Re: [algogeeks] Re: Ever growing sorted linked list
i want to ask one thing...the way some are saying first check with 2 then 4 and then 16to reach at that place we are suppose to traverse it and also hav eto put a condition say like countn or something...in this case also we are comparing so whats the usecorrect me if im wrong. On Wed, Jul 20, 2011 at 6:58 PM, bittu shashank7andr...@gmail.com wrote: can be done in O(n) find tow nodes from starting position find two nodes p,q such that p k k q as linked list is sorted we have to keep going on in right direction complexity will no less then O(N) as its linked list there is no notion of binary search sorted linked list think out why ? only think we can apply some logic to reduce the comparisons that's i also think will be gr8 improvement but approach sounds good if start comparing the nodes value using multiple of 2 fact .e.g. take an integer i=2^j from j=0 to start comparing 2^0th node, 2^1th node, 2^2th node2^jth node might be we are able to reduce the number of comparisons do notify me via gmail if i am wrong if u find difficulty in TC ? else happy learning if this would have been sorted array then we could have been solved it O(logn) suing same approach. Thanks Shashank Mani Computer Science Birla Institute of Technology, Mesra -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Ever growing sorted linked list
One a not so standard way of doing that is while creating the link list, keep an extra pointer in the node to point to the jump location. Just hold the previous node in a temp say address of 2 node and when i/p reaches 4, point the jumper pointer to 4 and make 4 the next jumper pointer. On Wed, Jul 20, 2011 at 7:27 PM, Ankur Khurana ankur.kkhur...@gmail.comwrote: @Popli : Bingo , that is what i was thinking and mentioned in my previous post.. On Wed, Jul 20, 2011 at 7:08 PM, Gaurav Popli abeygau...@gmail.comwrote: i want to ask one thing...the way some are saying first check with 2 then 4 and then 16to reach at that place we are suppose to traverse it and also hav eto put a condition say like countn or something...in this case also we are comparing so whats the usecorrect me if im wrong. On Wed, Jul 20, 2011 at 6:58 PM, bittu shashank7andr...@gmail.com wrote: can be done in O(n) find tow nodes from starting position find two nodes p,q such that p k k q as linked list is sorted we have to keep going on in right direction complexity will no less then O(N) as its linked list there is no notion of binary search sorted linked list think out why ? only think we can apply some logic to reduce the comparisons that's i also think will be gr8 improvement but approach sounds good if start comparing the nodes value using multiple of 2 fact .e.g. take an integer i=2^j from j=0 to start comparing 2^0th node, 2^1th node, 2^2th node2^jth node might be we are able to reduce the number of comparisons do notify me via gmail if i am wrong if u find difficulty in TC ? else happy learning if this would have been sorted array then we could have been solved it O(logn) suing same approach. Thanks Shashank Mani Computer Science Birla Institute of Technology, Mesra -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Ankur Khurana Computer Science , 4th year Netaji Subhas Institute Of Technology Delhi. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Ever growing sorted linked list
I think the novelty of this is that we are avoiding the comparison of new element with all the members of data in LL On Wed, Jul 20, 2011 at 7:27 PM, Ankur Khurana ankur.kkhur...@gmail.comwrote: @Popli : Bingo , that is what i was thinking and mentioned in my previous post.. On Wed, Jul 20, 2011 at 7:08 PM, Gaurav Popli abeygau...@gmail.comwrote: i want to ask one thing...the way some are saying first check with 2 then 4 and then 16to reach at that place we are suppose to traverse it and also hav eto put a condition say like countn or something...in this case also we are comparing so whats the usecorrect me if im wrong. On Wed, Jul 20, 2011 at 6:58 PM, bittu shashank7andr...@gmail.com wrote: can be done in O(n) find tow nodes from starting position find two nodes p,q such that p k k q as linked list is sorted we have to keep going on in right direction complexity will no less then O(N) as its linked list there is no notion of binary search sorted linked list think out why ? only think we can apply some logic to reduce the comparisons that's i also think will be gr8 improvement but approach sounds good if start comparing the nodes value using multiple of 2 fact .e.g. take an integer i=2^j from j=0 to start comparing 2^0th node, 2^1th node, 2^2th node2^jth node might be we are able to reduce the number of comparisons do notify me via gmail if i am wrong if u find difficulty in TC ? else happy learning if this would have been sorted array then we could have been solved it O(logn) suing same approach. Thanks Shashank Mani Computer Science Birla Institute of Technology, Mesra -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Ankur Khurana Computer Science , 4th year Netaji Subhas Institute Of Technology Delhi. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Thanks Regards, Saurabh -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Ever growing sorted linked list
hey
check my algo...i mentioned all the possible cases :)
On Tue, Jul 19, 2011 at 11:31 AM, oppilas . jatka.oppimi...@gmail.comwrote:
Yes we can avoid integer comparison.
But for jumping we will need to check whether the next node is null or
not.
So, depending upon the jump size, the number of comparison in worst case
can be very large if our jump size is increasing exponentially. So, why not
just compare with the next node instead of jumping.
On Tue, Jul 19, 2011 at 10:47 AM, sunny agrawal
sunny816.i...@gmail.comwrote:
yes Alok u r right that in any case we will traverse k times if k is the
position if the element that need to searched
but by jumping we can save the time of avoiding unnecessary comparisions
On Tue, Jul 19, 2011 at 10:44 AM, Alok Prakash alok251...@gmail.comwrote:
If i am not wrong, in a linked list to move to any number of position,
say n, we have to traverse all intermediate nodes of the linked list.
So, it does not matter if we are moving pointer by 2,4,8,... positions,
we have to scan all intermediate nodes.
Is it not a simple searching
On Tue, Jul 19, 2011 at 9:17 AM, sumit sumitispar...@gmail.com wrote:
Here is the idea I was thinking of:
- start from the root: if(root-data k) move to position 2 in the
list.
- in this way every time move the pointer to the position double the
current position and compare th eelement of node with k( moving here
is form 1 - 2 , 2-4, 4-8 ,8-16 ..)
- suppose you found a certain node at position n whose node-data
k , then now u only have to search for k between index n/2 to n (i.e.
if you found 16th element k then search for k between 8th and 16ht
element)..
correct me if any flaws.
On Jul 19, 3:38 am, Dumanshu duman...@gmail.com wrote:
@Gaurav: Ever Increasing means that you don't know the length of the
list. So you have to assume some index n, traverse the list upto that
point and check the results. If not found increment the n to some
higher value.
We are basically trying to improve the complexity by taking higher and
higher jumps for n.
On Jul 19, 2:03 am, Gaurav Popli abeygau...@gmail.com wrote:
yeah...im saying to reach position n at which k is placed we have to
trverse n positions from head or not...??
and i didnt understand the use of ever increasing...please elaborate
on it..
On Tue, Jul 19, 2011 at 2:08 AM, Dumanshu duman...@gmail.com
wrote:
@Swathi: plz give the TC of ur algo (exponential plus log)
On Jul 18, 10:14 pm, Swathi chukka.swa...@gmail.com wrote:
The solution to this problem will be a combination of exponential
increase
and the binary search..
start = 0;
end = 0;
index =0;
middle = 0;
while (1)
{
if(a[start] == data)
return start;
if(a[end] == data)
return end;
if(data end)
{
start = end;
end = pow(start,2); // here we are consider exponential for
faster
search.
// no need to check any boundary conditions as the array is
infinite
continue;
}
else
{
// do binary search between start index and end index because
data is
inbetween a[start] and a[end]
}
}
Hope this helps...
On Mon, Jul 18, 2011 at 10:32 PM, Gaurav Popli
gpgaurav.n...@gmail.comwrote:
i dont understand it..if k is at position an let supposeso
to
check at that position dont you have to traverse till that
position
...is thr anything else than the head of list...??or i
understood
wrongly...??
On Mon, Jul 18, 2011 at 9:55 PM, sagar pareek
sagarpar...@gmail.com
wrote:
Update on 2nd line
2.if( ptr2=ptr1-next-next...(5 or 10 times) ) goto
3.
else
make linear search till NULL encounter and exit with the
solution
On Mon, Jul 18, 2011 at 7:41 PM, sagar pareek
sagarpar...@gmail.com
wrote:
i have one approach :-
first compare root-data and k
if k is very much greater than root-data then set
next=5or10 ur choice
else set next 2or3 ur choice
take two pointers ptr1 and ptr2
now lets take k is much greater than root-data
then
1. set ptr1=root //initially
2. if( ptr2=ptr1-next-next...(5 or 10 times) ) else
make linear
search till NULL encounter
3. if ptr2-data==k return its position
4. else if (ptr2-datak) set ptr1=ptr2 goto 2
5. else traverse the ptr1 upto ptr2, if found return its
position else
return fail
if anyone has more efficient solution then pls tell :)
On Mon, Jul 18, 2011 at 6:53 PM, Dumanshu
duman...@gmail.com wrote:
You have a sorted linked list of integers of some length
you don't
know and it keeps on increasing. You are given a number k.
Print the
position of the number k.
Basically, you have to search for number k in the ever
growing sorted
list
Re: [algogeeks] Re: Ever growing sorted linked list
yeah...im saying to reach position n at which k is placed we have to
trverse n positions from head or not...??
and i didnt understand the use of ever increasing...please elaborate on it..
On Tue, Jul 19, 2011 at 2:08 AM, Dumanshu duman...@gmail.com wrote:
@Swathi: plz give the TC of ur algo (exponential plus log)
On Jul 18, 10:14 pm, Swathi chukka.swa...@gmail.com wrote:
The solution to this problem will be a combination of exponential increase
and the binary search..
start = 0;
end = 0;
index =0;
middle = 0;
while (1)
{
if(a[start] == data)
return start;
if(a[end] == data)
return end;
if(data end)
{
start = end;
end = pow(start,2); // here we are consider exponential for faster
search.
// no need to check any boundary conditions as the array is infinite
continue;
}
else
{
// do binary search between start index and end index because data is
inbetween a[start] and a[end]
}
}
Hope this helps...
On Mon, Jul 18, 2011 at 10:32 PM, Gaurav Popli
gpgaurav.n...@gmail.comwrote:
i dont understand it..if k is at position an let supposeso to
check at that position dont you have to traverse till that position
...is thr anything else than the head of list...??or i understood
wrongly...??
On Mon, Jul 18, 2011 at 9:55 PM, sagar pareek sagarpar...@gmail.com
wrote:
Update on 2nd line
2. if( ptr2=ptr1-next-next...(5 or 10 times) ) goto 3.
else
make linear search till NULL encounter and exit with the solution
On Mon, Jul 18, 2011 at 7:41 PM, sagar pareek sagarpar...@gmail.com
wrote:
i have one approach :-
first compare root-data and k
if k is very much greater than root-data then set next=5or10 ur choice
else set next 2or3 ur choice
take two pointers ptr1 and ptr2
now lets take k is much greater than root-data
then
1. set ptr1=root //initially
2. if( ptr2=ptr1-next-next...(5 or 10 times) ) else make linear
search till NULL encounter
3. if ptr2-data==k return its position
4. else if (ptr2-datak) set ptr1=ptr2 goto 2
5. else traverse the ptr1 upto ptr2, if found return its position else
return fail
if anyone has more efficient solution then pls tell :)
On Mon, Jul 18, 2011 at 6:53 PM, Dumanshu duman...@gmail.com wrote:
You have a sorted linked list of integers of some length you don't
know and it keeps on increasing. You are given a number k. Print the
position of the number k.
Basically, you have to search for number k in the ever growing sorted
list and print its position.
Please write the complexity of whatever solution you propose.
--
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--
Regards
SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD
--
Regards
SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD
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Re: [algogeeks] Re: Ever growing sorted linked list
If i am not wrong, in a linked list to move to any number of position, say
n, we have to traverse all intermediate nodes of the linked list.
So, it does not matter if we are moving pointer by 2,4,8,... positions, we
have to scan all intermediate nodes.
Is it not a simple searching
On Tue, Jul 19, 2011 at 9:17 AM, sumit sumitispar...@gmail.com wrote:
Here is the idea I was thinking of:
- start from the root: if(root-data k) move to position 2 in the
list.
- in this way every time move the pointer to the position double the
current position and compare th eelement of node with k( moving here
is form 1 - 2 , 2-4, 4-8 ,8-16 ..)
- suppose you found a certain node at position n whose node-data
k , then now u only have to search for k between index n/2 to n (i.e.
if you found 16th element k then search for k between 8th and 16ht
element)..
correct me if any flaws.
On Jul 19, 3:38 am, Dumanshu duman...@gmail.com wrote:
@Gaurav: Ever Increasing means that you don't know the length of the
list. So you have to assume some index n, traverse the list upto that
point and check the results. If not found increment the n to some
higher value.
We are basically trying to improve the complexity by taking higher and
higher jumps for n.
On Jul 19, 2:03 am, Gaurav Popli abeygau...@gmail.com wrote:
yeah...im saying to reach position n at which k is placed we have to
trverse n positions from head or not...??
and i didnt understand the use of ever increasing...please elaborate on
it..
On Tue, Jul 19, 2011 at 2:08 AM, Dumanshu duman...@gmail.com wrote:
@Swathi: plz give the TC of ur algo (exponential plus log)
On Jul 18, 10:14 pm, Swathi chukka.swa...@gmail.com wrote:
The solution to this problem will be a combination of exponential
increase
and the binary search..
start = 0;
end = 0;
index =0;
middle = 0;
while (1)
{
if(a[start] == data)
return start;
if(a[end] == data)
return end;
if(data end)
{
start = end;
end = pow(start,2); // here we are consider exponential for
faster
search.
// no need to check any boundary conditions as the array is
infinite
continue;
}
else
{
// do binary search between start index and end index because
data is
inbetween a[start] and a[end]
}
}
Hope this helps...
On Mon, Jul 18, 2011 at 10:32 PM, Gaurav Popli
gpgaurav.n...@gmail.comwrote:
i dont understand it..if k is at position an let supposeso to
check at that position dont you have to traverse till that
position
...is thr anything else than the head of list...??or i understood
wrongly...??
On Mon, Jul 18, 2011 at 9:55 PM, sagar pareek
sagarpar...@gmail.com
wrote:
Update on 2nd line
2.if( ptr2=ptr1-next-next...(5 or 10 times) ) goto 3.
else
make linear search till NULL encounter and exit with the
solution
On Mon, Jul 18, 2011 at 7:41 PM, sagar pareek
sagarpar...@gmail.com
wrote:
i have one approach :-
first compare root-data and k
if k is very much greater than root-data then set next=5or10
ur choice
else set next 2or3 ur choice
take two pointers ptr1 and ptr2
now lets take k is much greater than root-data
then
1. set ptr1=root //initially
2. if( ptr2=ptr1-next-next...(5 or 10 times) ) else make
linear
search till NULL encounter
3. if ptr2-data==k return its position
4. else if (ptr2-datak) set ptr1=ptr2 goto 2
5. else traverse the ptr1 upto ptr2, if found return its
position else
return fail
if anyone has more efficient solution then pls tell :)
On Mon, Jul 18, 2011 at 6:53 PM, Dumanshu duman...@gmail.com
wrote:
You have a sorted linked list of integers of some length you
don't
know and it keeps on increasing. You are given a number k.
Print the
position of the number k.
Basically, you have to search for number k in the ever growing
sorted
list and print its position.
Please write the complexity of whatever solution you propose.
--
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Groups
Algorithm Geeks group.
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algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
--
Regards
SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD
--
Regards
SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD
--
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To post to
Re: [algogeeks] Re: Ever growing sorted linked list
yes Alok u r right that in any case we will traverse k times if k is the
position if the element that need to searched
but by jumping we can save the time of avoiding unnecessary comparisions
On Tue, Jul 19, 2011 at 10:44 AM, Alok Prakash alok251...@gmail.com wrote:
If i am not wrong, in a linked list to move to any number of position, say
n, we have to traverse all intermediate nodes of the linked list.
So, it does not matter if we are moving pointer by 2,4,8,... positions, we
have to scan all intermediate nodes.
Is it not a simple searching
On Tue, Jul 19, 2011 at 9:17 AM, sumit sumitispar...@gmail.com wrote:
Here is the idea I was thinking of:
- start from the root: if(root-data k) move to position 2 in the
list.
- in this way every time move the pointer to the position double the
current position and compare th eelement of node with k( moving here
is form 1 - 2 , 2-4, 4-8 ,8-16 ..)
- suppose you found a certain node at position n whose node-data
k , then now u only have to search for k between index n/2 to n (i.e.
if you found 16th element k then search for k between 8th and 16ht
element)..
correct me if any flaws.
On Jul 19, 3:38 am, Dumanshu duman...@gmail.com wrote:
@Gaurav: Ever Increasing means that you don't know the length of the
list. So you have to assume some index n, traverse the list upto that
point and check the results. If not found increment the n to some
higher value.
We are basically trying to improve the complexity by taking higher and
higher jumps for n.
On Jul 19, 2:03 am, Gaurav Popli abeygau...@gmail.com wrote:
yeah...im saying to reach position n at which k is placed we have to
trverse n positions from head or not...??
and i didnt understand the use of ever increasing...please elaborate
on it..
On Tue, Jul 19, 2011 at 2:08 AM, Dumanshu duman...@gmail.com wrote:
@Swathi: plz give the TC of ur algo (exponential plus log)
On Jul 18, 10:14 pm, Swathi chukka.swa...@gmail.com wrote:
The solution to this problem will be a combination of exponential
increase
and the binary search..
start = 0;
end = 0;
index =0;
middle = 0;
while (1)
{
if(a[start] == data)
return start;
if(a[end] == data)
return end;
if(data end)
{
start = end;
end = pow(start,2); // here we are consider exponential for
faster
search.
// no need to check any boundary conditions as the array is
infinite
continue;
}
else
{
// do binary search between start index and end index because
data is
inbetween a[start] and a[end]
}
}
Hope this helps...
On Mon, Jul 18, 2011 at 10:32 PM, Gaurav Popli
gpgaurav.n...@gmail.comwrote:
i dont understand it..if k is at position an let supposeso to
check at that position dont you have to traverse till that
position
...is thr anything else than the head of list...??or i understood
wrongly...??
On Mon, Jul 18, 2011 at 9:55 PM, sagar pareek
sagarpar...@gmail.com
wrote:
Update on 2nd line
2.if( ptr2=ptr1-next-next...(5 or 10 times) ) goto 3.
else
make linear search till NULL encounter and exit with the
solution
On Mon, Jul 18, 2011 at 7:41 PM, sagar pareek
sagarpar...@gmail.com
wrote:
i have one approach :-
first compare root-data and k
if k is very much greater than root-data then set next=5or10
ur choice
else set next 2or3 ur choice
take two pointers ptr1 and ptr2
now lets take k is much greater than root-data
then
1. set ptr1=root //initially
2. if( ptr2=ptr1-next-next...(5 or 10 times) ) else make
linear
search till NULL encounter
3. if ptr2-data==k return its position
4. else if (ptr2-datak) set ptr1=ptr2 goto 2
5. else traverse the ptr1 upto ptr2, if found return its
position else
return fail
if anyone has more efficient solution then pls tell :)
On Mon, Jul 18, 2011 at 6:53 PM, Dumanshu duman...@gmail.com
wrote:
You have a sorted linked list of integers of some length you
don't
know and it keeps on increasing. You are given a number k.
Print the
position of the number k.
Basically, you have to search for number k in the ever
growing sorted
list and print its position.
Please write the complexity of whatever solution you propose.
--
You received this message because you are subscribed to the
Google
Groups
Algorithm Geeks group.
To post to this group, send email to
algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
--
Regards
SAGAR PAREEK
COMPUTER SCIENCE
