# UGC NET COMPUTER SCIENCE SOLVED PAPERS 2017-19 - UGC NET Computer Science November 2017 Paper 2

>>>>>>>>UGC NET Computer Science November 2017 Paper 2

• A

Every minimum spanning tree of G must contain Emin  • B

If Emax is in minimum spanning tree, then its removal must disconnect G.  • C

No minimum spanning tree contains Emax  • D

G has a unique minimum spanning tree.  • A

O(n log n)  • B

O(n2log n)  • C

O(n2 + log n)  • D

O(n3)  • Option : B
• Explanation :
When we are sorting an array of n integers, Recurrence relation for Total number of comparisons involved will be,
T(n) = 2T(n/2) + (n) where (n) is the number of comparisons in order to merge 2 sorted subarrays of size n/2. = (nlog2n)
Instead of integers whose comparison take O(1) time, we are given n strings. We can compare 2 strings in O(n) worst case.
Therefore, the total number of comparisons now will be (n2log2n) where each comparison takes O(n) time now.
In general, merge sort makes (nlog2n) comparisons, and runs in (nlog2n) time if each comparison can be done in O(1) time.

• A

3, 4, 5, 7, 9, 14, 20, 18, 17, 16, 15  • B

20, 18, 17, 16, 15, 14, 3, 4, 5, 7, 9  • C

20, 18, 17, 16, 15, 14, 9, 7, 5, 4, 3  • D

3, 4, 5, 7, 9, 14, 15, 16, 17, 18, 20  • Option : D
• Explanation :
Since Inorder traversal of a BST always gives elements in increasing order. For this question sorted order of keys will always be the in-order tree traversal for post-order traversal. So, option (D) is correct.

• A

Hub  • B

Modem  • C

Switch  • D

Gateway  • Option : C
• Explanation :
Switch takes data from one network and forward it to other network based on MAC address. So option (C) is correct.

• A

Static algorithms  • B

Adaptive algorithms  • C

Non - adaptive algorithms  • D

Recursive algorithms  • Option : C
• Explanation :
Non-adoptive algorithms don't take their decisions on measurements or estimates of the current traffic and topology. So, option (C) is correct.