[pgrouting-users] algo which allow negative cost
Stephen Woodbridge
woodbri at swoodbridge.com
Mon Jul 8 12:43:50 PDT 2013
Right, what you are describing is known as "VRP with profits". The idea
is that you have some number of locations that have a profit associated
to them and typically you have some other constraints like the maximum
time for the tour. Then you have to optimize the selection of locations
based on the maximum time and maximize profit for the tour.
This is not a simple shortest path algorithm. If you have no time
constraint, then you can just use TSP with the POI's that you want to
visit. Specify a start point and the list of POIs as a distance matrix
and TSP will optimize the order to visit the POIs based on making it the
shortest tour length. As you mention, this forces the use of all points.
Otherwise, we do not have anything like this yet. Dijkstra with a
negative cost does not solve this problem.
Thanks for taking the time to explain your needs. Feel free to add a
ticket for a future enhancement. Please refer to "VRP with profits" in
the ticket so we know what you are looking for.
Thanks,
-Steve
On 7/8/2013 12:49 PM, Yaron Lev wrote:
> Stephan i appreciate your help,
>
> this is not a game problem, this is a routing problem.
>
> i have some poi, such museums etc. i want to offer a route to my users
> which will be the shortest, and will pass through as many poi's as possible.
>
> if i just +1 other edges, nothing forces the algo, to go through those
> poi's.
>
> imagine the following where:
> F - poi
> E - end point
> S- start point
>
>
> FxxxxE
> xxxxxx
> xxxxxx
> xxxxxS
>
> using +1 technique the route will not pass through F, trust me i tried :)
>
> using the TSP approach, which i also tried, it forces the user go
> through all of the poi, regardless of there distance, which might be
> counter productive.
>
> thanks.
>
>
>
> On Mon, Jul 8, 2013 at 7:25 PM, Stephen Woodbridge
> <woodbri at swoodbridge.com <mailto:woodbri at swoodbridge.com>> wrote:
>
> Why not just add 1.0 to all you edge costs so you minimum edge cost
> is 0 for fruits and 1+ for all other edges?
>
> I assume that a -1 cost just means that the edge is desirable to
> capture but is your goal is to collect some number of POI in an
> area, than it might be better to get the location of all the POI and
> the cost to get from each POI to each of the other POI then you can
> build a distance matrix to compute TSP solution which is the minimum
> cost to hit every POI.
>
> It is not clear to be what you want to achieve here. We do not have
> algorithms targeted at game theory or solving game like problems. So
> you have to re-map the problem into point to point shortest distance
> and multiple point order optimization problems.
>
> -Steve
>
>
> On 7/8/2013 12:19 PM, Yaron Lev wrote:
>
> Hi,
>
> i am wondering if there is any algo within pgrouting 2, which
> will allow
> edges to have negative costs (without negative circles)?
>
> i was not able to find such, and in this event, is there some clever
> way, to achieve a similar effect?
>
> in specific, i am trying to solve a pac-man like problem.
>
> trying to find best route, given graph, where on some edges
> there are
> "fruits" to collect, collecting fruits will have a positive effect.
>
> your goal to collect as many fruits as possible, fruit can have a -1
> cost for example while each edge has it length as a positive cost.
>
> the outcome should be a route, where at any distance(this is
> flexible,
> as long as it achieve a similar goal) from route, the outcome
> route will
> have the best score out of all other routes.
>
> i tried using shortest path algo, with out negative costs, but i am
> unable to come up with a solution which will indeed provide such or
> similar solution. (in specfic un able to come up with costs,
> which will
> indeed "force" the route to collect those "fruits")
>
> i will appreciate any insight or direction for solving this problem.
>
>
> * the fruits are actually poi, such as bus stations, museums etc.
> * if negative edge costs are allowed, the problem is solved.
>
> thanks for your help!
>
>
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