You got a time machine that can travell fordward in time, the plausable MAX equals 5245181978721581550. You want to test if this maximum is valid, but you only got 12 machines, so you can use only 11 for testing, to define the maximum. The testing goes by this: you send the machine forward then it comes back. If it's unharmed, the test was successfull, but it can also come back damaged meaning valid time limit exceeded. What is the minimum number of time travels to determine the valid travel maximum?
That picture reminds me that I have big boobs and they are heavy :(
Lincoln Bell
>You want to test if this maximum is valid It's useless. Max supposedly equals 5 245 181 978 721 581 550 The Sun will turn into a red giant in 5 000 000 000 000 Sun basically dead in 10 000 000 000 000
Joseph Carter
I have 4 solutions, but they are 3-4-6-10 digit
Bentley Bell
bout tree fiddy
Jonathan Evans
>solve your own homework is the only correct answer
Jose Thompson
>You want to test if this maximum is valid That specific maximum? You could just send one machine to that specific time and see if it comes back damaged.
Anyways the answer is pretty simple. You just divide the interval you want to search in to nth root sized chunks, where n is the number of trials you are able to perform. Then you proceed to try at the end of every interval. If the machine is damaged in any interval, you just do the same thing inside that interval, divide it into nth root sized chunks and then search.
This means that if you had 11 machines, you would, in the worst case need to do at least 11*(5245181978721581550^(1/11)) trials or roughly 550 trials.
Juan Thomas
>do a test >if the machine comes back damaged then just go back in time to before you did the test Infinite tests
Sebastian Morgan
It can only go forward in time so it'll never get back
Jeremiah Foster
The minimum is 2 times with luck: You send one machine forward to for example 123215325 and it comes back damaged. Then you send one to 123215325 - 1 and if it comes back unharmed you know the max is 123215325.
Josiah Adams
OP here that's a paradox, let's say you have 2 machines, 1 is a proxy, 1 for travel. (since we are going back in time, i will use negatives, and "_t" to mark the machines timeline) time0: you send out travel0_t0, machine comes back damaged. you take proxy0_t0, go to time-1. time-1: send out undamaged travel0_t-1, machine comes back damaged but how will you send out travel0_t0, if you damaged it when it was travel0_t-1 on the first place, not to mention, how did you acknowledge it came back damaged, and use proxy0_t0 to go back to time-1, and do the whole cycle.
Let's say the time travel takes the machine out of time, and you can close the loop by replacing travel0_t0 with travel0_tX aat the end of the test, so it get's destroyed, and also closes the loop, BUT we don't care about infinite travels, we want to know the lowest number of travels
I was thinking about that, but 550 is not on the list I think I need to use 10 machines to find the smallest batch, then use the 11th to determine the max 1by1, but my math is off
Landon Roberts
can I carry dem?
Jacob Harris
Input 6666666666666666666 and hope this is the end
I know nothin about all these shiets, homies. I be bangin hoes and pussies all night. I’m black but ain’t nobody gone be calling me nigga. I’m be countin cash more than you white nerds. WE WUZ KANGZ !
Nathaniel Hall
Fuck that I'd go back in time and win the Mega Millions jackpot
well with that method 550 is the worst case number of searches. Conceivably you could send it to the first batch, and it would come back broken, then send it to the first sub-batch and it would come back broken, and that would continue until you use every machine only to discover that the very first year is the max, and you would have used 11 machines, sending them a total of 11 times.
Liam Campbell
these always focus on worst case scenarios. Conceivably you could do that
>impossible to solve.jpg Maybe if you are one of those whatitwascalled "natural math" or some other faggotry idiots who don't even know if + or x comes first. Otherwise ezpz.
John Morris
15 any other answer is retarded
Lincoln Gomez
>68252815 >68252826 (bait, don't give them (you))
Owen Lewis
It actually preys on the people who don't notice that there's only 1 packet of fries in the last equation as opposed to two. Like
Charles Myers
i'd go back to my conception and re-roll my gender
Carter Hughes
well, you need to make an easy to miss little thing for people to get tribalistic over some fucking numbers and desperately try to chime in with "you are all wrong, I'm smart," "no, YOU are wrong, I'm smart" and get 12 thousand likes and comments on whatever facebook group you're in
Ryder Morris
$5.99 plus tip
Connor Collins
The symbol of a single set of fries has an undefined value :^)
Ryder Price
mcdonald's menu equation > optimization problem, time travel thought this is more of the Jow Forums I know
Austin Long
>>INB4 binary search What do you mean? Binary search does give the min number of tests required, doesn't it? Is the problem that the 11 machines could all be destroyed before bsearch gives you the correct value?
Brandon Sanders
Yes that's exactly the problem. There's a similar problem in algorithm design (kleinberg and tardos) about designing jars to be dropped from specific heights.
That's not the only thing people miss. The box of fries in the fourth "equation" is also short a single fry, which means you need to determine the value of 1 fry from the third "equation." Since it's assumed the values are currency, this makes a single fry worth 1/7th of a dollar, which isn't a clean decimal resulting in this entire thing being dumb as fuck. >Respecting PEMDAS, the answer is 95/7 dollars or $13.57142857143. This is why it's a masterful shitpost device.
