1

Take a single person A out of the 10 people. The remaining 9 people has to be either friends or enemy with A. Therefore acc to pigeon hole principle there atleast⎾9/2⏋(i.e., 5) friends or enemy of A. Lets assume that A has five friends.Now, if at ... are four mutual enemies). similarly we can prove that there are either three mutual enemies or four mutual friends by assuming A has 5 enemies.

2

You are correct. For Dijkstra's algorithm, binomial heap and binary heap should give the same complexity.

3

this is one of the best solution for this type of question when by putting limit of x in given question it forms 1infinity form then we apply formula given below such type of question is solved by elim x->0(f(x)-1)g(x) in this question g(x)=1/x and f(x)=1-x so , according to above formula elim x->0 (1-x-1)*1/x =e-1

4

Describe a Θ(nlgn)-time algorithm that, given a set S of n integers and another integer x, determines whether or not there exists two elements of S whose sum is exactly x.

5

atleast 6 states are required in FA construction for this language, as shown below:

8

$\textbf{Pythagorean Identities:}$

$\sin^{2}\theta + \cos^{2}\theta = 1$ $1+\tan^{2}\theta = \sec^{2}\theta$ $1+\cot^{2}\theta = \csc^{2}\theta$$\textbf{Reciprocal Identities:}$

$\sin\theta = \dfrac{1}{\csc\theta}$ $\cos\theta = \dfrac{1}{\sec\theta}$ $\tan\theta = \dfrac{1}{\cot\theta}$ $\csc\theta = \dfrac{1}{\sin\theta}$ $\sec\theta = \dfrac{1}{\cos\theta}$ $\cot\theta = \dfrac{1}{\tan\theta}$$\textbf{Tangent and Cotangent Identities:}$

$\tan\theta =...9

$\textbf{1.Properties of determinants:-}$

The determinant is only valid for the square matrix.

$\mid A^{T}\mid = \mid A \mid $ $\mid AB \mid = \mid A \mid \mid B \mid $ $\mid A^{n} \mid = \big(\mid A \mid\big)^{n}$ $\mid kA\mid = k^{n} \mid A \mid$, here $A$ is the $n\times n$ matrix. If two rows (or two columns) of a determinant are interchanged, the sign of the value of the determinant changes. If in determinant any row or column is completely zero, the value of the determinant is zero. If two rows (or two columns) of a determinant are identical, the value...10

NP-Hard is atleast as hard as hardest problem in class NP. All problems in NP are reducible to NP-Hard. NP-Hard may or may not belong to class NP. If a problem Z is NP-Hard and it belongs to NP then Z is NP-Complete which is decidable using nondeterministic polynomial time turing machine.

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$$\textbf{Contents}$$

**Syllabus:**

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$$\textbf{Contents}$$

**Syllabus:** Connectivity, Matching, Coloring.

$$\small{\overset{{\large{\textbf{Mark Distribution in Previous...

14

Arjun sir . We use register renaming to over come anti dependency. Anti dependency in a instruction itself always causes a stall. So in order to overcome that we use register renaming during the programming phase itself. And the option has asked within a pipeline ... not any possibility of register renaming here as it is done before the program is executed. Please correct me if am wrong. Thanks

15

@Pragy Agarwal Shouldn't algorithm stop shifting once it has encountered an infinity. Because after that, all shifting will be of infinity elements only. Once it is clear that only infinty element can be shifted in empty cell, just fill that cell with infinity and stop. This method because it's been asked to minimize the shifting of elements.

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$$\textbf{Contents}$$

**Syllabus:**

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$$\textbf{Contents}$$

$$\textbf{1.Verbal Aptitude }$$

**Syllabus:**

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A parser works on the basis of given grammar. It takes the grammar as it is. Parser does not work on the basis of the yield of the grammar. Also, while constructing the LL(1) parser table, that entry for terminal 'c' will contain multiple entries. So, LL(1) parser ... grammar cannot be LL(1). Therefore, LL(1) parser cannot parse it. Hence, the answer should be option (D). Neither S1 nor S2.