In principle, SI is choosing fairly from a space of infinite programs. It’s only practical to see some programs as finite with weight proportional to the weight of all the infinite programs this finite program can be extended into. But no program knows its length, unless it explicitly counts to when to stop.
The wasteful encoding you propose does not make a difference to SI. What the wasteful encoding does is that the arithmetic encoding programs will be 10 times longer and thus 2^10 times more penalized, but there will be 2^10 times more of them. So in the sum-over-all-programs, arithmetic coding programs will take over the direct output programs just the same as before.
Programs that are 10 bits longer are penalized by 2^10. Programs that are 10 times longer are penalized by 2^(10n), where n is the size of the original program. The penalty isn’t washed out over time… it gets significantly worse.
Yes, you’re right, I was sloppy. Still, the programs are exactly that much more numerous, so their weight ends up being the same in your wasteful encoding as in a sane encoding.
Hmmm, right. It’s not enough to ignore the intermediate bits. Have to make them break the program unless they are all zero. Like if any of them are 1 then the program had no output except “syntax error” (but the raw output still allows them).
In principle, SI is choosing fairly from a space of infinite programs. It’s only practical to see some programs as finite with weight proportional to the weight of all the infinite programs this finite program can be extended into. But no program knows its length, unless it explicitly counts to when to stop.
The wasteful encoding you propose does not make a difference to SI. What the wasteful encoding does is that the arithmetic encoding programs will be 10 times longer and thus 2^10 times more penalized, but there will be 2^10 times more of them. So in the sum-over-all-programs, arithmetic coding programs will take over the direct output programs just the same as before.
Programs that are 10 bits longer are penalized by 2^10. Programs that are 10 times longer are penalized by 2^(10n), where n is the size of the original program. The penalty isn’t washed out over time… it gets significantly worse.
Yes, you’re right, I was sloppy. Still, the programs are exactly that much more numerous, so their weight ends up being the same in your wasteful encoding as in a sane encoding.
Hmmm, right. It’s not enough to ignore the intermediate bits. Have to make them break the program unless they are all zero. Like if any of them are 1 then the program had no output except “syntax error” (but the raw output still allows them).
I see. And don’t know the answer. I’m curious how SI fends off this one.