I am I not wrong, the problem asks for fib numbers which are even, not fib
numbers with even index...?
On Sat, Oct 1, 2011 at 10:16 PM, Wladimir Tavares wladimir...@gmail.comwrote:
F(0)+F(2)+...+F(2n) = F(2n+1) - 1
Wladimir Araujo Tavares
*Federal University of Ceará
What is the maximum number of edges possible in a graph with N nodes, and
where any three nodes can have at most two edges between them. 1=N=10.
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Shuaib
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I don't have a formal proof yet, but can anyone give a counter test case to
following:
Let f(n) be our function:
if n is even: f(n) = (n^2)/4
else: f(n) = ((n-1)^2)/4 + (n-1)/2
On Wed, Sep 7, 2011 at 5:36 AM, Shuaib Khan aries.shu...@gmail.com wrote:
What is the maximum number of edges
You don't need trial and error. Start by setting x = 1, 2, 3... and find
values for y. If the y you get is a positive integer, you have a solution.
On Sun, Aug 28, 2011 at 5:46 PM, sivaviknesh s sivavikne...@gmail.comwrote:
*Find the number of solutions for 3x+4y=60, if x and y are positive
Dave, I agree. :)
On Tue, Aug 16, 2011 at 6:33 PM, Dave dave_and_da...@juno.com wrote:
@Shuaib: We are talking about an array of numbers, arent't we? It is
natural to assume that the numbers fall into one of the defined data
types.
Dave
On Aug 16, 4:23 am, Shuaib aries.shu...@gmail.com
On Wed, Aug 10, 2011 at 1:45 AM, Brijesh Upadhyay
brijeshupadhyay...@gmail.com wrote:
No thers is not.. someone has asked me this., dont know anything else about
the question :|
Seems like you got half of the question there.
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Man, I feel so stupid. Yes, it is a case of conditional probability. We have
to calculate the probability of six heads, given that 5 heads have
occured. So answer is 17/18.
On Tue, Aug 9, 2011 at 1:47 AM, Arun Vishwanathan aaron.nar...@gmail.comwrote:
@shady: 3/5 can be the answer to such a
On Sun, Aug 7, 2011 at 11:06 PM, Dave dave_and_da...@juno.com wrote:
@Yasir: Sort the numbers. Then
int i = 0, j = 1, m, p = 0;
while( j N )
{
m = a[j] - a[i];
if( m = K )
++p;
else if( m K )
++j;
else
++i;
}
// answer = p
Looks like an infinite
On Mon, Aug 8, 2011 at 12:40 AM, Puneet Gautam puneet.nsi...@gmail.comwrote:
Sixth toss is independent of previous tosses and dependent only on
coin selection...!
1/5 + 4/5(1/2)= 3/5
is the correct answer
we want to calc. probability of getting heads the sixth time only
even if
is the
probability that heads will show up the first five times, plus a sixth time.
Not just the sixth time. The first five times head showing up is part of the
question.
On Mon, Aug 8, 2011 at 1:12 AM, Shuaib Khan aries.shu...@gmail.comwrote:
On Mon, Aug 8, 2011 at 12:40 AM, Puneet Gautam
On Mon, Aug 8, 2011 at 12:51 AM, aseem garg ase.as...@gmail.com wrote:
@Shuaib: **What is the probability that you toss *next time, heads turns
up***.
Well if you interpret it your way, then you are right. Otherwise, not.
Aseem
On Mon, Aug 8, 2011 at 1:19 AM, Shuaib Khan aries.shu
How about iterating over all the values and hashing them to a dict/map. And
then iterate once more, checking that for each element 'e', |K-e| exists in
the map or not. O(N)?
On Sun, Aug 7, 2011 at 10:53 PM, Yasir yasir@gmail.com wrote:
Guys,
Let's try to find out an efficient solution for
On Mon, Aug 8, 2011 at 1:07 AM, Yasir Imteyaz yasir@gmail.com wrote:
@Shuaib,
You are right, this approach will work! :)
But for each element 'e'
instead of checking whether *|K-e| *exists, *you should check for either
(e+K) or (e-K).*
You are right. Mistake. ;)
But here the
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