Any idea when are the t-shirts being sent out?
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Bartholomew, I can confirm this. I was 5xxth at the end of the round 1C and now
I am 644th.
On Monday, 12 May 2014 09:53:08 UTC-7, Bartholomew Furrow wrote:
srikkbhat, are you referring to Round 1B? That scoreboard changed after the
end of the contest when we identified contestants who had
Amit:
The insight is: imagine that contestant i gets a fraction Yi of the
public votes, the worst case scenario is if the rest of the public
votes are divided perfectly to allow all contestants to get the same
final score as our contestant (if there are any public votes left we
can distribute a
Upgrade to C++ and use the STL =)
On Apr 16, 3:49 pm, Oussama Bounaim o.boun...@gmail.com wrote:
Hi,
I'm looking for a C library that implements (stack, hashing,...) if you
know one please let me now about it.
Thanks.
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That's what happens in every rating system as far as I'm aware. If 10 people
get the exact same score and the same time and you take one second more, you
are the 11th, not the 2nd.
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*ranking system
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Hi,
I have used the ttmath http://www.ttmath.org/ library to solve problem A
of round 1B but I didn't upload it and I forgot to comment the code with a
link to the library... what should I do?
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@Brats:
I didn't write E[n=a+1] = 1 + E(n=a), I wrote for the specific case of a=3
(which is true for all a1) that:
E[n=a] = 1+ sum( p(n=0) * E[n=0] + p(n=1) * E[n=1] + ... + p(n=a) * E[n=a])
Where the n on the LHS is the number of elements in the wrong position
before hitting and on the RHS
You start at the southwest intersection (r=1, c=0) in the beginning.
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From the problem statement:
The *i*th of those lines contains information about intersections on the *
i*th row, where the 0th row is the northmost.
So, since we have N rows, the southmost row (where you start) is the
(N-1)ith row.
For rows=3, that is 2, yes.
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It's quite interesting that you choose the example of two coins, which
exactly resembles the game of goro for N=2, as in that case, p(hits=2) =
0.75.
The problem in your reasoning is that you are computing the cumulative
distribution function
I am not offended at all. I'm however starting to get a bit annoyed that you
don't pay attention to what I write.
I have explicitly defined X in that post as elements that go into the right
position after one hit for the array [2 1] (because that's what you were
asking about in your last
Ahah... nice joke =P (I hope...)
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For
@ Bharath
Linearity of expectation:
http://en.wikipedia.org/wiki/Expected_value#Linearity
Note that the second result is valid even if X is not statistically
independent of Y.
On May 10, 3:22 pm, Bharath Raghavendran rbharat...@gmail.com wrote:
I am not an expert in probability, but I don't
It's called MATLAB - The Language Of Technical Computing. So, yeah...
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Actually the proof has created a lot of confusion among some contestants.
Specifically in the first Lemma, the statement:
x't = t for t ≤ N - 1. This is true by the inductive hypothesis.
Includes the case x1, which doesn't make sense (it is impossible to have
only 1 element out of order).
@Eagle
Have you even bothered to read what I wrote about the same case, just
before you?...
On May 9, 7:38 am, Eagle khirwad...@gmail.com wrote:
I fail to understand the logic of the proof given in the 'contest
analysis'.
For the three unsorted numbers 3 1 2, the sample set of arrangements
of the equation?
Eagle
On May 9, 12:28 am, Pedro Osório mebm.pedroso...@gmail.com wrote:
Hi Ricardo,
For your example regarding 3, you have the following possibilities:
1 2 3 - perfect
1 3 2 - one correct
2 1 3 - one correct
2 3 1 - wrong
3 1 2 - wrong
3 2 1 - one correct
As you
@meir
It was my favorite problem too, but the problem is not easy in the
sense that proving that the solution is correct is far from trivial.
However since in these contests one normally uses their intuition and
pattern recognition to come up with a solution fast and not full
mathematical proofs,
Another reason may be that, regardless of n, the small input doesn't
include a certain tricky case where your solution fails, but the
large input does.
On May 9, 8:51 pm, Leopoldo Taravilse ltaravi...@gmail.com wrote:
I'd like to make a discussion about possible reasons for failing the Large
(a) is each element that is in the wrong position
(b) is each element that is not being held when Goro hits the table.
So T is the number of elements not being held - #set b + number of
elements being held which are in the wrong position - #(set a\set b).
Let us define #setb = Tb and #(set a\set
What is the mystery? They are intelligent so they can solve the
problems fast, they are familiar with the language they're using so
they can implement them fast and without mistakes. Practice helps in
both areas, and that's it.
On May 8, 7:38 pm, vivek dhiman vivek4dhi...@gmail.com wrote:
yeah!
Hi Ricardo,
For your example regarding 3, you have the following possibilities:
1 2 3 - perfect
1 3 2 - one correct
2 1 3 - one correct
2 3 1 - wrong
3 1 2 - wrong
3 2 1 - one correct
As you can see, if goro hits without holding anything, 1/6 of the time
it will be sorted, 1/2 of the time there
vivek:
I don't say it is only practice that matters the most.. Practice is
crucial, and none of the top guys get there without it, but the
latent ability that you have is the most important factor in my
opinion.
On May 8, 8:09 pm, vivek dhiman vivek4dhi...@gmail.com wrote:
So most of you say
You have a good point but I think they're excluding operating systems
from this one. That is, you can use a non-free operating system as
long as the environment running on that OS is free.
This is for practical reasons obviously, as everyone will use Windows/
Linux/OS X, when it comes to
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