Hi, This problem is same as a Travelling salesman problem........In 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 <[email protected]> wrote: > Hi, > Still waiting for solution........... > > On Wed, Oct 7, 2009 at 3:18 PM, monty 1987 <[email protected]> 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 <[email protected]> wrote: >> >>> The min. distance between two points i.e. the euclidean distance between >>> two points. >>> >>> >>> On Tue, Oct 6, 2009 at 5:52 PM, MrM <[email protected]> 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) ! >>>> 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, 5:22 pm, 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. >>>> >>>> >>>> >>> >> > > > > --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
