GATE Insights Version: CSE
GATE Insights Version: CSE
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GATE CSE 2015 SET 1
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MCQ (Single Correct Answer)
GATE CSE 2015 SET 1
A variable x is said to be live at a statement Si in a program if the following three conditions hold simultaneously:
There exists a statement Sj that uses x
2. There is a path from Si to Sj in the flow graph corresponding to the program
3. The path has no intervening assignment to x including at Si and Sj
The variables which are live both at the statement in basic block 2 and at the statement in basic block 3 of the above control flow graph are(A) p, s, u(B) r, s, u(C) r, u(D) q, v
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Відео

GATE CSE 2015 SET 1
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GATE CSE 2006
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GATE CSE 2005
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GATE CSE 2003
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GATE CSE 2020
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GATE CSE 2015 SET 2
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GATE CSE 2014 SET 1
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GATE CSE 2010
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GATE CSE 2008
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GATE CSE 2004
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GATE CSE 2004
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GATE CSE 2001
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GATE CSE 1995
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GATE CSE 1990
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GATE CSE 2017 SET 1

КОМЕНТАРІ

  • @rakeshsahu27
    @rakeshsahu27 8 годин тому

  • @FAIZANKHAN-jo9bb
    @FAIZANKHAN-jo9bb День тому

    Thank you bhai great explanation, but please use hindi language as well during the time of question explanation and answer explanatiom. Because people from belongs to north india he/she more comfortable with hindi language so please use hindi language as well. Thank you for making such grear video.

  • @FAIZANKHAN-jo9bb
    @FAIZANKHAN-jo9bb 2 дні тому

    Wrong answer dear 64, correct answer is 35

  • @prabhakargupta6596
    @prabhakargupta6596 3 дні тому

    Nonsence man

  • @PankajKumar-zx8kb
    @PankajKumar-zx8kb 4 дні тому

    Muh se gutka nikal ke bola kr bhai tu

  • @swatidubey9133
    @swatidubey9133 8 днів тому

    Good one

  • @gopiazad677
    @gopiazad677 10 днів тому

    Tatti explanation......

  • @ChetanMahale-d5x
    @ChetanMahale-d5x 11 днів тому

    2:54 projection eliminates duplicacy so it will be 1 only once

  • @shivam.sinha810
    @shivam.sinha810 11 днів тому

    Thank you

  • @FlyBee07
    @FlyBee07 12 днів тому

    Last option is wrong, if we consider tree where d is added after c, so when we visited vertex a-c-d and want to visit b from d Parents of d and b are different But They both r in leaf node So option c true

  • @Prerana-w8l
    @Prerana-w8l 13 днів тому

    Thank You so much sir

  • @FlyBee07
    @FlyBee07 14 днів тому

    Good job❤

  • @Rits2203
    @Rits2203 14 днів тому

    Your explanations are very clear. Thank you!

  • @bhaktithakur2065
    @bhaktithakur2065 19 днів тому

    Correct ans. Is D

  • @preambagga2527
    @preambagga2527 19 днів тому

    Very nicely explained !!

  • @GoldyGenda
    @GoldyGenda 19 днів тому

    totally bullshit.correct ans is 7!/4!*3!=35

  • @RishmithaChowdaryChimbili
    @RishmithaChowdaryChimbili 19 днів тому

    Super 👍

  • @papai4316
    @papai4316 22 дні тому

    How Calculate extarnal fragmentation

  • @FlyBee07
    @FlyBee07 22 дні тому

    Woow thank you for explaining this question, so smoothly ❤

  • @rajeevsankhla5028
    @rajeevsankhla5028 23 дні тому

    You're wrong..... right answer is B

  • @GatePrep-dl5kp
    @GatePrep-dl5kp 24 дні тому

    Wrong explanation

  • @anikdas517
    @anikdas517 24 дні тому

    poor explanation

  • @Rock28099
    @Rock28099 26 днів тому

    Thanks bro, I was solving PYQ and I marked rear node but in answer key of the book it was C. I started looking for answers. I found yours thanks!

  • @simranmansuri6778
    @simranmansuri6778 26 днів тому

    How 7 can lost? Why 7 can lost

  • @TankutTalu
    @TankutTalu 28 днів тому

    nice explanation

  • @akhileshyadav9862
    @akhileshyadav9862 Місяць тому

    Bro can you let me know minimum number of tables for c,d

  • @viki8815
    @viki8815 Місяць тому

    Totally wrong solution? How can you run P1 and P2 more than 1 times without while loop

  • @surenderreddy6294
    @surenderreddy6294 Місяць тому

    wow descent explanation

  • @anupamdas8067
    @anupamdas8067 Місяць тому

    kuch bhi

  • @manishachaurasia3405
    @manishachaurasia3405 Місяць тому

    Tum Gadhe ho Kya ,km se km gyaan dene se Phle seekh to lo. Kya time pass karaate ho ok-ok Bol ke.

  • @factresearchcentre8040
    @factresearchcentre8040 Місяць тому

    Good explanation sir ji❤

  • @VishalSharma-fi2mp
    @VishalSharma-fi2mp Місяць тому

    but how 😂

  • @itz_me_imraan02
    @itz_me_imraan02 Місяць тому

    Its wrong....plz dont provide wrong solutions

  • @theweakmaybegoodbuttheys-oh3zo
    @theweakmaybegoodbuttheys-oh3zo Місяць тому

    are u sure this is right ?

  • @amaanmohd8228
    @amaanmohd8228 Місяць тому

    Nice

  • @narainkarthii
    @narainkarthii Місяць тому

    Sir they asking Many to one supporting relation for the weak entity here you given weak entity as answer but still suporting then dimond with double border

  • @offaryaan
    @offaryaan Місяць тому

    N=1 was given not S

  • @rbasantkumarpradhan3568
    @rbasantkumarpradhan3568 Місяць тому

  • @harshilrathod7387
    @harshilrathod7387 Місяць тому

    Kindly explain clearly, nowadays msq are asked then how are you suppose to eliminate

  • @twinklesonkar3465
    @twinklesonkar3465 2 місяці тому

    Very helpful 😃

  • @aceyourdreams7529
    @aceyourdreams7529 2 місяці тому

    Thanks ❤

  • @gulshanmht
    @gulshanmht 2 місяці тому

    The option is correct, but the solution is wrong. If you are telling that option 'A' is wrong, then two processes can never enter critical section at the same time. For Option A) Theoretically this may be true, but practically it can't be true. [Explanation] Let's suppose the size of unsigned integer on your machine is 4 Byte(32 bit) then the max integer value it can store is ((2^32) - 1). If first process checks while condition it will store the value 0+1 = 1 to L because of "Fetch_And_Add" then enters Critical Section without executing "while" body. Now if every next process, only enters to "while" body by executing condition and then preempts just before executing its body, then the L will get values like 1, 2, 3, 4, 5, 6, and when it reaches (2^32) - 1 then next process (2^32th process) will try to store value 2^32 which is 1 greater than your machine can store in L. So, L can overflow and because of this L will get the value 0 by 2^32th process, that means it can also enter critical section. Now 1st process and 2^32th process is inside critical section at the same time (2 process in cs). As we know that having 2^32 processes is not practical. So, this option can't be true. For Option B) True [Explanation] Let's suppose if: 1) Process P1 enters C.S and preempts. Now the value of L is 1. 2) Process P2 enters while loop but preempts before executing its body. Now the value of L is 2 3) Process P1 leaves C.S. Now the value of L is 0. -> Preempts and Scheduled P2 4) Process P2 executes its next line that is L = 1 and then preempts, Now the value of L is 1 => At this point no process is inside the Critical Section and the value of L is 1. So, no process can enter Critical Section now. Every process will get stuck in while loop. It means deadlock. As we can see critical section is free(lock is available) but no one can enter critical section because L is non-zero(Showing Lock is not available). Option C and D are clearly false.

  • @satyam9408
    @satyam9408 2 місяці тому

    1 to the power any number 1 only, so option B is not correct, option A is correct.

  • @RanjeetKumar-kh9pj
    @RanjeetKumar-kh9pj 2 місяці тому

    Nice explanation,, thanks 🙏

  • @varshav.h3638
    @varshav.h3638 2 місяці тому

    If option has like course ids in which all female students are not enrolled?? Will it be correct.

  • @UJJWALKUMAR-dg9mw
    @UJJWALKUMAR-dg9mw 2 місяці тому

    Consider the following sequence of instructions, which is to be a 5 stage pipeline: IF, ID, OF, EX and WB. Each stage takes one cycle instruction. Total cycles require to execute this sequence ? ADD R2, R1, R0 MUL R4, R3, R2 SUB R6, R5, R4 ADD R6, R7, R8 MUL R7, R1, R2 SUB R1, R3, R4 Sir can you make a table for this one to get clear idea of total cycle needed ?

  • @harshilrathod7387
    @harshilrathod7387 2 місяці тому

    Kindly explain clearly

  • @nagendrabommireddi8437
    @nagendrabommireddi8437 2 місяці тому

    thanks for this question

  • @nagendrabommireddi8437
    @nagendrabommireddi8437 2 місяці тому

    A depth-first search is performed on a directed acyclic graph. Let d[u] denote the time at which vertex u is visited for the first time and f[u] the time at which the dfs call to the vertex u terminates. Which of the following statements is always true for all edges (u, v) in the graph ? (A) d[u] < d[v] (B) d[u] < f[v] (C) flu] < f[v] (D) f[u] > f[v] can you solve this question please ? if you solve it please comment it here i will get notified

  • @TankutTalu
    @TankutTalu 2 місяці тому

    take 104