Conflict serializability testing without precedence graph?

# Conflicting operations

Two operations I and J in a schedule are said to be conflict if they are operations which:

- are done by different transactions

AND

- are on the same data item

AND

- at least one of these instructions is write operation.

# Conflict equivalent

Two schedules are conflict equivalent if they can be transformed into each other by a sequence of swaps of non-conflicting, adjacent actions.

# Conflict serializable

A schedule S is conflict serializable if it is conflict equivalent to a serial schedule.

# Question

Check whether the schedule is conflict serializable.

read(t1,balx),read(t2,balx),write(t1,balx),write(t2,balx),commit(t1),commit(t2)

Here First t1 reads balx, then t2. Then t1 writes then t2.

This cannot be conflict equivalent to a serial schedule is what I believe. Because of w-w conflict probably.