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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.
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50 Marks, 90 Minutes, 30 Questions
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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 $5$ ... $A_{1} A_{2} A_{2}' A_{3} A_{4}$ $A_{1} A_{2}' A_{3} A_{4} A_{5}'$
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explain the recurrence relation???
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Fourier series of the periodic function (period 2π) defined by $f(x) = \begin{cases} 0, -p < x < 0\\x, 0 < x < p \end{cases} \text { is }\\ \frac{\pi}{4} + \sum \left [ \frac{1}{\pi n^2} \left(\cos n\pi - 1 \right) \cos nx - \frac{1}{n} \cos n\pi \sin nx \right ]$ ... $\frac{{\pi }^2 }{4}$ $\frac{{\pi }^2 }{6}$ $\frac{{\pi }^2 }{8}$ $\frac{{\pi }^2 }{12}$
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Many microprocessors have a specified lower limit on clock frequency (apart from the maximum clock frequency limit) because ?
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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.
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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 $5$
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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 $O( (\log \log n)^{0.5} )$
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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: $A$
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Which of the following predicate calculus statements is/are valid? $(\forall (x)) P(x) \vee (\forall(x))Q(x) \implies (\forall (x)) (P(x) \vee Q(x))$ $(\exists (x)) P(x) \wedge (\exists (x))Q(x) \implies (\exists (x)) (P(x) \wedge Q(x))$ ... $(\exists (x)) (P(x) \vee Q(x)) \implies \sim (\forall (x)) P(x) \vee (\exists (x)) Q(x)$
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so D or E ?
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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: $D$
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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 $\Theta(nk)\\ =\Theta(n\log \log n)$
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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 } } }
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no such constant exists for n^n, whereas for the nlogn I think it is 1/2
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I believe it should not be in Easy category :)...Medium would suffice..just my observation
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Yeap, that is what I meant...thanks
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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.
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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.
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60 Marks, 60 Minutes, 60 Questions
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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
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@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.
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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?
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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
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.
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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
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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.
30
all options are right
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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.
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60 Marks, 180 Minutes, 30 Questions
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60 Marks, 180 Minutes, 30 Questions
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60 Marks, 180 Minutes, 30 Questions
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60 Marks, 180 Minutes, 30 Questions
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80 Marks, 180 Minutes, 40 Questions
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80 Marks, 180 Minutes, 40 Questions
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80 Marks, 180 Minutes, 40 Questions
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50 Marks, 90 Minutes, 30 Questions
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50 Marks, 100 Minutes, 30 Questions
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50 Marks, 90 Minutes, 30 Questions
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50 Marks, 90 Minutes, 30 Questions
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50 Marks, 90 Minutes, 30 Questions
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50 Marks, 90 Minutes, 30 Questions
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50 Marks, 90 Minutes, 30 Questions
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100 Marks, 180 Minutes, 65 Questions
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50 Marks, 90 Minutes, 30 Questions
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50 Marks, 90 Minutes, 30 Questions
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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.
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No ..just keep ur basics clear in c and try to be strong in logic thats enough
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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
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Why not option d is correct ?
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150 Marks, 150 Minutes, 75 Questions
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okay. You may use <> in the toolbar to output code properly :)
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yes,,the line upto "count=count+1" is in the loop...
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Increment of count must be inside the loop rt?
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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?
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60 Marks, 108 Minutes, 40 Questions
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60 Marks, 108 Minutes, 40 Questions
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how can we eliminate option B unless we draw it actually...just because of argument of 'stronger answer' ?
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60 Marks, 108 Minutes, 40 Questions
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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.
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It is true for big O. But for $\Theta$ notation can we say $n! = \Theta (n^n)$?
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70 Marks, 126 Minutes, 40 Questions
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Yes, result. I will correct the statement. Thanks.
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60 Marks, 108 Minutes, 40 Questions
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60 Marks, 108 Minutes, 40 Questions
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Why you say it is a condition? It is the result of pumping lemma rt?
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That's a perfect explanation.
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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 $\Theta(n \log n)$.
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45 Marks, 81 Minutes, 30 Questions
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(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
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(C) $\forall i \in N, xy^iz \in L$ This is the result of Pumping Lemma for regular language
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ok
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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
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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
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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) $\exists i \in N, xy^iz \notin L$
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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: $B$
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45 Marks, 85 Minutes, 29 Questions
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46 Marks, 85 Minutes, 30 Questions
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Yes. That was a bad mistake. I have corrected it :)
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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 ... $2^{n + 1}.$
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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 $\left\{1, 2,...,n\right\}$ ... $2^{n+1}$ $2^{n}$ $2^{2n}/n!$ $2^{2n}$ $n!$
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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
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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
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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.
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44 Marks, 80 Minutes, 29 Questions
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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. $(n-1)/2 + 2\times (n-1)/2$ $=3(n-1)/2$
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51 Marks, 95 Minutes, 34 Questions
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50 Marks, 90 Minutes, 33 Questions
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45 Marks, 80 Minutes, 30 Questions
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45 Marks, 80 Minutes, 30 Questions
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55 Marks, 100 Minutes, 34 Questions
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61 Marks, 110 Minutes, 37 Questions
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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 :)
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