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Mohamed G. Gouda                                        CS 356
Summer      2007                                        Homework 2 Solution
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1. (3 Points):

process p

const  v   
inp    m      : integer     {m>0}
var    conp   : array [0..v-1] of boolean   {init. false}
       cansnd : array [0..v-1] of 0..m
par    c      : 0..v-1

begin
     timeout ~conp[c] ^ (crqst(c)#ch.p.q + data(c)#ch.p.q + drqst(c)#ch.p.q
	                + crply(c)#ch.q.p + drply(c)#ch.q.p = 0)
	                -> send crqst(c) to q
  		
[]   rcv crply(c) from q -> conp[c], cansnd[c] := true, m

[]   conp[c]             -> if cansnd[c] > 0 -> send data(c) to q;
					        cansnd[c] := cansnd[c] - 1
			    [] true          -> send drqst(c) to q;
					        cansnd[c] := 0
			    fi
[]   rcv drply(c) from q -> conp[c] := false
end

process q

const   v
var     conq : array [0..v-1] of boolean   {init. false}
par     c    : 0..v-1

begin
     rcv crqst(c) from p -> if true          -> conq[c] := true;
						send crply(c) to p
			    [] true          -> conq[c] := false;
					        send drply(c) to p
			    fi 
        
[]   conq[c]             -> conq[c] := false

[]   rcv data(c) from p  -> if ~conq[c]      -> {discard data message}
				                send drply(c) to p
			    [] conq[c]       -> {store data message}
						skip
			    fi
[]   rcv drqst(c) from p -> conq[c] := false; send drply(c) to p  
end

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2. (4 Points):

Only process q needs to be changed as follows. First, a new integer variable 
named sp is added to process q. Second, the first action in process q needs
to be modified as shown.

     rcv crqst(x) from p    -> 
	 if  conq ^ sp = x  -> y := sq; send crply(x,y) to p
	 []  conq ^ sp != x -> y := sq; send drply(x,y) to p
	 [] ~conq           -> conq, sp, y := true, x, sq;
			       send crply(x,y) to q
         [] ~conq           -> conq, sq, y := false, sq+1, sq
			       send drply(x,y) to q
         fi

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3. (3 Points):

In order to be able to compare these two protocols, we need to ensure that the
numbers of parity bits used in both protocols are equal. Thus for a data block
of m*n data bits, we use n parity bits, and for a data block of (m*n)/2 data
bits, we use n/2 parity bits. Now there is one-to-one correspondence between 
the data bits in both protocols and between the parity bits in both protocols.
If the corruption of some bits can be detected in one protocol, then corruption
of the corresponding bits can be detected in the other protocol. In other
words, the two protocols have the overhead and the same effectiveness in 
corruption detection. However in many cases once corruption is detected the
first protocol discards a large block of data bits while the second protocol 
discards only a small block of data bits. Thus the second protocol is better.

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