Map Routing, a la Google Maps?

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生来不讨喜
生来不讨喜 2021-02-06 23:59

I\'ve always been intrigued by Map Routing, but I\'ve never found any good introductory (or even advanced!) level tutorials on it. Does anybody have any pointers, hints, etc?

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  •  深忆病人
    2021-02-07 00:17

    From a conceptual point of view, imagine dropping a stone into a pond and watching the ripples. The routes would represent the pond and the stone your starting position.

    Of course the algorithm would have to search some proportion of n^2 paths as the distance n increases. You would take you starting position and check all available paths from that point. Then recursively call for the points at the end of those paths and so on.

    You can increase performance, by not double-backing on a path, by not re-checking the routes at a point if it has already been covered and by giving up on paths that are taking too long.

    An alternative way is to use the ant pheromone approach, where ants crawl randomly from a start point and leave a scent trail, which builds up the more ants cross over a given path. If you send (enough) ants from both the start point and the end points then eventually the path with the strongest scent will be the shortest. This is because the shortest path will have been visited more times in a given time period, given that the ants walk at a uniform pace.

    EDIT @ Spikie

    As a further explanation of how to implement the pond algorithm - potential data structures needed are highlighted:

    You'll need to store the map as a network. This is simply a set of nodes and edges between them. A set of nodes constitute a route. An edge joins two nodes (possibly both the same node), and has an associated cost such as distance or time to traverse the edge. An edge can either either be bi-directional or uni-directional. Probably simplest to just have uni-directional ones and double up for two way travel between nodes (i.e. one edge from A to B and a different one for B to A).

    By way of example imagine three railway stations arranged in an equilateral triangle pointing upwards. There are also a further three stations each halfway between them. Edges join all adjacent stations together, the final diagram will have an inverted triangle sitting inside the larger triangle.

    Label nodes starting from bottom left, going left to right and up, as A,B,C,D,E,F (F at the top).

    Assume the edges can be traversed in either direction. Each edge has a cost of 1 km.

    Ok, so we wish to route from the bottom left A to the top station F. There are many possible routes, including those that double back on themselves, e.g. ABCEBDEF.

    We have a routine say, NextNode, that accepts a node and a cost and calls itself for each node it can travel to.

    Clearly if we let this routine run it will eventually discover all routes, including ones that are potentially infinite in length (eg ABABABAB etc). We stop this from happening by checking against the cost. Whenever we visit a node that hasn't been visited before, we put both the cost and the node we came from against that node. If a node has been visited before we check against the existing cost and if we're cheaper then we update the node and carry on (recursing). If we're more expensive, then we skip the node. If all nodes are skipped then we exit the routine.

    If we hit our target node then we exit the routine too.

    This way all viable routes are checked, but crucially only those with the lowest cost. By the end of the process each node will have the lowest cost for getting to that node, including our target node.

    To get the route we work backwards from our target node. Since we stored the node we came from along with the cost, we just hop backwards building up the route. For our example we would end up with something like:

    Node A - (Total) Cost 0 - From Node None
    Node B - Cost 1 - From Node A
    Node C - Cost 2 - From Node B
    Node D - Cost 1 - From Node A
    Node E - Cost 2 - From Node D / Cost 2 - From Node B (this is an exception as there is equal cost)
    Node F - Cost 2 - From Node D

    So the shortest route is ADF.

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