And wby would this be optimal? What is to stop the greedy procastinating to consider a node as infected, but not to procastinate for long enough so it gets to consider a node that is "forbidden" by this principle? And even assuming we know the answer to this question, how can the greedy, knowing this, make an optimal local decision in a node when there are multiple other (non-local) similar scenarios (i.e. brother-ish nodes) in which choosing the alien node might eliminate the apparent need for choosing said node? (like, such issues break the necessarily local nature of greedy algorithms, and just this sort of band-aids dont seem to make for very compelling arguments, worse yet, I'd call them defective, else I am missing something :))

To the first question, I think it should be some data structure work to find the highest choose-able vertex within distance $$$d$$$.

To the second question, The proof just shows that if several vertices with same depth meet the condition, they are both optimal since they infects an identical subset of vertices need to be infected, so any one can be chosen as an optimal local choice.

Why is the last part true? If I have to choose between two (brothered) vertices, how do I know which one is better? Like, it seems as though if one has a more shallow afferent subtree, while the other has smth like height exactly d, it is more optimal to choose the latter, and it is still unclear to me what criteria can set these two apart :))-- like, it seems as though it cant be just height, because this characteristic can obviously not tell the minimum number of needed nodes to infect on its own

As we are assuming that the vertex $$$u$$$ we are going to infect is the lowest one, it does not matter which subtree is shallower or deeper. If a chosen vertex infects $$$u$$$, it infects all vertice need to be infected in a whole subtree, and no vertice lower than $$$u$$$ need to be considered.

My friend let me write on his account as a joke, but the algorithm is correct.

If all cows start infected, it means any cow can be a source node that starts sick. So then the idea of choosing nodes is this: We root the tree arbitrarily, and we will loop through all nodes from greatest depth to smallest depth and select nodes to start infected in this process.

Let the current node be $$$i$$$. If $$$i$$$ is within $$$d$$$ distance from a some node we have infected, we need not consider it. So now assume $$$i$$$ is not infected. Note that it is currently an uninfected node with largest possible depth, as all previous nodes are infected and all later nodes have shorter depth. That means that, if we have two nodes $$$x$$$ and $$$y$$$ that are both valid we can choose to infect the node with smaller depth. fdironia provided a proof above.

So now hopefully it makes sense how this greedy works if all nodes are infected. We return to the original problem. The issue now is that some nodes can't start infected. This is when there exists some node within $$$d$$$ distance that is not sick at the end of all days. Any node that satisfies the condition such that all nodes within $$$d$$$ distance are sick at the end of the process can start the disease. So we have turned this into the original problem; we only consider nodes that have to be sick, and out of the available nodes to make him sick we pick the one with smallest depth.

Please direct the downvotes to me rather than my friend, I apologize ;w;. We were next to each other after school and we both thought it was funny.