Thread: lets see how smart miscers are
05-16-2018, 08:09 AM
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#91
Originally Posted By numberguy12⏩
Correct. I changed my post because it was too easy.5 races.
( 4 heats in the beginning to cover all 9164, then take top 58 of each, and lump them all in the 5th race to determine overall top 58)
Say I have 208,382,207,121 horses. There are 456489 horses per race. I want to find the top 955 horses. What is the minimum number of races required?
( 4 heats in the beginning to cover all 9164, then take top 58 of each, and lump them all in the 5th race to determine overall top 58)
Say I have 208,382,207,121 horses. There are 456489 horses per race. I want to find the top 955 horses. What is the minimum number of races required?
05-16-2018, 08:24 AM
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#92
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Originally Posted By Globally⏩
This is changing the problem and adding assumptions. Most importantly, adding such technologychanges the intended nature of this problem as a logic problem and reduces it to a trivial problemIf you have photo finish and a way to measure distance, you can do it in 6 by having one horse run all 6 races against the other 24. That one horse is your standard metric.
The right answe is 7 using the process of elimination though.
The right answe is 7 using the process of elimination though.
If you are familiar with the Monty Hall problem, this is like saying.......what if the contestant just has x-ray vision and can see through the doors? Figuring out which door has the car behind it is no problem! The entire point of the problem is to solve it as a problem of logic. The answer to this problem of the 25 horses is 7 races.
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05-16-2018, 08:29 AM
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#93
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Originally Posted By BetaThanU⏩
And then you can only compare the UoM of a given race, but not against other races.Define the winners lap time has 1 horse unit of time, therefore every other horses time is a percentage of the horse unit. It’s as arbitrary as a second or minute.
Problem solved
Problem solved
As others have said, I was nitpicking and realize that the intention of the puzzle was that some of these variables are removed/defined.
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05-16-2018, 08:30 AM
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#94
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6 races. First 5 races you take the winner of each. On the 6th race you choose the top 3 finishers. Why is this so hard for some?
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05-16-2018, 08:32 AM
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#95
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Originally Posted By ChristianMR⏩
Winnar.Zero races.
Kill all but three horses. Those are now the three fastest.
Kill all but three horses. Those are now the three fastest.
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05-16-2018, 08:35 AM
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#96
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Originally Posted By eXistenceLies⏩
the second finisher in one heat could be faster than the whole field in another heat ... try again.6 races. First 5 races you take the winner of each. On the 6th race you choose the top 3 finishers. Why is this so hard for some?
05-16-2018, 08:36 AM
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#97
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Originally Posted By wincel⏩
Ah. Post is still showing those numbers for me, not sure what is going on there.Correct. I changed my post because it was too easy.
This is still an unanswered question if anyone wants to try (perhaps a true "test of IQ"- saying this jokingly because IQ is a rather silly concept- since you can't just google the answer to be 7 as in OP's problem):
Say I have 208,382,207,121 horses. There are 456489 horses per race. I want to find the top 955 horses. What is the minimum number of races required?
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05-16-2018, 08:44 AM
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#98
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Originally Posted By ifisthole⏩
No....That is going to in depth. Make it easy. Though I like killing 23 horses and then the last 3 are the fastest.the second finisher in one heat could be faster than the whole field in another heat ... try again.
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05-16-2018, 09:31 AM
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#99
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I'm getting 7 , maybe I have a mistake but this is my reasoning :
First, race every group of 5 horses, then race the winners. ( 6 races so far), then for a 7th heat:
- the losers disqualify their respective group (horses 16 and 21)
- the third, can only be the global third so he also disqualifies his group (horse 11)
- the second place (horse 6) can only qualify himself and the third of his group who can compete for global third (horse 7)
- in the winner bracket, second and third place also qualify as they can all be the global fastest.
- No need to race the fastest of the fastest (horse 1)
So you only need a 7th race to find 2nd and 3rd
First, race every group of 5 horses, then race the winners. ( 6 races so far), then for a 7th heat:
- the losers disqualify their respective group (horses 16 and 21)
- the third, can only be the global third so he also disqualifies his group (horse 11)
- the second place (horse 6) can only qualify himself and the third of his group who can compete for global third (horse 7)
- in the winner bracket, second and third place also qualify as they can all be the global fastest.
- No need to race the fastest of the fastest (horse 1)
So you only need a 7th race to find 2nd and 3rd
05-16-2018, 09:34 AM
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#100
Originally Posted By numberguy12⏩
I got 456489+1+1 races required.Ah. Post is still showing those numbers for me, not sure what is going on there.
This is still an unanswered question if anyone wants to try (perhaps a true "test of IQ"- saying this jokingly because IQ is a rather silly concept- since you can't just google the answer to be 7 as in OP's problem):
Say I have 208,382,207,121 horses. There are 456489 horses per race. I want to find the top 955 horses. What is the minimum number of races required?
This is still an unanswered question if anyone wants to try (perhaps a true "test of IQ"- saying this jokingly because IQ is a rather silly concept- since you can't just google the answer to be 7 as in OP's problem):
Say I have 208,382,207,121 horses. There are 456489 horses per race. I want to find the top 955 horses. What is the minimum number of races required?
You run 456489 races, and then you race the winners. We have now eliminated 456489*(456489-955)+(456489-955)*955 horses. Of the remaining 955^2 horses, we can again eliminate 954*955/2 horses to obtain 456490 total horses remaining, 1 of which we know to be the fastest. Race the remaining 456489 and you will be able to determine the top 955 by taking the top 954 of those and throwing in the fastest horse from the 456490th race.
When the number of horses per race squared is the number of horses and the sum of positive integers up to the number of top horses we want to find -1 is the number of horses per race, we can create simple schemes to determine the minimum number of races to obtain that it is always the number of horses per race+2, but we haven't really proved this is the minimum. I am also wondering how you do this for arbitrary numbers of horses, races, and top finishers. This seems to be a complicated combinatorics problem, and I'm not even sure anyone has solved it in its generality.
05-16-2018, 10:04 AM
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#101
05-16-2018, 10:19 AM
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#102
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Originally Posted By wincel⏩
Awesome solution here. Yes 456489+1+1 is correct.I got 456489+1+1 races required.
You run 456489 races, and then you race the winners. We have now eliminated 456489*(456489-955)+(456489-955)*955 horses. Of the remaining 955^2 horses, we can again eliminate 954*955/2 horses to obtain 456490 total horses remaining, 1 of which we know to be the fastest. Race the remaining 456489 and you will be able to determine the top 955 by taking the top 954 of those and throwing in the fastest horse from the 456490th race.
When the number of horses per race squared is the number of horses and the sum of integers up to the number of top horses we want to find -1 is the number of horses per race, we can create simple schemes to determine the minimum number of races, but we haven't really proved this is the minimum. I am also wondering how you do this for arbitrary numbers of horses, races, and top finishers. This seems to be a complicated combinatorics problem, and I'm not even sure anyone has solved it in its generality.
You run 456489 races, and then you race the winners. We have now eliminated 456489*(456489-955)+(456489-955)*955 horses. Of the remaining 955^2 horses, we can again eliminate 954*955/2 horses to obtain 456490 total horses remaining, 1 of which we know to be the fastest. Race the remaining 456489 and you will be able to determine the top 955 by taking the top 954 of those and throwing in the fastest horse from the 456490th race.
When the number of horses per race squared is the number of horses and the sum of integers up to the number of top horses we want to find -1 is the number of horses per race, we can create simple schemes to determine the minimum number of races, but we haven't really proved this is the minimum. I am also wondering how you do this for arbitrary numbers of horses, races, and top finishers. This seems to be a complicated combinatorics problem, and I'm not even sure anyone has solved it in its generality.
Exactly as you noted, numbers were chosen to make this problem analogous to OP's problem and therefore not tedious .....we have x^2 horses, x horses per race, (a-1)(a+2)/2 = x, where a is the number of top horses we want to determine. We will simply get x+2 as the answer. You always just add 2 to the number of horses per race....so in with OPs problem, you just add 5 and 2 to get 7.
The general solution with arbitrary everything is interesting, and I suspect it might be difficult (not eloquent). I asked this very question a little bit ago in the Math thread, maybe someone there might chime in
Edit: on spread
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05-16-2018, 11:52 AM
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#103
Originally Posted By numberguy12⏩
das it maneAwesome solution here. Yes 456489+1+1 is correct.
Exactly as you noted, numbers were chosen to make this problem analogous to OP's problem and therefore not tedious .....we have x^2 horses, x horses per race, (a-1)(a+2)/2 = x, where a is the number of top horses we want to determine. We will simply get x+2 as the answer. You always just add 2 to the number of horses per race....so in with OPs problem, you just add 5 and 2 to get 7.
The general solution with arbitrary everything is interesting, and I suspect it might be difficult (not eloquent). I asked this very question a little bit ago in the Math thread, maybe someone there might chime in
Edit: on spread
Exactly as you noted, numbers were chosen to make this problem analogous to OP's problem and therefore not tedious .....we have x^2 horses, x horses per race, (a-1)(a+2)/2 = x, where a is the number of top horses we want to determine. We will simply get x+2 as the answer. You always just add 2 to the number of horses per race....so in with OPs problem, you just add 5 and 2 to get 7.
The general solution with arbitrary everything is interesting, and I suspect it might be difficult (not eloquent). I asked this very question a little bit ago in the Math thread, maybe someone there might chime in
Edit: on spread
05-16-2018, 06:25 PM
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#104
Originally Posted By numberguy12⏩
Like I said, I agree that 7 is the answer using the method the problem most likely intended.This is changing the problem and adding assumptions. Most importantly, adding such technologychanges the intended nature of this problem as a logic problem and reduces it to a trivial problem
If you are familiar with the Monty Hall problem, this is like saying.......what if the contestant just has x-ray vision and can see through the doors? Figuring out which door has the car behind it is no problem! The entire point of the problem is to solve it as a problem of logic. The answer to this problem of the 25 horses is 7 races.
If you are familiar with the Monty Hall problem, this is like saying.......what if the contestant just has x-ray vision and can see through the doors? Figuring out which door has the car behind it is no problem! The entire point of the problem is to solve it as a problem of logic. The answer to this problem of the 25 horses is 7 races.
However I disagree that the logic for 6 can be compared to your X ray vision cheap answer. It is a pretty logical method to use one horse as a metric for time/speed to evaluate the others. The photo finish really isn’t even necessary, you just need to observe and remember the races.
If the races are so close that you need photo finish, you would need it in both cases anyway to determine #1,2,3 of each race. It’s not about adding technology.
Both answers are perfectly reasonably and logical.
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05-16-2018, 06:29 PM
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#105
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Originally Posted By numberguy12⏩
then those horses are unlucky and have bad juju with them.What if the top 3 overall horses are all in the very first heat of 5? Then the true 2nd and 3rd would not be represented in the final race, and the process would not determine the top 3.
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i dont need unlucky horses. i'll take the 3 winners and go.
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05-19-2018, 10:50 AM
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#106
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So is there a clear answer?
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05-19-2018, 11:01 AM
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#107
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Originally Posted By MajorTendonitis⏩
Yes, 7So is there a clear answer?
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05-20-2018, 12:36 AM
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#108
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Originally Posted By numberguy12⏩
Thanks .Yes, 7
But I swore it would be 6 . First five races with five horses each would leave you with one in each race that was the fastest .
So now your left with the 5 horses .
Now when you race those five together , the three that get to the finish line first eliminate the other two, leaving you with the three fastest horses?
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05-20-2018, 12:50 AM
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#109
Originally Posted By MajorTendonitis⏩
You aren't considering all the possibilities. It is possible that the 3 fastest horses were all in the same race from one of the original 5 races.Thanks .
But I swore it would be 6 . First five races with five horses each would leave you with one in each race that was the fastest .
So now your left with the 5 horses .
Now when you race those five together , the three that get to the finish line first eliminate the other two, leaving you with the three fastest horses?
But I swore it would be 6 . First five races with five horses each would leave you with one in each race that was the fastest .
So now your left with the 5 horses .
Now when you race those five together , the three that get to the finish line first eliminate the other two, leaving you with the three fastest horses?
05-21-2018, 01:23 PM
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#110
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Originally Posted By wincel⏩
My bad , good pointYou aren't considering all the possibilities. It is possible that the 3 fastest horses were all in the same race from one of the original 5 races.
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05-21-2018, 01:35 PM
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#111
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It's a typical trick question used in job interviews, and any answer other than pulling out your phone and googling the answer is wrong.
Don't waste time calculating chit that's already been calculated by others.
Don't waste time calculating chit that's already been calculated by others.
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05-21-2018, 01:50 PM
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#112
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05-21-2018, 01:52 PM
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#113
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Three races. Use a regular watch and not a stop watch to time the horses. Or a video camera. I work on a ranch breaking horses so I k ow this is the answer.
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05-21-2018, 02:02 PM
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#114
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Did it quickly in my head and got 7
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05-21-2018, 03:40 PM
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#115
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Originally Posted By stfudonny⏩
stfu donnythe correct answer is:
Spoiler!
Spoiler!
7
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05-21-2018, 03:58 PM
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#116
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What is 9?
05-21-2018, 04:00 PM
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#117
Originally Posted By cheeseNmills⏩
LOL...except eventually one encounters problems which actually need to be solved by someone who knows what they are doing...It's a typical trick question used in job interviews, and any answer other than pulling out your phone and googling the answer is wrong.
Don't waste time calculating chit that's already been calculated by others.
Don't waste time calculating chit that's already been calculated by others.
Not everything is on the internet. If you've trained to be a mathematician, these types of problems are pretty easy to construct, and it's easy to ask people something they have never seen before. Don't count on always being able to look up the answer. The Putnam exam is a nice example of problems rooted in basic concepts that is balls hard for people to solve.
05-21-2018, 04:17 PM
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#118
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Spoiler!
I think it's 7...
So you race the initial 5 (let's say race a, b, c, d, e), and take their winners. Let's say a1 wins race a, a2 was second place, etc.. Call this the initial races.
For race #6, you race the winners of the initial 5 races (a1-e1), call this the winner's race. Then, the winner of that is obviously the fastest. The horse that came in 3-5 in the winner's race came from initial heat sets where the 2nd and 3rd place horses couldn't be 2nd or 3rd fastest, since the winner of their initial race came in beyond that. However, the 3rd place horse of the winner's race could still be the third fastest and the 2nd place horse of the winner's race (and 2nd place of their initial race) could be the 2nd/3rd fastest.
So you then race the 2nd and 3rd place winner of the initial heat with the winner of the winner's race. So the 7th and final race would be the 2nd and 3rd place of winner's race winner's initial race + 2nd place of winner's race + 2nd place of 2nd place of winner's initial race + 3rd place of the winner's race. The top 2 of that are the other 2 fastest.
I think it's 7...
So you race the initial 5 (let's say race a, b, c, d, e), and take their winners. Let's say a1 wins race a, a2 was second place, etc.. Call this the initial races.
For race #6, you race the winners of the initial 5 races (a1-e1), call this the winner's race. Then, the winner of that is obviously the fastest. The horse that came in 3-5 in the winner's race came from initial heat sets where the 2nd and 3rd place horses couldn't be 2nd or 3rd fastest, since the winner of their initial race came in beyond that. However, the 3rd place horse of the winner's race could still be the third fastest and the 2nd place horse of the winner's race (and 2nd place of their initial race) could be the 2nd/3rd fastest.
So you then race the 2nd and 3rd place winner of the initial heat with the winner of the winner's race. So the 7th and final race would be the 2nd and 3rd place of winner's race winner's initial race + 2nd place of winner's race + 2nd place of 2nd place of winner's initial race + 3rd place of the winner's race. The top 2 of that are the other 2 fastest.
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