Prove that the basic two-phase locking protocol guarantees conflict serializability of schedules. (Hint: Show that, if a serializability graph for a schedule has a cycle, then at least one of the transactions participating in the schedule does not obey the two-phase locking protocol.)

Question

What will be an ideal response?

Answer 1

(This proof is by contradiction, and assumes binary locks for simplicity. A similar
proof can be made for shared/exclusive locks.)
Suppose we have n transactions T1, T2, ..., Tn such that they all obey the basic
two-phase locking rule (i.e. no transaction has an unlock operation followed by a lock
operation). Suppose that a non-(conflict)-serializable schedule S for T1, T2, ..., Tn
does occur; then, according to Section 17.5.2, the precedence (serialization) graph for
S must have a cycle. Hence, there must be some sequence within the schedule of the form:
S: ...; [o1(X); ...; o2(X);] ...; [ o2(Y); ...; o3(Y);] ... ; [on(Z); ...; o1(Z);]...
where each pair of operations between square brackets [o,o] are conflicting (either [w,
w], or [w, r], or [r,w]) in order to create an arc in the serialization graph. This
implies that in transaction T1, a sequence of the following form occurs:
T1: ...; o1(X); ... ; o1(Z); ...
Furthermore, T1 has to unlock item X (so T2 can lock it before applying o2(X) to follow
the rules of locking) and has to lock item Z (before applying o1(Z), but this must occur
after Tn has unlocked it). Hence, a sequence in T1 of the following form occurs:
T1: ...; o1(X); ...; unlock(X); ... ; lock(Z); ...; o1(Z); ...
This implies that T1 does not obey the two-phase locking protocol (since lock(Z) follows
unlock(X)), contradicting our assumption that all transactions in S follow the two-phase
locking protocol.