11 comments

  • foota5 days ago
    I've gone down a bit of a rabbit hole on path finding in the last week or two (most recently, this isn't the first time). When you have some knowledge of the topology of the graph you can use different techniques to do better than djikstra's.

    Of course, if you have lots of time and space and a completely static graph, you can run all pairs shortest paths and simply store all the results for O(1) path lookup, but there are intermediates for varying types of graphs. This stack exchange article is a good overview: https://cstheory.stackexchange.com/questions/11855/how-do-th....

    I've been wondering about how well D* lite would perform in practice with a somewhat varying graph. I read some suggestions that if the graph is changing even a bit on occasion, then it will mostly degrade to A*, since many changed paths would need to be re-evaluated.

    In the context of games, I've also been thinking about a technique called true distance heurustics (TDH), where you essentially precompute the distances between some fixed set of nodes, and then use those as a part of the heurustic for A* (or D* lite in this case), but it seems like updating these TDH in the case of a changing graph might introduce just as much overhead as not having them in the first place. It might be an interesting trade off though, if you have some "lines" (e.g., think train lines) that are much faster than roadways, you could handle each of these specially via the TDH, and in exchange you would be able to assume a lower "max speed" for use with the A* heurustic, allowing you to explore fewer paths (since with a lower "max speed" paths will more rapidly increase in cost), whereas if you had to assume all car based paths could move as fast as a train, you would have to explore more paths.

    • kevinwang5 days ago
      > When you have some knowledge of the topology of the graph you can use different techniques to do better than djikstra's.

      But that statement doesn't apply to the version of Dijkstra's developed in this paper right?

      > Universal optimality is a powerful beyond-worst-case performance guarantee for graph algorithms that informally states that a single algorithm performs as well as possible for every single graph topology.

      • foota5 days ago
        No, I don't believe so.

        It clarifies specifically what problem it is optimal for "We prove that our working-set property is sufficient to guarantee universal optimality, specifically, for the problem of ordering vertices by their distance from the source vertex", but A* only explores a subset of vertices based on the heuristic, so it can be more efficient.

      • mlyle5 days ago
        > > When you have some knowledge of the topology of the graph you can use different techniques to do better than djikstra's.

        > But that statement doesn't apply to the version of Dijkstra's developed in this paper right?

        Simple thought experiment: I have an algorithm that has path costs for one graph memorized and first compares to that graph; it's O(V+E) to compare and return the value for that memorized graph. It obviously beats this algorithm for that graph.

        • foota5 days ago
          Hm... technically they're saying that their algorithm performs as well as possible "for every single graph topology". Your special cased algorithm would only work for one particular graph in a family of isomorphic graphs (unless you have a linear time algorithm for the graph isomorphism problem, in which case please share), but would fail for the rest of the graphs in the family. So I think you could still say that their algorithm performs as well as possible for that topology.
    • telgareith5 days ago
      FYI, TDH sounds like what Skyrim or one of the previous games uses.

      I can't remeber exactly, but $thing was supposed to be hard to find, but because of how they baked in pathing costs- if you followed something it'd lead you there.

    • bee_rider5 days ago
      TDH seems like it would tend to coincidentally reproduce the tendency of animals (humans included) to create paths by having lots of people travel through somewhere. The cause and effect is flipped, but the player doesn’t have to know that, right?
    • Jadrago5 days ago
      Does google maps use something like this? I imagine all pairs is too expensive, but the topology should be fairly consistent over time
      • ahoka5 days ago
        One trick we used when I was in the transportation business is to pre-calculate the distances between border crossings and gas stations.
      • DennisL1235 days ago
        Storing the distances for all pairs is prohibitively expensive. Also, you'd need to store the path information as well. Fortunately, road networks exhibit a lot of hierarchical structure. For example, when going far away, you will almost certainly use the long-distance sub-network, i.e. highways. It is possible to exploit this in a preprocessing step that adds a linear amount of information, which is in turn used to speed up queries.
      • urbandw311er5 days ago
        Google maps uses something called contraction hierarchies
        • DennisL1235 days ago
          Without knowing what they actually use, I feel comfortable to state that the industry has moved on from Contraction Hierarchies to somewhat more flexible techniques. These allow you to take traffic information and road closures, and user preferences, and whatnot into account without requiring a full re-processing of the input data with each traffic update. The state of the art is a two-step preprocessing that first decomposes the road network into cells, and then processes these cells independently. Sometimes it goes by the name of customisable route planning, sometimes it is called multi-level Dijkstra.
  • blt5 days ago
    The paper's name is shorter than this post title, and summarizes the result much better.
    • mikestew5 days ago
      It took me a few minutes before I realized that putting “n” at the end of “prove” makes the HN title readable.

      But yeah, should have just used the original title.

  • vanderZwan5 days ago
    > Our universal optimality result reveals a surprisingly clean interplay between this property and Dijkstra’s algorithm: Any heap with the working set property enables the algorithm to efficiently leverage every structural attribute of the graph it operates on, to the fullest extent that any comparison-based algorithm possibly can.

    That last bit makes me wonder: what would a shortest path algorithm without comparisons look like? Are there also "radix sort" like approaches to shortest-path algorithms that surpass comparison-based algorithms or something?

    • akoboldfrying5 days ago
      A sibling post answers your question, but I found your quote interesting for a different reason: This working set property seems like something that would be very useful in practice for quite a few problems, even if it can't be pushed all the way to proving universal optimality. We often have some freedom in choosing what order to supply inputs to a problem we're trying to solve; if we can order things so that, 99% of the time, the minimum item that we're looking for turns out to be within the last 1000 items considered instead of the complete set of 1000000, that's a nearly 10x constant factor speed up right there.
      • zeroonetwothree4 days ago
        Well not exactly because it’s the log of the number. So it’s log 1000 vs log 1000000 which is a much smaller difference.

        Also if you actually know the minimum item it’s much better for it to be first because it means you can skip almost all the work. The paper is focused on doing the best for a worst case input. In real life if you can guess you can improve the average case substantially. For example, this is what A* tries to do.

        • akoboldfrying4 days ago
          log 1000 is about 10 times smaller than log 1000000, when logs are taken to base 2.
          • akoboldfrying4 days ago
            Too late to edit so I'll self- reply: My claim was completely wrong, log 1000 is only about half of log 1000000, and this is true regardless of base. Whoops...
    • Sesse__5 days ago
      Yes. If your distances are dense integers, you can use a simple array as the priority queue in Dijkstra, and it will be faster than a heap (Dial’s algorithm).
    • noctune5 days ago
      You can use a radix heap rather than a binary heap. I have an implementation here, with benchmarks using pathfinding: https://github.com/mpdn/radix-heap

      It has the nice property that the amortized cost of pushing/popping an element is independent of the number of other elements in the heap.

  • westurner5 days ago
    "Universal Optimality of Dijkstra via Beyond-Worst-Case Heaps" (2024) https://arxiv.org/abs/2311.11793 :

    > Abstract: This paper proves that Dijkstra's shortest-path algorithm is universally optimal in both its running time and number of comparisons when combined with a sufficiently efficient heap data structure.

    Dijkstra's algorithm: https://en.wikipedia.org/wiki/Dijkstra%27s_algorithm

    NetworkX docs > Reference > Algorithms > Shortest Paths: https://networkx.org/documentation/stable/reference/algorith...

    networkX.algorithms.shortest_path.dijkstra_path: https://networkx.org/documentation/stable/reference/algorith... https://github.com/networkx/networkx/blob/main/networkx/algo...

    /? Dijkstra manim: https://www.google.com/search?q=dijkstra%20manim

  • akoboldfrying5 days ago
    Robert Tarjan's name is on a simply astounding number of breakthrough papers in graph algorithms, spanning decades.
    • joshhug5 days ago
      I had him as a teaching assistant when I was teaching data structures at Princeton back in Fall 2013. Princeton CS has their professors rotate through as TAs every so often through their courses.

      That semester, I made a slight mistake on the final exam where I asked students to create an algorithm that could find the second shortest path from s to every other vertex in a graph. I forgot to specify that the second shortest path should be simple (i.e. should not reuse any vertex twice). Having to deal with non-simple paths makes the problem much much harder.

      None of the students figured it out in the time available, and I'm sure I would also have been stumped if I had tried to solve the problem. Bob figured it out though. And then I remember he graded all 150 solutions to the problem himself, having as a blast as he went through students attempts at an effectively impossible problem.

      • UltraSane4 days ago
        Being that damn smart must feel amazing, almost like having a superpower.
  • m0llusk5 days ago
    In most real situations a graph is likely to be a model with some expected characteristics or perhaps data regarding real situations. Either way with modern computing it seems like in many cases using machine learning to predict the path or next steps on the path might actually end up being a more optimal method. The issue is how much data and modeling is available and how any processing of that would best be accounted for in final results that make use of any analysis.
  • fiddlerwoaroof5 days ago
    Does this mean that Dijkstra’s algorithm can perform better than something like A*?
    • Jtsummers5 days ago
      The two algorithms solve different (but related) problems. A* finds the shortest path from a source to a single target node. Dijkstra's finds the shortest paths from a source to all other nodes. If you're using Dijkstra's as a search algorithm then it may be slower than A* (often will be, but it depends on the heuristic), but you'll be terminating the algorithm early (once your target has been found you don't need to continue the algorithm).

      The algorithm under discussion is not that search-use of Dijkstra's, but the original all shortest paths use, so it's not directly comparable here to A*.

    • devit5 days ago
      A* with a consistent heuristic is Dijkstra on a modified graph whose edge weights are the original edge weights plus f(target) - f(source) where f is the A* "heuristic".

      If the heuristic is not consistent, the edge weights aren't necessarily nonnegative, but you can still use the "hybrid Bellman–Ford–Dijkstra algorithm", which is a generalization of Dijkstra that works for all graphs, and should be asymptotically better than naive A*.

    • jprete5 days ago
      A* is faster in practice if the heuristics used by the specific implementation are accurate and if the graph is "general" for the problem space. I'm being very loose with the word "general" but essentially it should have typical structure for the problem space it represents.

      There's almost certainly a paper somewhere proving that A* with a given heuristic can always be made O(large) by choosing the right adversarial inputs.

    • mvkg5 days ago
      The paper's claim for Dijkstra's is it's "a single algorithm performs as well as possible for every single graph topology". A* is an augmented version of Dijkstra's only applicable when there is a priori knowledge of a good heuristic for the topology (e.g. manhattan distance in a cartesian plane). Since there is almost certainly no heuristic that is universally optimal for all topologies, A* shouldn't be more universally optimal than Dijkstra's (and can probably perform worse given a bad heuristic).
    • red75prime5 days ago
      Others pointed that A* and Dijkstra's algorithm solve different problems. But there's another possibility: less general but more efficient algorithm. For example, there are faster algorithms for planar graphs.
    • gcr4 days ago
      The paper studies "... the problem of ordering vertices by their distance from the source vertex."

      If all you need is shortest path between just one pair of points, this result doesn't necessarily apply.

    • foota5 days ago
      I think A* is solving a different problem than dijkstra's, since it requires an admissible heuristic to do any better than dijkstra's.

      As long as you have an admissible heurustic, A* won't ever perform worse than dijkstra's.

      • superjan5 days ago
        An example for those not in the know: to find a shortest route on a realworld map, an admissible heuristic would be that the minimum travel distance between two nodes will be a straight line. While examining options, A* takes this into account, Dijkstra does not.
      • jvanderbot5 days ago
        A* is not solving a different problem. What happens if h(x)=0 for all x in A*?
        • Jtsummers5 days ago
          > A* is not solving a different problem.

          A* finds the shortest path from a node to a single other node. Dijkstra's finds the shortest paths from a node to all other nodes. If you use it as a search algorithm to find the shortest path to a single target, then yes, it's equivalent to A* with h(x)=0, but you're terminating Dijkstra's early (once your target is found) and not running the full algorithm.

        • foota5 days ago
          A different problem in the sense that A* is useless (it degrades to dijkstra's) when there is no admissible heuristic. So I think it's reasonable to say that A* solves a different problem (namely, path finding when there is an admissible heuristic), since when there's no admissible heuristic it is identical to dijkstra's.
    • 5 days ago
      undefined
    • entropicdrifter5 days ago
      There's a notable exception:

      >when combined with a sufficiently efficient heap data structure

      So it depends on the circumstances a bit.

  • moron4hire5 days ago
    This came up for me not long ago. A* is a specialization of Dijkstra's that is the canonical "path finding algorithm" for game development. A* is good for finding how to get from a specific point A to a specific point B. But I wanted to know how to get from any point A to a list of point Bs. And so it turned out that the extra work that Dijkstra's does that A* skips is exactly the work you want when doing such a thing. It's also cacheable, which is incredible in the modern era of having basically infinite memory for this sort of thing.
    • o11c5 days ago
      That's wrong, A* can trivially handle a set of points at one end (you might have to "reverse" the direction depending on which end has the set).
  • vtodekl5 days ago
    [dead]
  • impure5 days ago
    A* has entered the chat
    • twojacobtwo5 days ago
      Several other commenters have now pointed out the differentiations, in case you weren't aware.
  • heraldgeezer5 days ago
    I recognize the name due to studying CCNA in the past. His name comes up with OSPF routing protocol.