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40 Single Revision Pages With Short Tricks : GATE CSE 2019 (in Hindi)
3,207 plays

This course will cover top 40 single Revision Pages for GATE Exam.

Sweta Kumari is teaching live on Unacademy Plus

Sweta Kumari
YouTuber (channel : GATE NoteBook) | Verified Educator | 2+ year Online Teaching Experience

Unacademy user
nice way of teaching
thank you mam u r great...
i was actually looking for this course of you mam so that i can make my single revision page too...thanks a lot mam
Sweta Kumari
6 months ago
I have shared the link with you ... great good going 👍👍👍
Excellent mam. I loved this. Add more SRP for most topics
Sweta Kumari
5 months ago
40 Completed now
thank you maam... these srp's are really helpful.... can you please provide an srp for serialisability of schedules
Jyotsna G
7 months ago
sorry maam....
Sweta Kumari
7 months ago
It's ok dear....your welcome
mam how to download from plus App m not finding link...
Sweta Kumari
6 months ago
There is option to download notes
  1. WELCOME TO unacademy unacademy Explore v Login Signup Join India's largest learning platform Enroll in courses and watch lessons from India's best educators Search Courses Topies&Educator

  2. Sweta KumariVERIFIED ABOUT ME Verified Educator I Software Engineer ISelected in 2 IT MNC(Java) IResearch Paper (NCCCS) IQuora Writer 12 yr Teaching Exp. FOLLOW M 154 courses 28.2k Followers 234 Following GATE CSE 2019 FULL LENGTH TARGET GATE CSE 201 UNACADEMY PLUS 10 DAYS MOCK TEST CRASH COURSE By Sweta Kumari By Sweta Kumari G Crash Course With Important Topics & PYQs for GATE CSE 2019 (Hindi)10 Days Crash Course- GATE CSE 2019 (Hindi) Full Length Mock Test with Answers - GATE CSE 2019 6 Lessons 20 Lessons View Details

  3. SHORT TRICKS SPECIAL 40 SINGLE REVISION PAGES GATE CSE 2019 S1 S2 SRP 6 S3 1 . T1 T2 TI NO FOR NOTES DOWNLOAD USING UNACADEMY LEARNING APP RIP WIP) T3 2. Blind Write Present : W, (A) Conflictin operations: 1. Different Transactions 2. Same item 3. At least 1 write operation (T2, T3) T1 TI T2 T1NO T2 YES YES T3 CONFLICT SERIALIZABLE? SERIALIZABLE

  4. S1 S2 S3 T1 T2 T3 T2 R(P W(a R(P) w(a) R(A) NO R(A) R(P) W(P) R(a) W(Q) W(P W(A) T3 R(Q) w(a) W(A) R(a) W(Q R(a) w(Q) 2. Blind Write Present : W3 (A) 3. T1-> T3 (T2, T3) T1. T1 T2 T3 R(A) Conflicting Operations 1. Different Transactions 2. Same item 3. At least 1 write operation R(A) W(A) W(A) T1 T2 T1 T2 T1NO T2 YES YES VIEW SERIALIZABLE? T3 CONFLICT SERIALIZABLE?

  5. IRRECOVERABLE & RECOVERABLE Schedules T1 T2T3 Cascading R(A) W(A) Transaction Transaction T2 T1 R(A) W(A) R(A) W(A) Cascadless R(A) R(A) // Dirty Transaction T1 Transaction T2 TI T2 W(A) Read W(A) Commit Failure R(A) W(A) W(A) Commit R(A) // Dirty Read Rollback R(A) W(A) Commit W(A) Strict Commit T1 W(A) Commit/ T2 Commit // Delayed Rollback R(A)/W(A)

  6. S1 S2 VIEW EQUIVALENT? CONFLICT EQUIVALENT? T1 R(A) W(A) T1 R(A) W(A) R(B) W(B) T2 Read(P) Write(P) R(A) Write(P) S1 R(B) W(B) R(A) W(A) Read(P) Write(P) R1(A), W2(A) W1(A), R2(A) + W1(A), W2(A) T1 R(X) X-X+5 W(x) T1 R(X) X-X+5 W(x) T2 T2 Read(P Let X = 2 and Y = 5 Write(P) Read(P) S2 Results by Schedule S1-X = 21 and Y = 10 Results by Schedule S2- X 21 and Y-10 Write(P) R(X) Write(P) Y-Y+5 W(Y) W(x) 1. Initial Readers same 2. Write Read Sequence must be same 3. Final writers are same R(X) R(Y) RESULT EQUIVALENT? Y = Y+5 W(X) W(Y)

  7. BCNF: For each Functional Dependency XY X must be a super key. R (A,B,C,D) A - BCD BC AD DB (Not BCNF) Key A, BC (NORMALIZATION) FIRST NORMAL FORM: 1. No Composite values 2. Same kind & unique entries 3. No 2 rows are identical SECOND NORMAL FORM: No partial dependency BC ADEF 1. One C.K 3NF & BCNF 2. Transitive Dependency 3NF 3. No Transitive Dependency BCNF 4. Lossless decomposition-2NF, 3NF, BCNF 5. No Lossy Decomposition 2NF, 3NF, BCNF 6. Dependency Preserving 2NF, 3NF 7. Not Always Dependency Preserving BCNF 8. R with only trivial FD- BCNF 9. BCNF 0 redundancy 10. If All are Prime Attributes 2NF, 3NF 11. Binary Relation BCNF 12. In Relational DB, Every Relation is in 1NF TRICKS where , BC C.K THIRD NORMAL FORM: X... NPA C.K A , BC NPA D, E NPA A - BCDE BCADE D E (NPA PROBLEM)

  8. FIFO First in First Out LRU Far From Left Optimal Far From Right SRP (PAGE 0Belady's Anomaly REPLACEMENT) Only in FIFO sometimes FIH) F F F F

  9. Maximum number of nodes in an AVL Tree of height H = 2H+1-1 SRP10 (TREE) Minimum number of nodes in a binary tree of height H H+1 Minimum number of nodes in an AVL Tree of height H is given by a recursive relation- N(H) = N(H-1) + N(H-2) + 1 Maximum number of nodes in a N(1)-2 binary tree of height H 2#*1 -1 leaf nodes in a Binary Tree- Degree-2 nodes + 1 Minimum height of an AVL Tree using N nodes floor(log2N) Maximum number of nodes at any level 'L' in a binary tree 2' Maximum height of an AVL Tree using N nodes is calculated using recursive relation- N(H) = N(H-1 ) + N(H-2) + 1 If there are n nodes in an AVL Tree, maximum height can not exceed 1.44log2n. In other words, Worst case height of an AVL Tree with n nodes 1.44log2n.

  10. Violate Affeet Stre ss Emphasi Ride Despite InveshgeteResist Resemble ene fit Comprise. Com bet Ob Next monh Tomorrow Last last wes Lest mon Last Yeant Te u Desenbe DemandEnsure Disunes Diveh Invade Join Preeede Conside r Affoy d Esuhen Suece ed Acuompn Ordet V e Hat To d Next Next VeaY Next uk Conhol Si Reah Board Petede Hack Enter POLICE AGENT CRIMINAL EPOSITION S.

  11. 5 GATE QUESTIONS ON TREE Here n-4, so, answer is 14. The number of structurally different possible binary trees with 4 nodes is? A complete n-ary tree is a tree in which each node has n children or no children. Let I be the number of internal nodes and L be the number of leaves in a complete n-ary tree. If L 41, and I 10, what is the value of n? L = (n-1)"I + 1 L-41,1 10 41 10* (n-1)+ 1 (n-1) 4 n 5 Degree 2 * e 18 Let T be a tree with 10 vertices. The sum of the degrees of all the vertices in T is Number of Leaf Nodes Number of Internal nodes with 2 children 1 Number of Leaf Nodes 10 1 11 In a binary tree, the number of internal nodes of degree 1 is 5, and the number of internal nodes of degree 2 is 10. The number of leaf nodes in the binary tree is Inorder traversal of a BST always gives elements in increasing order.

  12. S.NO PROBLEM 1.Is w E L? Membership Problem 2. | Is L- ? Emptiness Problem 3. Is L Finite or not? Finiteness Problem 4. Is L1-L2? Equivalence Problem 5. | ls L1 ^ L2-d ? intersection Empty Problem REGUL | DCFL | CFL | CSL | RECU | REL UD UD | UD | UD UD | UD | UD AR RSIVE UD UD 6. Is *? Totality Completeness Problem UD | UD | UD | UD UD UD 7. Is L1 C L2? Subset roblem 8. | L1 ^ L2-Finite ? Intersection finitenes Problem | D 9 Is (*-L) Finite? Co-Finiteness Problem 10. Is L Regular? Regularity Problem 11. Ambiguity Problem 12. Is complement of a language is of same type or DD UD | UD | UD | UD UD | UD | UD | UD UD D D UD not?

  13. Addressing Modes Applications M/M Operand Address Field Replaced by Immediate Addressing Mode To initialize registers to a constant value data | Direct (Add Data) Direct Addressing Mode and Register Direct Addressing Mode Only 1 NO RD (Register Data) Indirect (Add1 Add2 RI (Register Add Data) To access static data , To implement variables Indirect Addressing Mode and Register Indirect Addressing Mode Only 2 nly 2 Data) To implement pointers, To pass array as a parameter Only 1 For program relocation at run time To change the normal sequence of execution of Relative Addressing Mode instructions EA = PC + Address field For branch type instructions since it directly updates the program counter For array implementation or array addressing For records implementation For writing relocatable code i.e. for relocation of Only 1 Address field EA Index Register + Index Addressing Mode EA Base Register + Address field Base Register Addressing Mode program in memory even at run time For handling recursive procedures Auto-increment Addressing Mode For implementing loops Only 1 Increment After Execution and For stepping through arrays in a loop Only 1 Decrement before execution

  14. 1. For Strong Entity Set With Only Simple Attributes- 2. For Strong Entity Set With Composite Attributes- 3. For Strong Entity Set With Multi Valued Attributes- 4. Translating Relationship Set into a Table- 5. Binary relationship with cardinality ratio m:n 6. Binary relationship with cardinality ratio 1:n 7. For Binary Relationship With Cardinality Ratio m:1 8. For Binary Relationship With Cardinality Ratio 1:1 9. For Binary Relationship With Cardinality Constraint and Total Participation Constraint From One Side 10. For Binary Relationship With Cardinality Constraint and Total Participation Constraint From Both Sides- 1 11. For Binary Relationship With Weak Entity Set- TABLES

  15. TRUNCATE DELETE DML Can use WHERE clause Cannot use WHERE DDL Clustered Index Non- Clustered Only 1 Stores data & Information Reading is faster Index More than 1 (250) Stores only Delete a row from table Slower Can rollback data after Cannot rollback data using DELETE Statement clause Delete all row from table Faster information Slower after using DELETE Statement DELETE DROP DELETE remove some or the tuples from a table DML DROP can remove entire schema from the database Having Clause Where Clause Can be used with Cannot be used aggregates After GROUP BY Before GROUP DDL WHERE clause can be used No clause is used with aggregates Can be roll-backed can not be roll- backed Even if you delete all the tuples of the table using DELETE, space occupied by the table in the memory is not freed Table deleted using DROP frees the table space from memory. BY Post filter Pre filter