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Andrew S. Tanenbaum Edition 5th Exercise 6 Question 10 (Page No. 607)

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if two-way handshake were there, then there would be no deadlock. Reason being the connection will be established in 2 handshakes (ie after SYN() and SYN() + ACKN()). After that the sender will simply send the segment but the receiver will send the negative acknowledgement as it didn’t get for the previous one (ie for  SYN()+ ACKN()).
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