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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
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.
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.
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
I’d be too confused to make a decision
Don’t be, that’s falling for their trap!!!
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.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
OH that makes sense, so I misread it.
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.
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.
Passengers: 4 > 2 > 1 > 4
So that’s better, I guess?
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.
7→22→11→34→17→52→26→13→40→20→10→5→16→8→4→2→1→4→2→1→4→2→1…
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 …
You misread it, too, did you? 😉
I am too tired to think too hard about this, so, SURE!
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?
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