I take no offense. It seems this topic is over-flowing in many
countries, many hearts are unhappy and have no peace over this. They
forget the major sins they do, and focus on others' short-comings.
Although I'm not a muslim, I may drop in to orkut.com and see what some
children of God are thinki
That's true Dhyanesh, thanks. Any prose explanation for SPX2's pseudo
code?
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To
So
if(opening_value < 1.0)
step_value == .1
else if(opening_value < 10)
step_value == 1
else
step_value == 5 //or whatever value you like for larger steps
Now for any graph, you include 5 step levels, two below the stock being
displayed, two above, and the displayed stock right smack i
This problem is actually NP-Complete and is same as set-cover problem.
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Thanks Rootbear for your valuable suggestions.
-Original Message-
From: algogeeks@googlegroups.com [mailto:[EMAIL PROTECTED] On
Behalf Of rootbeer
Sent: Friday, March 10, 2006 6:19 AM
To: Algorithm Geeks
Subject: [algogeeks] Re: XML Document Matching
Hmmm. Although I have not played wit
Yes, this is NP-Complete and is called Partition problem or Subset-sum
problem. But I still fail to understand how Karthik's algo works.
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To po
Yes, I do accept the statement of Mayur. There are many other such forums and social groups do discuss and debate on such sensitive topics. Please do not fiddle with the focus and the intention of the group.
Adak, please do not take this is an offence. I am a muslim myself. So, hope U understan
Your solution is perfectly right. But this is NP hard problem and O(n L
). So for exponential L, the program runs into exponential time.
-DhyaneshOn 3/8/06, Karthik Singaram L <[EMAIL PROTECTED]> wrote:
This is the problem of dividing a
set into two such that the sum of their differences is minimu
This can be solved by dynamic programming. The recurrence would be something like this :
cost [ i , pi ] = min ( cost[ i+1 , i ] + Ci + i - pi ) ( i > pi )
( cost [i+1 , pi ] + i - pi ) ( i > pi )
cost [ n , j ] = Cn for 1<= j <=n
Least cost would be found
@pramod
not quite ... the equivalence tester does not give a "less than"
operation which is needed for sorting. It only tells whether they are
equal or not. You cannot order them on the basis of that.
-Dhyanesh
On 3/9/06, pramod <[EMAIL PROTECTED]> wrote:
Isn't third problem solved by sorting where
hi guys
new problem for uthink abt it and help me :)
Suppose we want to replicate a file over a collection of n servers,
labeled S1,S2,,Sn. To place a copy of the file at server Si results
in a placement cost of Ci, for an integer Ci>0. If a user requests the
file at server Si, and no
Have you found a solution yet?
Have you looked at an olap solution or a cross join with heavy filters.
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Yup.
There are ideas oozing out everywhere. If you need some advise I am
sure there is someone out there.
I get all excited when somene comes up with something brilliant. Did
you have something in mind?
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Hmmm. Although I have not played with the big documents yet, you will
want to do the following:
1. Have your targets as identifyable as you can.
2. You want to minimize your interation count.
I suppose after this you are still going slow. This is the fun part.
Now you have to look at different w
scratchProjects.com is a new programming site where you can learn to
code using real-life examples of programming projects. We will be
releasing a couple new articles each week during March, and hope to
grow even faster in the near future.
Writers not only have a chance to win a prize each month,
SPX2 wrote:
> And what you have to do after you have found path a1...an with a1 input
> and an output
> is to try to remember the cost of the path in a vector.
> that is all i think
> Depth path will assure different pathways.
> If you want the fastest path you can use dijkstra/royfloyd.
I think
And to add to Adak's philiosophy - this list is strictly (forgive my
impudence) for technical discussion (and in my personal opinion, it is
just plain evil to attempt to cause sparks by hurting others' religious
sentiments). Please keep such topics for lists that allow religious
propaganda.On 3/9/0
Hey. I'm making a app that draws a graph wiht the given data fromt he
stock market.
As of now i have it scaled genericaly for Y.
What i want to be ale to do is have the Y axis sclaed based on the
value taken form the data base.
Meaning I want to know if soemone can help me with an algorithim tha
one linea^=b^=a^=bOn 3/9/06, hemu <[EMAIL PROTECTED]> wrote:
A different way ( or operators) but underlined logic is same as that ofXORA= ( A & ~ B ) | ( ~A & B)B= ( A & ~ B ) | ( ~A & B)A= ( A & ~ B ) | ( ~A & B)
Don't send me any attachment in Micro$oft (.DOC, .PPT)
A different way ( or operators) but underlined logic is same as that of
XOR
A= ( A & ~ B ) | ( ~A & B)
B= ( A & ~ B ) | ( ~A & B)
A= ( A & ~ B ) | ( ~A & B)
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Isn't third problem solved by sorting where comparison function is
replaced by the equivalence tester? After sorting we just run through
the array to see if there n/2 repetitions.
ajay mishra wrote:
> @stefan , ur idea seems correct to me.
>
> On 3/8/06, SPX2 <[EMAIL PROTECTED]> wrote:
> >
> >
How about dynamic programming solution to this problem?
The nth entry may be chosen or may not be chosen. If it is chosen then
the sub problem to be solved is that of all those entries whose start
times are after the end time of the nth entry.
C(N,S) = 1+C(N',S') where S' = S U {n} and N' = tho
I am afraid, I don't understand your solution Karthik. When the loop is
run for program P1, Arr[0] = 1 and Arr[P1] = 1, when program P2 is
considered Arr[P0]=1, Arr[P2]=1 and Arr[P1+P2]=1 etc
After considering Pn, we have Arr[P1+...+Pn]=1, so how does this help
solve the problem?
Can you explain y
The problem (the original one) is what is called the upper-envelope
problem. Determining the upper-envelope in O(nlogn) requires one to use
what is called a duality transform.
A line in the plane y = ax + b , is mapped to the dual plane as a point
(a, -b) in the dual plane. And a point (h, k)
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