Lamport-Shostak-Peas Algorithm There is an algorithm, agreement protocol, known as Lamport’s Stockpiece Algorithm. So, Lamport’s Stockpiece Algorithm, right? This algorithm is also referred to as the Oral Message Algorithm (OM Algorithm), right? And it solves the Byzantine Agreement Problem for 3m + 1 or more processors. The total number of processors in the system must be 3 times the number of malicious processors plus one. Where there are total number of m faulty processors allowed, okay, where m is greater than zero. Now the source processor sends its value to every other processor. Yes, as usual, byzantine protocol, the source processor sends its value, it takes an initial value and sends the initial value to every other processor. For each processor i, let va be the value that processor i receives from the source, right? And if i does not receive any value, then it uses the default value 0. When the processor in the valley may receive panel or winner, the default is zero. C minimum on receiving bi, processor i acts as the new source and initiates OM of m minus 1 second time. It requires a value from the source processor, so every other processor which receives is executing one more time the OM algorithm and sends va to each of the n minus two other processors. For each i and each j, let vj be the value of the processor j received instead of two. Then the processor i uses the majority value, right, in the numerator. We can see the example and then come back to the algorithm. Now, the example is,There are four processors: P naught, P one, P two, and P three. P two is the faulty processor. Okay, P naught sends initial values to all three processors: P one, P two, and P three, similarly. Right, so P two, P f, P one, P two, P three, and a pyramidal. So, here, P naught is the source. Source P naught. Okay, there is one faulty processor. What is this? Is here OM f. This is called OM of one. Right, m is equal to one. Another one, OM of one. So, P naught becomes the source and sends its initial value 1 to all other processors. Okay, next, people OM of 0. People, all processors: P one, P two, P three, on receiving the initial value one from the source, have to initiate one more time called Ymilliam. So, this is OM of zero, n minus m minus one layout. OMF n minus one second column. So, why am I zero? Happens, upper zero happens. Left, P one sends the received value to P two. P two is a faulty processor. It received one but wrongly communicated zero to C1 and P three. And take P three, it received one from source processor at the initial first run, so it communicated one to P two, as well to P one. Character communications because they are non-faulting. Right, in the case titration here, and we can take up the consensus hippo at every node: P naught, P naught, linear value game is equal to one. Values: one is received from P naught, another is received from P two, another is received from P three. Okay, from P one, received one from P naught, zero from P two, and one from P three. And P two, P two received one from P naught, one from P one, one from P three, and P three has received one from P naught, zero from P two. Sorry, P one, one from P naught, one from P one, zero from P two. In a maximum of it, you look at all the other processors, agreed upon the same value, which is initiated by the source processor, P naught. Okay, and I’m assuming here the next case: suppose P naught is faulty. Source processor itself is faulty so that it sends 1 to P one, 0 to P two, over 1 to P two, P three. So, that is, P naught is faulty here. Right, so volume of one minute. Foreign one, zero, one. Either volume of zero. Okay, m minus one man. Okay, so this is one, two, three, seven, zero, one, two, three. Okay, so for P one, received one from P naught and sends the same to other processors, P two and P three. Similarly, P two received 0 from P naught and sends the same to [Music] zero, one from P naught, one from P three. And P two received zero from P naught, one from P one, one from P three. And P three, one from P naught, one from P two, sorry, zero from P two, sorry, one from P one, and 0 from P two. So, in the case law, what is maximum likelihood? Even though P one is faulty, all other processors assume the common same value, one. Right, so this is what actually is what the algorithm says.