COMP207 神秘错题合集
The procedure is as follows: You lock say X in transaction 1, Y in transaction 2 and Z in transaction 3. The fact that X is locked in transaction 1 does not mean that any of the other transactions can not move and similar for the other locks. Thus, any concurrent schedule satisfies this! Since there are concurrent schedules with none of these properties the right answer is "None of the above are correct".
仔细看中间……
我擦moive moive 居然可以……
If you read from something your timestamp must be higher than its write timestamp. Similar, if you write to something, your timestamp must be higher than the read or write timestamp (some require only restart on the read timestamp being higher, but it would not change anything for this question)