Conflict serializability in DBMS Theory - Step-by-Step Execution

Concept Flow - Conflict serializability

Start with schedule of transactions

↓

Identify conflicting operations

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Build precedence graph

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Check for cycles in graph

The flow shows how to check if a schedule is conflict serializable by identifying conflicts, building a graph, and checking for cycles.

Execution Sample

DBMS Theory

T1: R(A) W(A)
T2: R(A) W(A)
Schedule: R1(A) R2(A) W1(A) W2(A)

This schedule shows two transactions reading and writing the same data item A in an interleaved way.

Analysis Table

Step	Operation	Conflict With	Precedence Graph Edge	Cycle Detected?
1	R1(A)	None	None	No
2	R2(A)	None	None	No
3	W1(A)	R2(A)	T2 -> T1	No
4	W2(A)	W1(A)	T1 -> T2	Yes

💡 Cycle detected between T1 and T2, so schedule is NOT conflict serializable

State Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4
Precedence Graph Edges	{}	{}	{}	{"T2->T1"}	{"T2->T1", "T1->T2"}
Cycle Detected	No	No	No	No	Yes

Key Insights - 2 Insights

Why does writing after reading cause a conflict?

Because writing changes data that was read, the order matters. In the table, step 3's W1(A) conflicts with step 2's R2(A), creating an edge T2->T1.

What does a cycle in the precedence graph mean?

A cycle means transactions depend on each other in a loop, so the schedule cannot be rearranged into a serial order without conflicts. See step 4 where cycle is detected.

Visual Quiz - 3 Questions

Look at the execution table, which step first adds an edge to the precedence graph?

A Step 1

B Step 2

C Step 3

D Step 4

💡 Hint

Check the 'Precedence Graph Edge' column in the execution table rows.

At which step does the cycle appear in the graph?

A Step 2

B Step 4

C Step 3

D No cycle appears

💡 Hint

Look at the 'Cycle Detected?' column in the execution table.

If the last operation W2(A) was moved before W1(A), what would happen to the cycle?

A Cycle would disappear

B Cycle would remain

C New cycle would form

D Graph would have no edges

💡 Hint

Think about how changing operation order affects edges in the precedence graph.

Concept Snapshot

Conflict serializability checks if a schedule can be reordered into a serial one by:
- Identifying conflicting operations (read/write on same data by different transactions)
- Building a precedence graph with edges for conflicts
- Detecting cycles in the graph
No cycles mean the schedule is conflict serializable.

Full Transcript

Conflict serializability is a way to check if a schedule of database transactions can be rearranged without conflicts. We look at operations that conflict, like reads and writes on the same data by different transactions. We build a graph showing which transaction must come before another. If this graph has no cycles, the schedule is conflict serializable. If there is a cycle, it means transactions depend on each other in a loop, so the schedule cannot be safely reordered.

Practice

1. What does conflict serializability ensure in database transactions?

easy

A. Transactions produce the same result as if they were executed one after another

B. Transactions run faster by skipping some operations

C. Transactions always lock all data to avoid conflicts

D. Transactions are executed only by one user at a time

Solution

1. Step 1: Understand the goal of conflict serializability

   It aims to keep the database consistent by ensuring concurrent transactions behave like they ran sequentially.

2. Step 2: Compare options with the definition

   Only Transactions produce the same result as if they were executed one after another correctly states that the final result is as if transactions ran one after another.

3. Final Answer:

   Transactions produce the same result as if they were executed one after another -> Option A

4. Quick Check:

   Conflict serializability = Sequential result [OK]

Hint: Think: final result matches some order of transactions [OK]

Common Mistakes:

• Confusing conflict serializability with locking all data
• Thinking it speeds up transactions by skipping steps
• Assuming only one user can run transactions at a time

2. Which of the following is a correct way to check conflict serializability?

easy

A. Build a precedence graph and check for cycles

B. Count the number of transactions running simultaneously

C. Verify if all transactions use the same isolation level

D. Check if transactions use locks on all data items

Solution

1. Step 1: Recall the method to test conflict serializability

   It involves creating a precedence graph where nodes are transactions and edges show conflicts.

2. Step 2: Understand the graph's role

   If the graph has no cycles, the schedule is conflict serializable.

3. Final Answer:

   Build a precedence graph and check for cycles -> Option A

4. Quick Check:

   Precedence graph + no cycles = conflict serializable [OK]

Hint: Use graph cycles to detect conflicts fast [OK]

Common Mistakes:

• Thinking counting transactions shows serializability
• Confusing isolation levels with conflict serializability
• Assuming locking all data guarantees serializability

3. Consider two transactions:
T1: Read(A), Write(A)
T2: Read(A), Write(A)
Which schedule is conflict serializable?

medium

A. None of the above

B. Read(A) by T1, Read(A) by T2, Write(A) by T1, Write(A) by T2

C. Read(A) by T2, Read(A) by T1, Write(A) by T2, Write(A) by T1

D. T1 followed by T2

Solution

1. Step 1: Analyze schedules for conflicts

   Schedules where one transaction runs fully before the other have no conflicts and are serializable.

2. Step 2: Check interleaved schedule

   Interleaving reads and writes on the same data causes conflicts and cycles in the precedence graph.

3. Final Answer:

   T1 followed by T2 -> Option D

4. Quick Check:

   Non-interleaved = serializable [OK]

Hint: Serial schedules are always conflict serializable [OK]

Common Mistakes:

• Assuming interleaved schedules are always serializable
• Ignoring write-write conflicts on same data
• Thinking any order of reads is conflict free

4. Given a schedule with transactions T1 and T2:
T1: Read(X), Write(X)
T2: Read(X), Write(X)
The schedule is:
Read(X) by T1, Read(X) by T2, Write(X) by T2, Write(X) by T1.
What is the problem with this schedule?

medium

A. It is a serial schedule with no conflicts

B. It causes a deadlock between T1 and T2

C. It is not conflict serializable due to a cycle in the precedence graph

D. It is conflict serializable because reads happen before writes

Solution

1. Step 1: Build the precedence graph

   T1 reads X before T2 writes X, so edge T1 -> T2; T2 writes X before T1 writes X, so edge T2 -> T1.

2. Step 2: Detect cycle in graph

   Edges T1 -> T2 and T2 -> T1 form a cycle, meaning schedule is not conflict serializable.

3. Final Answer:

   It is not conflict serializable due to a cycle in the precedence graph -> Option C

4. Quick Check:

   Cycle in graph = not conflict serializable [OK]

Hint: Look for cycles in transaction dependency graph [OK]

Common Mistakes:

• Confusing deadlock with conflict serializability
• Assuming reads before writes always mean serializable
• Thinking schedule is serial when it is interleaved

5. You have three transactions:
T1: Read(A), Write(A)
T2: Read(B), Write(B)
T3: Read(A), Write(B)
If the schedule is:
Read(A) by T1, Read(B) by T2, Read(A) by T3, Write(A) by T1, Write(B) by T3, Write(B) by T2,
Is this schedule conflict serializable? If yes, what is a possible serial order?

hard

A. Yes, serial order: T2 -> T1 -> T3

B. No, schedule is not conflict serializable due to cycles

C. Yes, serial order: T1 -> T3 -> T2

D. Yes, serial order: T3 -> T1 -> T2

Solution

1. Step 1: Identify conflicts and build precedence graph

   T3 reads A before T1 writes A -> T3 -> T1.
   T2 reads B before T3 writes B -> T2 -> T3.
   T3 writes B before T2 writes B -> T3 -> T2.

2. Step 2: Check for cycles and find serial order

   Cycle T2 -> T3 -> T2 exists, so the schedule is not conflict serializable.

3. Final Answer:

   No, schedule is not conflict serializable due to cycles -> Option B

4. Quick Check:

   Precedence graph has cycle = not serializable [OK]

Hint: Build graph, check for cycles: cycle = not serializable [OK]

Common Mistakes:

• Missing edges like T2 read B before T3 write B
• Missing edges in the precedence graph
• Confusing read-write order causing edges