--- title: "View serializability check of schedule" subject: "Database Management System" chapter: 7 key: "semester_6/database_management_system:view-serializable-schedule-q" source: https://www.bctengineer.com/q/semester_6/database_management_system/view-serializable-schedule-q site: BCT Engineering --- # View serializability check of schedule Database Management System · Chapter 7: Transaction Processing - Asked in: year not given - Video: [View Serializability: A Tabular Form Example - Database](https://www.youtube.com/watch?v=Vb3G17vaug8) — Ritu Kapur Classes - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/database_management_system/view-serializable-schedule-q - Whole chapter: https://www.bctengineer.com/md/semester_6/database_management_system/chapter-7 ## Question Consider the transactions T1: r1(Q), w1(Q); T2: w2(Q); T3: w3(Q) and the schedule S: r1(Q), w2(Q), w1(Q), w3(Q). Check whether S is view serializable (compare with the serial schedule T1, T2, T3). ## Short answer S is view serializable, equivalent to serial T1,T2,T3. T1 reads initial Q, no other reads, final write by T3. Blind writes allow view equivalence despite conflict cycle. ## Answer Two schedules are view equivalent if (1) each transaction reads the initial value of an item in both or in neither, (2) each read reads the value written by the same transaction in both, and (3) the final write of each item is done by the same transaction in both. A schedule is view serializable if it is view equivalent to some serial schedule. T1: r1(Q), w1(Q); T2: w2(Q); T3: w3(Q) S: r1(Q), w2(Q), w1(Q), w3(Q) #### Step 1: conflict serializability (for comparison) Conflicts: r1(Q) before w2(Q) gives T1 → T2; w2(Q) before w1(Q) gives T2 → T1. This is a cycle, so S is **not conflict serializable**. We must test view serializability directly. #### Step 2: view properties of S - Initial read: T1 reads the initial value of Q in r1(Q). - Reads-from: there is only one read, and it reads the initial value, so there is no "reads from" relation. - Final write: w3(Q) is the last write on Q, so T3 does the final write. #### Step 3: compare with the serial schedule T1, T2, T3 Serial: r1(Q), w1(Q), w2(Q), w3(Q) - Initial read: T1 reads the initial Q. Same. - Reads-from: no other reads. Same. - Final write: T3. Same. All three conditions are met. **Conclusion: S is view serializable**, view equivalent to the serial schedule , even though it is not conflict serializable. This is possible because T2 and T3 perform blind writes (writes without reading). The final write w3(Q) overwrites w2(Q) and w1(Q), and no transaction reads those values, so their order does not matter. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is S = r1(Q), w2(Q), w1(Q), w3(Q) not conflict serializable? A. r1(Q) before w2(Q) gives T1 -> T2 and w2(Q) before w1(Q) gives T2 -> T1, a cycle B. T3 writes last C. There are only writes D. T1 reads Q Answer: A. The two opposite edges between T1 and T2 form a cycle in the precedence graph. (Key point: Not conflict serializable) 2. Which three conditions define view equivalence? A. Same initial reads, same reads-from relations, same final writes B. Same number of operations, same order, same locks C. Same commits, same aborts, same timestamps D. Same reads only Answer: A. View equivalence needs matching initial reads, matching reads-from, and matching final writes for each item. (Key point: View equivalence conditions) 3. Why is S view equivalent to the serial schedule T1, T2, T3? A. T1 reads the initial Q and T3 does the final write in both B. T2 writes last in both C. T1 reads from T2 in both D. Both have the same cycle Answer: A. In both, the only read is by T1 on the initial value and the final write is w3(Q). (Key point: Comparison with serial schedule) ## Related in Database Management System - [Conflict serializability precedence graph](https://www.bctengineer.com/question.md?key=semester_6%2Fdatabase_management_system%3Aprecedence-graph-three-transactions) — Database Management System, chapter 7 ## Related in other subjects - [p.u. impedance diagram: G, T1, line, T2, M](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Apu-impedance-diagram-20mva) — Power System Analysis I, chapter 3 - [Reactance diagram: G, T1, j18 Ω line, T2, M](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-g-t1-t2-m) — Power System Analysis I, chapter 3 - [Corrected bearings, closed link traverse T1-T2](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aclosed-link-traverse-t1-t2-corrected-bearings) — Surveying I, chapter 4 - [FP-tree on t1-t5 data](https://www.bctengineer.com/question.md?key=semester_7%2Fdata_mining%3Afp-tree-t1-t5) — Data Mining, chapter 4