In plain words: Gradient descent on a bounded convex shape solves exactly the problems in two known classes. Finding a spot where a smooth function on a square cannot be locally improved is the first problem shown fully hard for that intersection, which equals continuous local search.
Abstract · The Complexity of Gradient Descent: CLS = PPAD $\cap$ PLS
We study search problems that can be solved by performing Gradient Descent on a bounded convex polytopal domain and show that this class is equal to the intersection of two well-known classes: PPAD and PLS. As our main underlying technical contribution, we show that computing a Karush-Kuhn-Tucker (KKT) point of a continuously differentiable function over the domain $[0,1]^2$ is PPAD $\cap$ PLS-complete. This is the first non-artificial problem to be shown complete for this class. Our results also imply that the class CLS (Continuous Local Search) - which was defined by Daskalakis and Papadimitriou as a more "natural" counterpart to PPAD $\cap$ PLS and contains many interesting problems - is itself equal to PPAD $\cap$ PLS.
John Fearnley, Paul W. Goldberg, Alexandros Hollender, Rahul Savani
arXiv:2011.01929 · cs.CC, cs.LG, math.OC · submitted Nov 3, 2020 · updated Mar 3, 2023
abstract · pdf · html · Journal version
"Our main result is to show that finding a point where Gradient Descent on a continuously differentiable function terminates—or equivalently a KKT point—is PPAD ∩ PLS-complete, when the domain is a bounded convex polytope. This continues to hold even when the domain is as simple as the unit square [0, 1]. The PPAD ∩ PLS-completeness result applies to the “white box” model, where functions are represented as arithmetic circuits.
"...As an immediate consequence, our result provides convincing evidence that the problem is computationally hard. First of all, there are reasons to believe that PPAD ∩ PLS is hard simply because PPAD and PLS are believed to be hard. Indeed, if PPAD ∩ PLS could be solved in polynomial time, then, given an instance of a PPADcomplete problem and an instance of a PLS-complete problem, we would be able to solve at least one of the two instances in polynomial time. Furthermore, since CLS ⊆ PPAD ∩ PLS, the above-mentioned cryptographic hardness of CLS applies automatically to PPAD ∩ PLS, and thus to our problem of interest..."
https://arxiv.org/pdf/2011.01929.pdf
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