Placement Papers - Infosys

Infosys Paper Pattern III
Posted by :
Ramakrishna
(6)
Infosys (Paper)
The Questions are follows:
1. Number of null pointers in any binary tree = n+1
2. max(t1,t2,...tn) = pipelining
3. 50% -DBETXXXXXX - density
4.
Print (Head(T))
Traverse(left(T))
Print (Head(T))
Traverse(right(T)) -
Ans: none of the above
5. Boolean expn Evalvate
6. Common subexpn : -
Ans : a + e
7. LRU : 1, 2, 3.
8. Tr. Delay - 10000 bits
Ans. 10.01
9. Grammar of Number of shift / reduce operator :
Ans. 4
10. CPU scheduling 9,8 ?
11.
if even x/2
else p(p(3x+1))
2^k + 1: 3 . 2^(k-1) clarify this with sans
12. allocation
Ans: (ii) only
13. swapping :
Ans: reference only
14. Compiler - related Qn.
15. LAN frames - ? related Qn.
16. parameter passing (35,20)
17. sliding window protocol - BUFFER SIZE large
18. kernel mode - deallocate resource
19. logic circuit
Ans . Minimum OR = 3
20. Combinatorics related
21. priority scheduling
22. cobegin
begin x = y; x= x+1; y= x
begin x =y; z= z+1; y= z
coend
Ans. Number of values possi = 2
23. 2 bits flip / 2 bits exchange
Ans : the word with one \'1\'
24.
any address
K^+ v(a) + 2I - 2a
The Questions are follows:
1. Number of null pointers in any binary tree = n+1
2. max(t1,t2,...tn) = pipelining
3. 50% -DBETXXXXXX - density
4.
Print (Head(T))
Traverse(left(T))
Print (Head(T))
Traverse(right(T)) -
Ans: none of the above
5. Boolean expn Evalvate
6. Common subexpn : -
Ans : a + e
7. LRU : 1, 2, 3.
8. Tr. Delay - 10000 bits
Ans. 10.01
9. Grammar of Number of shift / reduce operator :
Ans. 4
10. CPU scheduling 9,8 ?
11.
if even x/2
else p(p(3x+1))
2^k + 1: 3 . 2^(k-1) clarify this with sans
12. allocation
Ans: (ii) only
13. swapping :
Ans: reference only
14. Compiler - related Qn.
15. LAN frames - ? related Qn.
16. parameter passing (35,20)
17. sliding window protocol - BUFFER SIZE large
18. kernel mode - deallocate resource
19. logic circuit
Ans . Minimum OR = 3
20. Combinatorics related
21. priority scheduling
22. cobegin
begin x = y; x= x+1; y= x
begin x =y; z= z+1; y= z
coend
Ans. Number of values possi = 2
23. 2 bits flip / 2 bits exchange
Ans : the word with one \'1\'
24.
any address
K^+ v(a) + 2I - 2a
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