When I did a similar job, it wasn't really the distance that made the difference. It would sometimes be quicker to drive 30 miles into the country for a drop than to try to cross the city at rush hour or during the school run. So it might be worth trying to work out the times between each drop and then find the route that has the shortest total time, rather than do it by distances on a map.
Here, rather clumsily derived, is a list of the different routes between, say, 8 places:
numberOfPlaces = 8;
edges = DeleteCases[
Map[#[[1]] \[DirectedEdge] #[[2]] &,
Tuples[Range[1, numberOfPlaces], {2}]],
x_ \[DirectedEdge] x_ /; x == x]
and we can assign times in minutes for travelling between each place, here just randomly. This is a bit of work initially, but most drivers would be able to estimate the times quite quickly.
randomweights = RandomInteger[{0, 45}, Length[edges]]
{17, 38, 25, 17, 9, 42, 7, 42, 11, 14, 40, 17, 4, 23, 4, ... 40}
Combine these with the edges:
edgeWeightRules =
MapThread[
Rule, {edges, randomweights}] /.
(1 \[DirectedEdge] 4 -> n_) -> 1 \[DirectedEdge] 4 -> 200
I've set the value for traveling between places 1 and 4 to 200, for use later on. Next, draw a graph:
cg = Graph[edges,
VertexLabels -> "Name",
VertexStyle -> Red,
EdgeWeight -> edgeWeightRules,
EdgeLabels -> edgeWeightRules,
ImagePadding -> 10]
This is not a physical map, but a picture showing the different travel times between different places. Then create a weighted adjacency graph:
wam = Normal[WeightedAdjacencyMatrix[cg]] /. 0 -> Infinity;
wag = WeightedAdjacencyGraph[wam,
EdgeShapeFunction ->
GraphElementData[{"DotLine", "ArrowSize" -> .01}],
VertexLabels -> "Name",
EdgeWeight -> edgeWeightRules,
EdgeLabels -> edgeWeightRules,
EdgeLabelStyle -> Directive[Gray, Italic, 12],
VertexLabelStyle -> Directive[Black, 16],
EdgeStyle -> Directive[LightGray],
ImagePadding -> 20]
Apparently, a Hamiltonian cycle is not a two-wheeled delivery device, but a path visting every vertex exactly once:
FindHamiltonianCycle[wag, All]
{{1 -> 2, 2 -> 3, 3 -> 4, 4 -> 5, 5 -> 6, 6 -> 7, 7 -> 1}
so we should find all of them, and then sort them by the total time each one takes:
routes = Sort[{Total[# /. edgeWeightRules], #} & /@
FindHamiltonianCycle[wag, All]]
This returns a list of all possible cycles, with the quickest one first. So the quickest route that turns up at all locations once takes 54 minutes (that's driving past and throwing the items out of the window...:).
{54, {1 -> 6, 6 -> 4, 4 -> 3, 3 -> 5, 5 -> 7, 7 -> 2, 2 -> 1}}
Then it's possible to investigate the different routes, using a Manipulate to step through the Hamiltonian cycles:
Manipulate[
Column[{
Row[{"length of ", Last[routes[[n]]], " is ", First@routes[[n]]}],
HighlightGraph[
wag,
PathGraph[
Last@routes[[n]],
DirectedEdges -> True],
GraphHighlightStyle -> "Thick",
ImageSize -> 400]
}], {n, 1, Length[routes], 1}]
The artificially slowed-down route shows up as being part of the longest cycles.
At least this approach avoids the problems of trying to find routes through the one-way systems of cities.
