Presentation by Aaron Bernstein (coauthor) on the paper https://www.youtube.com/watch?v=Bpw3yqWT_d0
Presentation by Danupon Nanongkai (coauthor) on the paper https://www.youtube.com/watch?v=awvBpvlbG1M
Implementation in Java https://github.com/nevingeorge/Negative-Weight-SSSP
https://algs4.cs.princeton.edu/44sp/
Read the top section then the first answer in the Q&A at the bottom.
Basically the problem is to find the least-cost (shortest) path across a graph of nodes from point a to point b, but the best algorithm we have (Djikstra’s) runs in exponential time in the worst case when there are some negative edge weights (see context below). This guy found an algorithm that runs must faster, in near-linear time.
Context that is helpful is to think of edge weights not as distances, but as time or cost (this allows negative weights). Imagine a process workflow or something.
Another helpful context that emerges from this is “negative cycle” which is an endless loop of negative weight edges (therefore asking an algorithm to find the “shortest” lowest weight path from one point to another that includes that cycle will fail). You can have negative weights without negative cycles though
Dijkstra doesn't handle negative weights so in that case you can use Ford-Bellman. It's slower than Dijkstra but far from exponential - O(VE) or something like that.
I took this straight from the Q&A I referenced in my link:
Q. Does Dijkstra's algorithm work with negative weights? A. Yes and no. There are two shortest paths algorithms known as Dijkstra's algorithm, depending on whether a vertex can be enqueued on the priority queue more than once. When the weights are nonnegative, the two versions coincide (as no vertex will be enqueued more than once). The version implemented in DijkstraSP.java (which allows a vertex to be enqueued more than once) is correct in the presence of negative edge weights (but no negative cycles) but its running time is exponential in the worst case.
This is a strange parenthetical. No shortest-path algorithm can be correct in the presence of a negative cycle; the concept of "shortest path" does not apply to such a graph.
I think @DropPanda gave a great example of regenerative charging in an EV.
In the video the author does use cost/price-function terminology.
Djikstra doesn't handle negative weights, Bellman-Ford is optimal for positive edge only. This apparently is also easier to parallelize.
The Quanta article mentioned below is excellent https://www.quantamagazine.org/finally-a-fast-algorithm-for-...
???
CLRS says it takes negative weight edges (just no negative cycles) and does it in O(VE). I have no idea where you're getting this from
What do you mean “optimal”? The Bellman-Ford algorithm gives you the shortest path, period.
In the simplest case, suppose you have a graph with exactly one negative cycle that is completely isolated from the rest of the graph. The shortest paths between vertices in the negative cycle are not well-defined, but the shortest paths in the rest of the graph are well-defined.
In general, look at the meta-DAG induced by the SCC decomposition of the graph. Mark any meta-vertices whose SCC contains some negative cycle. For s, t in the original graph, the distance from s to t is well-defined iff there exists no path from s's meta-vertex to t's meta-vertex in the meta-graph that traverses through a marked meta-vertex. (Proof left to the reader.)
When have a negative weight cycle, you can drive down the cost to be arbitrarily low by running through that cycle as many times as you'd like. What does "shortest path" or "lowest cost path" mean in this context then?
More info: https://stackoverflow.com/questions/13159337/why-doesnt-dijk...
Dijkstra's is effectively a greedy algorithm for finding shortest paths. By their very nature greedy algorithms cannot "look ahead." If there is a negative weight edge locked behind a expensive edge, Dijkstra's won't be able to see the negative weight edge until it's too late.
10
A -----------------> C
\ ^
\ 11 -15 /
\-----> B -----/For example, suppose you cannot exchange the US dollar for the Russian ruble. You have to go through some path such as USD -> EUR -> RUB. Of all possible paths, which one gives you the most favorable exchange? You can use Bellman-Ford to find out.
Why is it necessary to assume I can't exchange dollars directly for rubles? If that is possible, wouldn't I still want whatever path has the most favorable exchange rate?
I don't see what this has to add over just using the weights as they are. Dijkstra's algorithm tracks the distance from a source node to a target node. There's nothing about the algorithm that says that distance must be the sum of all the edge weights along a path; if it's the product, the algorithm will work without any changes.[1] The only requirement on the distance function is that you know how to calculate d(s,t) from the two values d(s, t') and d(t', t).
[1] You'll still have problems [with Dijkstra's algorithm] if some weights are greater than 1 and some aren't, which is the exact analog of the problems that occur when distance is additive and some weights differ in sign. But that has no bearing on whether you can use multiplication directly or you need to convert it to addition.
I think it relies on the assumption that d(a, c) >= d(a, b) + d(b, c). You could relax this of course, but then you're just back to bellman-ford algorithm.
When your distance aggregation is instead multiplicative and you allow multipliers < 1, this no longer holds.
Edit: I see your footnote already mentions this. I agree, the key point here isn't really whether things add or not, it's whether your aggregate distances are monotonically increasing or not. If they are, you can use dijkstra. If not, you have to use bellman ford or the fancier algorithms like the one in the arxiv.
A naïve attempt at adapting Dijkstra's to graphs with negative weight edges is to add a constant bias to every edge weight to make all weights positive, then run Dijkstra's. But that doesn't work because it changes the ordering of paths (wrt. cost) if their lengths differ. A path with length one will have its cost increased by B, while a path with length 100 will have its length increased by 100 * B. If the length-100 path was cheaper before adding the bias, after adding it bias it can be more expensive. So path ordering is not preserved.
But that idea, modified slightly, is not too far off a technique that works. Instead of adding a weight to each edge, the technique is to add a weight to each vertex (call this phi(v)) and modify each edge (u, v)'s weight to w(u, v) + phi(u) - phi(v). It turns out that modifying each edge's weight in this way preserves the ordering of paths that start from s and end at t. Consider a path s -> a -> b -> t; its new weight under w_phi is:
w(s, a) + phi(s) - phi(a) +
w(a, b) + phi(a) - phi(b) +
w(b, t) + phi(b) - phi(t)
= phi(s) + w(s, a) + w(a, b) + w(b, t) - phi(t)
This technique, called "price functions" is not novel -- it has been known from the 70's. The paper's main contribution is how to discover price functions very efficiently. The main way the paper does this as follows: first, decompose the graph into smaller strongly connected components (SCCs) whose paths have a "low" number of negative edge weight graphs by removing some small subset of edges (low diameter decomposition, or LDD). Second, recursively find a phi function that makes all edges in the interiors of these SCCs have positive weight. Third, modify that phi (since phis compose additively) for the DAG induced by the SCC decomposition: "fixup" weights of edges that cross from one SCC into another. Finally, modify that phi again to account for the edges that were removed from the graph. (This last part is usually "slow," but it turns out that it is fast because of the LDD algorithm tends to only remove a small number of edges in any path.)
Glossed over a lot of things, and probably not completely accurate -- see the lecture notes or the paper for the real analysis.
In RL we use Bellman's equations as a core component and implementations of Dijkstra's are embedded in many software suites as a useful tools for navigation (navmesh) etc....
This algorithm is faster than all of those, so you know: The same but faster
TL;DR: Everything will continue to speed up