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 <[email protected]> > > 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 <[email protected]> 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 --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---
