bonanova 77 Report post Posted January 26, 2008 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? Share this post Link to post Share on other sites

0 Guest Report post Posted January 26, 2008 Seven heats Share this post Link to post Share on other sites

0 Guest Report post Posted January 26, 2008 when I read brhan's spoiler I didn't believe it - after all - you need 5 heats just to make sure each horse races once. After much reflection, I now see how it's done - good job brhan! Share this post Link to post Share on other sites

0 unreality 1 Report post Posted January 26, 2008 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? * 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. Share this post Link to post Share on other sites

0 Guest Report post Posted January 26, 2008 (edited) * 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. 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. Edited January 26, 2008 by brhan Share this post Link to post Share on other sites

0 Guest Report post Posted January 26, 2008 (edited) top horse races twice, 2nd & 3rd race 2 or 3 times max...Heat 1 - horses 1-5 Heat 2 - horses 6-10, Heat 3 - horses 11-15, Heat 4 - horses 16-20, Heat 5 - horses 21-25 Heat 6 - winners of each heat 1-5 Heat 7 - place 2 & 3 from heat 6, plus places 2 & 3 from the first heat that winner of heat 6 raced, plus place 2 from the first heat that 2nd place in heat 6 originally raced fasted is winner of heat 6, 2nd & 3rd fastest are the 1st and 2nd place in heat 7 for example: each row is a heat, for the example first 3 win each heat, but logic will work as long as you follow the rule for how heat 6 & 7 are selected 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 1 6 11 16 21 6 11 2 3 7 example's result is 1 6 11 Edited January 26, 2008 by PDR Share this post Link to post Share on other sites

0 roolstar 0 Report post Posted January 26, 2008 Here's how I see it: The 5 first heats will be called A, B, C, D & E The outcomes of these heats are: A1, A2, A3, A4 & A5 (By rank) B1, B2, B3... Well you get the idea... The 6th heat is between A1, B1, C1, D1 & E1 Let's take as an example the outcome to be: First Place: A1 ==> Garanteed to be The Fastest! Second Place: B1 ==> B3 is automatically eliminated from being in the top 3 (best he can do is 4th place behind A1, B1 & B2) Third Place: C1 ==> C2 & C3 are automatically eliminated from being in the top 3 Fourth Place: D1 ==> D2 & D3 are automatically eliminated from being in the top 3 Fifth Place E1 ==> E2 & E3 are automatically eliminated from being in the top 3 The 7th and last heat will be among A2, A3, B1, B2 & C1 ==> Fastest 2 will join A1 in the Top 3! Fastest horse: 2 runs Second best: 3 runs Third best: 3 runs Of course we need to overlook the fact that the horses get tired after each heat, or we have to give them time enough to rest after each heat... Share this post Link to post Share on other sites

0 Guest Report post Posted February 22, 2008 Here's how I see it: The 5 first heats will be called A, B, C, D & E The outcomes of these heats are: A1, A2, A3, A4 & A5 (By rank) B1, B2, B3... Well you get the idea... The 6th heat is between A1, B1, C1, D1 & E1 Let's take as an example the outcome to be: First Place: A1 ==> Garanteed to be The Fastest! Second Place: B1 ==> B3 is automatically eliminated from being in the top 3 (best he can do is 4th place behind A1, B1 & B2) Third Place: C1 ==> C2 & C3 are automatically eliminated from being in the top 3 Fourth Place: D1 ==> D2 & D3 are automatically eliminated from being in the top 3 Fifth Place E1 ==> E2 & E3 are automatically eliminated from being in the top 3 The 7th and last heat will be among A2, A3, B1, B2 & C1 ==> Fastest 2 will join A1 in the Top 3! Fastest horse: 2 runs Second best: 3 runs Third best: 3 runs Of course we need to overlook the fact that the horses get tired after each heat, or we have to give them time enough to rest after each heat... why is the answer not SIX heats? Race 5 heats to get the top 5 horses. Then race the top five horses in the sixth heat. Your fastest three are the horses that finishes first, second and third in the sixth heat. What am i not seeing? Share this post Link to post Share on other sites

0 Guest Report post Posted February 22, 2008 why is the answer not SIX heats? Race 5 heats to get the top 5 horses. Then race the top five horses in the sixth heat. Your fastest three are the horses that finishes first, second and third in the sixth heat. What am i not seeing? It is possible that the 3 fastest horses could be in one group. If you only do SIX races, you will leave out the 2nd and 3rd fastest horses because you are only considering (in the sixth heat) the horses that stood first from each group. Share this post Link to post Share on other sites

0 Guest Report post Posted February 22, 2008 For more discussion, also check this topic Horse Race, started in June 2007. Share this post Link to post Share on other sites

0 unreality 1 Report post Posted February 28, 2008 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 Share this post Link to post Share on other sites

0 Guest Report post Posted March 17, 2008 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: "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!! Share this post Link to post Share on other sites

0 Guest Report post Posted March 17, 2008 (edited) 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 March 17, 2008 by brhan Share this post Link to post Share on other sites

0 Guest Report post Posted March 17, 2008 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. Share this post Link to post Share on other sites

0 Guest Report post Posted November 1, 2008 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!! Share this post Link to post Share on other sites

0 Guest Report post Posted November 13, 2008 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 Share this post Link to post Share on other sites

0 bonanova 77 Report post Posted November 13, 2008 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? Share this post Link to post Share on other sites

0 NAPALM71 0 Report post Posted August 31, 2012 Seven heats ummm this is not a solution it is an answer in which you gave no explanation meaning it is NOT a solution Share this post Link to post Share on other sites

0 mmiguel 1 Report post Posted September 2, 2012 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? 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 Share this post Link to post Share on other sites

0 mmiguel 1 Report post Posted September 2, 2012 * 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). Share this post Link to post Share on other sites

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?

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