i think that can’t really be answered bc there’s no hard rules on who specifically gets off.
if it’s first-on, first-off then all the original riders would cycle out in as little as 2 cycles. but if it’s first-on, LAST-off then at least 1 person from the original bunch would always be on the train.
if it’s random, who knows! someone who took probability and statistics can work that one out lmao
If the trolley is moving at light speed by the time it hits the station, it is impossible for anyone to get on or off because—from the trolley’s perspective—no time passes between stops. Ergo, the number of passengers on it must be the same every stop.
If the initial number of passengers is odd or a non-zero integer, this inability to board/unboard would contradict the rules.
Thus, in order to satisfy all the conditions, the initial number of people on the trolley must be 0. As an even number it will be subject to halving, but 0/2=0, so the rules are satisfied.
Hence, pulling the lever is the optimal solution as 0 people will die. QED.
Also, the trolley going in a loop the speed of light would be immediately deformed and destroyed. If inner points of the trolley go c, then outer points would be going faster, so the tram would forcefully deform. Same with the people inside
Not to be the 🤓 but technically that only applies to Euclidean spacetime. It is possible to have spaces in which loops occur without there being a localized curvature gradient. The manifold might loop but at a small enough scale all manifolds are locally Euclidean. There are also just weird things that happen in hyperbolic geometry where you can have infinite nested concentric circles that are all technically the same size and are centered at infinity (Horocycles).
Anyway, point is that we don’t necessarily know the topology of the space in which the loop resides, so we can’t make the assumption that the trolley would be destroyed.
It only accelerates to light speed, therefore it will need infinite time to complete the loops. Thus the risk is not the killing but getting stuck.
If the conjecture holds, naturally there is a small cycle so people can get on and off and use the train as a form of teleporting to the future.
If there are different loops, then still people can take turns.
Even if there are values that diverge, if it can be shown that at least one event of division occurs with a certain average frequency in the infinite divergence, then at any such point all previous guests can exit and the train can be ridden for one such span.
Only if there are no cases of division and endless steps of 3n+1 in the limit, would people be trapped on the train at no subjective time passing, and in essence time travel into the infinitely far future where they are killed.
Wouldn't that be 3n + 1? n passengers already present, another 2n + 1 enter, resulting in a total of 3n + 1. Doing it in my head, we seem to always end up in a cycle of 4 -> 2 -> 1 -> 4. All of these are < 5, so once we enter that cycle, the number of possible passengers killed is always less than five.
One is: would you bet human lives on a conjecture being true? The collatz conjecture does hold for every number we have tried, but there have been conjectures that were disproven with a very large counterexample. You could kill countless humans if wrong, so even if you think the chance of a counterexample is low, is it low enough to outweigh that potentially very hight value counterexample?
The second one is: Let's say the collatz conjecture holds, and the number of passsengers just loops 4 -> 2 -> 1 -> 4 -> 2 -> 1 eventually. What is the 'final' number, when the trolley is done with the infinite loops? It can't be 1, because that is always followed by a 4. And it can't be 4 because it's always followed by 2 and so on. But it has to be one of those, because any other number is not possible. It reminds me of the Vsauce Video Supertasks, which comes to the conclusion that we can't know the answer to these type of questions.
So in conclusion, flipping the switch will either give you an arbitrarily large number of deaths, or an unknown number of deaths. Fun!
In the spirit of supertasks, I think no matter what you should take the collatz loop. Assuming the people who get free are chosen randomly, it results in merely an infinite number of people taking a short train ride and then going about their day; a unifying rite of passage for humanity (or a rare opportunity for the blessed few).
It really depends.
For example, if you walk 1m, then 0.5m, then 0.25m and continue infinitely, then "after infinity" you will have walked exactly 2m. This is the classic 'Achilles and turtle' example and works fine if the value converges. It's just mathematics.
There is only a problem if the value diverges. Imagine the step example, but on even steps, you raise a blue flag, and on odd steps you raise a red flag. Now the question what flag is raised "after infinity" is impossible to answer. It clearly should be either red or blue, but it also can't really be either, because that would mean infinity is either even or odd, which makes no sense.
Flipping the switch gives you 1, 2, or 4 deaths. It will always end up looping those three numbers, so after an infinite amount of time it has to be one of those three options.
All three of those options are less than 5, and they occur after an infinite amount of time instead of (relatively speaking) immediately.
From both a pure numbers perspective and a theoretical minimizing or delaying harm perspective, pulling the lever is the right move.
You're assuming the collatz conjecture holds, which is unknown.
But even if it does hold, you do understand the second problem, right? 1 can not possibly be the outcome, because whenever there is a 1 in that infinite loop, it is followed by a 4. And if 1 is the outcome, then it wasn't done infinitely, because otherwise there must have been a 4 afterwards. The same argument holds for 4 and 2 as well. So we're stuck in the reality that it would have to be one of those numbers, but it also can't really be one of those numbers. It's paradoxical.
It's a lyric from a Rush song, so I don't know what you're on about.
But yes, in the original trolley problem "not deciding" means letting it run over the people on the initial track, which is still viewed as a moral decision.
So not only is your comment irrelevant, it's also wrong.
(Oh, and I have a degree in philosophy, so your condescension is unwarranted).
No matter what wouldn't this grow to infinite passengers? Is that supposed to be the point?
Because any even number is going to halve itself down to 1, which is odd an odd number, then double plus one will always make another odd number so it would grow to infinity.
Edit: Misread the problem, read replies for explanations
It's stated wrongly. The Collatz Conjecture is about the case where you triple an odd number and add one, that way you enter a loop if you get down to 1 (1 -> 4 -> 2 -> 1)
I think everyone inside would die when it accelerated to light speed. And since most numbers are larger than 5 I'd say don't pull the lever. The only question is what happens to the people inside the trolley if you don't pull the lever?
It'll always reduce to a cycle of 4→2→1→4→2→1 etc.
Which means people can get on and off so no one is trapped, and since they don't die until after an infinite number of stops it means no one will get killed
If the number of passengers is unknown, we can't guarantee it is strictly positive or even an integer.
Sure, accidentally killing lim {-2,-1,-2,-1...} people is good compared to lim {4,2,1,...} or 5, but only if it's an integer that's a 50/50 on being negative. If it's not an integer, which is infinitely more likely if we truly have nothing to go off and have to assume it was randomly chosen, then reality might break upon reaching the station.
Doesn't "2 times, plus 1" mean from that point forward, there will always be an odd number of people on the trolley? Meaning, after not too long, ALL people will be on the trolley? And, since there is no "after" infinity, they'll all be stuck on the trolley until they die? So, throwing the lever kills all of humanity, but it will, for them, happen in the distant, distant future from when they started?
How long does it take to get an infinite number of loops in? Well, it's going at a finite speed, so it must be an infinite amount of time. Maybe you can argue that at the speed of light causes the inside of the trolley to not experience time past that point, but there's still all the time spent at sub-light speed accelerating. So at least an astronomical amount of time.
And the rules as stated result in an arbitrarily large number of people on the trolley. So these people after a point aren't being pulled from Earth, they must be being created wholesale. And then living a life out on the trolley, unless they exit.
So the choices are really 1. Kill 5 people or 2. Create an unknown but large number of people that will live out some sort of lives on the trolley, or get shunted out into the real world, and some smaller but still large number of people that will die prematurely.
If someone dies due to extraneous circumstances, they would die either way so it doesn't have to factor into your considerations on whether or not to pull the lever.
And while I haven't done a geometric proof to show that for all odd numbers, it will eventually reduce to one, I've worked out the sets for every odd number up to twenty and the pattern holds. While that's not rigorous enough for a theorem, it's good enough for me.
Wouldn't the tram require all the energy in the universe to accelerate to light speed whilst the object would create a massive wave of radiation and shock waves thus destroying everything in its path?
Thus no more station and peoples on the rail meaning no one can get on or off.
66 Comments
slacktoid@lemmy.ml · 41 pts · 180d
Motherfucker here making a math problem for infinite series from a philosophical question. Is nothing sacred /s
jdr@lemmy.ml · 34 pts · 180d
After an infinite number of loops?
After an infinite number of loops I'd want to be killed.
hakase@lemmy.zip · 13 pts · 180d
After an infinite number of loops are any of the original passengers still on the trolley?
Delta_V@lemmy.world · 13 pts · 180d
Anything moving at light speed does not experience the passage of time, so yes. Nobody can actually get off the trolley.
threelonmusketeers@sh.itjust.works · 3 pts · 180d
If time stops for people on the trolley, wouldn't their subjective experience be of immediately getting off the trolley?
jdr@lemmy.ml · 5 pts · 180d
Without solving the collatz conjecture I think you can see it always stays above zero.
hakase@lemmy.zip · 7 pts · 180d
Sure, the total number of passengers does, but do any of the original passengers stay on the entire time as new passengers cycle on and off?
yuri@pawb.social · 8 pts · 180d
i think that can’t really be answered bc there’s no hard rules on who specifically gets off.
if it’s first-on, first-off then all the original riders would cycle out in as little as 2 cycles. but if it’s first-on, LAST-off then at least 1 person from the original bunch would always be on the train.
if it’s random, who knows! someone who took probability and statistics can work that one out lmao
AnarchoEngineer@lemmy.dbzer0.com · 26 pts · 180d
If the trolley is moving at light speed by the time it hits the station, it is impossible for anyone to get on or off because—from the trolley’s perspective—no time passes between stops. Ergo, the number of passengers on it must be the same every stop.
If the initial number of passengers is odd or a non-zero integer, this inability to board/unboard would contradict the rules.
Thus, in order to satisfy all the conditions, the initial number of people on the trolley must be 0. As an even number it will be subject to halving, but 0/2=0, so the rules are satisfied.
Hence, pulling the lever is the optimal solution as 0 people will die. QED.
reabsorbthelight@lemmy.world · 5 pts · 180d
Also, the trolley going in a loop the speed of light would be immediately deformed and destroyed. If inner points of the trolley go c, then outer points would be going faster, so the tram would forcefully deform. Same with the people inside
AnarchoEngineer@lemmy.dbzer0.com · 8 pts · 180d
Not to be the 🤓 but technically that only applies to Euclidean spacetime. It is possible to have spaces in which loops occur without there being a localized curvature gradient. The manifold might loop but at a small enough scale all manifolds are locally Euclidean. There are also just weird things that happen in hyperbolic geometry where you can have infinite nested concentric circles that are all technically the same size and are centered at infinity (Horocycles).
Anyway, point is that we don’t necessarily know the topology of the space in which the loop resides, so we can’t make the assumption that the trolley would be destroyed.
Agent641@lemmy.world · 1 pts · 180d
No, I don't pull the lever. I don't want to wait for it to do infinite loops to see the gore and carnage of running over a bunch of people.
Redjard@lemmy.dbzer0.com · 10 pts · 180d
It only accelerates to light speed, therefore it will need infinite time to complete the loops. Thus the risk is not the killing but getting stuck.
If the conjecture holds, naturally there is a small cycle so people can get on and off and use the train as a form of teleporting to the future.
If there are different loops, then still people can take turns.
Even if there are values that diverge, if it can be shown that at least one event of division occurs with a certain average frequency in the infinite divergence, then at any such point all previous guests can exit and the train can be ridden for one such span.
Only if there are no cases of division and endless steps of 3n+1 in the limit, would people be trapped on the train at no subjective time passing, and in essence time travel into the infinitely far future where they are killed.
wonderingwanderer@sopuli.xyz · 3 pts · 180d
That's impossible because if n is an odd number, then 3n+1 is even. The number can never increase twice in a row.
Chais@sh.itjust.works · 9 pts · 180d
Passengers: 4 > 2 > 1 > 4
So that's better, I guess?
Scipitie@lemmy.dbzer0.com · 3 pts · 180d
Edit: misread the prompt, below original post is wrong, I apologize!
Noe start with three or any odd number :p 3 , 7, 15, 31, 33.....
I think only 2^x has the outcome you describe, please correct me if I'm mistaken.
addison@piefed.social · 10 pts · 180d
Odd numbers cause double plus one additional passengers to board. So a cycle starting with 3 is a bit different.
3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 -> 4 ...
Scipitie@lemmy.dbzer0.com · 2 pts · 180d
Edited my response, I misread it. Thank you!
jbrains@sh.itjust.works · 5 pts · 180d
You misread it, too, did you? 😉
Scipitie@lemmy.dbzer0.com · 2 pts · 180d
Yes! :/
wonderingwanderer@sopuli.xyz · 2 pts · 180d
7→22→11→34→17→52→26→13→40→20→10→5→16→8→4→2→1→4→2→1→4→2→1...
Edit: also, 9→28→14→7...
And, 15→46→23→70→35→106→53→160→80→40→20→10→5→16→8→4→2→1...
Or even, 19→58→29→88→44→22→11→...
And lest we forget, 3→10→5→16→8→4→2→1...
That covers every odd number below 20. Want me to do 21 and 25, too? Perhaps 27?
Scipitie@lemmy.dbzer0.com · 1 pts · 180d
The trick is to read the prompt. I had 2n+1 are the new number of passengers.
Had to reconstruct it from your answer though.
Anyway, thanks!
MousePotatoDoesStuff@lemmy.world · 8 pts · 180d
The conjecture has been checked by computer for all starting values up to 2^71 ≈ 2.36×10^21.
So you're probably good to go.
Gladaed@feddit.org · 3 pts · 179d
Escape your special characters, people.
Also I find your approximation to be overly precise. 2^70^ is ca. 10^21^
MousePotatoDoesStuff@lemmy.world · 1 pts · 179d
Whoops. Fixed it lmao
Gladaed@feddit.org · 2 pts · 179d
Not exactly. Jerboa shows them as and <\sub>
Please check the source code as I can't be fucked to escape myself
ShinkanTrain@lemmy.ml · 3 pts · 179d
Yeah but the amount of numbers they haven't checked is even higher!
MousePotatoDoesStuff@lemmy.world · 1 pts · 179d
But those numbers are too high to fit into a trolley (or maybe even the Solar System/galaxy/observable universe?)
jbrains@sh.itjust.works · 7 pts · 180d
2n+1 is not in the Collatz conjecture.
Mathematics is not ready for such carelessness.
And I did a dumb. Withdrawn.
SpaceNoodle@lemmy.world · 3 pts · 180d
Wouldn't that be 3n + 1? n passengers already present, another 2n + 1 enter, resulting in a total of 3n + 1. Doing it in my head, we seem to always end up in a cycle of 4 -> 2 -> 1 -> 4. All of these are < 5, so once we enter that cycle, the number of possible passengers killed is always less than five.
Honytawk@feddit.nl · 6 pts · 180d
This one is easy.
Either you kill 5 people, or you pull the lever and kill none. Because an infinite loop never ends.
All the rest is just fluff with no bearing on the question.
pyre@lemmy.world · 5 pts · 180d
yeah i don't get what the point is. the answer is easy, you switch tracks unless there is at least one billionaire among the five.
xxd@discuss.tchncs.de · 6 pts · 179d
There are two super interesting problems in here.
One is: would you bet human lives on a conjecture being true? The collatz conjecture does hold for every number we have tried, but there have been conjectures that were disproven with a very large counterexample. You could kill countless humans if wrong, so even if you think the chance of a counterexample is low, is it low enough to outweigh that potentially very hight value counterexample?
The second one is: Let's say the collatz conjecture holds, and the number of passsengers just loops
4 -> 2 -> 1 -> 4 -> 2 -> 1eventually. What is the 'final' number, when the trolley is done with the infinite loops? It can't be1, because that is always followed by a4. And it can't be4because it's always followed by2and so on. But it has to be one of those, because any other number is not possible. It reminds me of the Vsauce Video Supertasks, which comes to the conclusion that we can't know the answer to these type of questions.So in conclusion, flipping the switch will either give you an arbitrarily large number of deaths, or an unknown number of deaths. Fun!
Artisian@lemmy.world · 2 pts · 179d
In the spirit of supertasks, I think no matter what you should take the collatz loop. Assuming the people who get free are chosen randomly, it results in merely an infinite number of people taking a short train ride and then going about their day; a unifying rite of passage for humanity (or a rare opportunity for the blessed few).
vithigar@lemmy.ca · 2 pts · 179d
You can't answer this kind of question because "after infinity" is meaningless nonsense.
xxd@discuss.tchncs.de · 2 pts · 179d
It really depends. For example, if you walk 1m, then 0.5m, then 0.25m and continue infinitely, then "after infinity" you will have walked exactly 2m. This is the classic 'Achilles and turtle' example and works fine if the value converges. It's just mathematics.
There is only a problem if the value diverges. Imagine the step example, but on even steps, you raise a blue flag, and on odd steps you raise a red flag. Now the question what flag is raised "after infinity" is impossible to answer. It clearly should be either red or blue, but it also can't really be either, because that would mean infinity is either even or odd, which makes no sense.
wizardbeard@lemmy.dbzer0.com · 1 pts · 179d
Flipping the switch gives you 1, 2, or 4 deaths. It will always end up looping those three numbers, so after an infinite amount of time it has to be one of those three options.
All three of those options are less than 5, and they occur after an infinite amount of time instead of (relatively speaking) immediately.
From both a pure numbers perspective and a theoretical minimizing or delaying harm perspective, pulling the lever is the right move.
xxd@discuss.tchncs.de · 1 pts · 179d
You're assuming the collatz conjecture holds, which is unknown.
But even if it does hold, you do understand the second problem, right? 1 can not possibly be the outcome, because whenever there is a 1 in that infinite loop, it is followed by a 4. And if 1 is the outcome, then it wasn't done infinitely, because otherwise there must have been a 4 afterwards. The same argument holds for 4 and 2 as well. So we're stuck in the reality that it would have to be one of those numbers, but it also can't really be one of those numbers. It's paradoxical.
Tudsamfa@lemmy.world · 2 pts · 179d
This is just the "Achilles and the turtle" paradox again, isn't it? You won't trick me into inventing calculus a second time!
carbon@piefed.social · 6 pts · 180d
I’d be too confused to make a decision
SubArcticTundra@lemmy.ml · 4 pts · 180d
Don't be, that's falling for their trap!!!!
wonderingwanderer@sopuli.xyz · 1 pts · 180d
If you choose not to decide, you still have made a choice
carbon@piefed.social · 1 pts · 179d
lol ok buddy, back to your philosophy 101 class, you have homework
wonderingwanderer@sopuli.xyz · 1 pts · 179d
It's a lyric from a Rush song, so I don't know what you're on about.
But yes, in the original trolley problem "not deciding" means letting it run over the people on the initial track, which is still viewed as a moral decision.
So not only is your comment irrelevant, it's also wrong.
(Oh, and I have a degree in philosophy, so your condescension is unwarranted).
Wirlocke@lemmy.blahaj.zone · 5 pts · 180d
No matter what wouldn't this grow to infinite passengers? Is that supposed to be the point?
Because any even number is going to halve itself down to
1, which is oddan odd number, then double plus one will always make another odd number so it would grow to infinity.Edit: Misread the problem, read replies for explanations
Lauchmelder@feddit.org · 8 pts · 180d
It's stated wrongly. The Collatz Conjecture is about the case where you triple an odd number and add one, that way you enter a loop if you get down to 1 (1 -> 4 -> 2 -> 1)
wonderingwanderer@sopuli.xyz · 5 pts · 180d
2n+1 enter the trolley, meaning 3n+1, which in the case of n being an odd number, will always equal an even.
tomiant@piefed.social · 4 pts · 180d
I am too tired to think too hard about this, so, SURE!
MegaMichelle@a2mi.social · 2 pts · 179d
@mech
In the amount of time it takes to describe the situation, I could go free the people tied to the tracks.
HeyThisIsntTheYMCA@lemmy.world · 2 pts · 179d
how fast is it going if i dont pull it? what kind of smear are we looking at?
pruwybn@discuss.tchncs.de · 2 pts · 180d
I think everyone inside would die when it accelerated to light speed. And since most numbers are larger than 5 I'd say don't pull the lever. The only question is what happens to the people inside the trolley if you don't pull the lever?
wonderingwanderer@sopuli.xyz · 2 pts · 180d
It'll always reduce to a cycle of 4→2→1→4→2→1 etc.
Which means people can get on and off so no one is trapped, and since they don't die until after an infinite number of stops it means no one will get killed
Tudsamfa@lemmy.world · 2 pts · 179d
If the number of passengers is unknown, we can't guarantee it is strictly positive or even an integer.
Sure, accidentally killing lim {-2,-1,-2,-1...} people is good compared to lim {4,2,1,...} or 5, but only if it's an integer that's a 50/50 on being negative. If it's not an integer, which is infinitely more likely if we truly have nothing to go off and have to assume it was randomly chosen, then reality might break upon reaching the station.
DarrinBrunner@lemmy.world · 1 pts · 180d
Trust me when I say I'm an idiot and a liar.
Doesn't "2 times, plus 1" mean from that point forward, there will always be an odd number of people on the trolley? Meaning, after not too long, ALL people will be on the trolley? And, since there is no "after" infinity, they'll all be stuck on the trolley until they die? So, throwing the lever kills all of humanity, but it will, for them, happen in the distant, distant future from when they started?
mech@feddit.org · 1 pts · 180d
2 times + 1 get on additionally, so there will be 3 times plus 1 on the trolley.
Which can be either odd or even.
Delta_V@lemmy.world · 1 pts · 180d
It eventually gets stuck in a loop.
5, 16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1 . . . etc
mumblerfish@lemmy.world · 1 pts · 180d
Yeah, it is suppose to be 3n+1: https://en.wikipedia.org/wiki/Collatz_conjecture
Tudsamfa@lemmy.world · 1 pts · 179d
n are already on the trolley, 2n +1 enter the trolley.
bacon_pdp@lemmy.world · 0 pts · 25d
Well it is not true for zero.
ttyybb@lemmy.world · 1 pts · 180d
I read it the same way at first but let's say theres seven people on the trolly, twice that plus one is 15 people getting on the trolly, 15+7 is 22.
wonderingwanderer@sopuli.xyz · 1 pts · 180d
Why does it have to accelerate to the speed of light? I don't understand what role that part plays in this...
Atlas_@lemmy.world · 1 pts · 180d
How long does it take to get an infinite number of loops in? Well, it's going at a finite speed, so it must be an infinite amount of time. Maybe you can argue that at the speed of light causes the inside of the trolley to not experience time past that point, but there's still all the time spent at sub-light speed accelerating. So at least an astronomical amount of time.
And the rules as stated result in an arbitrarily large number of people on the trolley. So these people after a point aren't being pulled from Earth, they must be being created wholesale. And then living a life out on the trolley, unless they exit.
So the choices are really 1. Kill 5 people or 2. Create an unknown but large number of people that will live out some sort of lives on the trolley, or get shunted out into the real world, and some smaller but still large number of people that will die prematurely.
I think life is worth living, so I prefer 2
wonderingwanderer@sopuli.xyz · 3 pts · 180d
It will always reduce to a cycle of 4→2→1→4→2→1, so you won't end up with a huge number.
But yeah, it would take infinite time to reach infinite loops, and meanwhile people can get on and off, so in reality nobody dies prematurely...
Atlas_@lemmy.world · 2 pts · 180d
People can die on the trolley,
Also you're assuming the collatz conjecture.
wonderingwanderer@sopuli.xyz · 2 pts · 180d
If someone dies due to extraneous circumstances, they would die either way so it doesn't have to factor into your considerations on whether or not to pull the lever.
And while I haven't done a geometric proof to show that for all odd numbers, it will eventually reduce to one, I've worked out the sets for every odd number up to twenty and the pattern holds. While that's not rigorous enough for a theorem, it's good enough for me.
how_we_burned@lemmy.zip · 1 pts · 180d
Wouldn't the tram require all the energy in the universe to accelerate to light speed whilst the object would create a massive wave of radiation and shock waves thus destroying everything in its path?
Thus no more station and peoples on the rail meaning no one can get on or off.
ttyybb@lemmy.world · 1 pts · 180d