private static Evidence checkShuffleMisbehavior(
Map<Integer, VerificationKey> players,
Map<VerificationKey, DecryptionKey> decryptionKeys,
Map<VerificationKey, Packet> shuffleMessages,
Map<VerificationKey, Packet> broadcastMessages)
throws FormatException {
if (players == null
|| decryptionKeys == null
|| shuffleMessages == null
|| broadcastMessages == null) throw new NullPointerException();
SortedSet<String> outputs = new TreeSet<>();
// Go through the steps of shuffling messages.
for (int i = 1; i < players.size(); i++) {
Packet packet = shuffleMessages.get(players.get(i + 1));
if (packet == null) {
// TODO Blame a player for lying.
return null;
}
Message message = packet.payload();
// Grab the correct number of addresses and decrypt them.
SortedSet<String> decrypted = new TreeSet<>();
for (int j = 0; j < i; j++) {
if (message.isEmpty()) {
return Evidence.ShuffleMisbehaviorDropAddress(
players.get(i), decryptionKeys, shuffleMessages, broadcastMessages);
}
String address = message.readString();
message = message.rest();
for (int k = i + 1; k <= players.size(); k++) {
address = decryptionKeys.get(players.get(k)).decrypt(address);
}
// There shouldn't be duplicates.
if (decrypted.contains(address)) {
return Evidence.ShuffleMisbehaviorDropAddress(
players.get(i), decryptionKeys, shuffleMessages, broadcastMessages);
}
decrypted.add(address);
}
// Does this contain all the previous addresses?
if (!decrypted.containsAll(outputs)) {
return Evidence.ShuffleMisbehaviorDropAddress(
players.get(i), decryptionKeys, shuffleMessages, broadcastMessages);
}
decrypted.removeAll(outputs);
// There should be one new address.
if (decrypted.size() != 1) {
return Evidence.ShuffleMisbehaviorDropAddress(
players.get(i), decryptionKeys, shuffleMessages, broadcastMessages);
}
outputs.add(decrypted.first());
}
// Now check the last set of messages from player N.
// All broadcast messages should have the same content and we should
// have already checked for this. Therefore we just look for the first
// one that is available.
// (The message for player 1 should always be available, so theoretically
// we don't need to loop through everybody, but who knows what might have happened.)
Packet packet = null;
for (int j = 1; j <= players.size(); j++) {
packet = broadcastMessages.get(players.get(j));
if (packet != null) {
break;
}
}
if (packet == null) {
// TODO blame someone.
return null;
}
Message message = packet.payload();
// Grab the correct number of addresses and decrypt them.
SortedSet<String> addresses = new TreeSet<>();
for (int j = 0; j < players.size(); j++) {
if (message.isEmpty()) {
return Evidence.ShuffleMisbehaviorDropAddress(
players.get(players.size()), decryptionKeys, shuffleMessages, broadcastMessages);
}
Address address = message.readAddress();
// There shouldn't be duplicates.
if (addresses.contains(address.toString())) {
return Evidence.ShuffleMisbehaviorDropAddress(
players.get(players.size()), decryptionKeys, shuffleMessages, broadcastMessages);
}
addresses.add(address.toString());
}
// Does this contain all the previous addresses?
if (!addresses.containsAll(outputs)) {
return Evidence.ShuffleMisbehaviorDropAddress(
players.get(players.size()), decryptionKeys, shuffleMessages, broadcastMessages);
}
addresses.removeAll(outputs);
// There should be one new address.
if (addresses.size() != 1) {
return Evidence.ShuffleMisbehaviorDropAddress(
players.get(players.size()), decryptionKeys, shuffleMessages, broadcastMessages);
}
return null;
}
