Fastest way to defeat the Kirby-Paris hydra? The Kirby-Paris hydra, previously seen on math.stack here and in this layperson-targeted video, is a one-player game that models the mythical hydra as a tree. Each node/vertex of the tree is considered to be a "head" of the hydra, except for the root which is its "body". A player "moves" by selecting a leaf node L and removing it. Then, the hydra "regrows" n copies what is left of L's parent's subtree, all rooted on L's grandparent. For our purposes, let n=2. The goal is to reduce the hydra to its body.
The surprising result is that, not only is the game winnable, it is impossible to lose. Any hydra will be "defeated" by a sufficiently long but finite sequence of moves. However, it seems that the sequence length grows very rapidly with increasingly complex hydra.
My question is whether the choice of moves has any effect on how many moves are required to defeat a particular hydra. If all sequences of moves have the same length, how is this proven? If there exist at least two sequences of different lengths, how may one find the shortest one?
 A: First, a bit of terminology. Define k heads, growing out of one head, growing out of the body, as a k-fork. Define k heads growing out of each other in sequence, without any branching, as a k-chain.
As stated in the question, assume each chop produces n=2 copies, not counting the one already present.
Proof: Order can matter
Consider this diagram in which you start with a hydra on the top left (the horizontal line is the body/root). Chopping the higher head labeled A produces four-fork. There is no meaningful choice; the second chop yields three three-forks.
Suppose instead you chopped the lower head labeled B. This yields three three-chains. Chopping the top head of a three-chain yields a three-fork. Doing this for each chain gets us to three three-forks, but this takes four moves when before it took only two.

Having gotten to the same hydra using a different number of moves, it follows that there are two complete sequences to kill this hydra that have different lengths. (Incidentally, three three-forks take 120 chops to kill, regardless of order.)
Conjecture: Kill the highest head
The shorter path in the example above was to kill the highest head, the one furthest from the body. I conjecture that this will always lead to defeating the hydra in the fewest number of chops. Conversely, I conjecture that the longest sequence is found by killing the leaf head closest to the body.
Imagine that the leaf labeled A in the diagram is replaced by a large subtree. Chopping B will then duplicate that subtree twice, creating a lot of heads. Chopping higher leafs causes the hydra to regrow smaller subtrees, adding fewer heads. Each head requires a chop to kill, so adding fewer heads is a good thing. The total number of chops equals the number of heads produced plus the initial number of heads.
However, I can't prove that greedily producing the fewest possible heads with each chop will result in a global minimum. It's possible that producing more heads now will allow one to produce fewer heads later.
