Hi ankur,
can you give some more elaboration of that problem becoz i think some
condition will be der. whether max time, min time avg time .. like of
ant can traverse each once like that
On Tue, Nov 3, 2009 at 9:44 AM, ankur aggarwal ankur.mast@gmail.comwrote:
An ant stays at
7.
Define L1 as being a hop away from the origin, L2 as being 2 hops away
from origin, L3 as being 3 hops away.
Thus we are looking for E[X|L1] since originally the ant is only a hop
away.
So, E[X|L1]=1/3(1)+2/3(E[X|L2]+1) since it may return to the origin
with chance 1/3 or go away another level
thanks to all of u ...
i think curve fitting fits here ... thanks..
On Oct 31, 10:39 am, nikhil garg nikhilgar...@gmail.com wrote:
This is a very common problem actually. And is most often solved using curve
fitting only. We choose the curve to be polynomial of minimum degree and
then use
firstly 5 races are required, because if even one horse is not used,
this horse maybe the fastest and so the result will
not be correct.
with the following steps, 7 races are enough and so the question
becomes will 6 races or even 5 races enough?
1. group the horses 5 by 5, and suppose the
old problem
7 races
first split 25 horses into five group, group A,B,C,D,E
Ai means the ith rank of group A, the same as Bi,Ci,Di,and Ei.
first 5 races, one race for one group.
the 6th race select A1,B1,C1,D1,E1,we get the fastest three horse.
assume it's A1,B1,C1, (A1 faster than B1,and B1 faster
Is there a way to find the sum of the Kth series ( Given below)
K=0 S={1,2,3,4,5,6,}
K=1 S={1,2,4,7,11,16..} common diff = 1,2,3,4 5 ...
K=2 S={1,2,4,8,15,26...} common diff = 1,2,4,7 11... (series with
K=1)
K=3 S={1,2,4,8,16,31...} common diff = 1,2,4,8 15... (series with
K=2)
Note
it has been discusse rite??
On Fri, Nov 6, 2009 at 7:44 AM, Shiqing Shen ssq...@gmail.com wrote:
old problem
7 races
first split 25 horses into five group, group A,B,C,D,E
Ai means the ith rank of group A, the same as Bi,Ci,Di,and Ei.
first 5 races, one race for one group.
the 6th race
This is a 'finite calculus' (differences summations) problem.
You can solve it using difference operator (actually its inverse which gives
you the discrete integration which is nothing but summation).
If you do not know finite calculus, Google for it (or refer Concrete
Mathematics by Knuth).
The
