Six turnings of the Turner at T=0 results in the same 1-hour segment being looped into 6 times. This allows six iterations—but those iterations do not recurse beyond the actual number of loops.
< is that you are gaining information as a result of some work, but that work is never performed.
Sorry, I’m really not following your pie argument. Harry would learn about the pies in the near future; since it is his style, he would think about throwing them to frighten the bullies. So, his observation of Harry+1 throwing the pies is not necessary for him to think to throw pies anyway.
What do you mean by “they don’t recurse”? Surely the fact that this procedure results in fast graph isomorphism testing shows that it is not a particularly “stable” solution? Or, do fast integer factorization by writing down the first digit for the least factor greater than 1, listen as future you says “no, no, maybe,” and change it to whatever, and then figure out the second digit, and so forth. The scenario you’ve outlined results in nearly instant integer factorization (or password guessing, or whatever), so it is probably illegal.
Note that both graph isomorphism and integer factorization are problems that may well lie in P, so these aren’t great examples. Traveling salesman is a bit better.
Sure, though my impression is that people don’t think graph isomorphism actually is in P. And integer factorization turned out to be a problem for Harry. But you’re right, we can actually just simulate a nondeterministic Turing machine this way: every time you have a choice for which state to visit next, just listen as future you tells you which ones not to visit.
So, his observation of Harry+1 throwing the pies is not necessary for him to think to throw pies anyway.
Harry+0′s actions or non-actions were radically transformed by the act of Harry+1′s throwing the pies.
The solutionspace for Harry+0′s problems were altered by the actions of Harry+1.
From this we must derive the answer that iteration can alter outcomes. However, from the factoring of primes we see that the TT resists allowing iteration to recurse beyond the actual number of iterations.
Where number-of-iterations = i, where i < 6, then Harry+0..i can perform as many recursive alterations of his own solution-seeking as can be achieved without exceeding the value of i.
No. That’s where Harry+1 got them. Harry did not.
Six turnings of the Turner at T=0 results in the same 1-hour segment being looped into 6 times. This allows six iterations—but those iterations do not recurse beyond the actual number of loops.
< is that you are gaining information as a result of some work, but that work is never performed.
That doesn’t follow. How do you figure?
Sorry, I’m really not following your pie argument. Harry would learn about the pies in the near future; since it is his style, he would think about throwing them to frighten the bullies. So, his observation of Harry+1 throwing the pies is not necessary for him to think to throw pies anyway.
What do you mean by “they don’t recurse”? Surely the fact that this procedure results in fast graph isomorphism testing shows that it is not a particularly “stable” solution? Or, do fast integer factorization by writing down the first digit for the least factor greater than 1, listen as future you says “no, no, maybe,” and change it to whatever, and then figure out the second digit, and so forth. The scenario you’ve outlined results in nearly instant integer factorization (or password guessing, or whatever), so it is probably illegal.
Note that both graph isomorphism and integer factorization are problems that may well lie in P, so these aren’t great examples. Traveling salesman is a bit better.
Sure, though my impression is that people don’t think graph isomorphism actually is in P. And integer factorization turned out to be a problem for Harry. But you’re right, we can actually just simulate a nondeterministic Turing machine this way: every time you have a choice for which state to visit next, just listen as future you tells you which ones not to visit.
Harry+0′s actions or non-actions were radically transformed by the act of Harry+1′s throwing the pies.
The solutionspace for Harry+0′s problems were altered by the actions of Harry+1.
From this we must derive the answer that iteration can alter outcomes. However, from the factoring of primes we see that the TT resists allowing iteration to recurse beyond the actual number of iterations.
Where number-of-iterations = i, where i < 6, then Harry+0..i can perform as many recursive alterations of his own solution-seeking as can be achieved without exceeding the value of i.