I think that's a great question, but I would guess the answer is no, there are no other solutions.

One important axiom for this type of puzzle is that everyone does deduce every piece of information that they can mathematically deduce with 100% certainty. That would eliminate the "just guess" and "guess because your friend will not play along" type of answers some are giving in the thread.

With this, my hand-wavy reasoning for not being any other solution is that when you try to model the knowledge the players possess after each turn (or equivalently, the state of the game), there are no branches or decisions they could make except for the STOP condition of the game. In each turn, they can (and because of the "axiom" above: they DO) eliminate some numbers from the set of possible number of statues the other player could be seeing if they passed in their round. Once the set goes down to one, they KNOW the answer. Since there are no choices [1] they could possibly make to alter the game before it ends, I don't see how there could be an alternate solution.

[1] I'm discounting the possibility that they insert a constant pattern of extra passes in the game. (E.g., "We will both pass 5 times and then deduce the number based on the strategy described above.")

I thought about that too, if he passes 6 times (and I've passed 6 times) I know that he has at least 6 (and he knows I have 10 or 12), so I pass a 7th time, and he passes a 7th time too. So now I know he can't have 6 therefore he has 8.

This, however isn't the quickest way so it would probably fail if your partner is optimizing.

Hmm. I was pretty convinced that my method would work, then I got convinced it couldn't work, now I'm confused again.

A few notes: You can see 12, your friend can see 8. So your friend knows you have either 12 or 10. You know your friend has either 8 or 6.

Step 1: you pass, friend knows you have >=1. Friend passes, you know they have >=1.

Step 2: you pass, friend knows you have >=2. Friend passes, you know they have >=2.

[...]

Step 7: you pass, friend knows you have >=7. Friend passes, you know they have >=7; you have 12, so there must be 20, you win.

I'm actually pretty convinced that it would work now... anyone want to poke a hole in it?

I thought of something along those lines at first as well. However, the problem is that the answer is not logically deducible using this method as it requires both of you to agree on a counting system and an end condition. And since the puzzle states that the two of you are not able to communicate, this method would not be viable. The given solution allows you to deduce the result with only the information and actions you have readily available to you.

Only if you use + 1 as a rule. What if your friend uses more efficient + 2, as you know the difference is by two?

(Then unless you know who has the lower number, you're out of luck.)

I thought that, when the keeper comes to ask you, you distract him and make him late for the other person. You do that for 12 days, and the next day you pass immediately, so your friend knows it was 12 by knowing that the keeper came early the 13th time.