It’s a nice result, but I doubt it’ll ever be practically relevant.
You could implement the algorithm today, although with a limit of less than 3^^^3. Rough guesstimates suggest that about a limit of 50 would be workable at home, 100 for a worldwide distributed computing effort and 500 for a universe-sized computer that does one operation per bit per plank time. In all cases, nowhere near 10000 (let alone 3^^^3).
Heuristics could help a bit, but making the algorithm more complex is likely to also make the proof that it works longer and so harder to find. The fallback to brute-force search is unhelpful because no computer that could actually exist would be able to find the proof that way.
If program A uses heuristics, program B doesn’t need to prove that these heuristics are correct. It only needs to see that A’s use of heuristics makes A either output 1 or move on to the brute-force search.
Agreed with the rest of your comment. An easier way to reach the same conclusion: if program B is implemented as “always defect”, program A has to run the brute-force search to the end, there’s no shortcut (because having any shortcut in the code would make the proofs invalid if B were a copy of A).
It’s a nice result, but I doubt it’ll ever be practically relevant.
You could implement the algorithm today, although with a limit of less than 3^^^3. Rough guesstimates suggest that about a limit of 50 would be workable at home, 100 for a worldwide distributed computing effort and 500 for a universe-sized computer that does one operation per bit per plank time. In all cases, nowhere near 10000 (let alone 3^^^3).
Heuristics could help a bit, but making the algorithm more complex is likely to also make the proof that it works longer and so harder to find. The fallback to brute-force search is unhelpful because no computer that could actually exist would be able to find the proof that way.
If program A uses heuristics, program B doesn’t need to prove that these heuristics are correct. It only needs to see that A’s use of heuristics makes A either output 1 or move on to the brute-force search.
Agreed with the rest of your comment. An easier way to reach the same conclusion: if program B is implemented as “always defect”, program A has to run the brute-force search to the end, there’s no shortcut (because having any shortcut in the code would make the proofs invalid if B were a copy of A).