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srn437
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supertasks
« on: Aug 30th, 2007, 10:07am » |
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Zeus gathers an infinite amount of demons and gives the following demands: demon 1 kill prometheus in an hour, demon 2 kill him in half an hour, demon 3 kill him in a quarter of an hour...
Prometheus was found dead in that hour. Zeus was taken to court. Why was he found innocent? You get the next supertask riddle after solving this one.
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Archae
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Re: supertasks
« Reply #1 on: Sep 1st, 2007, 8:19am » |
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I would say, to answer this 'supertask', that Zeus was found innocent because he could argue that there is no way that a demon could have killed Prometheus.
Assume that all demons act in the same manner - they either all obey Zeus or all disobey him. If the latter is the case, it is clear to see that Prometheus' death was not caused by Zeus at all. In the former case, Zeus can argue that no demon killed Prometheus: if you assign blame to a certain demon, then Zeus can show that the next demon he created would have killed Prometheus first, so the demon accused could not have done it. And since he created an infinite amount of these demons, there is no first demon who killed Prometheus.
However, Zeus should still be charged for premeditated murder.
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srn437
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Re: supertasks
« Reply #2 on: Sep 1st, 2007, 9:45am » |
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Correct. Next supertask. There is a hotel with an infinite amount of people. If its full, how do you fit another person? Or another infinite amount of people? Or another infinite amount of infinite amounts of people?
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srn437
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Re: supertasks
« Reply #4 on: Sep 1st, 2007, 4:02pm » |
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Correct. Next supertask. There is a light that is on. A person turns the switch off in a minute, on in 30 seconds, off in 15 seconds...
In 2 minutes(which he will be done by that time), will the light be on or off and will the switch be on or off(at that speed, the light might not keep up with the switch)?
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denis
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Re: supertasks
« Reply #5 on: Sep 1st, 2007, 4:51pm » |
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The light will be off because the switch will be broken from overuse 
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| « Last Edit: Sep 1st, 2007, 4:52pm by denis » |
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srn437
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Re: supertasks
« Reply #6 on: Sep 1st, 2007, 7:16pm » |
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Good answer, but no.
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mikedagr8
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Re: supertasks
« Reply #7 on: Sep 2nd, 2007, 5:27am » |
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You can't tell, as you near the two minutes, you will be going so fast, that it wont change. So I'll go with on. By what I have said, I mean that 1/2+1/4+1/8+.... never actually reaches 1.
Do these puzzles have something in common? They seem to have so far.
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| « Last Edit: Sep 2nd, 2007, 5:30am by mikedagr8 » |
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denis
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Re: supertasks
« Reply #8 on: Sep 2nd, 2007, 7:00am » |
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Supposing the switch is still working, we won't know what position it's in at the two minute mark because if you say it's on, there exists a time interval (half of the last time interval used to set it to on) whereby it will be set to off. If you say it's off, there exists a time interval (half of the last time interval used to set it to off) whereby it will be set to on. So you really can't tell.
Whether the light is on or off in 2 minutes depends on the type of lamp. I think, an incandescant bulb would have its filament still hot enough to emit a sufficient amount of light to be considered on (but dimmer than regular on). A fluorescent light would probably be off because its latency time to turn on exceeds the intervals times between the quick on/off switching near the 2 minute mark.
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| « Last Edit: Sep 2nd, 2007, 8:07am by denis » |
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srn437
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Re: supertasks
« Reply #9 on: Sep 2nd, 2007, 8:09am » |
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Should I just give you the answer?
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denis
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Re: supertasks
« Reply #10 on: Sep 2nd, 2007, 8:14am » |
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on Sep 2nd, 2007, 8:09am, srn347 wrote:| Should I just give you the answer? |
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That's really up to you but generally, you give a little bit of time for others to chime in with their answers or give hints.
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| « Last Edit: Sep 2nd, 2007, 8:15am by denis » |
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srn437
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Re: supertasks
« Reply #11 on: Sep 2nd, 2007, 10:10am » |
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Ok. The light switch won't break or anything like that.
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Archae
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Re: supertasks
« Reply #12 on: Sep 3rd, 2007, 10:41pm » |
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Well, unless I am misreading the question, it should be off, along with the switch in the 'off' position. Reread the problem:
on Sep 1st, 2007, 4:02pm, srn347 wrote:There is a light that is on. A person turns the switch off in a minute, on in 30 seconds, off in 15 seconds...
In 2 minutes(which he will be done by that time), will the light be on or off and will the switch be on or off(at that speed, the light might not keep up with the switch)? |
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The way I interpret the riddle, we start at time t=0 and proceed to two minutes (t=2). By the rules, at t=1/4, the light will be off; at t=1/2, the light will be on; at t=1, the light will be off again.
Pretty much, what's happening here is that the infinite series is constructed to be approaching the two minute mark, but when the flicking actually occurs in practice, that order will be reversed. Then the problem of running off to infinity occurs in the millisecond (and smaller) right after the clock starts, and in another instant, we do not need to worry about it anymore. Then, the last flick will be at t=1, which will turn the switch and the light off, and will stay like that until t=2. (Unless, of course, there is another flick precisely at t=2, in which case - which seems more true as I write it - that light should be on.)
But, if I have completely misinterpreted the question, I would say that the light should be both on and off simultaneously, so the position of the switch is both correct and incorrect.
Just my two cents.
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srn437
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Re: supertasks
« Reply #13 on: Sep 4th, 2007, 8:51pm » |
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Misenterpretation. Maybe I should give the answer after one more close guess.
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Whiskey Tango Foxtrot
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Re: supertasks
« Reply #14 on: Sep 5th, 2007, 11:13am » |
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I also think this is a trick involving the order in which you stated the actions. By stating them backwards you made us think the light is first modified after a minute, then again after thirty seconds, etc. In reality, the first change takes place after an infinitesimally small amount of time. The changes continue taking place until fifteen seconds, when the light is turned off. It is turned on at thirty seconds, back off at the one minute mark, and will be turned on at two minutes.
I'm pretty sure this is what Archae was saying but I found it a bit confusing on the first read-through.
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| « Last Edit: Sep 5th, 2007, 11:17am by Whiskey Tango Foxtrot » |
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srn437
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Re: supertasks
« Reply #15 on: Sep 5th, 2007, 4:47pm » |
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That's the same answer as stated(incorrectly) previously.
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Whiskey Tango Foxtrot
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Re: supertasks
« Reply #16 on: Sep 5th, 2007, 6:26pm » |
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Okay, then I say it will be off. The reasoning is the same as above except there is no switching at the 2 minute mark.
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"I do not feel obliged to believe that the same God who has endowed us with sense, reason, and intellect has intended us to forgo their use." - Galileo Galilei
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Archae
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Re: supertasks
« Reply #17 on: Sep 5th, 2007, 7:29pm » |
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Sorry for the poor wording, there. I was kinda rambling on I realize (saying things as they hit me), so it didn't make much sense.
I'll keep on the problem.
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srn437
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Re: supertasks
« Reply #18 on: Sep 5th, 2007, 10:59pm » |
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It will be off, but not why you said. If a switch is in the middle(which it will be), the light will be off. And for the next one, a machine can in half an hour make a machine half its size, but twice as fast. In an hour, how many will there be? What size(in terms of the original machines size)?
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| « Last Edit: Sep 5th, 2007, 11:01pm by srn437 » |
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towr
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Re: supertasks
« Reply #19 on: Sep 6th, 2007, 12:21am » |
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on Sep 5th, 2007, 10:59pm, srn347 wrote:| If a switch is in the middle(which it will be) |
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No it won't.
Just try it
(Mathematically speaking its state is undefined; not in the middle. And physically, it's impossible.)
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SMQ
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Re: supertasks
« Reply #20 on: Sep 6th, 2007, 7:46am » |
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on Sep 5th, 2007, 10:59pm, srn347 wrote:| a machine can in half an hour make a machine half its size, but twice as fast. In an hour, how many will there be? What size(in terms of the original machines size)? |
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"Just before" an hour there will be a countable infinity of machines: one of each size 1, 1/2, 1/4, 1/8, etc. "Just after" an hour there will be a countable infinity of machines, but "almost twice as many": one of size 1 and two each of sizes 1/2, 1/4, 1/8, etc. (The terms in quotes are not rigorously defined in the reals, but could be with hyperreals.)
The first machine makes a machine of 1/2 size at 1/2 hour and at 1 hour. The second machine makes a machine of 1/4 size as 3/4 hours and at 1 hour. The third machine makes a machine of 1/8 size at 7/8 hours and at 1 hour, etc.
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| « Last Edit: Sep 6th, 2007, 7:47am by SMQ » |
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srn437
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Re: supertasks
« Reply #21 on: Sep 6th, 2007, 4:19pm » |
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I have tried it, have you?
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Archae
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Re: supertasks
« Reply #22 on: Sep 6th, 2007, 11:01pm » |
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It depends on the switch I believe: I have had switches that will stay in the middle with the light off, but also switches that will rest in the middle with the light on.
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JiNbOtAk
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Re: supertasks
« Reply #23 on: Sep 7th, 2007, 1:51am » |
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on Sep 6th, 2007, 12:21am, towr wrote:| (Mathematically speaking its state is undefined; not in the middle. And physically, it's impossible.) |
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I think what towr meant is that switches are discrete, it can only be either on, or off. To say that the switch in the middle position, from between on and off, would result in the bulb being turned on ( or off ), is pointless, as that would mean the on ( or off ) position would just need to be shifted to the current middle position. ( Am I making any sense ? )
Anyway, if switches aren't discrete, one would expect it to act like dimmers, where the middle position would cause the bulb to still be on, but the light is somewhat dimmer.
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towr
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Re: supertasks
« Reply #24 on: Sep 7th, 2007, 2:47am » |
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No, what I meant was that physically toggling a switch with infinite speeds makes it go back in time; trying to accelerate it to an infinite speed increases its mass to the point you form a black hole; also once it reached light speed, time for the switch stops and it can't go faster; and even at much lower speeds the switch will already be destroyed due to internal material stresses. 
And mathematically speaking the end state is indeterminate. It's equivalent to asking "is the last natural number odd or even", ignoring that there is no such number.
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