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1
Clcok frequency becomes low means time period of clock becomes high. When this time period increses beyond the time period in which the non-volatile memory contents must be refreshed, we loose those contents. So, clock frequence can't go below this value. The ... or else will loose their contents. Non-volatile memory are being developed and computers in future should be tailor made for that.
posted Jun 6 in GATE Subject Wise Arjun 1 take
2
50 Marks, 90 Minutes, 30 Questions
posted Jun 6 in GATE Overflow Tests Subject Wise Arjun 1 take
3
Consider the following circuit involving a positive edge triggered D FF. Consider the following timing diagram. Let $A_{i}$ represents the logic level on the line a in the i-th clock period. Let $A'$ represent the compliment of $A$. The correct output sequence on $Y$ over the clock periods $1$ through $5$ ... $A_{1} A_{2} A_{2}' A_{3} A_{4}$ $A_{1} A_{2}' A_{3} A_{4} A_{5}'$
posted Feb 21 in GATE Full Length soujanyareddy13 14 takes
4
explain the recurrence relation???
posted Feb 21 in GATE Full Length 16 takes
5
Fourier series of the periodic function (period 2π) defined by $f(x) = \begin{cases} 0, -p < x < 0\\x, 0 < x < p \end{cases} \text { is }\\ \frac{\pi}{4} + \sum \left [ \frac{1}{\pi n^2} \left(\cos n\pi - 1 \right) \cos nx - \frac{1}{n} \cos n\pi \sin nx \right ]$ ... $\frac{{\pi }^2 }{4}$ $\frac{{\pi }^2 }{6}$ $\frac{{\pi }^2 }{8}$ $\frac{{\pi }^2 }{12}$
posted Feb 8 in GATE Overflow Tests Full Length soujanyareddy13 352 takes
6
Many microprocessors have a specified lower limit on clock frequency (apart from the maximum clock frequency limit) because ?
posted Feb 1 in GATE Overflow Tests Full Length soujanyareddy13 79 takes
7
We can simply do a binary search in the array of natural numbers from $1..n$ and check if the cube of the number matches $n$ (i.e., check if $a[i] * a[i] * a[i] == n$). This check takes $O(\log n)$ ... cannot be lower than $\log n$. (It must be $\Omega \left( \log n \right)$). So, (D) is also false and (C) is the correct answer.
posted Jan 26 in GATE Overflow Tests Full Length soujanyareddy13 74 takes
8
Consider a file of $16384$ records. Each record is $32$ $bytes$ long and its key field is of size $6$ $bytes$. The file is ordered on a non-key field, and the file organization is unspanned. The file is stored in a file system with block size $1024$ $bytes$, and the size of ... and second-level blocks in the multi-level index are respectively $8$ and $0$ $128$ and $6$ $256$ and $4$ $512$ and $5$
posted Jan 17 in GATE Overflow Tests Full Length soujanyareddy13 98 takes
9
The cube root of a natural number $n$ is defined as the largest natural number $m$ such that $(m^3 \leq n)$ . The complexity of computing the cube root of $n$ ($n$ is represented by binary notation) is $O(n)$ but not $O(n^{0.5})$ $O(n^{0.5})$ but not $O((\log n)^k)$ for any constant ... $m>0$ $O( (\log \log n)^k )$ for some constant $k > 0.5$, but not $O( (\log \log n)^{0.5} )$
posted Jan 4 in GATE Overflow Tests Full Length soujanyareddy13 135 takes
10
The corresponding English meaning: If $P(x)$ is true for all $x$, or if $Q(x)$ is true for all $x$, then for all $x$, either $P(x)$ is true or $Q(x)$ is true. This is always true and hence valid. To understand deeply, consider $X = \{3,6,9,12\}$. For ... in proving validity of many statements as is its converse given below: $\exists(x)(P(x)) \equiv \neg \forall (x)(\neg P(x))$ Correct Answer: $A$
posted Dec 28, 2020 in GATE Overflow Tests Subject Wise soujanyareddy13 40 takes
11
Which of the following predicate calculus statements is/are valid? $(\forall (x)) P(x) \vee (\forall(x))Q(x) \implies (\forall (x)) (P(x) \vee Q(x))$ $(\exists (x)) P(x) \wedge (\exists (x))Q(x) \implies (\exists (x)) (P(x) \wedge Q(x))$ ... $(\exists (x)) (P(x) \vee Q(x)) \implies \sim (\forall (x)) P(x) \vee (\exists (x)) Q(x)$
posted Dec 24, 2020 in GATE Overflow Tests Subject Wise soujanyareddy13 63 takes
12
D should be the answer.
posted Dec 14, 2020 in GATE Overflow Tests Subject Wise soujanyareddy13 85 takes
14
Both the programs are equivalent in the sense that the output will be the same at the end of execution. Q just writes 1 to u but this will be overwritten by the following write of 0. So, in any computer both P and Q should produce the same result at the end of execution. Correct Answer: $D$
posted Nov 27, 2020 in GATE Overflow Tests Random soujanyareddy13 32 takes
15
I assume 2^2^k is evaluated as 2^(2^k) In each iteration of while loop j is becomeing 2^2^l (where l is the loop iteration count) and loop terminates when j = 2^2^k. So, l must be equal to k when loop terminates. So, complexity of while loop is $\Theta(k)$. Now, the outher for loop runs for $n$ iterations and hence the complexity of the entire code is $\Theta(nk)\\ =\Theta(n\log \log n)$
posted Nov 15, 2020 in GATE Overflow Tests Subject Wise soujanyareddy13 117 takes
16
What is the time complexity? main() { n=2^2^k, k>0 for(i = 1 to n) { j=2 while(j ≤ n) { j=j^2 } } }
posted Oct 30, 2020 in GATE Overflow Tests Subject Wise soujanyareddy13 65 takes
17
no such constant exists for n^n, whereas for the nlogn I think it is 1/2
posted Oct 23, 2020 in TIFR Full Length soujanyareddy13 2 takes
18
I believe it should not be in Easy category :)...Medium would suffice..just my observation
posted Oct 23, 2020 in TIFR Full Length soujanyareddy13 4 takes
19
Yeap, that is what I meant...thanks
posted Oct 23, 2020 in GATE Overflow Tests Subject Wise soujanyareddy13 107 takes
20
their is the second option is true. Option 2 : Code A uses lesser memory and is faster than Code B If we are define 3 variable then low memory used. and if go with the 4 variable then every time d=(a-b) firstly. so the code becomes slow. so the I m go with Option 2 : Code A uses lesser memory and is faster than Code B.
posted Oct 15, 2020 in GATE Overflow Tests Random soujanyareddy13 42 takes
21
Lets take the English meaning Government will not be unpopular $\implies$ People will not suffer $\implies$ Either no inflation or government regulates it $\implies$ If no regulation then no inflation $\implies$ if no regulation then no wage or price rise In your statement, (W+P)->G ... rt? So, a, b and d are answers. I misread the first sentence earlier. That is why I posted only (a) as answer.
posted Oct 8, 2020 in GATE Overflow Tests Subject Wise soujanyareddy13 126 takes
22
60 Marks, 60 Minutes, 60 Questions
posted Oct 2, 2020 in ISRO Full Length gatecse 50 takes
23
To put these things in English sentences is the mistake I commited..to ammend it, W=wages raised,P=price raised,I=Inflation occured,G=government regulates it,PS=people suffer,GU=government becomes unpopular....here are relations...(W+P)->I , I->(G+Pe) , Pe-> GU....with the ... now, as given in the option B if we make value of G =0(not regulated) then both W and P has to be zero......thanks
posted Sep 29, 2020 in GATE Overflow Tests Subject Wise soujanyareddy13 136 takes
24
@Shaun Patel : In option (b), if inflation is not regulated, then it is not necessary that wages are not raised, i.e. wages might have been raised, because then inf;ation would have occured, and still we could have said that inflation is not regulated, because then people will suffer. Similar case for option (d). So option (b) and (d) are incorrect.
posted Sep 22, 2020 in GATE Overflow Tests Subject Wise soujanyareddy13 69 takes
25
A is surely corrrect but what about B and D...For B,D, as people will not suffer(concluded) and government has not regulated the inflation(given in options) then it can be concluded that there is no inflation and hence no price/wages are raised....isn't it?
posted Sep 14, 2020 in GATE Overflow Tests Subject Wise soujanyareddy13 80 takes
26
Pankaj and Mythili were both asked to write the code to evaluate the following expression: $a - b + c/(a-b) + (a-b)^2 $ Pankaj writes the following code statements (Code A): print (a-b) + c/(a-b) + (a-b)*(a-b) Mythili writes the following code statements (Code ... Code B Option 3 : Code A uses more memory and is faster than Code B Option 4 : Code A uses more memory and is slower than Code B Like
posted Sep 7, 2020 in GATE Overflow Tests Subject Wise soujanyareddy13 101 takes
27
It is told in the question "If the people suffer, the government will be unpopular". And "government will not be unpopular" means, people will not suffer. It is like $A \rightarrow B$ is true and ~B is given. So, ~A must be true. So, (A) is valid (always true). Lets take ... if no regulation then no wage or price rise So, (B) and (D) are valid (always true) and (C) and (E) are not valid.
posted Aug 31, 2020 in GATE Overflow Tests Random soujanyareddy13 45 takes
28
If either wages or prices are raised, there will be inflation. If there is inflation, then either the government must regulate it or the people will suffer. If the people suffer, the government will be unpopular. Government will not be unpopular. Which of the ... wages are not raised Prices are not raised If the inflation is not regulated, then the prices are not raised Wages are not raised
posted Aug 19, 2020 in GATE Overflow Tests Subject Wise gatecse 124 takes
29
No. Pumping lemma cannot say if a language is regular. If pumping lemma is violated, it can say a language is not regular. But if pumping lemma is satisfied, it doesn't mean language is regular.
posted Aug 15, 2020 in TIFR Full Length soujanyareddy13 3 takes
30
all options are right
posted Aug 15, 2020 in TIFR Full Length soujanyareddy13 22 takes
31
Please read all the topics. If you have any doubt in question you can post here. We shall be posting questions in this topic by November and we shall try to cover all types of problems.
posted Aug 15, 2020 in TIFR Full Length soujanyareddy13 11 takes
32
60 Marks, 180 Minutes, 30 Questions
posted Aug 15, 2020 in TIFR Full Length soujanyareddy13 7 takes
33
60 Marks, 180 Minutes, 30 Questions
posted Aug 15, 2020 in TIFR Full Length soujanyareddy13 5 takes
34
60 Marks, 180 Minutes, 30 Questions
posted Aug 15, 2020 in TIFR Full Length soujanyareddy13 6 takes
35
60 Marks, 180 Minutes, 30 Questions
posted Aug 15, 2020 in TIFR Full Length soujanyareddy13 5 takes
36
80 Marks, 180 Minutes, 40 Questions
posted Aug 15, 2020 in TIFR Full Length soujanyareddy13 3 takes
37
80 Marks, 180 Minutes, 40 Questions
posted Aug 15, 2020 in TIFR Full Length soujanyareddy13 7 takes
38
80 Marks, 180 Minutes, 40 Questions
posted Aug 15, 2020 in TIFR Full Length soujanyareddy13 11 takes
39
50 Marks, 90 Minutes, 30 Questions
posted Aug 10, 2020 in GATE Overflow Tests Subject Wise gatecse 120 takes
40
50 Marks, 100 Minutes, 30 Questions
posted Aug 3, 2020 in GATE Overflow Tests Subject Wise gatecse 110 takes
41
50 Marks, 90 Minutes, 30 Questions
posted Jul 26, 2020 in GATE Overflow Tests Subject Wise gatecse 145 takes
42
50 Marks, 90 Minutes, 30 Questions
posted Jul 19, 2020 in GATE Overflow Tests Subject Wise gatecse 135 takes
43
50 Marks, 90 Minutes, 30 Questions
posted Jul 12, 2020 in GATE Overflow Tests Random gatecse 40 takes
44
50 Marks, 90 Minutes, 30 Questions
posted Jul 5, 2020 in GATE Overflow Tests Subject Wise gatecse 80 takes
45
50 Marks, 90 Minutes, 30 Questions
posted Jun 28, 2020 in GATE Overflow Tests Subject Wise gatecse 86 takes
46
100 Marks, 180 Minutes, 65 Questions
posted Jun 23, 2020 in GATE Full Length gatecse 403 takes
47
50 Marks, 90 Minutes, 30 Questions
posted Jun 21, 2020 in GATE Overflow Tests Subject Wise gatecse 121 takes
48
50 Marks, 90 Minutes, 30 Questions
posted Jun 14, 2020 in GATE Overflow Tests Subject Wise gatecse 68 takes
49
jp
From Wikipedia: When a connection is requested by an application, the application indicates to the network The Type of Service required The Traffic Parameters of each data flow in both directions The Quality of Service (QoS) Parameters requested in each direction These parameters form the traffic descriptor for the connection.
posted Jun 7, 2020 in GATE Overflow Tests Subject Wise gatecse 105 takes
50
No ..just keep ur basics clear in c and try to be strong in logic thats enough
posted May 29, 2020 in UGC NET Full Length soujanyareddy13 24 takes
51
Dis maths:kenneth rosen really its a bible and you wl also learn many new things apart from syllabus Aptitude:r.s agarwal standard book
posted May 29, 2020 in UGC NET Full Length soujanyareddy13 9 takes
52
Why not option d is correct ?
posted May 26, 2020 in UGC NET Full Length soujanyareddy13 1 take
53
150 Marks, 150 Minutes, 75 Questions
posted May 26, 2020 in UGC NET Full Length soujanyareddy13 2 takes
54
okay. You may use <> in the toolbar to output code properly :)
posted Jan 16, 2020 in GATE Subject Wise Lakshman Patel RJIT 118 takes
55
yes,,the line upto "count=count+1" is in the loop...
posted Jan 16, 2020 in GATE Subject Wise Lakshman Patel RJIT 143 takes
56
Increment of count must be inside the loop rt?
posted Jan 15, 2020 in GATE Subject Wise Lakshman Patel RJIT 60 takes
57
def brian(n): count = 0 while ( n != 0 ) n = n & ( n-1 ) count = count + 1 return count Here n is meant to be an unsigned integer. The operator & considers its arguments in binary and computes their bit wise AND. For example, 22 & 15 gives 6 ... of 15 is 00001111, and the bit-wise AND of these binary strings is 00000110, which is the binary representation of 6. What does the function brian return?
posted Jan 15, 2020 in GATE Subject Wise Lakshman Patel RJIT 57 takes
58
60 Marks, 108 Minutes, 40 Questions
posted Jan 15, 2020 in GATE Subject Wise Lakshman Patel RJIT 30 takes
59
60 Marks, 108 Minutes, 40 Questions
posted Jan 15, 2020 in GATE Subject Wise Lakshman Patel RJIT 54 takes
60
how can we eliminate option B unless we draw it actually...just because of argument of 'stronger answer' ?
posted Jan 13, 2020 in GATE Subject Wise Lakshman Patel RJIT 47 takes
61
60 Marks, 108 Minutes, 40 Questions
posted Jan 13, 2020 in GATE Subject Wise Lakshman Patel RJIT 24 takes
62
Let $n$ be the number of vertices. Total number of incoming edges $= 7 \times n$. This should be equal to the total number of outgoing edges. So, either all vertices must have 7 edges leaving or some should have more than 7 leaving while others could have less than 7 ... leaving it. So, (c) is the correct answer. Only if we restrict $n$ to 8, exactly seven edges need to leave every vertex.
posted Jan 13, 2020 in GATE Subject Wise Lakshman Patel RJIT 23 takes
63
It is true for big O. But for $\Theta$ notation can we say $n! = \Theta (n^n)$?
posted Jan 13, 2020 in GATE Subject Wise Lakshman Patel RJIT 64 takes
64
70 Marks, 126 Minutes, 40 Questions
posted Jan 12, 2020 in GATE Subject Wise Lakshman Patel RJIT 40 takes
65
Yes, result. I will correct the statement. Thanks.
posted Jan 11, 2020 in GATE Subject Wise Lakshman Patel RJIT 32 takes
66
60 Marks, 108 Minutes, 40 Questions
posted Jan 11, 2020 in GATE Subject Wise Lakshman Patel RJIT 29 takes
67
60 Marks, 108 Minutes, 40 Questions
posted Jan 11, 2020 in GATE Subject Wise Lakshman Patel RJIT 79 takes
68
Why you say it is a condition? It is the result of pumping lemma rt?
posted Jan 9, 2020 in GATE Subject Wise Lakshman Patel RJIT 130 takes
69
That's a perfect explanation.
posted Jan 9, 2020 in GATE Subject Wise Lakshman Patel RJIT 180 takes
70
n! corresponds to n*(n-1)*(n-2)*....*1 which is $\Theta (n^n)$. So by taking $\log$ to both terms, the answer comes out to be $\Theta(n \log n)$.
posted Jan 9, 2020 in GATE Subject Wise Lakshman Patel RJIT 124 takes
71
45 Marks, 81 Minutes, 30 Questions
posted Jan 9, 2020 in GATE Subject Wise Lakshman Patel RJIT 107 takes
72
(B)1, 3 and 4 only As from the given options, Myhill-Nerode theorem is useful by providing necessary and sufficient condition for proving that a language regular. Pumping lemma is often used to prove that a language is non-regular. Drawing an NFA can be useful to prove a language is regular. Forming a regular expression can also help us prove if it is a regular language
posted Jan 9, 2020 in GATE Subject Wise Lakshman Patel RJIT 71 takes
73
(C) $\forall i \in N, xy^iz \in L$ This is the result of Pumping Lemma for regular language
posted Jan 9, 2020 in GATE Subject Wise Lakshman Patel RJIT 135 takes
74
ok
posted Dec 31, 2019 in ISRO Full Length Arjun 282 takes
75
General aptitude is a vast area but for GATE they ask only from a small section which is clearly mentioned in GATE syllabus. Many people won't prepare for these but it is advisable to just read the sections mentioned in GATE syllabus from a recommended book ... from one of the recommended books such as Kenneth H. Rosen. For more books: http://gatecse.in/gatecse/index.php?title=Best_books_for_CSE
posted Dec 22, 2019 in GATE Subject Wise Counsellor 130 takes
76
Which of the following are useful in proving a language to be regular? Myhill-Nerode theorem Pumping lemma Drawing an NFA Forming a regular expression (A) All of these (B) 1, 3 and 4 only (C) 2, 3 and 4 only (D) 3 and 4 only
posted Dec 22, 2019 in GATE Subject Wise Counsellor 79 takes
77
Let $L$ be a regular language and $w$ be a string in $L$. If $w$ can be split into $x, y$ and $z$ such that $|xy| \leq$ number of states in the minimal DFA for $L$, and $|y| > 0$ then, (A) $\forall i \in N, xy^iz \notin L$ (B) $\exists i \in N, xy^iz \in L$ (C) $\forall i \in N, xy^iz \in L$ (D) $\exists i \in N, xy^iz \notin L$
posted Dec 19, 2019 in GATE Subject Wise Counsellor 122 takes
78
is wrong. Microprogrammed CU can never be faster than hardwired CU. Microprogrammed CU it has an extra layer on top of hardwired CU and hence can only be slower than hardwired CU. is a suitable answer as we can add new instruction by changing the content of control ... control makes more sense. control unit can also be hardwired, so this is also not correct. Reference: Slides Correct Answer: $B$
posted Dec 19, 2019 in GATE Subject Wise Counsellor 94 takes
79
45 Marks, 85 Minutes, 29 Questions
posted Dec 17, 2019 in GATE Subject Wise Counsellor 142 takes
80
46 Marks, 85 Minutes, 30 Questions
posted Dec 17, 2019 in GATE Subject Wise Counsellor 101 takes
81
Yes. That was a bad mistake. I have corrected it :)
posted Dec 16, 2019 in GATE Subject Wise Counsellor 82 takes
82
First we need to find different number of inputs for $f$ given that permutations are equivalent. Now given two binary string $x,y$: $x$ is a permutation of $y$ iff $x$ has same number of ones and zeros as $y$. So the total number of different inputs is n+1 ( namely input with ... $2^{n + 1}.$
posted Dec 16, 2019 in GATE Subject Wise Counsellor 74 takes
83
A function $f:\left\{0, 1\right\}^{n}\rightarrow \left\{0, 1\right\}$ is called symmetric if for every $x_{1}, x_{2},....,x_{n} \in \left\{0, 1\right\}$ and every permutation $\sigma$ of $\left\{1, 2,...,n\right\}$ ... $2^{n+1}$ $2^{n}$ $2^{2n}/n!$ $2^{2n}$ $n!$
posted Dec 16, 2019 in GATE Subject Wise Counsellor 93 takes
84
Microprogrammed control unit: is faster than a hardwired control unit facilitates easy implementation of new instructions is useful when very small programs are to be run usually refers to control unit of a microprocessor
posted Dec 16, 2019 in GATE Counsellor 30 takes
85
yeah I thought the same, but for the second part adding only the height difference from A to P wont suffice coz we need to add all those right trees also, going up from A to P...thanks
posted Dec 16, 2019 in GATE Counsellor 14 takes
86
Answer should be (D). We first find $a$ in the BST in $O(\log n)$ time. Now there are two possibilities, b can be in the right subtree of $a$ or b can be in the right subtree of any of the parents of $a$. For the first case, we simply search for $b$, in the right ... $N(x)$ denote the no. of elements in the subtree rooted at $x$ and if LEFT(a), RIGHT(b) returning 0 for NULL.
posted Dec 16, 2019 in GATE Subject Wise Counsellor 98 takes
87
44 Marks, 80 Minutes, 29 Questions
posted Dec 8, 2019 in GATE Subject Wise Counsellor 82 takes
88
Complexity of the algorithm is $O(n)$ and is irrespective of the success of if case as both if as well as else is $O(1)$ operation. If you say exactly, the complexity in terms of comparisons will be 1. $n-n/2-1$ (number of elements for which first if succeeds) $+ 2 \times (n/2) $ (number of elements for which first if fails) $= 3n/2 -1$ 2. $ (n-1)/2 + 2\times (n-1)/2$ $=3(n-1)/2$
posted Dec 8, 2019 in GATE Subject Wise Counsellor 105 takes
89
51 Marks, 95 Minutes, 34 Questions
posted Dec 6, 2019 in GATE Subject Wise Counsellor 42 takes
90
50 Marks, 90 Minutes, 33 Questions
posted Dec 6, 2019 in GATE Subject Wise Counsellor 78 takes
91
45 Marks, 80 Minutes, 30 Questions
posted Dec 2, 2019 in GATE Subject Wise Counsellor 82 takes
92
45 Marks, 80 Minutes, 30 Questions
posted Dec 2, 2019 in GATE Subject Wise Counsellor 132 takes
93
55 Marks, 100 Minutes, 34 Questions
posted Dec 2, 2019 in GATE Subject Wise Counsellor 76 takes
94
61 Marks, 110 Minutes, 37 Questions
posted Dec 2, 2019 in GATE Subject Wise Counsellor 123 takes
95
Np. Anyway good to highlight the "polynomial time" :) Reduction of an NPC problem to a P problem (in polynomial time) would imply 2 things: 1) All problems in NP are reducible to this problem in P because all NP problems are reducible to NPC problems. 2) We can ... by definition of class P. This implies that all problems in NP are solvable in polynomial time, or NP = P Thats the exact answer :)
posted Nov 28, 2019 in GATE Subject Wise Counsellor 40 takes
96
My bad..didn't notice that....
posted Nov 28, 2019 in GATE Subject Wise Counsellor 64 takes
97
57 Marks, 100 Minutes, 37 Questions
posted Nov 27, 2019 in GATE Subject Wise Counsellor 79 takes
98
Simple linear search to find max min algo maxmin(a,n,max,min) { max=min=a[1]; for i=2 to n do { if a[i]>max then max:=a[i]; else if a[i]<min then min:=a[i]; } } 1.Average case complexity of the above algo given that the first if conditions fails for n/2 elements 2.Average case complexity of the above algo if the first condition fails 1/2 times plz xplain
posted Nov 27, 2019 in GATE Subject Wise Counsellor 109 takes
99
(All reductions in polynomial time) is mentioned in the question :)
posted Nov 26, 2019 in GATE Subject Wise Counsellor 75 takes
100
Option B is the most correct answer. (To know why it is not THE correct answer read the tail section of this answer) The reasons are as follows. A problem p in NP is NP-complete if every other problem in NP can be transformed into p in ... The answer is partially correct because when we say reduction, we usually mean polynomial time reduction, but neverthless it doesnt hurt to be precise.
posted Nov 26, 2019 in GATE Subject Wise Counsellor 138 takes
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