Thread: lets see how smart miscers are
05-15-2018, 08:01 PM
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#61
05-15-2018, 08:23 PM
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#62
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1412 as far as i can tell not sure how you'd do it with less
im probably retarded
im probably retarded
05-15-2018, 08:33 PM
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#63
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11 races.
05-15-2018, 08:57 PM
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#64
- StevieMe
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Can confirm it's 7, worked it out on Excel.
1) Run 5 heats of 5 horses, the 5 winners go into their champions race.
2) Eliminate the bottom two of that champions race which means eliminating every horse beaten by those two (the 4 horses that each of them beat in the first race, 10 horses in total are eliminated at this point)
3) Eliminate the bottom two of the fastest horse's group, they can't win. (12 horses eliminated so far)
4) Eliminate the bottom four of the 3rd fastest horse's group (16 horses eliminated so far)
5) Eliminate the bottom three of 2nd fastest horse's group (19 horses eliminated so far)
6) Of the 6 horses left, let the fastest watch the other 5 battle for 2nd and 3rd place.
1) Run 5 heats of 5 horses, the 5 winners go into their champions race.
2) Eliminate the bottom two of that champions race which means eliminating every horse beaten by those two (the 4 horses that each of them beat in the first race, 10 horses in total are eliminated at this point)
3) Eliminate the bottom two of the fastest horse's group, they can't win. (12 horses eliminated so far)
4) Eliminate the bottom four of the 3rd fastest horse's group (16 horses eliminated so far)
5) Eliminate the bottom three of 2nd fastest horse's group (19 horses eliminated so far)
6) Of the 6 horses left, let the fastest watch the other 5 battle for 2nd and 3rd place.
05-15-2018, 09:00 PM
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#65
- PeaceWithin
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Originally Posted By PeaceWithin⏩
Seven races. Can't be bothered to type the whole thing out but...
You run five horses for five races (5 races)
You take the top five horses and run them for a final race (1 race)
There are still five more horses you need to race... 2nd/3rd place from the "finals" race... and the 2nd/3rd place from whichever heat the overall(absolute) winner won during the first set of five races, and most importantly... the 2nd place winner in whichever heat the 2nd place horse from the "finals" race ran in.
7 races.
You run five horses for five races (5 races)
You take the top five horses and run them for a final race (1 race)
There are still five more horses you need to race... 2nd/3rd place from the "finals" race... and the 2nd/3rd place from whichever heat the overall(absolute) winner won during the first set of five races, and most importantly... the 2nd place winner in whichever heat the 2nd place horse from the "finals" race ran in.
7 races.
Originally Posted By StevieMe⏩
You laid that out so much better than I did. Basically what I was trying to say, haha.Can confirm it's 7, worked it out on Excel.
1) Run 5 heats of 5 horses, the 5 winners go into their champions race.
2) Eliminate the bottom two of that champions race which means eliminating every horse beaten by those two (the 4 horses that each of them beat in the first race, 10 horses in total are eliminated at this point)
3) Eliminate the bottom two of the fastest horse's group, they can't win. (12 horses eliminated so far)
4) Eliminate the bottom four of the 3rd fastest horse's group (16 horses eliminated so far)
5) Eliminate the bottom three of 2nd fastest horse's group (19 horses eliminated so far)
6) Of the 6 horses left, let the fastest watch the other 5 battle for 2nd and 3rd place.
1) Run 5 heats of 5 horses, the 5 winners go into their champions race.
2) Eliminate the bottom two of that champions race which means eliminating every horse beaten by those two (the 4 horses that each of them beat in the first race, 10 horses in total are eliminated at this point)
3) Eliminate the bottom two of the fastest horse's group, they can't win. (12 horses eliminated so far)
4) Eliminate the bottom four of the 3rd fastest horse's group (16 horses eliminated so far)
5) Eliminate the bottom three of 2nd fastest horse's group (19 horses eliminated so far)
6) Of the 6 horses left, let the fastest watch the other 5 battle for 2nd and 3rd place.
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05-15-2018, 09:48 PM
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#66
05-15-2018, 10:11 PM
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#67
- litle_dan
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has anyone factored in that after the first heats (meaning after every horse runs once) the second tier of races may not be reliable information due to exhaustion.
or am i just thinking too deeply here
op more info
or am i just thinking too deeply here
op more info
05-15-2018, 10:40 PM
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#68
Edit- I’m wrong it’s 7. I made that a lot more complicated than it needed to be.
Solutionhttp://puzzles.nigelcoldwell.co.uk/fiftynine.htm
Off the top of my head 10 or 11
25 Horses
5 races of 5 and take the top 3 winners of each group since the 3rd place of one group could possibly be faster than any other group.
(5 races)
15 Horses
3 races of 5. I’m guessing here you have to take the top 3 as well which leaves you with 9 horses.
(8 races)
9 Horses
1 Race of 5 (Group A), take #1 and race with remaining 4 (Group B)
If #1 is 3rd or below we have our answer. If #1 is 1st or 2nd we’ll have to race group B’s top 3 with Group A’s #2 and #3
(10 or 11 races)
Another method I thought of is 1 race of 5
(1 race)
Take these top 3 and race them against every horse adding two horses each race and eliminating the slowest two.
With 20 Horses left after the initial race it will take 10 more races
(11 races)
Assuming each horse is completely rested.
Solutionhttp://puzzles.nigelcoldwell.co.uk/fiftynine.htm
Off the top of my head 10 or 11
25 Horses
5 races of 5 and take the top 3 winners of each group since the 3rd place of one group could possibly be faster than any other group.
(5 races)
15 Horses
3 races of 5. I’m guessing here you have to take the top 3 as well which leaves you with 9 horses.
(8 races)
9 Horses
1 Race of 5 (Group A), take #1 and race with remaining 4 (Group B)
If #1 is 3rd or below we have our answer. If #1 is 1st or 2nd we’ll have to race group B’s top 3 with Group A’s #2 and #3
(10 or 11 races)
Another method I thought of is 1 race of 5
(1 race)
Take these top 3 and race them against every horse adding two horses each race and eliminating the slowest two.
With 20 Horses left after the initial race it will take 10 more races
(11 races)
Assuming each horse is completely rested.
05-15-2018, 10:45 PM
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#69
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It's 11.
25 horses, 5 runs you eliminate the bottom 2 from each.
Leaves you with 15 horses, and 5 trials so far
Take top 3 of more groups of 5, eliminating 6 horses. Total horses = 9, total runs = 8
Take a group of 5 and run them, eliminate 2 more horses. Total horses = 7, runs = 9
Take a group of 5 horses, eliminate 2 more. Total horses = 5, total runs = 10
Do one more run, eliminate bottom 2. Total runs = 11.
25 horses, 5 runs you eliminate the bottom 2 from each.
Leaves you with 15 horses, and 5 trials so far
Take top 3 of more groups of 5, eliminating 6 horses. Total horses = 9, total runs = 8
Take a group of 5 and run them, eliminate 2 more horses. Total horses = 7, runs = 9
Take a group of 5 horses, eliminate 2 more. Total horses = 5, total runs = 10
Do one more run, eliminate bottom 2. Total runs = 11.
05-15-2018, 10:59 PM
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#70
- 1hardgainer
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My guess is 8 races: 5 races w/ 5 horses in each race to determine the top 5, then 3 more races to determine the fastest of the 5:
5 races to determine top 5
1 race to determine fastest of top 5
1 race to determine fastest of remaining 4
1 race to determine fastest of remaining 3
= 8 total races
EDIT: After reading some of the other responses my answer is totally wrong. Never even thought about the possibility of having the top 3 fastest horses unknowingly put in the same group of 5 during the first round of races. Now I'm really curious what the answer is.
5 races to determine top 5
1 race to determine fastest of top 5
1 race to determine fastest of remaining 4
1 race to determine fastest of remaining 3
= 8 total races
EDIT: After reading some of the other responses my answer is totally wrong. Never even thought about the possibility of having the top 3 fastest horses unknowingly put in the same group of 5 during the first round of races. Now I'm really curious what the answer is.
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05-15-2018, 11:13 PM
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#71
05-15-2018, 11:45 PM
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#72
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You're out of your element.
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It is a disgrace for a man to grow old without seeing the beauty and strength of which his body is capable.
05-16-2018, 12:16 AM
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#73
- aal04
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1st race... 2 eliminated
get 3rd fastest horse. race in second heat.
best case scenario he wins (minimum is the word used by op). 4 horses eliminated
next race. 3rd horse races. he wins again (minimum). 4 horses eliminated.
etc
It will take minimum 2,4,4,4,4,4 - 6 races minimum to find the 3 fastest horses (minimum is not a brute force)
get 3rd fastest horse. race in second heat.
best case scenario he wins (minimum is the word used by op). 4 horses eliminated
next race. 3rd horse races. he wins again (minimum). 4 horses eliminated.
etc
It will take minimum 2,4,4,4,4,4 - 6 races minimum to find the 3 fastest horses (minimum is not a brute force)
05-16-2018, 12:25 AM
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#74
- numberguy12
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People are giving scenerios where the fastest 3 can be determined, but it must be remembered to find the minimum # of races required, you must also show that no lesser amount of races (than the number you figured out) will suffice.
For example if you conclude 10 races would work....you must also show 9 or less cannot work.
The answer is 7 races. It was argued correctly in a post above. You must also show that 6 or less doesn't work. This was also touched upon (must first do 5 races to cover all 25 horses.....then must do another race of the top 5 because you can't select overall top 3 otherwise.....even after the 6 races, the hypothetical scenario of overall top 3 being in the very first heat of 5 means you can't conclude you have the top 3 from the 6th race....thus 6 races or less are insufficient).
For example if you conclude 10 races would work....you must also show 9 or less cannot work.
The answer is 7 races. It was argued correctly in a post above. You must also show that 6 or less doesn't work. This was also touched upon (must first do 5 races to cover all 25 horses.....then must do another race of the top 5 because you can't select overall top 3 otherwise.....even after the 6 races, the hypothetical scenario of overall top 3 being in the very first heat of 5 means you can't conclude you have the top 3 from the 6th race....thus 6 races or less are insufficient).
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05-16-2018, 12:39 AM
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#75
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Originally Posted By aal04⏩
Interesting take. The issue here is that you don't know the 3rd fastest horse in the very first heat will win all the remaining heats. That particular horse very well might not win. So in most interpretations of this problem (i.e what is the minimum required races for any random group of 25 horses), this would not work.1st race... 2 eliminated
get 3rd fastest horse. race in second heat.
best case scenario he wins (minimum is the word used by op). 4 horses eliminated
next race. 3rd horse races. he wins again (minimum). 4 horses eliminated.
etc
It will take minimum 2,4,4,4,4,4 - 6 races minimum to find the 3 fastest horses (minimum is not a brute force)
get 3rd fastest horse. race in second heat.
best case scenario he wins (minimum is the word used by op). 4 horses eliminated
next race. 3rd horse races. he wins again (minimum). 4 horses eliminated.
etc
It will take minimum 2,4,4,4,4,4 - 6 races minimum to find the 3 fastest horses (minimum is not a brute force)
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05-16-2018, 12:58 AM
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#76
05-16-2018, 01:11 AM
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#77
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Misc Hunting and Fishing crew
Age:28 (idk the Age thing is glitched on here)
05-16-2018, 01:43 AM
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#78
05-16-2018, 01:52 AM
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#79
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Originally Posted By wincel⏩
5 races.What if there are 9164 horses, and there are 2291 horses in each race, and I want to determine the top 58 horses?
( 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?
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05-16-2018, 02:02 AM
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#80
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Originally Posted By aal04⏩
Being stubborn does not make you right. The fault in your interpretation of the case was already explained in the thread:I still stand by my answer of 6.
It is still *possible to find the 3 fastest horses with 6.
The question was the bare "minimum", not the most efficient way.
It is still *possible to find the 3 fastest horses with 6.
The question was the bare "minimum", not the most efficient way.
Originally Posted By numberguy12⏩
You are being asked for the minimum number of reasons to determine the fastest 3 horses out of 25. You cannot inject the assumption that the third place horse from the first heat will win as a 'best case'. That is injecting data into the example to produce a desired outcome.Interesting take. The issue here is thatyou don't know the 3rd fastest horse in the very first heat will win all the remaining heats.That particular horse very well might not win. So in most interpretations of this problem (i.e what is the minimum required races for any random group of 25 horses), this would not work.
7 is the minimum number, and it is a classic 'process of elimination' technique. As numberguy12 previously stated, the minimum number of races required to find the fastest horse simpliciter is 6, and it's important to demonstrate that as part of finding the fastest 3 horses.
05-16-2018, 02:07 AM
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#81
- janglingjack
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I have no fukin idea how to even begin to answer this
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05-16-2018, 02:10 AM
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#82
- numberguy12
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Originally Posted By aal04⏩
Again, 6 only works with a very specific outcome of the races. The whole point of the problem is to find a method that works in general....i.e. whatever the outcome of the races is. This is so you are not dependent on a particular event occurring, and can still be assured that 7 races will suffice. With many particular outcomes, 6 will not suffice.I still stand by my answer of 6.
It is still *possible to find the 3 fastest horses with 6.
The question was the bare "minimum", not the most efficient way.
It is still *possible to find the 3 fastest horses with 6.
The question was the bare "minimum", not the most efficient way.
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05-16-2018, 02:14 AM
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#83
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9 races
5 races: determine fastest 3 of each race
6rd race: take slowest of each race and race them
7rd race: take second-fastest of each race and race them
8rd race: take fastest and race them
9rd race: race the slowest, second-fastest and fastest of each race
edit: and this is how we know its 7 races, because 9 is out of the question.
5 races: determine fastest 3 of each race
6rd race: take slowest of each race and race them
7rd race: take second-fastest of each race and race them
8rd race: take fastest and race them
9rd race: race the slowest, second-fastest and fastest of each race
edit: and this is how we know its 7 races, because 9 is out of the question.
05-16-2018, 02:42 AM
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#84
Originally Posted By PeaceWithin⏩
While I agree with you overall, this exercise assumes that the horses are machines and run identical times in each of their races. Therefore if one horse ran multiple races, you could use it as a metric for comparison of horses that never raced each other.Wut? The whole point of the OP is a puzzle based on ranking winners, excluding losers, and seeing how efficiently you can determine the top three without timing them.
I guess I don't understand what you mean by consistency... Like I said earlier, the OP presupposes that you can determine who won each race. Even if you have photo finishes, comparing photo finishes from multiple different races against the semi-final race is an exercise in futility if you don't have a stopwatch to determine how fast each of those different races was...
I guess I don't understand what you mean by consistency... Like I said earlier, the OP presupposes that you can determine who won each race. Even if you have photo finishes, comparing photo finishes from multiple different races against the semi-final race is an exercise in futility if you don't have a stopwatch to determine how fast each of those different races was...
ie if horse one blows everyone out of the water by 3 lengths in race one, then wins by a half length to 2 different horses in race 2, you could eliminate the bottom 4 of race one instantly.
You don’t need a stopwatch.
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05-16-2018, 03:00 AM
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#85
- likeawashboard
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This seems like a good IQ question.
05-16-2018, 03:01 AM
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#86
- Furybox
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Race a random 5
Take the third, race with another random 4. Drop any which place lower than max(where the original 3rd place ends up, 3rd place in the current race) [i.e. If it places first, you can drop all four, if it places forth, you cancel it and the one below it]
Repeat
^ feel like that's a more efficient algorithm for it, but it doesn't get me to a final number!
Take the third, race with another random 4. Drop any which place lower than max(where the original 3rd place ends up, 3rd place in the current race) [i.e. If it places first, you can drop all four, if it places forth, you cancel it and the one below it]
Repeat
^ feel like that's a more efficient algorithm for it, but it doesn't get me to a final number!
05-16-2018, 03:29 AM
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#87
- numberguy12
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Originally Posted By Globally⏩
see what you are saying here. Taking an example.....if we observe in each of the opening 5 heats, that the winner utterly blows out 2-5 (make it extreme: say the 2-5 are still close to the starting line while the winner crosses the finish line). Then you take the 5 super fast winners and put them in the 6th race, where all 5 finish super close together......any rational person would say a 7th race is not needed. You just take the top 3 from the 6th.While I agree with you overall, this exercise assumes that the horses are machines and run identical times in each of their races. Therefore if one horse ran multiple races, you could use it as a metric for comparison of horses that never raced each other.
ie if horse one blows everyone out of the water by 3 lengths in race one, then wins by a half length to 2 different horses in race 2, you could eliminate the bottom 4 of race one instantly.
You don’t need a stopwatch.
ie if horse one blows everyone out of the water by 3 lengths in race one, then wins by a half length to 2 different horses in race 2, you could eliminate the bottom 4 of race one instantly.
You don’t need a stopwatch.
The problem here is the same kind of issue mentioned a few posts up with aal04: this is dependent on a certain outcome happening (the 6 races only works because the race outcomes happened a certain way- super different finishing times, followed by close). What if the outcome just happened not to be that? The problem is asking for a general solution, so that no matter what happens, 7 races will be the minimum required. I'm probably wording this terribly though. Someone else may be able to explain better
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05-16-2018, 03:33 AM
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#88
05-16-2018, 03:44 AM
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#89
- BetaThanU
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Originally Posted By Mortar34⏩
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.Trick question.
If you don’t use a stopwatch you aren’t finding the “fastest,” you’re just discovering who won the most races. “Fastest” denotes time. Without time, it’s only a unilateral accomplishment.
If the answer was 6, 11, 12, etc..the stupid beasts could be crawling by the last race and you’d still crown a winner, but it wouldn't necessarily be the “fastest.”
If you don’t use a stopwatch you aren’t finding the “fastest,” you’re just discovering who won the most races. “Fastest” denotes time. Without time, it’s only a unilateral accomplishment.
If the answer was 6, 11, 12, etc..the stupid beasts could be crawling by the last race and you’d still crown a winner, but it wouldn't necessarily be the “fastest.”
Problem solved
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05-16-2018, 03:44 AM
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#90
Originally Posted By CopeyDan⏩
If 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.Not sure, but anyone who answers 6 is a simpleton.
The right answe is 7 using the process of elimination though.
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