Hello
For the data flow graph G= (V,E),I have to determine a partition - in
such a way that
sub graphs created could be computed in parallel and the sub graphs
are approximately of the same size
(sqrt(|V|))
For example as in graph attached ( http://i55.tinypic.com/35lvu6x.png -
the graph should be
Hello
For the data flow graph G= (V,E),I have to determine a partition - in
such a way that
sub graphs created could be computed in parallel and the sub graphs
are approximately of the same size
(sqrt(|V|))
For example as in graph attached ( http://i55.tinypic.com/35lvu6x.png -
the graph should be
Trivial algorithm : (Assuming it is in ascending order in both columns and
rows)
logarithmic time in max(n,m)
Divide the 2-d table to 4 parts, the -right-bottom-most and the
left-bottom-most are the smallest and largest values in the subtable.
It should be clear that atleast two subtables can be
http://fgiesen.wordpress.com/2009/12/13/decoding-morton-codes/
On Sep 28, 2:52 pm, yako...@gmail.com yako...@gmail.com wrote:
Hello
For the data flow graph G= (V,E),I have to determine a partition - in
such a way that
sub graphs created could be computed in parallel and the sub graphs
are
#includestdio.h
#includestdlib.h
int main()
{
int a[20],i,n,max,t,j,k;
printf(Enter the no. of elements\n);
scanf(%d,n);
for(i=0;in;i++)
scanf(%d,a[i]);
for(i=0;in-1;i++)
{
j=n-1;
max=0;
k=i;
while(ij)
{
Maximum Sized Binary Search Tree in a Binary Tree:
http://www.rawkam.com/?p=822
On Mon, Sep 27, 2010 at 10:34 AM, Chonku cho...@gmail.com wrote:
@Prodigy
As per your example, 8 15 20 25 which is the is indeed the maximum binary
search tree in this binary tree is only a solution to smaller
Hi Friends
This is an interesting problem and request you all to give a brief
intro to your code before putting the code. Or you may just mention
the under lying concept in the code. This will be of great help and
others will find it more ready to improve by adding there approach.
Coming back to
In my opinion also, this is a Majority vote algorithm as mentioned by
Navin and as coded by Dave. Only point I would add to @Dave's code is
that it wont be possible to find if the majority element has 2n/3
occurance as majority element keeps changing during the run and as the
majority algorithm
Power set : http://www.rawkam.com/?p=330
On Sun, Sep 19, 2010 at 2:45 AM, Gene gene.ress...@gmail.com wrote:
The power set of a set with N elements has size O(2^N). You can't do
anything with it in polynomial time.
On Sep 18, 5:03 pm, bittu shashank7andr...@gmail.com wrote:
can we solve
see a nice solution and related puzzle at
http://www.rawkam.com/?p=1034
On Tue, Sep 28, 2010 at 6:31 AM, sharad kumar aryansmit3...@gmail.comwrote:
hey isn't it suppposed to be tournament problem..
On Fri, Sep 24, 2010 at 12:06 AM, Divesh Dixit
dixit.coolfrog.div...@gmail.com wrote:
still there is some problem related to numbers encoding like..
22333101 here how will u going to
encode it???
On Tue, Sep 28, 2010 at 1:38 AM, ligerdave david.c...@gmail.com wrote:
it's a compression problem. using hex instead of oct saves more space
Given a singly-linked, find out (just give the basic logic) the mid
point of a single linked list in a single parse of the list. Assume
the program would be loaded in read-only memory so no manipulation of
the list is allowed.
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@ sorurav
yes , the basic logic is required so that the code can be understood in
single Run..
i also request the same to all dear friends.
Regards..
On Tue, Sep 28, 2010 at 8:11 PM, sourav souravs...@gmail.com wrote:
Hi Friends
This is an interesting problem and request you all
You are given an unlimited number of each of n different types of
envelopes. The dimensions of envelope type i are xi × yi.(i is in sub
script) In nesting envelopes inside one another, you can place
envelope A inside envelope B if and only if the dimensions A are
strictly smaller than the
that will be
12 23 34 1 0 1c1
what's the some related problem?
only last char in the group represent the char, leading chars
represent the number of the repeated char. space(or whatever you like
it to be) is the separator of groups. when you see space, replace w/
'\t'.
On Sep 28, 2:58 am,
A possible solution i can think is create a directed graph where each
vertex is a envelope and edges are from a bigger envelope to smaller
envelope ( one which can fit in bigger envelope ) . Now the problem is
reduce to finding longest path in the graph .
Regards
Rahul
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You received this
Can u elaborate more?
On Sep 28, 5:41 am, saurabh singh saurabh.n...@gmail.com wrote:
u can use log(n)+1 space to do that by using bit string
On Mon, Sep 27, 2010 at 10:37 PM, AdrianW atw...@gmail.com wrote:
Suppose you have N strings that can be generated on-the-fly, and you
wanted to
May be this helps
http://ds-gyan.blogspot.com/2009/12/two-numbers-with-minimum-difference.html
On Sep 28, 10:28 am, coolfrog$ dixit.coolfrog.div...@gmail.com
wrote:
@ sorurav
yes , the basic logic is required so that the code can be understood in
single Run..
i also request the same
http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-087-practical-programming-in-c-january-iap-2010/lecture-notes/MIT6_087IAP10_lec05.pdf
Thanking In Advance
--
Rahul K Rai
And The Geek Shall Inherit The Earth
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You received this message because you are subscribed to the
i think it is similar to finding max in a list O(n) or sorting algorithm
O(n log n)
-- Prashant Kulkarni
On Tue, Sep 28, 2010 at 11:33 PM, Rahul Singal rahulsinga...@gmail.comwrote:
A possible solution i can think is create a directed graph where each
vertex is a envelope and edges are
maintain two pointer
1) slow pointer : it will point to next node for each iteration (ie
node=node-next)
2) fast pointer : it will point to two ahead node for each iteration ( ie
node=node-next-next)
when fast pointer reaches the end of linked list (NULL), slow pionter will
point to middle of
Hi Rahul,
All it is saying is that you can do simple arithmetic on pointers
directly which modified the address it is pointing to.
Suppose you have an array of int type (int arr[10];)then you can
access elements of the array in following ways:
1. arr[1], arr[4] etc.
2. arr+1, arr+4 etc. as arr is
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