Race condition Example
Duration: 4 min
This video lesson is available to enrolled students.
Inside: a video lesson and guided study material.
Module outline
- Quantitative Ability: Percentage, Profit and Loss, Ratio and Proportion (Ratios), Average, Time and Work, Speed Time and Distance, Simple Interest and Compound Interest, Age Problems, Number System, Divisibility Rules, HCF LCM, Algebra, Permutation and Combination, Probability, Data Interpretation, Logarithm, Sequence and Series (Series), Powers and Exponents (Surds and Indices), Speed Mathematics and Basic Mathematics
- Logical Reasoning: Seating Arrangements, Puzzle Solving (Puzzles), Data Arrangement, Coding Decoding, Blood Relations, Directions (Direction Test), Ranking, Order and Sequence, Data Sufficiency, Venn Diagram, Inequality, Series (Number and Letter Series) (Numerical Relations and Reasoning), Logical Sequence Of Words, Deductive and Inductive Reasoning (Logical Deduction and Induction) (Prepositional Reasoning), Selection Decision Table (Decision Making), Analogy, Non Verbal Reasoning (Spatial Aptitude) (Spatial Reasoning) (Visual Reasoning)
- Verbal Ability: Comprehension / Reading Comprehension / Unseen Passages (Critical Reasoning) (Paragraph Questions), Subject Verb Agreement (Verb Noun Agreement), Tenses, Articles, Preposition, Conjunction, Modals, Active Passive Voice (Voice), Direct and Indirect Speech, Vocabulary, Sentence Re-arrangements (Para Jumbles) (Narrative Sequencing), Sentence Completion (Fill in the blanks), Sentence Correction (Error Correction), Cloze Test (Cloze Passage), Facts, Inference and Judgements
- Programming Fundamentals (Technical): C Fundamentals
- Data Structures (Technical): Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs, Hashing
- Algorithms (Technical): Algorithm Analysis, Sorting Algorithms, Greedy Algorithms, Dynamic Programming, Minimum Spanning Trees, Shortest Path Algos, Time Complexity Analysis, Advanced Algorithms
- OOPs (Technical): Object Oriented Programming (OOPS)
- DBMS (Technical): Basics of DBMS, ER Diagram, Relational Model & Functional Dependencies, Keys & Integrity Constraints, Normalization (1NF - BCNF), Decomposition Properties & 4NF, File Organization & Indexing, Relational Algebra, Relational Calculus, Transaction Management, Concurrency Control, Database Recovery, ORDBMS, Database Security & Authorization, Query Processing & Optimization, Enhanced Data Models, Data Warehousing & Mining, Big Data Systems, NoSQL
- SQL (Technical): SQL
- Operating Systems (Technical): Introduction to OS, Process Management, CPU Scheduling, Process Synchronization, Deadlock, Threads & Process Creation, Memory Management, Virtual Memory, Disc Scheduling, File Management, Windows OS, Linux OS, Security, Distributed Systems, Virtual Machines
- Computer Networks (Technical): Introduction to CN, Data Communication, DLL: Access Control, DLL: Flow Control, DLL: Error Control, DLL: Framing, Data Link Layer - Ethernet, Net Layer: IPv4 & Proto, Net Layer: IP Addressing, Net Layer:Routing Protocol, Transport Layer Services, TL: Congestion & UDP, Application Layer, Hardware basics, Network Security, Mobile Technology, Cloud Computing and IoT, Cyber and Info Security
- Software Engineering (Technical): Fundamentals of Software Engineering, Software Requirements and Quality Assurance, Software Design, Estimation and Metrics, Software Testing, Software Maintenance and Configuration Management
- Coding/Automata Round: Coding Questions by Yash Sir, C Programming by Yash Sir, Programming in C++, Java, Python for Placements by Yash Sir, Pseudo Codes
- Personality / Behavioral Assessment: Behavioral Assessment (Behavioral Competency Profiling) (Psychometric Test)
- Writing Assessment: English Writing Ability
AI summary & chapters
AI Summary
An AI-generated summary of this video lecture.
This video solves a GATE-2015 operating-systems concurrency question. Two functions, P1 and P2, share a variable B with initial value 2. P1 executes C = B - 1; then B = 2 * C;, while P2 executes D = 2 * B; then B = D - 1;. The question asks for the number of distinct values that B can possibly take after execution. The instructor labels the four instructions I11, I12 for P1 and I21, I22 for P2. Because each function's two instructions must preserve their internal order, the possible interleavings are the six permutations of choosing which P1 instruction and which P2 instruction occur at each step. A table with Case 1 through Case 6 lists the six interleavings of I11, I12, I21, and I22. The instructor evaluates each case by tracking the shared variable B (and local variables C and D) through the sequence. For example, one interleaving gives I11: C = 2 - 1 = 1; then I21: D = 2 * 2 = 4; then I12: B = 2 * 1 = 2; then I22: B = 4 - 1 = 3. Other interleavings yield final values such as 2 and 3 (and in some orderings, other intermediate results). By circling the final B values across all six cases and removing duplicates, the instructor concludes that B can take 3 distinct final values. The answer '3' is circled at the bottom of the question slide.
Chapters
0:00 – 2:00 00:00-02:00
The slide presents the GATE-2015 question in an orange-and-white two-column table labeled P1() and P2(). The top line states that the two functions share a variable B with an initial value of 2 and execute concurrently. P1 contains 'C = B - 1;' then 'B = 2 * C;'; P2 contains 'D = 2 * B;' then 'B = D - 1;'. The red bottom line asks for the number of distinct values B can possibly take after execution. Handwritten blue annotations progressively label the P1 steps as I1, J1 and the P2 steps as I2, J2 (later refined to I11, I12 and I21, I22), establishing the instruction labels used for interleaving analysis.
2:00 – 3:58 02:00-03:58
The slide now shows two code blocks headed P1() and P2() with labeled instructions (I11) C = B - 1;, (I12) B = 2 * C; and (I21) D = 2 * B;, (I22) B = D - 1;. Below, a green-headed table lists six columns Case1 through Case6, each containing a vertical sequence of the four instruction labels in different orders. Blue vertical line marks highlight selected rows in Case1-Case4 as the instructor computes each interleaving. On-screen values such as C=1, D=4, B=2 appear during the step-by-step evaluation. The instructor circles key intermediate values (e.g., B=3) and crosses out redundant steps, then returns to the original question slide where the final answer '3' is circled at the bottom, indicating three distinct possible values for B.
The core teaching point is systematic enumeration of legal interleavings in a concurrent execution where each process's internal instruction order must be preserved. With two instructions per process, there are C(4,2) = 6 valid interleavings, which the instructor tabulates as Case1-Case6. The method is: (1) label each instruction, (2) list all order-preserving interleavings, (3) simulate the shared variable B through each sequence while tracking local variables C and D, (4) collect the final B values and count distinct ones. The worked example shows how one interleaving produces C=1, D=4, and a final B=3, while others produce different finals. The conclusion is that the number of distinct values B can take is 3, which is circled as the answer. This case-by-case table method is the key exam technique for GATE concurrency questions of this type.