Ranita works as an Assistant Professor at the Computer Science and Engineering Department at IIT Roorkee. She has completed her PhD from the Computer Science and Engineering Department at IIT Kharagpur. Before that, she has done her MTech from IIEST Shibpur, and got job offers from companies like Amazon, Samsung Labs etc. However, she preferred to pursue higher studies to stay in the field of academics and teaching. She believes in the motto of free education and contribution.


Alma Mater:

IIT Kharagpur
2012 to 2016
Master of Engineering
Indian Institute of Engineering Science and Technology, Shibpur
2010 to 2012
Bachelor of Technology
Kalyani Government Engineering College
2005 to 2009


Assistant Professor
Indian Institute of Technology Roorkee
Project Linked Personnel
Indian Statistical Institute, Kolkata
2009 to 2010
7 months 1 week ago

The correct answer is (b) remains same. To understand this, let us revise the informal algorithm for Timestamp based protocol for concurrency control (copied from Wikipedia, you can also refer to the formal version of the algorithm there):

1) If a transaction wants to read an object,

a) but the transaction started before the object's write timestamp it means that something changed the object's data after the transaction started. In this case, the transaction is canceled and must be restarted.
b) and the transaction started after the object's write timestamp, it means that it is safe to read the object. In this case, if the transaction timestamp is after the object's read timestamp, the read timestamp is set to the transaction timestamp.
2) If a transaction wants to write to an object,

a) but the transaction started before the object's read timestamp it means that something has had a look at the object, and we assume it took a copy of the object's data. So we can't write to the object as that would make any copied data invalid, so the transaction is aborted and must be restarted.
b) and the transaction started before the object's write timestamp it means that something has changed the object since we started our transaction. In this case, we use the Thomas write rule and simply skip our write operation and continue as normal; the transaction does not have to be aborted or restarted
c) otherwise, the transaction writes to the object, and the object's write timestamp is set to the transaction's timestamp.

Now consider the following example schedule:

T1      T2      T3      T4

Consider that after the execution of the last R(A), T4 needs to rollback (as given in the question). So, RTS(A) was 4 and now we need to decide what should be the new RTS for A. Note that, the schedule at hand now is something like the following:

T1      T2      T3

Independent of T4, this schedule will have RTS(A) = 3 when T2 tries to write and hence T2 should be aborted by rule 2(a).

Now, after rollback of T4, consider option (a) i.e. changing the RTS(A) to 0. This will certainly allow W(A) by T2 to execute without any problem as it falls under rule 2(c) and hence violates concurrency by allowing T2 to write a data already read by T3.

Consider option (c) i.e. RTS(A) becomes equal to the timestamp of the transaction which read 'A' just before T4, in this example then RTS(A) becomes 1. And hence the same problem arises as choosing option (a).

Therefore, if we choose option (b) and keep the RTS(A) same as before i.e. 4, then W(A) by T2 will fall under rule 2(a) and hence will be aborted. So, option (b) is the correct answer.

This is why RTS is kept as the largest of the timestamps of the transactions which has read the data, and not the recent timestamp.

Example text
8 months 3 weeks ago

This is an example of fan trap and chasm trap and how to resolve these scenarios. The following brief explanation may help you in understanding the concept which is usually found to be dubious across the web. This explanation is made in parity with the concept explained in Wikipedia (https://en.wikipedia.org/wiki/Entity–relationship_model#Model_usability_issues). You can also refer to https://db.grussell.org/section005.html to understand in detail.

Example text
10 months 1 week ago

You are very close to the answer. But, there is one catch; the first row of pixels (at y = 0) and the last row of pixels (at y = 37) are not full length. You may try to modify the approach accordingly to get to the correct answer. Very good presentation, by the way.

Example text
10 months 2 weeks ago

Consider the attribute set ABCDEG and the FD set
AB → C, AC → B, AD → E, B → D, BC → A, E → G
Is the following decomposition of R(ABCDEG)
(a) dependency-preserving?
(b) lossless-join?
Give proper justification for your answer.

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@raviraaja By recurrence relation, do you mean the function for the output values? Those are just decremented by 3 at each step.

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15 Jan 2016 - 1:25pm

This code will give: "lvalue required as increment operand" Compilation ERROR
The result of pre-increment operator is just a value, it does not point to any memory location; therefore, the post-increment operator cannot be used over that.

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Yes, if you hard code the values in the for loop, then yes. Putting any non-zero constant value in the second statement i.e. condition checking part, will make it work like an infinite loop.

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I have named the different calls to function fun() for ease of understanding.
First point to note that variable num inside function is static, so its value will persist between function calls.
Next, "return num--" uses post-decrement operator, therefore first time return value will be 16, not 15.
Now, The sequence of function calls from main function in the loop will be as follows:
A, B, D, C, B, D, C, B, D, C.... until function call in B gets 0 as return value.
Now as you trace the calls you get the function call statement and its associated return values as follows:
A(16), B(15), D(14-printed), C(13), B(12), D(11-printed), C(10), B(9), D(8-printed), C(7), B(6), D(5-printed), C(4), B(3), D(2-printed), C(1), B(0-loop terminates).
Therefore, the answer is Option (C) 14 11 8 5 2
Please let me know if this is clear to you now.

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10 Nov 2015 - 12:54pm

Yes, possibly you are right. We will correct it as necessary.

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Yes, unless the system is preemptive, number of context switches will be 4.

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Optimal partitioning based on Time.
Go through this: https://en.wikipedia.org/wiki/Quicksort#Choice_of_pivot
The constant multiplication factor also matters when you are dealing with large value of n. Choosing median ensures that you cannot end up with worst case complexity which is O(n2).

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I said exactly the same, linear time means O(n). Sorting takes O(n log n) time, never less than that. May I know which question you are exactly referring too? You have also said "space" complexity, which is completely different from "time" complexity.

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