msg
      nc   : integer,       /nonce of next data msg, init. nc.p=nc.q 
      h    : integer,       /received data
      c,c' : integer        /received nonces

begin
   rcv data(h) from p	-->
        (t, c, c') := DCR(sk, h);
        if nc = c   -->     /deliver t
                            nc  := c';
                            send ack(nc) to q

        [] nc <> c  -->     send ack(nc) to q
        fi
end

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3.

Let s be the source of the long chain and  
    d be the destination of the long chain.

for every certificate (u, v) that occurs only in the
    long chain do the following to disperse (u, v):

    add (u, v) to D.s or D.d

for every certificate (u, v) that occurs only in the
    short chains do the following to disperse (u, v):

    if there is a node x such that x is the source 
       or destination of every chain that has (u,v),
    then add (u, v) to D.x
    else add (u, v) to both D.u and D.v.

for every certificate (u, v) that occurs in the long
    and some short chains do the following to disperse (u, v):

    Let D'(u,v) be the location of certificate (u,v) given
    a dispersal D' where (u,v) is assigned to the source s
    of the long chain.

        1. remove from the set Y of short chains that contain
           (u,v) every chain whose source or destination is s

        2. if Y is empty, 
           then D'(u,v) = {s}
           elseif there is a node x such that x is the source or
                  destination of every remaining chain in Y, 
               then D'(u,v) = {s, x}
               else D'(u,v) = {s, u, v}  

    Let D"(u,v) be the location of certificate (u,v) given
    a dispersal D" where (u,v) is assigned to the destination d 
    of the long chain.

        1. remove from the set Y of short chains that contain
           (u,v) every chain whose source or destination is d

        2. if Y is empty, 
           then D"(u,v) = {d}
           elseif there is a node x such that x is the source or
                  destination of every remaining chain in Y, 
               then D"(u,v) = {d, x}
               else D"(u,v) = {d, u, v}  

    If |D'(u,v)| =< |D"(u,v)|,
    then disperse (u,v) to every D.z where z is in D'(u,v)
    else disperse (u,v) to every D.z where z is in D"(u,v)