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16 bits is not the size of the IP datagram; it is the amount of bits required to encode its length. Hence, you can have datagrams of length up to 65536 octets (i.e., bytes), as described in RFC 760: Total Length: 16 bits Total Length is the length of the datagram, measured in octets, including internet header and data. This field allows the ...


3

is a protocol a pack of algorithms and programming source codes? (in both the computer and network) No. A protocol is an abstract description of how systems communicate. Source code could be an implementation of a protocol, but the protocol is not the same as its implementation. Basically, a protocol tells a programmer "the messages should be formatted like ...


3

If you're feeling up for something particularly geeky, you might enjoy a fair exchange protocol, either one based on gradual release (see e.g., this paper) or one based on Bitcoin (e.g., this). The number of steps is very large though! Probably not suitable to carry out by hand. You could try a simpler protocol: Alice picks a message $m_A$, computes a ...


2

In this context, the rendezvous would mean "a bunch of threads meeting together" instead of people. Rendezvous is a method of synchronisation in which at least two threads "meet". In other words, each thread that reaches the rendezvous point waits until all other threads have reached the same point before proceeding. In the context of Distributed ...


2

I believe the answer is "yes", when $Q$ is square and nonsingular, and "not necessarily", otherwise. See "On the Capacity of a Discrete Channel. I", by Saburo Muroga, Journal of the Physical Society of Japan, Vol. 8, No. 4, July - August 1953, pp 484-494. You might also be interested in my note on Muroga's paper, in which I attempt to provide an alternate ...


2

Your channel is known as the Z-channel. Take a look at Marco Mondelli, S. Hamed Hassani, and Rüdiger Urbanke, How to Achieve the Capacity of Asymmetric Channels.


1

Yes. Consider the following protocol. Alice flips a coin, and sets $X$ and $Y$ to be equal to the outcome of the coin flip. She then sends $X,Y$ to Bob. Note that the transcript will include the value of $X,Y$. Now unconditionally, $X,Y$ are dependent (there is probability $1/2$ that they are Heads,Heads and probability $1/2$ that they are Tails,Tails). ...


1

You haven't defined when a randomized protocol is declared to be successful, so I will assume that at the end of the protocol, Bob tells a judge what he thinks the message is, and we want that for each message $x$, the probability that Bob is correct is at least $1-\epsilon$. Let $X$ be a random $n$-bit string, and let $\Pi$ be the corresponding transcript. ...


1

According to the paper 1 Generally speaking, Byzantine consensus considers the problem of reaching agreement among a group of n parties, among which up to f can have Byzantine faults and perform arbitrarily. There exist a few variant formulations for the Byzantine consensus problem. Two theoretical formulations are "Byzantine broadcast" and "Byzantine ...


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