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, in which the endpoints of the intervals
become constraints on the selection of the x(i), and may prevent you
from selecting equally-spaced points.
Dave
On Oct 21, 10:21 am, Vivek S <s.vivek.ra...@gmail.com> wrote:
> 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 <saltycoo...@gmail.com>
>
>
>
>
>
> > 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 <dave_and_da...@juno.com>
>
> >> 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 necessary, this can be written as two inequalities:
> >> x(i) - x(j) <= c and
> >> x(j) - x(i) <= c.
>
> >> Since the relationship is "and," we can just use the two inequalities
> >> as part of the constraint conditions.
>
> >> Dave
>
> >> On Oct 21, 12:02 am, 林夏祥 <saltycoo...@gmail.com> wrote:
> >> > 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 two is
> >> "or",
> >> > and we cannot put them together to get a system of linear inequalities.
> >> > 2009/10/21 Dave <dave_and_da...@juno.com>
>
> >> > > 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, 9:22 am, monty 1987 <1986mo...@gmail.com> wrote:
> >> > > > We have to locate n points on the x-axis
> >> > > > For each point xi
> >> > > > the x co-ordinate of it lies between a
> >> range
> >> > > > [r1i,r2i]
> >> > > > Now we have to decide the location of points such that
> >> > > > minimum { distance between any two points } is maximum.
>
> >> > > > Any answer is welcomed.
>
> >> > --
> >> > 此致
> >> > 敬礼！
>
> >> > 林夏祥- Hide quoted text -
>
> >> > - Show quoted text -
>
> > --
> > 此致
> > 敬礼！
>
> > 林夏祥
>
> --
> "Reduce, Reuse and Recycle"
> Regards,
> Vivek.S- Hide quoted text -
>
> - Show quoted text -
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