Is Veeky Forums Smart Enough to Solve This?

Assume you have 25 horses and you want to find the fastest 3 of the 25. What is the minimal number of races you can perform assuming you can only observe 5 horses racing at a time. You have no external tools and you only know the order of the horses when they finish their race.

7 races

this is a fucking easy question you dumb Amerilard

you just time them dumbass

It actually not.

You're assuming that the horses in the first 5 races who didn't place first aren't faster than the horses who placed first in other groups.

This is a famous fucking riddle given to all French students, I know what the fucking answer is you fucking retarded Amerinigger. If you're going to pull the retarded nigger stunt like you are because education fucking sucks in your nigger riddled country the answer is fucking 11. Holy fuck I hope you fucking kill yourself or get cancer and die you fucking hopeless brainlet

What part of no external tools don't you understand you fucking imbecile

>says question is easy
>gets it wrong
>wrong answer is pointed out
>gets mad and complains while correcting himself
Okai

>6
5 races to pick the fastest of the 25 in 5 groups, then a final race among the 5 winners.

Because you're using an apeish interpretation of the problem. This is an extremely famous problem and there is the 7 race answer and an 11 race answer. Get this through your ape skull: you are a dumb fat nigger

1+(20/4)+1=7 races
amerilard confirmed dumb

so not only did you get the answer wrong, but you have known of this question since you were a child, so you didn't figure it out

that doesn't work. What if the fastest 3 horses are all in your first group

No because there's two different answers depending on how you interpret the problem, which Mr Frenchy clearly stated. If you could read English you'd know that

samefagging this hard

Yeah, I noticed after posting, what if you do 5 races in groups of 5 and then you make 5 races more in which all of those you got in their first place compete with each other, the ones in second and so on and then you make a final race among the 5 finalists giving you a total of 11 races. I'm just a brainlet and I don't study a math oriented career :(

How can it be interpreted differently?
Seems like an obvious 7 to me

idk if that works either, but here's what you can definitely do: in any race, eliminate the bottom 2. so you can definitely do it in 11 races, because then you eliminate 22 leaving only the top 3.

That seems trivial though. It seems you should be able to do it with fewer than 11 racdes

But, how could you know that the two slowest in any race aren't faster than those in other group?

The slowest two in a race can't possibly be in the top three of all the horses, since we know there are three faster horses than them.

Since everyone seems to acknowledge that this is a quick google away, how about 729 horses, 9 horses can be raced at a time, and we still want the top 3 horses.

is it 121 races?

Nope, its a little lower. How are you racing the horses for 121?

not him, but I'd assume he just makes a race of 5 horses, keeps the top 3 and makes them race with other 2, keeps the top 3 and so on...
this way every race excludes 2 horses

but that's not the optimal strat

I believe you can generalize the method used to obtain 7 to any N horses, for races of n horses, for the top x

To my knowledge there isn't a perfect generalization, as ranking the top t objects in a set of n with a k-sorter is an important unsolved computer science problem. However, there is a generalization of the 7 race solution to the numbers I gave.

just keep racing and eliminating the bottom 6 of every race.

LOL obviously I'm a retarded

93?

How did you arrive at 93?

are we trying to rank the top 3, or just find which three horses are the top 3?

First groups of 9, race all of them for a total of 81 races

Now you have 243 horses left since the 4th to 9th placers are trash

You can imagine 81 groups of 3 horses left

Now race all the horses who placed first in both their races which will take 9 races for a total of 90 races now

If the horse took first place, all 3 horses in his group survived
If he took second, 2 horses survive
If he took third, only he survives
4th to 9th places are eliminated again

So in a race we were actually racing 27 horses (each 1st placed horse belonged in a group) and 6 of them survived from the 27 horses, which leaves us 54 horses now

These horses are grouped as 9 groups of 6

x1 x2 x3 y1 y2 z1 where x is the first place finisher's group, y the second place finisher's and z the third

Now since we have 9 groups of 6 left, we can race all the first placers again, this is our 91st race

The 1st placer has all his group survive
2nd placer has x1 x2 and y1 survive
3rd placer only himself

Which leaves us a total of 10 horse
Sadly I don't think 1 race is enough here, so we can hold 2 more races and finally crown the winners in 93

But does that 1st placer need to ever be raced again? Isn't it the 5 others in his group that we should concern ourselves with?

>But does that 1st placer need to ever be raced again?
To determine whether he is top 3, of course we do. You have a better strategy?

I'm saying he is the winner of every race, and is the fastest horse. If I'm understanding your method right, that is. If that is the case then we only need to race the other 9 for 2nd and 3rd.

OOH FUCK ME
you are right, so it IS 92

Yup yup! And that can be shown to be the best.

nice