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Congestion games are attractive because they can model many concrete situations where some competing entities interact through the use of some shared resources, and also because they always admit pure Nash equilibria which correspond to the local minima of a potential function. We explore the problem of computing a state of minimum potential in this setting. Using the maximum number of resources that a player can use at a time, and the possible symmetry in the players' strategy spaces, we settle the complexity of the problem for instances having monotone (i.e., either non-decreasing or non-increasing) latency functions on their resources. The picture, delineating polynomial and NP-hard cases, is complemented with tight approximation algorithms.

In this work, we study the Stochastic Budgeted Multi-round Submodular Maximization (SBMSm) problem, where we aim to adaptively maximize the sum, over multiple rounds, of a monotone and submodular objective function defined on subsets of items. The objective function also depends on the realization of stochastic events, and the total number of items we can select over all rounds is bounded by a limited budget. This problem extends, and generalizes to multiple round settings, well-studied problems such as (adaptive) influence maximization and stochastic probing. We show that, if the number of items and stochastic events is somehow bounded, there is a polynomial time dynamic programming algorithm for SBMSm. Then, we provide a simple greedy $1/2(1-1/e-\epsilon)\approx 0.316$-approximation algorithm for SBMSm, that first non-adaptively allocates the budget to be spent at each round, and then greedily and adaptively maximizes the objective function by using the budget assigned at each round. Finally, we introduce the {\em budget-adaptivity gap}, by which we measure how much an adaptive policy for SBMSm is better than an optimal partially adaptive one that, as in our greedy algorithm, determines the budget allocation in advance. We show that the budget-adaptivity gap lies between $e/(e-1)\approx 1.582$ and $2$.