@Betlist: That is a neat diagram you posted. What tool did you use to
generate it?
On 4 May 2013 23:33, Joseph DeVincentis dev...@gmail.com wrote:
In your cases with 10 diamonds, you have probabilities of 1/16, 4/16,
6/16, 4/16, and 1/16 of being in the respective cases. To get to the first
Atleast the response can be a list of testcases that failed.
Then we can debug at out end.
In my case Question1 had 6 incorrect test cases of 100. Very low
probability that i will pick the incorrect test case to debug
On Sun, May 5, 2013 at 4:56 AM, Andres Felipe Ruiz andresfelip...@gmail.com
Dear Bartholomew,
Is it possible to have a virtual contest mode in the future that we can
compete with the contestants in the past contests? I may not be eligible to
play in that round but I could join that after the contest. The scoreboard will
be updated from time to time according to the
A brief outline:
Problem A:
If A=1, no motes can ever be eaten, so the answer is you have to remove all N.
On the other hand, if A1, suppose that in the optimal solution you remove k
motes. It seemed intuitive to me that the k motes you remove would be the
largest k and that for the others
Problem B:
For the small case, there are only at most 20 diamonds. So even if you need to
decide left/right all 20 times, there are at most 2^20 scenarios. We all know
that 2^20 is a million and a bit. So we can just play out all the scenarios and
work out the probability the hard way.
For
On Sunday, May 5, 2013 6:48:17 AM UTC+3, CoderBaba wrote:
Well contrary to what Bartholomew Furrow said I think the violators should
have been disqualified from Qualification round. In that case these emails
would not have come in one after another.
In the qualification round it's not
On Saturday, May 4, 2013 3:07:37 PM UTC-4, ICY wrote:
samarth3692 and HARSH94 have identical B-small solutions (ranks 981, 984).
prakhar120793 (rank 950) had a wrong code attached to B-small solution.
I am sure Google have better ways to detect cheaters... like a problem with a
very big input
But that would say there are 32 possibilities, but as drawn, there are only 30.
From first case for N=10, you go to the first case in N=11 in all cases,
because the diamond have to slide to the right...
Dňa nedeľa, 5. mája 2013 5:33:09 UTC+2 /dev/joe napísal(-a):
In your cases with 10
@Jugesh: I have to admit, it was created in mspaint O:-)
Dňa nedeľa, 5. mája 2013 8:11:46 UTC+2 Jugesh Sundram napísal(-a):
@Betlist: That is a neat diagram you posted. What tool did you use to
generate it?
On 4 May 2013 23:33, Joseph DeVincentis dev...@gmail.com wrote:
In your
I'd like to ask, why accepted solution returns for input
1
11 1 3
value 0.1875 (6/32)?
I believe that correct answer is 0.16667 (5/30).
The following picture shows all possibilities for N=11:
http://i.imgur.com/DF6QL2u.png
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There are 30 possibilities, but they are not all equally likely, because of
the result that after 4 diamonds are stacked on one side, the next one is
forced to the other side.
On Sun, May 5, 2013 at 3:48 AM, Betlista betli...@gmail.com wrote:
But that would say there are 32 possibilities, but
I don't get it that just checking all additions + removals record for a
mode is just enough.
Could you please help me understand that?
Please check mine.
http://ideone.com/gPoM1g
On Sun, May 5, 2013 at 12:27 PM, Luke Pebody luke.peb...@gmail.com wrote:
A brief outline:
Problem A:
If A=1,
The point is that removing a mote out of the middle of the sorted sequence
doesn't help. If you can already eat that mote then you are better off not
removing it. If you can't eat that mote without adding other motes, then
you will have to add other motes to be able to eat the next larger
*So the only cases for removing motes that you have to consider are cases
where all of the n largest motes are removed, for some n between 0 and all
of them.*
I didn't understand this.
On Sun, May 5, 2013 at 8:30 PM, Joseph DeVincentis dev...@gmail.com wrote:
The point is that removing a mote
In the previous step, I showed that removing a mote while not removing a
larger mote is not helpful. You will still need to add at least as many
motes as you would have needed to add if you have not removed the mote, in
order to get large enough to eat the larger mote. So your score will be
worse
* for each moteSize in motes:
* while currentSize = moteSize:
* currentSize += currentSize - 1
* additions += 1
* currentSize += moteSize
* removals -= 1
* if additions + removals record:
* Set record = additions + removals
Here, shouldn't you check whether any other
At most you will have to remove all motes (it doesn't make sense to add
more motes than n, since we can just remove all motes instead).
So, the lower limit is 0 (if you can already eat all the motes) and the
upper limit is n (all of them, since you can just remove all).
On Sun, May 5, 2013 at
There is 5/32 chance that exactly 1 out of 5 diamonds will roll to fall left,
and 1/32 chance that all will roll to fall right, even if the 5th one will
end up falling on left side.
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To
Thank you, I think I understood it ;-)
Dňa nedeľa, 5. mája 2013 15:40:37 UTC+2 /dev/joe napísal(-a):
There are 30 possibilities, but they are not all equally likely, because of
the result that after 4 diamonds are stacked on one side, the next one is
forced to the other side.
On
That gets checked in the next pass through the loop. If the next mote can
be eaten already, then in that next pass:
* The while loop is skipped, since currentSize moteSize
* The mote gets eaten
* The number of removals is reduced by one (since we ate the mote instead
of removing it
* And we again
Wow, this is all over the blogosphere!
https://twitter.com/ichattopadhyaya/status/331173658879983616
(yes, that's my twitter)
On Sun, May 5, 2013 at 7:06 AM, bigonion haibren...@gmail.com wrote:
On Sunday, May 5, 2013 6:48:17 AM UTC+3, CoderBaba wrote:
Well contrary to what Bartholomew
Hi.
Many times the Contest Analysis doesn't contain a working solution.
I know that we can see the solutions from other coders, but many times is hard
to understand their codes: no comments, variable's names ...
It would help a lot if Code Jam provide us with a working solution with a lot
of
a) That would entail question - In which language?
If you say C++ I will ask - why not Python. If you say All major
languages. Somebody would ask for LOLCODE.
So on one hand it is a question of resources - just like you said it is not as
simple as just copying some solution from contestant. If
Good points Stanislav! But there's one point here. They already provide
solution sometimes in Python! So they could keep posting solutions in this
language or C++ or Java, cause it's what the majority of people here uses,
and because one that programs in one of these languages is able to
undestand
I agree with Stanislav. I believe the Code Jam organizers would like the
participants to develop a level of thinking such that giving a brief
outline would be sufficient to understand the solution.
If you observed, the contest analysis gives us an insight into the essence
of a problem. Assuming
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