Dave has correctly understood the problem.
I take it from the problem description that you are given n intervals
[r1(i),r2(i)]_i=1^n and are to choose an x(i) in each interval so as
to minimize the maximum value of |x(i)-x(j)|. The intervals may
overlap, in which case you may be able to pick equal
It is not a solution for minimize maximum distance either, since |x(i)-x(j)|
<= c does not hold for every pair of points, only for adjacent points.
2009/10/21 saltycookie
> Yes, you are quite right. If I am not mistaken, you give a good solution
> for finding the minimum maximum distance.
>
> Bu
I take it from the problem description that you are given n intervals
[r1(i),r2(i)]_i=1^n and are to choose an x(i) in each interval so as
to minimize the maximum value of |x(i)-x(j)|. The intervals may
overlap, in which case you may be able to pick equally-spaced points,
or they may not overlap,
to maximize the minimum distance between any two points:-> to maximize the
minimum distance between adjacent points
-> for this all points must be equally spaced.
hence, choose 'n' equally spaced points in the range (r1, r2) starting from
r1 and ending at r2.
2009/10/21 saltycookie
> Yes, you
Yes, you are quite right. If I am not mistaken, you give a good solution for
finding the minimum maximum distance.
But what about the original problem where we want to find the maximum
minimum distance? I am not clear about the connection between the two
problems.
Thanks.
2009/10/21 Dave
>
> 林
林夏祥 , think again. If we are trying to minimize the maximum distance,
then we want to minimize the upper bound. That is what I specified:
letting c be the upper bound, find the smallest c such that all of the
distances do not exceed c. That gives rise to the inequalities
|x(i)-x(j)| <= c.
If neces
I guess choosing n equally spaced points withing the given interval will be
the desired solution.
2009/10/20 Mithun Kumar Singh
> Hi,
>This problem is same as a Travelling salesman problemIn
> travelling salesman we need to cover points in Min distance...here we need
> to just opposi
I don't think LP can solve it. We are to maximize c, not minimize c.
The formulas we have are:
|x(i)-x(j)| >= c for all i and j
r1(i) <= x(i) <= r2(i) for all i
The first inequality actually is combination of two linear equalities: x(i)
- x(j) >= c or x(i) - x(j) <= -c. Notice the relation of the
This is a linear programming problem. The way you formulate the
problem depends on the capabilities of the linear programming software
you have.
Basically, you want to
minimize c
by finding x(1) to x(n) such that
|x(i)-x(j)| <= c for all i and j
r1(i) <= x(i) <= r2(i) for all i
Dave
On Oct 5,
Hi,
This problem is same as a Travelling salesman problemIn
travelling salesman we need to cover points in Min distance...here we need
to just opposite..
PS: Answer may be misleading ...if so Pls correct me... :)
-Mithun
On Tue, Oct 20, 2009 at 6:16 PM, monty 1987 <1986mo...@gmail.
Hi,
Still waiting for solution...
On Wed, Oct 7, 2009 at 3:18 PM, monty 1987 <1986mo...@gmail.com> wrote:
> The important thing is all the points do not lie in same range i.e.
> x1 ,x2 ,x3 each of them have their own range.
>
>
> On Wed, Oct 7, 2009 at 3:15 PM, monty 1987 <1986mo...
The important thing is all the points do not lie in same range i.e.
x1 ,x2 ,x3 each of them have their own range.
On Wed, Oct 7, 2009 at 3:15 PM, monty 1987 <1986mo...@gmail.com> wrote:
> The min. distance between two points i.e. the euclidean distance between
> two points.
>
>
> On Tue, Oct 6, 2
The min. distance between two points i.e. the euclidean distance between two
points.
On Tue, Oct 6, 2009 at 5:52 PM, MrM wrote:
>
> you can arrange them with equal distances !
> if n=1 then, it does not matter where you put the point !
> if n>1 then, put them with distances = (r2i-r1i) / (n-1) !
you can arrange them with equal distances !
if n=1 then, it does not matter where you put the point !
if n>1 then, put them with distances = (r2i-r1i) / (n-1) !
it means ou put the first point on r1i and the last point on r2i, the
remaining point are distributed with equal distances !
On Oct 5, 5
Can you please make question a bit clear. What does minimum distance between
any two points. Do u mean sum of minimum distance?
On Mon, Oct 5, 2009 at 7:52 PM, monty 1987 <1986mo...@gmail.com> wrote:
> We have to locate n points on the x-axis
> For each point xi
> the
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