Math Puzzle

[rj144]

Here's the question I found on the internet although I won't say where so you can figure it out:

"There are 25 mechanical horses and a single racetrack. Each horse completes the track in a pre-programmed time, and the horses all have different finishing times, unknown to you. You can race 5 horses at a time. After a race is over, you get a printout with the order the horses finished, but not the finishing times of the horses. What is the minimum number of races you need to identify the fastest 3 horses?"

I'll post the answer later. Try to figure it out without cheating.

 

[NorwegianWood]

 

[ToxicShocker]

There should only be one race

 

[rj144]

Those equations are not useful here. Do you have a more applicable gif?

 

[rj144]

There should only be one race

A little off.

 

[AGGAMEMNON66]

@Lubaolong

 

[BEER]

7

 

[Fox by the Sea]

i am tagging @Karnage because i dont want to be bored of this problem all by myself.

 

[Lubaolong]

No, I won't help you with your middle school algebra homework, @rj144

 

[milliniar]

Bout tree fiddy

 

[Fox by the Sea]

1. gaussian conjugate heat equation
2. ricci flow manifold for 4plus dimensions
3. divide by 7
4. HORSES~

 

[Fedorgasm]

Not reading other responses cuz i want to try it myself

I'm thinking you have to start with 5 races and eliminate the slowest 2 horses in each. That's the only way to know that you don't accidentally eliminate one of the fastest 3.

So after 5 races you have 15 horses left.

Then you run 3 more races, and eliminate 6, so you have 9 horses left after a total of 8 races.

Then you run 2 races. The first race eliminates 2 horses, but the 2nd race can only eliminate 1. So after 10 total races, you have 6 horses left.

Run 2 more races. Eliminate 2 in the first and 1 in the 2nd and you're left with the fastest 3 horses after 12 races.


Now that I took the time to write that all out, I'm thinking there is a faster way, if I recorded which horses faced each other along the way I could eliminate some races. Maybe I'll take the time to figure that out if nobody else has yet.

 

[Lubaolong]

It's better to post puzzles that aren't easily found on Google. Then we will see how long it takes to solve. Wheat from chaff.

 

[MillingMeSoftly]

Fedorgasm has it right

 

[Literally Bricked]

I worked out seven, not sure if there is a path that allows you to confirm in six.

 

[odog]

Ok, I came to my own conclusion, then looked up the real answer. So here is my question.

Let’s say in the first race you do you just happen to have the 5 fastest through bad luck. Then aren’t you eliminating some of the fastest horses right away?

 

[Lubaolong]

Now that I took the time to write that all out, I'm thinking there is a faster way, if I recorded which horses faced each other along the way I could eliminate some races. Maybe I'll take the time to figure that out if nobody else has yet.

Break them into 5 groups and race them. Top finishers from each group race.

That gives you 1st place horse overall.

To find 2nd and 3rd you have to do more races, as the group that the 1st place horse was in, for example, could have all the top 5 potentially.

The horses that got 4th and 5th in the 2nd race were the fastest of their group. Their group can't possibly have a top 3, so all of them can be eliminated.

You can also likewise eliminate all horses from the 3rd place finisher group, because they are all slower than at best the 3rd place overall horse.

This leaves 1st place horse group and 2nd place horse group of the 2nd race to sort through.

You know 1st place horse is the fastest, so it doesn't need to race again.

So that leaves 9 horses left. You can eliminate 4th and 5th from first place horse group, because you know they are slower than 1st, 2nd, 3rd in that group. Likewise, you can eliminate 3rd, 4th, and 5th from the 2nd place horse group because they are slower that 2nd place horse group winner, 2nd place group 2nd place, and 1st place group winner.

So that leaves 5 horses left.

#2 and #3 from the 1st place horse group, 2nd place horse group winner and runner up, and 3rd place horse group winner.

Race these last 5 to get the 2nd and 3rd fastest overall.

So 5 initial races, +1 race of those group winners, +1 final race.

 

[Fedorgasm]

Not reading other responses cuz i want to try it myself

I'm thinking you have to start with 5 races and eliminate the slowest 2 horses in each. That's the only way to know that you don't accidentally eliminate one of the fastest 3.

So after 5 races you have 15 horses left.

Then you run 3 more races, and eliminate 6, so you have 9 horses left after a total of 8 races.

Then you run 2 races. The first race eliminates 2 horses, but the 2nd race can only eliminate 1. So after 10 total races, you have 6 horses left.

Run 2 more races. Eliminate 2 in the first and 1 in the 2nd and you're left with the fastest 3 horses after 12 races.


Now that I took the time to write that all out, I'm thinking there is a faster way, if I recorded which horses faced each other along the way I could eliminate some races. Maybe I'll take the time to figure that out if nobody else has yet.

Ok my new answer is 9. I used the same method as I used previously, but when you finish the first 8 races you have enough info to determine 1st place, and the only 2 candidates for 2nd, and the 3 candidates for 3rd, so you only need one more race at that point.


Edit:
Ah crap, Luba is right. I started eliminating the slowest a round too late. The method was right but I did it wrong. The answer is 7 races.

 

[rj144]

Ok, I came to my own conclusion, then looked up the real answer. So here is my question.

Let’s say in the first race you do you just happen to have the 5 fastest through bad luck. Then aren’t you eliminating some of the fastest horses right away?

No, you're not.

 

[rj144]

Ok my new answer is 9. I used the same method as I used previously, but when you finish the first 8 races you have enough info to determine 1st place, and the only 2 candidates for 2nd, and the 3 candidates for 3rd, so you only need one more race at that point.


Edit:
Ah crap, Luba is right. I started eliminating the slowest a round too late. The method was right but I did it wrong. The answer is 7 races.

Good try, but a little wrong.