Why does slaying the hydra require infinite ordinals?

Question

I have recently come across the problem of the mathematical hydra. More information can be found here or a video here.

I have to say that I don't fully understand why infinite ordinals are required to prove the hydra can always be killed. For example, here is how I would go about it.

Represent the hydra as a decorated tree so that leaf nodes are grouped together and represented by a single number on an edge leading to a representative node as shown below. [a bit awkward to represent in ASCII, but here goes]

(r denotes the root, 1-multiplicity labels are omitted so o-[1]-o = o-o)

r
o-o-o
   \
    o

= 

r
o-o-[2]-o

since there are 2 leaf nodes in that branch.

Then by repeatedly applying the cut move:

$o-[A]-o-[B]-o \ \ \ \ \ \rightarrow \ \ \ \ \ o-[A(m+1)^B]-o + A((m+1)^B - 1)/m$

where the first term on the RHS changes the tree, the second term counts the number of cuts needed to perform the cut move, and m is the number of new copies spawned,

the height of each branch is decreasing and the height of the tree is always non-increasing as the number of cuts is being tallied. Eventually this leaves us with just the root and the total number of cuts required to kill the hydra.

Since the hydra is finite and the number of cuts added with each cut move, $A((m+1)^B - 1)/m$, is finite, then the total number of cuts is therefore always finite and so the hydra will always be killed.

So where does this strategy either go wrong or invoke infinite ordinals?

Answer 1

Your idea works only for a certain very special class of strategies: strategies that preserve the symmetry of the hydra inherent in your labelling scheme. That is, by labelling "repeated" edges with numbers, each "head" in your graph actually represents many different heads of the hydra, one for each possible choice of a single edge from every repeated edge below it. You then assume that every time you cut off one head of the hydra, you also immediately cut off all other heads which are represented by the same point in your graph. In this way, by just changing the numerical labels, you guarantee that the graph only shrinks as you cut.

However, an arbitrary strategy does not need to do this. It can cut off one head but not also cut off all the other "copies" of it, destroying the symmetry of the graph. In order to correctly represent the hydra, you would then need to add new branches to your graph, and so your argument falls apart.

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[Old response, based on a misunderstanding of your strategy, preserved since it still may be helpful to some readers:]

I am somewhat confused by your notation and don't know how you got the specific numbers you wrote, but I believe you have in mind a correct simple strategy for defeating any hydra. To be a bit more precise, here's the strategy. Pick a head of maximal height, and repeatedly cut off all maximal height heads from that head's grandfather until you have eliminated all such heads (it is not difficult to show this will happen eventually). Doing this to all the grandfather nodes of heads of maximal height, you will reduce the height of the tree. Now repeat until the height of the tree is reduced to $0$.

This strategy requires nothing involving ordinals, and indeed can be proven to be a successful strategy in PA. What requires something like ordinals (or at least, some technique which cannot be performed in PA) is proving that every strategy is successful. That is, no matter what strategy you use for cutting off heads, you will eventually defeat the hydra. This is a much stronger statement than just finding a single strategy that works!