like how to "assign remaining customers to this and other ads so that $\frac{\bar{v}-v_i}{m-1}$ becomes a new upper bound for the spread." ?

Thanks I understand all other cases except for the case $\frac{\bar{v}-v_i}{m-1} < \frac{\bar{v}}{m}$. Could you explain more in detail how to deal with this case ?

@BernardoSubercaseaux $1,2,4$ are indices of customers, they're not values ie. $A_1$ is shown to customers with values $v[1] = 3, v[2] = 4, v[4] = 2$ totals to $9$. The value vector is $v[] = \{3,4,12,2,4,6\}$.

Since you already have 2 configured networks, why can't you do $\pi_1(\pi_0(x))$ in linear time by piping the outputs of network 0 as inputs to network 1 ?

@ManosMastronikolas just note that you cannot $\bf first$ find the subset that sum to $T$ $\bf then$ construct the rectangles because it's not known it can be done in polynomial time.

@ManosMastronikolas yeah you can give a try, given an instance of subset sum $(S,T)$ convert it to an instance of rectangle packing $(P,\mathcal R)$ in polynomial time such that there's a way to fit all rectangles of $\mathcal R$ into $P$ iff there's a subset of $S$ that sums to $T$.

@AinsleyH. Am I understanding correctly ? consider the residual graph $R$ at the end of the previous iteration, it gives us a min-cut $X_{i-1}$ which partitions the vertices into $(S_{i-1}, T_{i-1})$. Any vertex $u \in G - X_{i-1}$ ie. has not been cut off must lie in $S_{i-1}$, so it must reachable from $s$ in $R$. If we now add an edge from $u$ to $t$, we have an augmenting path in $R$, thus the flow increases.

thanks I believe the algo is correct but I'll appreciate if you could comment on the following situation : fix $u\in \mathcal{F}$ but $u \notin F_{i-1}$, there're $2$ minimum $(s,F_{i-1})-$cuts $X$ and $X'$ where $X'$ is actually also a $(s,F_{i-1}\cup \{u\})-$cut but $X$ is not. Unfortunately, the algo only finds $X$ (the "bad choice") and branches into it.

thanks, so we're using max-flow to find min-cut thus we can just set all other edges to have capacity 1 ?

@AinsleyH. thanks but in the first iteration couldn't $X$ just be all out-edges of $s$ ? then $R$ is just $\{s\}$ so minimized, $X$ is also $(s,F)$ seperator since $F = \emptyset$ ...

Could you explain more on what "let 𝑋 be (𝑠,𝐹)-separator that minimizes "a reachable set" 𝑅" means ? like what is an (𝑠,𝐹)-separator, what is "a reachable set" 𝑅 ?

thanks I see now, because $a_{k+1} - a_{k} = a_k$ so you have $a_1,...,a_k$ again , upvoted

$2^k - (\text{any odd positive integers up to} 2^{k-1})$ doesn't give you all odd positive integers up to $2^{k}$, in particular how can you get $1$ ?