Common use of Byzantine Agreement Clause in Contracts

Byzantine Agreement. decide on x in the first round already. We have established all-same validity! If the correct nodes have different (binary) input values, the validity condition becomes trivial as any result is fine. What about agreement? Let u be the first node to decide on value x (in Line 8). Due to asynchrony another node v received messages from a different subset of the nodes, however, at most f senders may be different. Taking into account that byzantine nodes may lie (send different propose messages to different nodes), f additional propose messages received by v may differ from those received by u. Since node u had at least n/2 + 3f + 1 propose messages with value x, node v has at least n/2 + f + 1 propose messages with value x. Hence every correct node will propose x in the next round, and then decide on x. So we only need to worry about termination: We have already seen that as soon as one correct node terminates (Line 8) everybody terminates in the next round. So what are the chances that some node u terminates in Line 8? Well, we can hope that all correct nodes randomly propose the same value (in Line 12). Maybe there are some nodes not choosing randomly (entering Line 10 instead of 12), but according to Lemma 11.22 they will all propose the same. − Thus, at worst all n f correct nodes need to randomly choose the same bit, which happens with probability 2−(n−f)+1. If so, all correct nodes will send the same propose message, and the algorithm terminates. So the expected running time is exponential in the number of nodes n in the worst case.

Byzantine Agreement. decide on x in the first round already. We have established all-same validity! If the correct nodes have different (binary) input values, the validity condition becomes trivial as any result is fine. − − What about agreement? Let u be the first node to decide on value x (in Line 8). Due to asynchrony another node v received messages from a different subset of the nodes, however, at most f senders may be different. Taking into account that byzantine nodes may lie (send different propose messages to different nodes), f additional propose messages received by v may differ from those received by u. Since node u had at least n/2 + 3f + 1 n 2f propose messages with value x, node v has at least n/2 + f + 1 n 4f propose messages with value x. Hence every correct node will propose x in the next round, and then decide on x. So we only need to worry about termination: We have already seen that as soon as one correct node terminates (Line 8) everybody terminates in the next round. So what are the chances that some node u terminates in Line 8? Well, we can hope that all correct nodes randomly propose the same value (in Line 12). Maybe there are some nodes not choosing randomly at random (entering Line 10 instead of 12), but according to Lemma 11.22 Xxxxx 17.22 they will all propose the same. − Thus, at worst all n f correct nodes need to randomly choose the same bit, which happens with probability 2−(n−f)+1. If so, all correct nodes will send the same propose message, and the algorithm terminates. So the expected running time is exponential in the number of nodes n in the worst case.n.

Byzantine Agreement. decide on x in the first round already. We have established all-same validity! If the correct nodes have different (binary) input values, the validity condition becomes trivial as any result is fine. − − What about agreement? Let u be the first node to decide on value x (in Line 8). Due to asynchrony another node v received messages from a different subset of the nodes, however, at most f senders may be different. Taking into account that byzantine nodes may lie (send different propose messages to different nodes), f additional propose messages received by v may differ from those received by u. Since node u had at least n/2 + 3f + 1 n 2f propose messages with value x, node v has at least n/2 + f + 1 n 4f propose messages with value x. Hence every correct node will propose x in the next round, and then decide on x. So we only need to worry about termination: We have already seen that as soon as one correct node terminates (Line 8) everybody terminates in the next round. So what are the chances that some node u terminates in Line 8? Well, we can hope that all correct nodes randomly propose the same value (in Line 12). Maybe there are some nodes not choosing randomly at random (entering Line 10 instead of 12), but according to Lemma 11.22 3.22 they will all propose the same. − Thus, at worst all n f correct nodes need to randomly choose the same bit, which happens with probability 2−(n−f)+1. If so, all correct nodes will send the same propose message, and the algorithm terminates. So the expected running time is exponential in the number of nodes n in the worst case.n.

Byzantine Agreement. decide on x in the first round already. We have established all-same validity! If the correct nodes have different (binary) input values, the validity condition becomes trivial as any result is fine. What about agreement? Let u be the first node to decide on value x (in Line 8). Due to asynchrony asynchrony, another node v received messages from a different subset of the nodes, however, at most f senders may be different. Taking into account that byzantine nodes may lie (send different propose messages to different nodes), f additional propose messages received by v may differ from those received by u. Since node u had at least n/2 + 3f + 1 propose messages with value x, node v has at least n/2 + f + 1 propose messages with value x. Hence every correct node will propose x in the next round, round and then decide on x. So we only need to worry about termination: We have already seen that as soon as one correct node terminates (Line 8) everybody terminates in the next round. So what are the chances that some node u terminates in Line 8? Well, we can hope that all correct nodes randomly propose the same value (in Line 12). Maybe there are some nodes not choosing randomly (entering Line 10 instead of 12), but according to Lemma 11.22 Xxxxx 17.22 they will all propose the same. − Thus, at worst all n f correct nodes need to randomly choose the same bit, which happens with probability 2−(n−f)+1. If so, all correct nodes will send the same propose message, and the algorithm terminates. So the expected running time is exponential in the number of nodes n in the worst case.