Balls & Bins: A punishment and reward game Consider a game where one has a set of bins $(b_1, b_2, ...)$, and each bin has an associated initial count of balls $(c_1, c_2, ...)$.  The rules of the game are as follows:

(1) Once a bin has a threshold number of balls $Q$ (the threshold is uniform for all bins), the bin may be capped, and capping is irreversible.  However, one is not obligated to cap the bin once this condition is met.  Capping must be SEQUENTIAL.  One cannot simultaneously cap two bins to avoid e.g. a situation where capping one bin drops the number of balls in the other bin below the threshold $Q$ necessary for capping.

(2) Once a bin is capped, a bin-specific set of "punishments" or "rewards" can be specified for at most $k$ other bins.  For example, we might say that, in capping bin $3$, we need to subtract two balls from bin $1$ and add seven balls to bin $5$.  So to each bin we assign a two $k$-tuples: the first, e.g. {1,5}, with values from the previous example, specifies the bins to be "punished" or "rewarded", and the second, e.g. {-2, 7}, again from the provided examples, species how many balls to add or remove from each of the bins in the first tuple.

(3) Negative values for bin ball counts are permissible, and a bin may initiate (i.e. have some $c_i$ value) $\leq 0$.

(4) We add the constraint that punishments or rewards must be symmetric.  If capping some bin $b_i$ means subtracting one ball from bin $b_j$, it must also be the case that capping bin $b_j$ means subtracting one ball from bin $b_i$.

(5) We add the constraint that the total number of punishments or rewards a bin may receive is at most $k$.  By the symmetry condition from (4), this correspondingly implies that the total number of punishments or rewards a bin may deliver is at most $k$.

Consider now the decision problem: provided some set of bins $(b_1, b_2, ...)$, initial ball counts $(c_1, c_2, ...)$, and access to all of the dependencies elaborated on in rule (2), can all of the bins be capped?

This problem is trivially in $NP$, but for what smallest value of $k$ is it $NP$-complete?  Can the decision problem be $NP$-complete for $k=2$?  Is there an easy reduction from an $NP$-complete problem or a problem solvable in polynomial time?

Update // Added further constraints (4) + (5), where (5) is implied by (4).  That should be it.
Here's an argument that it's polynomial for $k=2$. 

If $k=2$, each bin can only be connected to two other bins, and thus the graph falls into paths and cycles. We can thus just worry about a cycle. In fact, we can guess which bin is capped first. Since there are only a polynomial number of guesses, we can try all of them, and reduce the problem to the case where the graph is a path.

Now, whether a bin can be capped only depends on whether or not you've capped its neighbors. That is, for bin $1$, we have one of the following constraints:


bin $1$ can be capped only before you've capped bin $2$,
bin $1$ can be capped only after you've capped bin $2$,
bin $1$ can be capped any time,
bin $1$ can never be capped.


For case 1, you might as well cap bin $1$ as the first thing you do. Now, you have a path of length $n-1$, and you can use recursion. 

For case 2, you might as well cap bin $1$ as the last thing you do. Now, you have a path of length $n-1$, and you can use recursion.

For case 3, capping bin $1$ will either add or subtract balls from bin $2$. If it adds them, make it the first bin you cap. If it subtracts them, make it the last bin you cap. In either case, you've reduced the problem to a path of length $n-1$.

For case 4, you've solved the problem: it's impossible to win.

This gives an algorithm of time $O(n^2)$. For a cycle of length $n$, we need to test $n$ different paths, breaking the cycle in every possible place. And for a path of length $n'$, in constant time, we can convert the problem to the same problem on a path of length $n'-1$.