I see. So I guess there is some benefit gained from this, but it is very minor. It seems to me that the simplest rule that explains why Harry’s integer factorization is not okay, but, for example, the “silver on the tree” password from the end of TSPE is okay, is the following: if you would gain information at time T, and send information from time T to any time S < T, then it must be the case that you would have gained that same information even if you hadn’t learned it at time S.
Now consider your “keying” scenario. We have clones 0-5, and at time 1 clone x goes back to time 0 and becomes clone x+1. When clones give a “time key,” it will be a number between 0 and 6, identifying a clone/wall-clock pair. Now suppose at wall-clock time T clone 1 says: “the thought I had at time P doesn’t work.” Assume for simplicity that time P refers to clone 0 at a wall-clock time S > T (though it would work out the same anyway). Now at time S, when clone 0 has the thought, he has two choices. On the one hand, he can continue working out why it doesn’t work, but in this case he gains only the minor benefit of knowing in advance that it will not work out. On the other hand, he can move on and not consider the thought, but at time T as clone 1 just repeat (without knowing why), the fact that it doesn’t work. In that case, he gained information that he would not have learned had he not told himself. Or, in your terminology, recursion was required in order to reach a stable state.
then it must be the case that you would have gained that same information even if you hadn’t learned it at time S.
that doesn’t follow. Where would Harry have gotten the pies if not from Harry+1?
In that case, he gained information that he would not have learned had he not told himself. Or, in your terminology, recursion was required in order to reach a stable state.
The recursion is non-iterative beyond the number of loops actually manifested, however. Each individual only adjusts the one previous, and only in immediately non-iterative manners. “Nope. Nope. Nope. Maybe. Nope. Nope. Maybe.”
That lets you prune out failed items but doesn’t recurse back to an instantaneous success.
that doesn’t follow. Where would Harry have gotten the pies if not from Harry+1?
He got them from the breakfast table. Where did he get the idea to get them? Well, he would have seen the pies later on anyway. Just like he would have learned about time turners later on in the day anyway, but a more stable scenario was obtained by learning about them earlier.
The recursion is non-iterative beyond the number of loops actually manifested, however.
I’m not quite sure how to parse this. If you would think about an idea at time T, but don’t because future you tells you it won’t work out, that means your whole thought process going forward has completely changed. But maybe the thing you start thinking about instead doesn’t work out, so someone warns you about the idea at time T+epsilon. And so on. So if you are proposing that Time works by iterating through a number of scenarios until you get to something stable, the situation you’ve pointed to “requires recursion.” (It’s worth pointing out that Harry, when he gets his Time Turner, doesn’t think this is a likely answer to how Time works.) But perhaps I am not understanding you correctly?
My main objection to the scenario you are proposing, though, is that you are gaining information as a result of some work, but that work is never performed. Try taking your scenario to its logical extreme. You sit in a room with one copy of future-you, and a large composite number N on a sheet of paper. On scrap work hidden from future-you, you write down an integer K. If you are not told that K does not divide N, you check. If it does, you keep track of the factor of N you have found. In any case, you then systematically select a new integer K’ to check for divisibility. Once you have a complete factorization, you sit quietly, and at the end of the hour you go back in time. Then, you let past-you know the “keys” for all of the integers that weren’t factors. Thus, you must have ended up only trying actual factors. So, you have a slightly more complicated version of Harry’s factorization algorithm.
Edit to add: I guess this situation actually still is still an exponential time algorithm, since you still have to consider every possibly integer. But you could do, for example, graph isomorphism testing in polynomial time with a similar method: try constructing a map by first finding what vertex 1 should map to. Future you says “nope, nope, maybe,” and on “maybe” you try to figure out where to put vertex 2, given that vertex 1 maps to whatever. If the graphs are not isomorphic, future you will say “no” to every choice for vertex 1, so you have your answer in polynomial time. Otherwise, you will construct the isomorphism in polynomial time.
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.
I see. So I guess there is some benefit gained from this, but it is very minor. It seems to me that the simplest rule that explains why Harry’s integer factorization is not okay, but, for example, the “silver on the tree” password from the end of TSPE is okay, is the following: if you would gain information at time T, and send information from time T to any time S < T, then it must be the case that you would have gained that same information even if you hadn’t learned it at time S.
Now consider your “keying” scenario. We have clones 0-5, and at time 1 clone x goes back to time 0 and becomes clone x+1. When clones give a “time key,” it will be a number between 0 and 6, identifying a clone/wall-clock pair. Now suppose at wall-clock time T clone 1 says: “the thought I had at time P doesn’t work.” Assume for simplicity that time P refers to clone 0 at a wall-clock time S > T (though it would work out the same anyway). Now at time S, when clone 0 has the thought, he has two choices. On the one hand, he can continue working out why it doesn’t work, but in this case he gains only the minor benefit of knowing in advance that it will not work out. On the other hand, he can move on and not consider the thought, but at time T as clone 1 just repeat (without knowing why), the fact that it doesn’t work. In that case, he gained information that he would not have learned had he not told himself. Or, in your terminology, recursion was required in order to reach a stable state.
that doesn’t follow. Where would Harry have gotten the pies if not from Harry+1?
The recursion is non-iterative beyond the number of loops actually manifested, however. Each individual only adjusts the one previous, and only in immediately non-iterative manners. “Nope. Nope. Nope. Maybe. Nope. Nope. Maybe.”
That lets you prune out failed items but doesn’t recurse back to an instantaneous success.
He got them from the breakfast table. Where did he get the idea to get them? Well, he would have seen the pies later on anyway. Just like he would have learned about time turners later on in the day anyway, but a more stable scenario was obtained by learning about them earlier.
I’m not quite sure how to parse this. If you would think about an idea at time T, but don’t because future you tells you it won’t work out, that means your whole thought process going forward has completely changed. But maybe the thing you start thinking about instead doesn’t work out, so someone warns you about the idea at time T+epsilon. And so on. So if you are proposing that Time works by iterating through a number of scenarios until you get to something stable, the situation you’ve pointed to “requires recursion.” (It’s worth pointing out that Harry, when he gets his Time Turner, doesn’t think this is a likely answer to how Time works.) But perhaps I am not understanding you correctly?
My main objection to the scenario you are proposing, though, is that you are gaining information as a result of some work, but that work is never performed. Try taking your scenario to its logical extreme. You sit in a room with one copy of future-you, and a large composite number N on a sheet of paper. On scrap work hidden from future-you, you write down an integer K. If you are not told that K does not divide N, you check. If it does, you keep track of the factor of N you have found. In any case, you then systematically select a new integer K’ to check for divisibility. Once you have a complete factorization, you sit quietly, and at the end of the hour you go back in time. Then, you let past-you know the “keys” for all of the integers that weren’t factors. Thus, you must have ended up only trying actual factors. So, you have a slightly more complicated version of Harry’s factorization algorithm.
Edit to add: I guess this situation actually still is still an exponential time algorithm, since you still have to consider every possibly integer. But you could do, for example, graph isomorphism testing in polynomial time with a similar method: try constructing a map by first finding what vertex 1 should map to. Future you says “nope, nope, maybe,” and on “maybe” you try to figure out where to put vertex 2, given that vertex 1 maps to whatever. If the graphs are not isomorphic, future you will say “no” to every choice for vertex 1, so you have your answer in polynomial time. Otherwise, you will construct the isomorphism in polynomial time.
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.