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Home/ Questions/Q 709019
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T04:27:24+00:00 2026-05-14T04:27:24+00:00

I’m working on a directed network problem and trying to compute all valid paths

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I’m working on a directed network problem and trying to compute all valid paths between two points. I need a way to look at paths up to 30 “trips” (represented by an [origin, destination] pair) in length. The full route is then composed of a series of these pairs:

route = [[start, city2], [city2, city3], [city3, city4], [city4, city5], [city5, city6], [city6, city7], [city7, city8], [city8, stop]]

So far my best solution is as follows:

    def numRoutes(graph, start, stop, minStops, maxStops):
    routes = []

    route = [[start, stop]]
    if distance(graph, route) != "NO SUCH ROUTE" and len(route) >= minStops and len(route) <= maxStops: 
            routes.append(route)

    if maxStops >= 2:
        for city2 in routesFromCity(graph, start):
            route = [[start, city2],[city2, stop]]
            if distance(graph, route) != "NO SUCH ROUTE" and len(route) >= minStops and len(route) <= maxStops: 
                routes.append(route)
    if maxStops >= 3:       
        for city2 in routesFromCity(graph, start):
            for city3 in routesFromCity(graph, city2):
                route = [[start, city2], [city2, city3], [city3, stop]]
                if distance(graph, route) != "NO SUCH ROUTE" and len(route) >= minStops and len(route) <= maxStops:
                    routes.append(route)

    if maxStops >= 4:
        for city2 in routesFromCity(graph, start):
            for city3 in routesFromCity(graph, city2):
                for city4 in routesFromCity(graph, city3):
                    route = [[start, city2], [city2, city3], [city3, city4], [city4, stop]]
                    if distance(graph, route) != "NO SUCH ROUTE" and len(route) >= minStops and len(route) <= maxStops:
                        routes.append(route)
    if maxStops >= 5:
        for city2 in routesFromCity(graph, start):
            for city3 in routesFromCity(graph, city2):
                for city4 in routesFromCity(graph, city3):
                    for city5 in routesFromCity(graph, city4):
                        route = [[start, city2], [city2, city3], [city3, city4], [city4, city5], [city5, stop]]
                        if distance(graph, route) != "NO SUCH ROUTE" and len(route) >= minStops and len(route) <= maxStops:
                            routes.append(route)
return routes

Where numRoutes is fed my network graph where numbers represent distances:

[[0, 5, 0, 5, 7], [0, 0, 4, 0, 0], [0, 0, 0, 8, 2], [0, 0, 8, 0, 6], [0, 3, 0, 0, 0]]

a start city, an end city and the parameters for the length of the routes.

distance checks if a route is viable and routesFromCity returns the attached nodes to each fed in city.

I have a feeling there’s a far more efficient way to generate all of the routes especially as I move toward many more steps, but I can’t seem to get anything else to work.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T04:27:25+00:00Added an answer on May 14, 2026 at 4:27 am

    You could use a recursive function. Your maxStops can be a parameter and each time you call you decrease this by 1. When minStops is 0, you yield a result, When the maxStops is 0 you stop recursing.

    Here is a code example:

    def routesFromCity(x):
        for i in range(2, 10):
            yield x * i
    
    def findRoutes(start, stop, minStops, maxStops):
        if start == stop:
            if minStops <= 0:
                yield []
        else:
            if maxStops > 0:
                for nextCity in routesFromCity(start):
                    for route in findRoutes(nextCity, stop, minStops - 1, maxStops - 1):
                        yield [(start, nextCity)] + route
    
    for route in findRoutes(1, 12, 2, 5):
        print route
    

    Output:

    [(1, 2), (2, 4), (4, 12)]
    [(1, 2), (2, 6), (6, 12)]
    [(1, 2), (2, 12)]
    [(1, 3), (3, 6), (6, 12)]
    [(1, 3), (3, 12)]
    [(1, 4), (4, 12)]
    [(1, 6), (6, 12)]
    
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