Algorithms And Data Structures

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Thursday, January 23, 2020

If the access time of a symbol table can be made logarithmic

January 23, 2020 |

Algorithms And Data Structures, PGCET CSE Exam (Karnataka) 2011, Technical Aptitude, UnSolved

If the access time of a symbol table can be made logarithmic, it greatly reduces the search time. This can be implemented by

Aa linear list

Ba search tree

Chashing mechanism

Dself-organisation chart

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The following graph is reduced to its minimum spanning tree

January 23, 2020 |

Algorithms And Data Structures, PGCET CSE Exam (Karnataka) 2011, UnSolved

The following graph is reduced to its minimum spanning tree using prim's method. Indicate the sequence in which the nodes get included

A1, 2, 5, 6, 4, 3

B1, 2, 6, 4, 5, 3

C1, 2,.3, 6, 4, 5

D1, 2, 6, 3, 4, 5

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Given the set of keys ( 40, 80, 35, 90, 45, 50, 70 ), which

January 23, 2020 |

Algorithms And Data Structures, PGCET CSE Exam (Karnataka) 2011, UnSolved

Given the set of keys ( 40, 80, 35, 90, 45, 50, 70 ), which of the following represents a heap ? ( Each pair of parentheses indicate nodes at one level, left child first)

A( ( 90 ) ( 40 - 80 ) ( ( 35 - 70 ) - ( 45 - 50 ) ) )

B( ( 80 ) ( 90 - 70 ) ( ( 40 - 45 ) ( 35 - 50 ) ) )

C( ( 90 ) ( 80 - 70 ) ( ( 40 - 45 ) ( 35 - 50 ) ) )

D( ( 70 ) ( 80 - 90 ) ( ( 50 - 45 ) ( 40 - 35 ) ) )

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Wednesday, January 22, 2020

Match the methodology of List-A with the application of List

January 22, 2020 |

Algorithms And Data Structures, PGCET CSE Exam (Karnataka) 2011, Solved

Match the methodology of List-A with the application of List-B :

Aa->III b->I c->IV d->II

Ba->IV b->III c->II d->I

Ca->III b->IV c->I d->II

Da->III b->II c->I d->IV

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Given two problems X and Y, Y is NP complete and X reduces

January 22, 2020 |

Algorithms And Data Structures, PGCET CSE Exam (Karnataka) 2011, Solved, Technical Aptitude

Given two problems X and Y, Y is NP complete and X reduces to Y in polynomial time. Which of the following is a valid statement ?

AX is NP hard

BX is NP complete

CX is an NP, but not necessarily NP complete

DIf X can be solved in polynomial time, so also can Y.

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