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Posts by soujanyareddy13

3
SQL query will be SELECT cno FROM Completed, Pre-Req WHERE student_no = '2310' GROUP BY cno HAVING pre-Cno IN ( SELECT C.cno FROM Completed AS C WHERE C.student_no = '2310'; )
posted Mar 18 in Interview Experience 100 views
4
Refer to the similar thread below to get answer for this question https://gateoverflow.in/4273/char-pointer-to-access-an-int
posted Mar 18 in Interview Experience 49 views
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sir that we do in dijkstra as well , there also we chose initially one vertex as source vertex , but there are only V passes .
posted Mar 17 in Interview Experience 68 views
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Sir, How is the loop nested when we have semicolon at the end of both for loops? Am i missing something ?
posted Mar 17 in Interview Experience 34 views
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i am not sure , i just give a try on this : { p | (&exist;q)(&exist;r) works(p q r) and ( (&forall;y) (works(x y z )^( y = small bank co-operation)) -> (&exist;z) (z<r) )}
posted Mar 17 in Interview Experience 64 views
11
in bellman ford algo v-1 times will give you the shortest path but if there is any -ve edge cycle to check you have to perform 1 more cycle . Why V-1times ? simple , a vertices is connected to atmost V-1 vertices to check them you need V-1 times ( there is no -ve edge cycle ) .
posted Mar 17 in Interview Experience 84 views
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My question: Should we include dirty bit and vaild-invalid bit in this calculation ? Please comment
posted Mar 15 in Job Openings 380 views
13
8 million = 8*2^10*2^10 =2^23 So address is 23 bits. Word offset its = 6 (2*6=64) No of bits to identify a Block= 8(2^8 = 256) No of bits to indentify tag= 23-6-8=9 1) Additional memory for tag=(256*9)/8= 288 bytes 2) Total size of the cache =Cache size +Tag memory + Dirty bit memory +Valid bit memory =256*64 +(256*9)/8 + 256/8 + 256/8
posted Mar 15 in From GO Admins 1,734 views
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Consider a 8 million word main memory and 256 block cache.Both partitioned into 64 words block.What is the size of additional memory for tags? Size of the cache? consider direct mapping is used and word size 1 byte.
posted Mar 15 in Interview Experience 226 views
17
A. Encapsulation http://stackoverflow.com/questions/24626/abstraction-vs-information-hiding-vs-encapsulation
posted Mar 13 in Interview Experience 741 views
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lg *n, (Inverse Ackermann function) is the number of times we can take log repeatedly until we get 1. This function can almost be considered constant for all practical purposes. There shouldn't be any confusion regarding options C and D as n! is asymptotically larger than lg n. A) lg (lg * n) B) lg * ... can write B as lg * (lg n) = lg * n - 1. So, A is asymptotically lower than B. A < B < D < C.
posted Mar 13 in Interview Experience 140 views
19
Surely a misprint in question, but don't follow that book for GATE.
posted Mar 13 in Interview Experience 86 views
20
In topological sort, in the inner loop we consider only "adjacent vertex". So, the complexity at max can be O(|V|2) which happens when |E| = |V|2. In Bellman Ford algorithm, the inner loop is run for each and every edge.
posted Mar 13 in Interview Experience 104 views
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