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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.
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!
You can’t answer this kind of question because “after infinity” is meaningless nonsense.
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.
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).
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.
This is just the “Achilles and the turtle” paradox again, isn’t it? You won’t trick me into inventing calculus a second time!
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.
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.
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.
Yeah but the amount of numbers they haven’t checked is even higher!
But those numbers are too high to fit into a trolley (or maybe even the Solar System/galaxy/observable universe?)
Escape your special characters, people.
Also I find your approximation to be overly precise. 270 is ca. 1021
Whoops. Fixed it lmao
Not exactly. Jerboa shows them as <sup> and <\sub>
Please check the source code as I can’t be fucked to escape myself
In the amount of time it takes to describe the situation, I could go free the people tied to the tracks.
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.
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.
Motherfucker here making a math problem for infinite series from a philosophical question. Is nothing sacred /s
After an infinite number of loops?
After an infinite number of loops I’d want to be killed.
After an infinite number of loops are any of the original passengers still on the trolley?
Anything moving at light speed does not experience the passage of time, so yes. Nobody can actually get off the trolley.
Anything moving at light speed does not experience the passage of time
Nobody can actually get off
If time stops for people on the trolley, wouldn’t their subjective experience be of immediately getting off the trolley?
Without solving the collatz conjecture I think you can see it always stays above zero.
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?
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
how fast is it going if i dont pull it? what kind of smear are we looking at?
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.
That’s impossible because if n is an odd number, then 3n+1 is even. The number can never increase twice in a row.
Passengers: 4 > 2 > 1 > 4
So that’s better, I guess?
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.
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 …
Edited my response, I misread it. Thank you!
You misread it, too, did you? 😉
Yes! :/
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?
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!
I’d be too confused to make a decision
Don’t be, that’s falling for their trap!!!
If you choose not to decide, you still have made a choice
lol ok buddy, back to your philosophy 101 class, you have homework
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).
2n+1 is not in the Collatz conjecture.
Mathematics is not ready for such carelessness.
And I did a dumb. Withdrawn.
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.
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
2n+1 enter the trolley, meaning 3n+1, which in the case of n being an odd number, will always equal an even.
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)
It isn’t stated wrongly: 2n+1 get on, which means 3n+1 are on board
touche
OH that makes sense, so I misread it.
I am too tired to think too hard about this, so, SURE!
