19 replies to this topic

[unreality]

Spoiler for my solution

A sketch to my solution of seven heats:

1. Divide the horses into five groups ==> 5 heats

2. Do another heat with the horses that stood first of each group ==> 1 heat
Obviously, the winner in this race is the fastest.

3. Do one more heat of the horses which stood 2nd and 3rd of the group of the fastest horse, 2nd and 3rd of the heat in step 2, and the horse that stood second from the group 2nd of the heat in step 2. ==> 1 heat.

From step 2, we know the fastest horse, and from step 3 the second and third fastest horses.

Just to comment on unreality's solution, some of the heats are redundant and hence cut them off.

No, yours doesn't work, my friend The fastest of each heat might not necessarily be the 5 fastest because the heats are random. Two of the five fastest could have run in the same initial heat. My solution is the best one so far, from what I can tell. Bonanova never did give his answer

[frog101858]

Okay, call me slow but I am just not getting this.. The numbers in red you can rule out due to at least 3 other horses being faster..

1: A1 A2 A3 A4 A5
2: B1 B2 B3 B4 B5
3: C1 C2 C3 C4 C5
4: D1 D2 D3 D4 D5
5: E1 E2 E3 E4 E5

6: A1 B1 C1 D1 E1 Winner of this heat is the fastest #1 (C1 and D1 eliminated due to 3 faster scores, based on arbitrary times)

So this is where I'm not following... Who's to say that E2 isn't faster than B1

Just to try to prove my point on this, I arbitrarily assigned times to these 15 horses

A1 (2) A2 (6) A3 (7)
B1 (6) B2 (7) B3 (9)
C1 (8) C2 (9) C3 (10)
D1 (10) D2 (11) D3 (12)
E1 (3) E2 (4) E3 (5)

And if C1 and D1 were the fastest of the horses in heat 3 and 4 then that means all other horses in their heats were slower than A1 B1 or E1
If B1 was 3rd fastest in this heat then B2 and B3 must be slower

So now we have 6 horses left
A2 A3
B1

E1 E2 E3


This is where I get lost... I don't understand this part: "[color="#FFFF00"]So, send home the horse who got third place against the second place horses from this race"

So if we did that we would take out E3

.......................................................

Alright it finally hit.. and even though I was tempted to delete this now since I figured it out, I thought I'd go ahead and post it anyways for other people who are as slow as I am... :0)


I was thinking that according to my numbers if we took out E3 that it was a faster time than the others remaining, and I couldn't figure out what justified taking it out. But if it came in 3rd place in heat 5 and E3 came in 2nd place in heat 6, then that means that E1, E2 and A1 are all faster, therefore justifying ruling this one out... Eureka!!!

A2 A3
B1

E1 E2


And now there are only 5 horses to race in HEAT 7

oh my.. only took me 2 hours.. Don't I feel stupid!!

[brhan]

No, yours doesn't work, my friend The fastest of each heat might not necessarily be the 5 fastest because the heats are random. Two of the five fastest could have run in the same initial heat. My solution is the best one so far, from what I can tell. Bonanova never did give his answer

... and your solution was ...

* Split them into five heats. Take the 3 fastest in each heat, meaning you now have the 15 fastest.
* So do 3 heats now. Take the 3 fastest in each heat. You have the 9 fastest horses.
* Do a single heat with 5 random horses of the 9, cutting out the 2 last-place horses, leaving you with 7.
* Take another random 5 horses and do it again, cuttnig out 2, leaving you with 5 horses.
* Do one final heat and take the top 3 from that race.

Those top 3 horses will only have to race 5 times each.

It works ... but a lot races ....
The top 3 horses need to run 3 times, and in total seven races. If you don't understand my approach, there are other posts with the same solution but different approach. You may need to optimize your procedure ...

Edited by brhan, 17 March 2008 - 11:50 AM.

[Duh Puck]

Nice problem. I was stuck on eight heats too for a bit, just like frog101858, for basically the same reason. Once I popped into Excel and started charting it out, it clicked. Seven heats indeed.

[NintendoMad]

Damn I thought it was 6, 9 or 11!!! AHHHHH!! I get it from the spoiler!!! 7!! I know why know!! I understand from PDR and roolstar!!

[tygar]

question?? y cant u just do 6 heats?? i havent read all the posts but it seams logical to me that u divide the 25 into 5 heats the 1st place winners of the 5 heats makes 5 horses that race again the fastest is duh the fastest the second is duh the second fastest and the third is duh the third fastest just 6 heats needed seems logical to me

[bonanova]

question?? y cant u just do 6 heats?? i havent read all the posts but it seams logical to me that u divide the 25 into 5 heats the 1st place winners of the 5 heats makes 5 horses that race again the fastest is duh the fastest the second is duh the second fastest and the third is duh the third fastest just 6 heats needed seems logical to me

Hi tygar, and welcome to the Den.

What if the 5 fastest horses were in the first heat?

[NAPALM71]

Spoiler for my solution

Seven heats

ummm this is not a solution it is an answer in which you gave no explanation meaning it is NOT a solution

[mmiguel]

The annual Horse racing tournament is a week away.
You own 25 horses, and you're allowed to enter up to three.
You decide to hold a series of races to find your three fastest horses.
You have access to a track that permits five horses to race against each other.

Here's the deal:

[1] No two of your horses are equally fast - there will be no ties.
[2] You have no stopwatch - you can't compare performances from different heats.
[3] You do not want to tire the horses needlessly - the fewer heats needed, the better.
[4] You want to know which horse is fastest, which is next fastest and which is third-fastest of the 25.

How many heats are needed?

Spoiler for try

7?
Randomly divide into 5 groups of 5 and race each group.

5 races so far
Let G(X,Y) be the Yth fastest horse from group X

Race G(x,1) for all x in 1..5 i.e. fastest horse from each group
i.e.
G(1,1),
G(2,1),
G(3,1),
G(4,1),
G(5,1)


Let H(Y) be the number of the original group (of the 5 original groups) of the Yth fastest horse from the 6th race.

Race the following 5 horses:
G(H(1),2)
G(H(1),3)
G(H(2),1)
G(H(2),2)
G(H(3),1)

The 2nd and 3rd fastest are the 1st and 2nd fastest in this 7th race respectively

[mmiguel]

* Split them into five heats. Take the 3 fastest in each heat, meaning you now have the 15 fastest.
* So do 3 heats now. Take the 3 fastest in each heat. You have the 9 fastest horses.
* Do a single heat with 5 random horses of the 9, cutting out the 2 last-place horses, leaving you with 7.
* Take another random 5 horses and do it again, cuttnig out 2, leaving you with 5 horses.
* Do one final heat and take the top 3 from that race.

Those top 3 horses will only have to race 5 times each.

In my solution, the top horse races twice, the 2nd and 3rd to best horses race 3 times each (the max for any horse).