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Suppose in one step of BFGS: compute $p_k$ such that $B_{k}p_{k} = -grad(f)$ $x_{k+1} = x_{k}+\alpha_{k}*p_{k}$ where $\alpha_k$ is determined by Wolfie Condtion. Maybe you can check it here:https://en.wikipedia.org/wiki/Wolfe_conditions
The problem is NP-hard even when all sets have at most (or exactly) one element. This can be seen by a reduction from (the decision version of) vertex cover. Given graph $H$, you can build the graph $G=(V,E)$ by starting with a graph containing a single vertex $s$ and doing the following for each $e=(u,v)$ of $H$: Add three new vertices to $G$ namely $x_e, ... 1 You didn't explain how your solution today works, so it's hard to give any concrete pointers. But more generally, in a Monte Carlo solution, you'll take random steps all the way to a terminal state (win/loss). As you explore more and more nodes, you can start to focus more on promising ones, by having them be more likely to be selected. In pandemic there are ... 1 I'm not sure this will answer your question "mathematically", but it will definitely give you some idea to why it is "advantageous". Usual RL requires to keep in memory a state-value vector. When the number of possible states is large, or even infinite - this is not practical to keep in memory. In addition, the usual RL requirements for ... 1 You could try expressing this as an integer linear expression: let$v_{x,b}$be 1 if item$x$is placed in box$b$, and let$w_{x,x',b,b'}$be 1 if$v_{x,b}=1$and$v_{x',b'}=1$; then your objective function is a linear function of these zero-or-one variables. You can also enforce the relationship between the$v$'s and$w$'s, and ensure that the$v\$'s ...