Re: [algogeeks] Re: google paper/...plz help..its urgent
thanks very much.
On Sun, Jan 16, 2011 at 5:04 PM, Lakhan Arya lakhan.a...@gmail.com wrote:
@pacific
Sets of size 2 can have 2 elements common with set of size greater
than 2. for example if set is (1,2) than it is adjacent to sets like
(1,2,3) (1,2,4), (1,2,3,4...n) etc.
So (1,2) is adjacent to (1,2,3), (1,2,4) etc.
On Jan 16, 1:04 pm, pacific pacific pacific4...@gmail.com wrote:
@Lakhan
Why are you not considering sets of size 2 ? Because two sets of size two
cannot have both of the elements as same.
On Sat, Jan 15, 2011 at 9:39 PM, Lakhan Arya lakhan.a...@gmail.com
wrote:
@bittu
I don't think answer of 6th question to be a)
No. of vertices of degree 0 will be those who didnot intersect with
any set i exactly 2 points. All sets of size greater than equal 2 must
intersect with any other set having exactly 2 common elements between
them in exactly 2 points. e.g if a set is (1,2) then it will be
adjacent to (1,2,3) , (1,2,3,4) etc..
The sets of size 0 and 1 cannot intersect in 2 points so they all will
be of degree 0.
Number of Sets of size 0 --- 1
Number of Sets of size 1 --- n
so Total number of vertices n+1.
In the similar way number of connected components will be n+2.
On Jan 15, 8:44 pm, bittu shashank7andr...@gmail.com wrote:
1.c U Can verify by putting n =I where I is positive integer value
say
n=5 try it out its so easy
2 a...what i have understood.
as we know that formal grammar is defined as (N, Σ, P, S)
so For instance, the grammar G with N = {S, A}, Σ = {a, b}, P
with start symbol S and rules
S → aA
A → Sb
S → ε
generates { a^ib^i : =0} so answer is A.
3 expected value doe discrete distributional is defined as
E(i)=sum(pi * xi); so from my points of view ans is 1/n ...Really
Gud
Question one has think..still thinking
4.b -Explaination
Informally the NP-complete problems are the toughest problems in NP
in the sense that they are the ones most likely not to be in P. NP-
complete problems are a set of problems that any other NP-problem can
be reduced to in polynomial time, but retain the ability to have
their
solution verified in polynomial time. In comparison, NP-hard problems
are those at least as hard as NP-complete problems, meaning all NP-
problems can be reduced to them, but not all NP-hard problems are in
NP, meaning not all of them have solutions verifiable in polynomial
time.
(A) is incorrect because set NP includes both P(Polynomial time
solvable) and NP-Complete .
(B) is incorrect because X may belong to P (same reason as (A))
(C) is correct because NP-Complete set is intersection of NP and NP-
Hard sets.
(D) is incorrect because all NP problems are decidable in finite set
of operations.
5. The Most Typical..Still Need Time
6 a zero degree means vertex is not connected from any other vertex
in graph
7.a
8.No Answer Answer Comes to Be 252
15c10,14c9,10c5,10*9*8*7*6 all are greater then from output so
say
No Answer
Correct Me if I am Wrong
Next Time I will Try to provide the solution of 2nd, 5th
problem ..explanations from-others are appreciated
Thanks Regards
Shashank Mani Don't B Evil U Can Earn while U Learn
Computer Science Engg.
BIT Mesra
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regards,
chinna.
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Re: [algogeeks] Re: google paper/...plz help..its urgent
@Lakhan
Why are you not considering sets of size 2 ? Because two sets of size two
cannot have both of the elements as same.
On Sat, Jan 15, 2011 at 9:39 PM, Lakhan Arya lakhan.a...@gmail.com wrote:
@bittu
I don't think answer of 6th question to be a)
No. of vertices of degree 0 will be those who didnot intersect with
any set i exactly 2 points. All sets of size greater than equal 2 must
intersect with any other set having exactly 2 common elements between
them in exactly 2 points. e.g if a set is (1,2) then it will be
adjacent to (1,2,3) , (1,2,3,4) etc..
The sets of size 0 and 1 cannot intersect in 2 points so they all will
be of degree 0.
Number of Sets of size 0 --- 1
Number of Sets of size 1 --- n
so Total number of vertices n+1.
In the similar way number of connected components will be n+2.
On Jan 15, 8:44 pm, bittu shashank7andr...@gmail.com wrote:
1.c U Can verify by putting n =I where I is positive integer value say
n=5 try it out its so easy
2 a...what i have understood.
as we know that formal grammar is defined as (N, Σ, P, S)
so For instance, the grammar G with N = {S, A}, Σ = {a, b}, P
with start symbol S and rules
S → aA
A → Sb
S → ε
generates { a^ib^i : =0} so answer is A.
3 expected value doe discrete distributional is defined as
E(i)=sum(pi * xi); so from my points of view ans is 1/n ...Really Gud
Question one has think..still thinking
4.b -Explaination
Informally the NP-complete problems are the toughest problems in NP
in the sense that they are the ones most likely not to be in P. NP-
complete problems are a set of problems that any other NP-problem can
be reduced to in polynomial time, but retain the ability to have their
solution verified in polynomial time. In comparison, NP-hard problems
are those at least as hard as NP-complete problems, meaning all NP-
problems can be reduced to them, but not all NP-hard problems are in
NP, meaning not all of them have solutions verifiable in polynomial
time.
(A) is incorrect because set NP includes both P(Polynomial time
solvable) and NP-Complete .
(B) is incorrect because X may belong to P (same reason as (A))
(C) is correct because NP-Complete set is intersection of NP and NP-
Hard sets.
(D) is incorrect because all NP problems are decidable in finite set
of operations.
5. The Most Typical..Still Need Time
6 a zero degree means vertex is not connected from any other vertex
in graph
7.a
8.No Answer Answer Comes to Be 252
15c10,14c9,10c5,10*9*8*7*6 all are greater then from output so say
No Answer
Correct Me if I am Wrong
Next Time I will Try to provide the solution of 2nd, 5th
problem ..explanations from-others are appreciated
Thanks Regards
Shashank Mani Don't B Evil U Can Earn while U Learn
Computer Science Engg.
BIT Mesra
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regards,
chinna.
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[algogeeks] Re: google paper/...plz help..its urgent
@pacific
Sets of size 2 can have 2 elements common with set of size greater
than 2. for example if set is (1,2) than it is adjacent to sets like
(1,2,3) (1,2,4), (1,2,3,4...n) etc.
So (1,2) is adjacent to (1,2,3), (1,2,4) etc.
On Jan 16, 1:04 pm, pacific pacific pacific4...@gmail.com wrote:
@Lakhan
Why are you not considering sets of size 2 ? Because two sets of size two
cannot have both of the elements as same.
On Sat, Jan 15, 2011 at 9:39 PM, Lakhan Arya lakhan.a...@gmail.com wrote:
@bittu
I don't think answer of 6th question to be a)
No. of vertices of degree 0 will be those who didnot intersect with
any set i exactly 2 points. All sets of size greater than equal 2 must
intersect with any other set having exactly 2 common elements between
them in exactly 2 points. e.g if a set is (1,2) then it will be
adjacent to (1,2,3) , (1,2,3,4) etc..
The sets of size 0 and 1 cannot intersect in 2 points so they all will
be of degree 0.
Number of Sets of size 0 --- 1
Number of Sets of size 1 --- n
so Total number of vertices n+1.
In the similar way number of connected components will be n+2.
On Jan 15, 8:44 pm, bittu shashank7andr...@gmail.com wrote:
1.c U Can verify by putting n =I where I is positive integer value say
n=5 try it out its so easy
2 a...what i have understood.
as we know that formal grammar is defined as (N, Σ, P, S)
so For instance, the grammar G with N = {S, A}, Σ = {a, b}, P
with start symbol S and rules
S → aA
A → Sb
S → ε
generates { a^ib^i : =0} so answer is A.
3 expected value doe discrete distributional is defined as
E(i)=sum(pi * xi); so from my points of view ans is 1/n ...Really Gud
Question one has think..still thinking
4.b -Explaination
Informally the NP-complete problems are the toughest problems in NP
in the sense that they are the ones most likely not to be in P. NP-
complete problems are a set of problems that any other NP-problem can
be reduced to in polynomial time, but retain the ability to have their
solution verified in polynomial time. In comparison, NP-hard problems
are those at least as hard as NP-complete problems, meaning all NP-
problems can be reduced to them, but not all NP-hard problems are in
NP, meaning not all of them have solutions verifiable in polynomial
time.
(A) is incorrect because set NP includes both P(Polynomial time
solvable) and NP-Complete .
(B) is incorrect because X may belong to P (same reason as (A))
(C) is correct because NP-Complete set is intersection of NP and NP-
Hard sets.
(D) is incorrect because all NP problems are decidable in finite set
of operations.
5. The Most Typical..Still Need Time
6 a zero degree means vertex is not connected from any other vertex
in graph
7.a
8.No Answer Answer Comes to Be 252
15c10,14c9,10c5,10*9*8*7*6 all are greater then from output so say
No Answer
Correct Me if I am Wrong
Next Time I will Try to provide the solution of 2nd, 5th
problem ..explanations from-others are appreciated
Thanks Regards
Shashank Mani Don't B Evil U Can Earn while U Learn
Computer Science Engg.
BIT Mesra
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regards,
chinna.
--
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[algogeeks] Re: google paper/...plz help..its urgent
1.c U Can verify by putting n =I where I is positive integer value say
n=5 try it out its so easy
2 a...what i have understood.
as we know that formal grammar is defined as (N, Σ, P, S)
so For instance, the grammar G with N = {S, A}, Σ = {a, b}, P
with start symbol S and rules
S → aA
A → Sb
S → ε
generates { a^ib^i : =0} so answer is A.
3 expected value doe discrete distributional is defined as
E(i)=sum(pi * xi); so from my points of view ans is 1/n ...Really Gud
Question one has think..still thinking
4.b -Explaination
Informally the NP-complete problems are the toughest problems in NP
in the sense that they are the ones most likely not to be in P. NP-
complete problems are a set of problems that any other NP-problem can
be reduced to in polynomial time, but retain the ability to have their
solution verified in polynomial time. In comparison, NP-hard problems
are those at least as hard as NP-complete problems, meaning all NP-
problems can be reduced to them, but not all NP-hard problems are in
NP, meaning not all of them have solutions verifiable in polynomial
time.
(A) is incorrect because set NP includes both P(Polynomial time
solvable) and NP-Complete .
(B) is incorrect because X may belong to P (same reason as (A))
(C) is correct because NP-Complete set is intersection of NP and NP-
Hard sets.
(D) is incorrect because all NP problems are decidable in finite set
of operations.
5. The Most Typical..Still Need Time
6 a zero degree means vertex is not connected from any other vertex
in graph
7.a
8.No Answer Answer Comes to Be 252
15c10,14c9,10c5,10*9*8*7*6 all are greater then from output so say
No Answer
Correct Me if I am Wrong
Next Time I will Try to provide the solution of 2nd, 5th
problem ..explanations from-others are appreciated
Thanks Regards
Shashank Mani Don't B Evil U Can Earn while U Learn
Computer Science Engg.
BIT Mesra
--
You received this message because you are subscribed to the Google Groups
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To post to this group, send email to algogeeks@googlegroups.com.
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For more options, visit this group at
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[algogeeks] Re: google paper/...plz help..its urgent
@bittu
I don't think answer of 6th question to be a)
No. of vertices of degree 0 will be those who didnot intersect with
any set i exactly 2 points. All sets of size greater than equal 2 must
intersect with any other set having exactly 2 common elements between
them in exactly 2 points. e.g if a set is (1,2) then it will be
adjacent to (1,2,3) , (1,2,3,4) etc..
The sets of size 0 and 1 cannot intersect in 2 points so they all will
be of degree 0.
Number of Sets of size 0 --- 1
Number of Sets of size 1 --- n
so Total number of vertices n+1.
In the similar way number of connected components will be n+2.
On Jan 15, 8:44 pm, bittu shashank7andr...@gmail.com wrote:
1.c U Can verify by putting n =I where I is positive integer value say
n=5 try it out its so easy
2 a...what i have understood.
as we know that formal grammar is defined as (N, Σ, P, S)
so For instance, the grammar G with N = {S, A}, Σ = {a, b}, P
with start symbol S and rules
S → aA
A → Sb
S → ε
generates { a^ib^i : =0} so answer is A.
3 expected value doe discrete distributional is defined as
E(i)=sum(pi * xi); so from my points of view ans is 1/n ...Really Gud
Question one has think..still thinking
4.b -Explaination
Informally the NP-complete problems are the toughest problems in NP
in the sense that they are the ones most likely not to be in P. NP-
complete problems are a set of problems that any other NP-problem can
be reduced to in polynomial time, but retain the ability to have their
solution verified in polynomial time. In comparison, NP-hard problems
are those at least as hard as NP-complete problems, meaning all NP-
problems can be reduced to them, but not all NP-hard problems are in
NP, meaning not all of them have solutions verifiable in polynomial
time.
(A) is incorrect because set NP includes both P(Polynomial time
solvable) and NP-Complete .
(B) is incorrect because X may belong to P (same reason as (A))
(C) is correct because NP-Complete set is intersection of NP and NP-
Hard sets.
(D) is incorrect because all NP problems are decidable in finite set
of operations.
5. The Most Typical..Still Need Time
6 a zero degree means vertex is not connected from any other vertex
in graph
7.a
8.No Answer Answer Comes to Be 252
15c10,14c9,10c5,10*9*8*7*6 all are greater then from output so say
No Answer
Correct Me if I am Wrong
Next Time I will Try to provide the solution of 2nd, 5th
problem ..explanations from-others are appreciated
Thanks Regards
Shashank Mani Don't B Evil U Can Earn while U Learn
Computer Science Engg.
BIT Mesra
--
You received this message because you are subscribed to the Google Groups
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[algogeeks] Re: google paper/...plz help..its urgent
Answer for question 6 maybe b)
also for question 7 maybe b)
On Jan 12, 2:14 pm, snehal jain learner@gmail.com wrote:
1. Quick-sort is run on two inputs shown below to sort in ascending
order
(i) 1,2,3, ….,n
(ii) n, n - 1, n - 2, …., 2, 1
Let C1 and C2 be the number of comparisons made for the inputs (i) and
(ii) respectively. Then,
a) C1 C2
b) C1 C2
c) C1 = C2
d) We cannot say anything for arbitrary n
2. Which of the following languages over {0, 1} is regular?
a) 0i1j such that i ≤ j
b) 0iw1j such that w ∈ {0, 1}∗ and i ≥ 0
c) All strings of 0s and 1s such that every pth character is 0 where p
is prime
d) None of the above
3. We are given a set X = {x1, x2, ..., xn} where xi = 2i. A sample S
(which is a subset of X) is
drawn by selecting each xi independently with probability pi = 1 / 2.
The expected value of the
smallest number in sample S is:
a) 1 / n
b) 2
c) sqrt(n)
d) n
4. Let S be an NP-complete problem and Q and R be two other problems
not known to be in
NP. Q is polynomial time reducible to S and S is polynomial-time
reducible to R. Which one of
the following statements is true?
a) R is NP-complete
b) R is NP-hard
c) Q is NP-complete
d) Q is NP-hard
5. For any string s ∈ (0 + 1)*, let d(s) denote the decimal value of s
(eg: d(101) = 5, d(011) = 3).
Let L = {s ∈ (0+1)* | d(s) mod 5 = 2 and d(s) mod 7 = 4}. Which of the
following statements is
true?
a) L is recursively enumerable, but not recursive
b) L is is recursive, but not context-free
c) L is context-free, but not regular
d) L is regular
Common data for questions 6 and 7
The 2n vertices of a graph G corresponds to all subsets of a set of
size n. Two vertices of G are
adjacent if and only if the corresponding sets intersect in exactly 2
elements
6. The number of vertices of degree zero in G is:
a) 1
b) n
c) 2n - 1
d) None
7. The number of connected components in G is:
a) 2n
b) n + 2
c) n C 2
d) None
8. There are 5 nested loops written as follows,
int counter = 0;
for (int loop_1=0; loop_1 10; loop_1++) {
for (int loop_2=loop_1 + 1; loop_2 10; loop_2++) {
for (int loop_3=loop_2 + 1; loop_3 10; loop_3++) {
for (int loop_4=loop_3 + 1; loop_4 10; loop_4++) {
for (int loop_5=loop_4 + 1; loop_5 10; loop_5++) {
counter++;}
}
}
}
}
What will be the value of counter in the end (after all the loops
finished running)?
a) 15C5
b) 14C5
c) 10C5
d) 10 * 9 * 8 * 7 * 6 * 5
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