7 million views! Thank you! Here's a Microsoft interview puzzle I think you will enjoy. A cat is hiding in one of five boxes that are lined up in a row. The boxes are numbered 1 to 5. Each night the cat hides in an adjacent box, exactly one number away. Each morning you can open a single box to try to find the cat. Can you win this game of hide and seek? What is your strategy to find the cat? What if there are n boxes? Watch the video for the solution! ruclips.net/video/yZyx9gHhRXM/видео.html
I know this is a little bit late but i got a different result and i want to know if it is right. My result was 6,because that in the same race that we used to discover the first fastest horse we could use to discover the second and third fastest horse. Sorry if it is hard to understand, it's because i'm not english.
@@MrJelle18 I was saying that,in the race that we discovered the fastest one,we could say that the second and third place would be the second and third fastest horses.
Depends on the race. Run them through an Army obstacle course. I'm guessing they'll get hung up on the wall climbing, rope swinging and belly crawl parts. It also depends on the length of the race. Native American runners could outrun horses in an endurance race. People are willing to keep going when animals would quit. So, if you're in shape for a marathon, you'll probably win.
@@flatebo1 Humans have an innate advantage in an endurance race vs most animals our species evolved to hunt critters by running them to exhaustion, that is to say we are naturally built for persistence hunting.
@@seraphina985 have you ever rode a horse? He is really enduring, why do you think he was used as a transport mean in the past? Even if its an endurance race he will leave a long distance with the human so he he could rest and start again, but don't worry he will not get tired before minimum 5 hours
what i dont get is why we need 7 races : we must do five races for the five groups and from that we have found the 5 fastest horses of them all secondly we just race those five horses and the first three to finish are the fastest three horses right ? so that a total of 6 races ! P.S : no where in the question does it state that you need to know their ranking aka whos fastest of the three ...
@@shadi3993 But if the 3 fastest horses are in the same group your method will only identify one of them. Here is how you can really solve this problem with 6 races: Let 5 horses race and label the 3 fastest horses from that group A, B, C. Assume that C is the 3rd fastest horse of all 25 horses and proove that by letting C race against the remaining 20 horses with 5 more races (6 in total by now). In the unlikely case that C is indeed the 3rd fastest horse, you've identified the 3 fastest horses (A, B, C) with 6 races. This is'nt an efficent method because it requires more than 7 races on average but sometimes it can solve the problem with only 6 races.
Yeah this was great. I was having a hard time until I saw you group them in to a,b,c,d,e based off of the rank of their race. I was getting stuck on the three fastest horses being in the same race initially but the re-grouping did the trick. The visual really helps.
Feel you, i was stuck at a minimum of 11 races , because i wasnt regrouping and i thought id always have to let the top 3 of every race continue in case the 3 fastest horses happened to be in the same race at the start.
Outcomes are dependent on competition. This is a software developer question written in a bubble. And he didn't specify time , vs, outcome. Race six runs the winner of each group, but it takes a minimum of two races to remove all the second and third place finishers. Therefore his solution is wrong. It's not right.
@@sasquatchrosefarts What do you mean? After race 6 you know no horse behind the third place horse can win, 1 horse behind the second place horse gets a chance for third place, 2 horses behind the first place horse get a chance for second and third place, the winner of the winners definitely is first place, leaving 5 horses to be checked so we finally know the second and third place. When you identify a horse is slower than a horse that turns out to be slower than a slow horse then you also learn something about the first horse even though you didn't let it race again. Horses can be identified to be slower after a later race. If I misunderstood can you explain why you consider the solution to be wrong?
@@esco8778 Well you order the groups based on the place their "winner horse" made at the "winners race". So you have the 5 races at the start, after that the 5 winner horses will race each other. After that you order the groups by the finishing order of the groups fastest horse in the "winners race". So if the winner horse of group C also wins the "winners race" group C will be the "fastest group".
@Dpg Dpg You didn't calculate it, you looked it up. Sorry you don't pass In order to calculate it, you must take drag a line from the edge of the earth to the edge of the sun
for people getting 6 races, we’re trying to find a method which guarantees the top 3 horses will be found. after the first 5 races, there are 15 horses which could still potentially be top 3 overall, and after the 6th race, there are still 5 in contention, including some of those 15 who didn’t win their initial race.
And it is open to interpretation, so some will interpret it in a way so you can label the horses and get 7 races, others will not label them and get 11 races. I would think a good answer would also be that the instructions are not complete
If you are looking to hire a programmer that's definitely not the answer you're looking for. If this were a simulation it would take less resources to program this solution in than to program in a clock. Programmers have to always think abstractly to find the most practical solutions.
@@bro748 That's also a very good answer. Also, if I hired a programmer, the only reason I'd have them out racing horses without proper equipment, would be as a punishment. Or maybe as a reward, depending on the programmer.
@@bro748 If you are looking to hire a programmer, you want to get a solution that does not blow up in your face when you get a 26th horse and the algorithm is just not working anymore. So you should phrase this as a more general problem with N horses. And then it gets interesting whether you want to find a generalizable algorithm that performs well on both low and high numbers or something that switches in between. Adding a clock scales nicely O(n), which is the best you can get with this, so this is actually one of the best answers. :P
I got the setup exactly the same up until the 6th round, but at the seventh round I couldn't quite get the final 2 horses. I thought c2 could be faster than a2 and b2 and got confused how I'd identify the 2nd and 3rd fastest horses with such complexity. What I missed is that c2 could infact be faster than a2 and b2, but it's irrelevant as in that case there is still a1, b1 and c1 faster. Looking at it from the other side and eliminating all the horses that can not possibly qualify for top 3 is by far the easiest and smartest solution. I didn't come up with that.
Exactly, also this implies how subjective is categorizing letters a to e in the same group as 2 to 5 since there is no timer, there's no way to know if A2 is actually faster than E1 because E1 could've for example finished in 5 seconds, A1 7, but that doesn't mean A2 couldn't have finished in 4.9 making it actually slower than E1
I still didn’t get it. C2 could in fact be faster than a1,a2, b1, b2. So we are missing that comparison. But you said that is irrelevant so just want to understand what I’m missing.
@@bhupendrasatpathy C2 can never be in the top 3 because it is already slower than at least 3 horses: A1, B1, and C1. It is slower than C1 because it lost the race with all the Cs. It’s slower than A1 and B1 because A1 and B1 are faster than C1 since they beat it in the race of the fastest among groups.
But all of B2 3 4 5 just means they slower than B1. All pf A2 3 4 5 just slower than A1. Theres no line where B2 3 4 5 faster than A2 3 4 5 because groups are ranked by heats. Say A1 is 10sec. B1 is 9sec. A2 3 4 5 could all be 4sec. B2 3 4 5 could all be 8sec.
ideal horses too, so individually they have no volume but occupy a finite space together. They also do not attract each other so little ponies won't be made
@@backwashjoe7864 I’d argue that being frictionless and in a vacuum were redundant, but… Instead I’m going to add that they should also be made up of non-baryonic matter so as not to be affected by virtual particles. Frictionless non-baryonic spherical horses in a vacuum.
I think it also depends on who is keeping track and actually watching the races. Alice or Bob would be able to figure it out. Charlie is either selling popcorn or shoveling the stalls.
the answer he gave is not correckt! The problem is that you dont have a time/ watch nor a distance! If you let 5 groups run, and you take the 1st. of each group, you can not be sure if these horses are one of the fastest 3. 1st horse in group (a) maybe can run just 2Kmh and the 1st of group (c) can run 40kmh while 2nd of group (E) can run 60 kmh! . you have to have a constant to really get the 3 fastest! My logic: So you let the first 5 race, then you let the first second and third place on the track and get 2 new ones in, every race the 4,5 will be replaced. with this method you have always a constant for first, second and third place without needing a watch ! each race, 11 in total , will naturaly eliminate 4 and 5 place ! imagine this: lets say the first group are the 5 slowest of all, so the slowest up to the 5th slowest of all. in the second race the 4th and 5th place gets replaced with (in this simplyfiyed case)faster ones, and maybe one is the fastest of all and the other one the 2nd fastest of all. they will be 1st and 2nd place of the 2nd race and they will stay on this position. the 3th will be found after all other races are finished. You can switch the start situation, that the first 5er group have two of the fastest of all in it, so the first and second place will be set after the first race, and again the other races and just to find the real 3th place.---- i think its 11 races ! 😃
@@crashoverwrite5196 > 1st horse in group (a) maybe can run just 2Kmh and the 1st of group (c) can run 40kmh while 2nd of group (E) can run 60 kmh! No. If you look carefully at the solution, the groups "a, b, c, d, e" are named *after* the 6th race. the winner of group a is the 1st place of the winners, the winner of group b is the 2nd place of the winners, so by definition a_1 > b_1 > c_1 > d_1 > e_1 - and then, you have a tree from the winner, you only want the top 3 so you can eliminate any that are 4th or more from the winner, and you are left with exactly 5 horses to race and find the 2nd, 3rd overall
@@sodiboo You dont know how fast they are! so my asumtion is valid. You need a fix point for the 1st,2nd and 3rd place. my comment was based on the crux, not the solution he gave.
I know I will never work at these kinds of places, but to see the thought processes in these puzzles is awesome. Training yourself to look at the same real world problem in different perspectives to solve real world engineering/ self reliance problems are invaluable. Great video
This was before logic and rational thought was deemed racist. Now states like NY are outlawing 'tests' like this in the pursuit of 'equity' and we're slowly starting to fall behind other countries like China.
I was getting 9 as a first effort but then as soon as you said 7 I was able to figure out how to get 7… It’s truly wonderful how much knowing that something can be done can influence your ability to do that something
Just like our full potential as a living being. Very few people know this and it then makes it impossible to achieve it until you are taught it by someone who has achieved their full potential. Most believe it is impossible only because they don't know how.
I just want to say, your question is exponentially better than the common interview question. Me sitting here with tired horses, one fell in a race....
@@fmobus you do understand what an exponential equation looks like right? Calling something exponentially better is in comparison to calling it linearly better, where something exponentially better reaches its limit faster than something that reaches its limit linearly. So “exponentially better” means “a lot better” and “exponentially more” means “a lot more”…
If I was doing a job interview at Google and they would ask me this question , explaining I dont have a watch , I'd be like "Smartphone DOH , you know howmany stopwatch apps there are ? Are you sure you are from Google , geez"
I realized pretty quickly that 6 races isn't enough, at which point I decided it's to complex and might require like 20 races or something. Glad to see such an elegant solution, thank you!
@@galacticlava1475 2nd place horse in 6th race might be slower than second place from the race the first place horse raced in. for all we know, that specific race might have consisted of the 5 fastest horses in the herd. thats why we take the 2nd and 3rd horse from that group, because even if 2nd place horse in race 6 is faster than 2nd place horse in race with the overall fastest horse, it may still be slower than 3rd place horse in that race. you consider the 2nd place horse in the race producing the 2nd fastest in the 6th race as u want to find the top 3 horses in speed, as 2nd place horse in 6th race might be faster than 2nd place horse in race that produced the fastest horse. Lastly, u include no3 from 6th race as all these extra horses included can still be slower than it, and if so, u still have ur fastest 3.
I don´t know if you are even allowed to label the horses, since it is not said you can. And if you can´t label the horses, 11 horses is the right answer, because every race exactly 2 horses can´t go on for further testing. It all depends on the interpretation of the questions.
That's sometimes a good approach - when the optimization is a complex process. But in this case the optimization is simple enough that there's no reason not to use it on small numbers of horses too.
@@a0flj0 Sure, but without knowing this answer beforehand, who is actually going to think about this? Google should want to see that you know how to solve a problem in front of you, not necessarily do it in the best and most accurate way possible. In real life you'll have resources to assist you in researching what the best way to do something is. You're not just given an issue and told to fix it immediately without any outside help.
The mathematically optimal answer shown in the video is satisfying to discover! The use of this method seems to be dependent on the specific numbers of total horses and horses per race presented in the problem, though... While trying to solve the problem, I came up with an algorithm which has the advantage of generalizing smoothly to any variation of the problem with M horses in the pool, N horses allowed per race and K spots at the top of the field, where K ≤ (N+1)/2. Here it is: Start: Have a group of untested horses, and a "leaderboard" with K spots. All M horses start in the untested group. Place N untested horses in the lineup for the next race. Run [Race Type A]. [Race Type A] Race the horses in the group; place the top K on the leaderboard, in order, and throw out the (N - K) horses at the bottom. Now add the horse in Kth place on the leaderboard, and (N - 1) untested horses, to the racing group. Run [Race Type B]. [Race Type B] Run the race with the current group. The next step depends on the outcome of the race: If the horse currently in Kth place on the leaderboard wins the race, eliminate all the other horses in the race and set up the next race with the Kth horse and (N - 1) new horses. No change to the leaderboard. Run [Race Type B]. If the Kth horse comes in 2nd or worse, eliminate the Kth horse and all horses in the race who were slower. If K or more horses beat the Kth horse, add the top K finishers to the next race; otherwise, add all horses who placed above the Kth horse. Add the other K-1 horses on the leaderboard to the race. Fill the remaining spots in the race, if any, with more untested horses. Run [Race Type A]. When no more untested horses remain, the leaderboard contains the K fastest horses. I ran a Monte Carlo simulation of this algorithm with 10,000 trials, and for the values given in the problem (M=25, N=5, K=3), the expected number of races is just under 8. Not bad!
This is interesting! What does a simulation find in the situation (M=25, N=5, K=2)? By the way, the approach of the video's method isn't necessarily the optimal one though. For example, in the case of (M=4, N=2, K=2), the video's approach would suggest that 4 races is optimal. However, it's relatively easy to find a method that takes 3 races in 1/3 of all possible scenarios, and 4 races in the other 2/3 of scenarios, thereby resulting in the optimal method that requires 11/3 = 3.6666... races on average; which is more efficient than 4 races. And maybe 7 races is optimal for the case finding the Top 3 out of 25 horses, but the video failed to give a solid proof of it (the "proof of minimality" in the video is flawed).
Google doesn't actually use these sorts of logic puzzles in interviews. At least they haven't for the last 20 years or so. They ask algorithmic/coding questions, system design questions, questions that check technical knowledge and the like. There are lots of videos posted by Google about how to prep for a Google interview. But tl;dr, re-read Introduction to Algorithms (AKA the mobile book).
So something we talked about in economics the other day is how if you take too much time to solve a problem in the most efficient way it can be more costly than just doing the inefficient way that instantly comes to mind. The answer I came up with was 11. I know it’s not even close to 7 but if I was able to have a system ready in 10 seconds that works I’m still proud of that answer Edit: very happy with the conclusion of the video though. Super cool thought process!!
But then again, if you spent a bit longer thinking more efficient solution you would save 4 races. That's 20 races you would have saved those mechanical horses from doing. If you do this more than once the maintenance costs are going to pass your thinking time.
well, this makes sense to me, but i think that if u create the best solutions u will be faster to create the best solutions next time and everytime, and maybe just at the start is harder, maybe
I think he is wrong, he would not be hired by Google. (About the answer in the video) I think he assumes that all horses finish at equal time intervals. The fastest horse, without any timing available, doesnt say anything aboit the slowest horse. The horse that finished 2nd to the fastest of all horses (the fastest group) could still be slower than the horse that finished 2nd to the horse that finished #5 in the 'race of winners' So without also determining which are the slowest horses this is just assumption. I'd say one need at least 8 races therefore. 11 is making sure anyway. Cross reference and you can make the whole top25.
@Moh Taw, - "What if the 5 horses in group "a" were actually faster than all other horses ... How you will solve it without a watch" That's exactly why horses a2 and a3 also participate in the seventh race (along with b1, b2 and c1), which they will win in your scenario.
@@davidhawley1132 ok thanks David. I’ve noted your assumptions. I’ll speak to the client in relation to your concerns of legacy constraints and find out if these are going to be problematic... I have liased with them and it turns out that we live in 2021 and we’re dealing with a startup. It’s also apparent that there are already timing tools available on the market, so I’ve conducted build vs buy analysis and it makes sense to procure the stop watch instead. $3 from Amazon. Since finding the fastest horse is going to be repetitive it also means we can make huge time savings in efficiency across the product life cycle and there is less risk involving the horses. So I’m grateful for your input as this led to a cost saving of tens of thousands. I’m giving you a pay rise and I’ll write a case study based on your input for lessons learned in the next knowledge sharing workshop so we can all benefit from your unrivalled wisdom. Lol obviously this was not a serious solution dude.
@@davidhawley1132 We have "stopwatches" that work just fine in distributed simulations. It's a very difficult problem with an embarrassingly simple solution. (But, I'm also talking software - getting circuits in time without a common clock is something I'm not going to mess with, unless performance is not a concern.) (And who wants to work on something where performance is not a concern?) :D
I had to answer this question to get accepted into university. I got the right answer by making a mistake that led to the same answer. I got accepted 🗿
I'm the interviewer, I'm now wishing I'd never asked this question after hearing 27 variations of how to solve it. It's late Friday and I want to finish work. . . . . . .First one to tell me, 'Sell 22 of them, keep the cash!' Gets the job! I don't care, I'm retiring next week.
What has happened about retiring next week? Are you so busy making the papers ready that you forgot you had announced your retirement publicly on RUclips?
I get it, but jumped to the initial conclusion of 6 races. I’ll say that I knew I was wrong and that I was missing something. Good lesson here, I’m glad I watched.
I was about to, but then I thought "what if all 3 fastest horses is in the same group? This can't be the solution." I couldn't figure out the right answer though.
Sorry, Mäx-Tick, but someone with a background in math posted a solid proof in this comment section 8 months ago, showing that it's impossible to design a method that is guaranteed to identify the three fastest horses every time in just 6 races.
@@writer_gupta_ji*(edit this method is flawed as pointed out by others below me)* I solved it in less then 2 minutes, tho I solved for 6-7 (ill explain that later) using a different method which is race 5 horses then race first place against 4 new repeat 5 times keeping track of the second place horses at the end of the sixth race if the fastest horse of race 5 isn't first or second then the top 3 are the 3 fastest else race 2nd place horse of race 5 against 3rd place of race 6 and first place of race 7 is 3rd fastest. This is a better solution because there is an 84.43% chance u will only need 6 races. *>>> AGAIN THIS METHOD IS FLAWED SO PLZ STOP COMMENTING THAT IM WRONG, I AM WELL AWARE.
@@nothingnothing1799 Maybe I am missing something, but how does your method work? Imagine if your very first race had all three fastest horses in it, and you basically eliminate the 2nd and 3rd fastest ones at the very beginning, and never return to them in your procedure above, how do you get to the top three?
0. Destroy 22 of the houses. The last 3 are now the fastest. (You can also sell them to be more humane but in a hypothetical scenario where you HAVE to know which ones are the fastest) This is the smallest amount of races required to determine the 3 faster horses.
There are ways to learn this in 0 races. Consider disassembling one horse, reverse engineering its software to find where the horses movement rate is located, then hack the other 24 horses and learn their rates without the horse moving. The question is "The three fastest horses" not "Rank the horses by the fastest" Another way is to destroy all but three of the horses, and then you automatically know the three fastest horses.
It's a programming problem, ie why Google would ask it to a potential employee. It doesn't really matter what horse is the 1st, 2nd, and 3rd fastest, they're trying to get you to come up with a problem-solving technique to identify what those horses are. Programming is all about developing these techniques, the actual numbers/results don't matter so long as the techniques are correct.
@@digiquo8143 Agreed. There are all kinds of videos on you tube (including several other of the 25 horses problem) that pose intriguing puzzles that challenge your brain to think, find possible solutions, look at possibilities, and come up with sound techniques that will work in any case. Someone else here mentioned six races, and when I worked through his logic, I found out that he was in fact correct, but it only worked in 2 possible cases, both of which were statistically insignificant. The correct technique, as shown in this video, will hold up rigorously for all cases.
@@DreadX10 race horses in sets of 5 and kill the ones not 1st place, 5 races. Race remaining horses and kill 5th and 4th place. 6 races. You are now certain that the 3 horses you picked are the fastest of the original 25.
@@ozzygilliam9194 ...but two or three of the fastest horses might (by chance) be in your group of 5, so then you would have shot one or two of the fastest horses. you could race them in groups of 5 and the two slowest are retired (see Blade Runner).
Horse C in 3rd place beats 4th and 5th place horses hes faster than the eliminated horses even tho he never raced them from the 4th place group and 5th place Group at the beginning because the 4th and 5th already beat the 8 horses who got eliminated its obvious that horse in 3rd place faster than 10 so far counting 4th n 5th place. Horse 1st and 2nd beat everyone in their group the eliminated ones from there group never raced the horse 3rd place horse 1st and 2nd could had beaten someone faster than the horse in 3rd place now which is the eliminated ones and couldn't race to know for sure. so we can't know for sure. All we know is 8 is left we can only pick 5 so 85% chance we need 7 try 12.5% chance we need a 6th try only.
I was wondering how you could find the 2nd and 3rd fastest overall. The other method I came up with was more of a slider and would require 11 races, which works but isn't minimal. (Line up the horses, race the first 5, keep the top 3, add the next two horses, race and repeat)
I would race 5 groups of 5 and aswell remove to two slowest in each. Leaves 15, repeat, leaves 9, then 5 and of those it's the fastest three. 11 races (5 x 5horses, 3x5, 2x 5, and last race with 5)
@@aramisortsbottcher8201 Wrote out a very similar thing about a day after you did. I just continued to take the top 3, even when it left we with 9 instead of 10. When I got to 6, I ran one race with 5 and left one behind. Then ran it again with the last horse eliminated and the spare 6th plugged in. For a small touch of extra certainty.
@@robertdavis5693 I first had this idea too, but when the point with nine horses was arrived, I raced the first five and after that the next four, but filled the fifth slot with one of the previous race. That way I could sort out two of five instead of one of four horses. This left me with 5 instead of 6 horses, so they all fit into one race :D
The smartest person seems to be the person who comes up with the question with the answer in mind. Honestly takes a lot of brain power to come up with a problem and solution a lot of people cant solve
Vsauce! Michael here. What you experienced is what we call "Jamais Vu". It's what you experience when a word has been spoken so many times that it loses its meaning. Thanks for watching, this has been Michael... or am I?
you can do it just in 6 races, 5+4+4+4+4+4=25 first race, 5 random second race, best of first race and 4 more do it in 6 races, you must mark where the horses are when the winner of the first race reaches the finish line.
I came up with 8 races and proved it using pennies as horses. I didn't consider the 7th race could determine both the 2nd and 3rd fastest horses by eliminating the entire field of horses that only had 1 winner (race 1 -5). Thanks for posting.
I got 8 as well, but then when I saw the actual solution was 7 I immediately noticed we repeat the 7th and 8th race with the bottom 2 horses that have absolutely no hope of ever winning a top spot, and so worked out the best solution from there. So I worked out how to do it in 7 after seeing the solution was 7, but before seeing how to do it.
I also got 8. It is easier to find a solution that is not optimal, but that is fine in computer science. After you found a solution that works, you should seek to find a better solution, if necessary. 👍
I got 6, but I was wrong. I assumed the first 3 winners of the 6th race were the 3 fastest, without considering that they hadn't raced yet against those of higher level
I'm not sure about all combinations of numbers of horses etc, but the method in the video works for ¼(k-1)²(k+2)² horses where you can race ½(k-1)(k+2) horses in each race.
I think it's more adequately written as (½(k-1)(k+2))² horses and ½(k-1)(k+2) horses per race. That makes it more obvious the former is the square of the latter.
+codebeard Is k the number of horses we're trying to find, or is it just a coincidence that k=3 for 25 horses? If k is the number of horse we're trying to find, the minimum number of races is ½(k-1)(k+2) + 2.
2:28 couldn't you skip the last race? If you race the top 5 horses you immediately know the top three since, like you described earlier, we get a list of the order that horses finished, not just the winning horse only.
@@30GBofnegativeluckso basically if you think about it the fastest horse in group a could be (for example) 25 mph and the second fastest in group a could be 20mph but it doesn’t get send to the fastest race bc it wasn’t the fastest in its group and we can say the fastest group e horse was only 19 mph but the fastest in its group
Something i think could be explained as i didnt initially think of it and other ppl dont seem to have noticed is the fastest horse of a given race could still be slower than the slowest horse in another race. Just depends on luck of the initial groups
I was thinking about this too, but with this method, it won't matter. Even if the horses are arranged strictly by performance (the slowest in group A is faster than the fastest in group B), it will just mean that the winners of race 7 (2nd and 3rd fastest overall) will all come from group A
There is another problem in every race the horses are randomly asigned and you don't have the time so by that definition in one of the groups you can still have the 1 fastes hourse and 2nd fastes from all 25 hourses, but by this method of elimination you have already decided that the secound fastes hourse is no longer competing and the 1st fastes hourse is compeeting to the once left from other groups.
I'm genuinely curious what percentage of people still have watches. They seem kind of redundant now TBH. Edit - and by have I mean "actually wear and use". Buried in the couch or the back of a junk drawer isn't what I'm talking about.
I understood how you got the three fastest horses in seven races (in a few minutes after I finished watching the video). I drew the diagrams you drew, and I began eliminating horses.
Its not correckt! The problem is that you dont have a time/ watch nor a distance! If you let 5 groups run, and you take the 1st. of each group, you can not be sure if these horses are one of the fastest 3. 1st horse in group (a) maybe can run just 2Kmh and the 1st of group (c) can run 40kmh while 2nd of group (E) can run 60 kmh! . you have to have a constant to really get the 3 fastest! My logic: So you let the first 5 race, then you let the first second and third place on the track and get 2 new ones in, every race the 4,5 will be replaced. with this method you have always a constant for first, second and third place without needing a watch ! each race, 11 in total , will naturaly eliminate 4 and 5 place ! imagine this: lets say the first group are the 5 slowest of all, so the slowest up to the 5th slowest of all. in the second race the 4th and 5th place gets replaced with (in this simplyfiyed case)faster ones, and maybe one is the fastest of all and the other one the 2nd fastest of all. they will be 1st and 2nd place of the 2nd race and they will stay on this position. the 3th will be found after all other races are finished. You can switch the start situation, that the first 5er group have two of the fastest of all in it, so the first and second place will be set after the first race, and again the other races and just to find the real 3th place.---- i think its 11 races ! 😃
@@crashoverwrite5196 Yeah. My first thought was to do what you described, but I abandoned that quickly because I knew it would require more races than necessary. I figured it out quickly enough. The answer is 7. Just pay attention to what is said in the video.
@@crashoverwrite5196 What about if you ran the first 6 races as described in the video but for the seventh, ran all 2nd place finishing horses from the first 5 races against one another and for the 8th (and final race) run the top two horses from both of the 6th and 7th races?
My first idea was taking the 2 slowest off each time as well. Ended up with 11 races however I appreciate it can be done in less. But upon further contemplation I figured if I'm at Google I'd just pick 3 random horses. The truth is what we say it is.
Same idea here. A way with less races may possibly exist, but designing and verifying its veracity likely takes longer than the few races that it might save us, so we better simply get those horses to run. One could ask the interviewer, "is the question here really about the best algorithm, or is it about figuring out which are the three fastest horses in the most efficient way?"
yes man. you got it with 11. The problem is that you dont have a time/ watch nor a distance! If you let 5 groups run, and you take the 1st. of each group, you can not be sure if these horses are one of the fastest 3. 1st horse in group (a) maybe can run just 2Kmh and the 1st of group (c) can run 40kmh while 2nd of group (E) can run 60 kmh! . you have to have a constant to really get the 3 fastest! My logic: So you let the first 5 race, then you let the first second and third place on the track and get 2 new ones in, every race the 4,5 will be replaced. with this method you have always a constant for first, second and third place without needing a watch ! each race, 11 in total , will naturaly eliminate 4 and 5 place ! imagine this: lets say the first group are the 5 slowest of all, so the slowest up to the 5th slowest of all. in the second race the 4th and 5th place gets replaced with (in this simplyfiyed case)faster ones, and maybe one is the fastest of all and the other one the 2nd fastest of all. they will be 1st and 2nd place of the 2nd race and they will stay on this position. the 3th will be found after all other races are finished. You can switch the start situation, that the first 5er group have two of the fastest of all in it, so the first and second place will be set after the first race, and again the other races and just to find the real 3th place.---- i think its 11 races ! 😃
I started out saying 10, then realized I could eliminate the 4th and 5th place horses from the first 5 races. Just as I was patting myself on the back and saying the answer was 8 races, the rest of the video played. Guess I'm not as smart as I thought I was.
@@johns9652 6 races, because reality isn't so comfortable; you have the winners run, and the winner of winners is the fastest with 2nd & 3rd place marked behind it. If UFC, or other events operated like y'all are suggesting, then all competition sports should be eliminated, and all champions should return their trophies.
I used that approach initially. 25 horses to start, 5 races, two slowest eliminated each race, 15 horses left. 3 races, two slowest eliminated each race, 9 horses left. 2 races, but with nine horses, you'll have one group of five and one group of four, and you need to keep the fastest three from each group, so eliminate the two slowest from the group of five, but only one from the group of four. This creates somewhat of a dilemma, as you have already run 10 races and you still have 6 horses. Run a race of five, with the sixth sitting out. Knock out the two slowest from that group, leaving the three fastest and with the one that sat out, that's four remaining. That was 11 races so far The final race will show who is 1-2-3 out of the original 25. But that was 12 races.
@@storm0fnova No, he means kill all 22 before even racing. His answer is 0 races, because it's a foregone conclusion that the 3 survivors would be fastest.
Actually it's 6 races. Your original question says; "What is the minimum number of races needed so you CAN identify the fastest 3 horses?" You didn't say "MUST". Therefore; (H3 = 3rd finishing horse from R1) R1: H1-5 R2: H3 + H6-H9 R3: H3 + H10-H13 R4: H3 + H14-H17 R5: H3 + H18-H21 R6: H3 + H22-H25 - H3 was faster than Horses 4-21, we now have our 3 fastest horses. gg
The word "can" implies an ability to perform a task with a (near) 100% likelihood of success. If I'd say "I _can_ multiply two two-digit numbers from the top of my head within 3 seconds", then my claim implies that I can (pretty much) _always_ perform that multiplication within 3 seconds, and not just in the unlikely "lucky" scenario that one of the two-digit numbers happens to be the number 10. Likewise, "number of races needed so you can identify the fastest three horses" implies a number of races in which you'd generally manage to successfully identify the fastest three horses. (The word "minimum" then asks that from all numbers of races in which you'd generally succeed to identify the fastest three horses, which number is the minimum number? So, for example, suppose you master three different methods that successfully identify the fastest three horses: the first method generally requires 9 races, the second method generally requires 8 races, and the third method generally requires 7 races; then the minimum number of races in which you can identify the three fastest horses, is 7 races.)
I made it seven, but it was a bit of a guess. I got as far as racing five groups of five and then racing the winners, giving six races. After that I figured one more race should do it, but more by intuition than reasoning. The step that eluded me was that any horse beaten by three others could be eliminated, although that seems obvious, now you mention it. I think I would have got there eventually, but it's easy to say that and my working would most likely have involved some extra unnecessary steps. Thanks for the clarification 🙂
Damn I feel like I got pretty close. I immediately had 2 ideas. The first method was to just keep the top 3 and add another 2 new horses until complete, which is 11 races. The other idea was to use a bracket and I figured out to get the fastest overall requires 6 races, but for second and third place I just did the same thing and compared the second fastest from the winner's group to the second fastest from the initial groups which gave me 8. Intuitively I could feel this was wrong though, and if I had actually wrote any of this down instead of trying to solve in my head I probably would have done a lot better, but even still I'm not sure I would have seen that last step. What I've learned here is knowing when and how to use grids can be extremely useful, and that I should note things down.
@@jordantyler148 actually Im not sure, imight have been right, but not the way i thought. You say it isnt a solution unless it works in all cases, but what if i had a method that gives the answer in 6 rounds if and only if all top 3 horses are in the first group, and gives it in 7 rounds if not? Im not sure if I have such a method, but if I did it would be a better answer than the video, would it not?
@@jasonbernard5468 It would be interesting if you had a method that sometimes worked in 6 rounds, but occasionally needed an additional round, but only if you knew if it was done after 6 rounds, not if you do the 6 rounds and you dont know if its right.
@@jordantyler148 There is a method that works for all cases and in best case is only 6 races, but worst case 11. It does not involve doing 5 races with all horses like this. You do an initial race which gives you an initial #1, #2, #3. You take you 3rd place and race it against 4 new horses. If any are faster than current #3 then you race the fastest 3 of those against your current #1 and #2, giving you a new set of finishers. And continue with whatever the current #3 is. Best case, they are in order and you do 1 initial race plus 5 additionals where none are faster that that first #3, so 6 total races. Worst case they are in reverse order. and you will have to re-race after each of the 5 races, so 11 max. The question is to find the minimum, not the most optimal solution. So this would be a "better" answer.
I get a paper and label the horses so and I based my time on the fastest one in each group so then I can tell which ones are the fastest Ones ( I will race the fastest of each group to see which one is the absolute fastest and based the speed on that one
My best answer would be run a horse race gambling game while doing the laziest solution done everyday making sure to keep horses always healthy, do it for 6 months to a year, you get money and you get the best horses trained after months of running 🥳. Since the fastest horse now wont necessarily be the consistently fastest horse, you need reliable statistics!
That's actually the correct answer, because it's said you don't have a watch, so you could never compare the times of each group's fastest horses, 2nd, 3rd, etc (one race at a time, remember?) The 2nd place from group A may be faster than group B's fastest horse. The guy who made this video may be good at math, but he surely lacks text interpretation...
I was started to write a long comment throughly explaining why I disagreed with your answer. I read through the comments before responding to see if others came up with the same conclusion. Upon reading those comments, others had realized what I had, and explained how they were confused and why your answer made sense. I see where I was wrong, after reading the comments. If the 4th, and 5th place horse from the the 6th race were the fastest in their group, that entire group is eliminated. Even the 3rd place horse would be the very minimum possibility. Excellent mind puzzle, made me think thank you.
I was starting to do the same - only when I scanned through comments I was only surprised that everyone except me was in agreement. If the horses are divided randomly into even groups of 5 - how do we know that all of the top 3 are not in the same group and therefore 2nd and 3rd place are not automically incorrectly eliminated?
No it actually is not. On placing yes, but with actual times it can be very different. For instance A1 can run 1min but A2 needed 5min and the whole group E can still lie between them
Horse C in 3rd place beats 4th and 5th place horses hes faster than the eliminated horses even tho he never raced them from the 4th place group and 5th place Group at the beginning because the 4th and 5th already beat the 8 horses who got eliminated its obvious that horse in 3rd place faster than 10 so far counting 4th n 5th place. Horse 1st and 2nd beat everyone in their group the eliminated ones from there group never raced the horse 3rd place horse 1st and 2nd could had beaten someone faster than the horse in 3rd place now but they got eliminated so can't race horse 3rd place so we can't know for sure. All we know is 8 is left we can only pick 5 so 85% chance we need 7 try 12.5% chance we need a 6th try only.
personally I thought it was 6, 25/5 for the 5 initial races and then 1 with the winners, surely first, second and third place meet all those requirements, 3 fastest horses, I hadn't thought about the fact that 2nd place in group A could be faster than 3rd place Group C in race 6, interesting, something to think about
Horse C in 3rd place beats 4th and 5th place horses hes faster than the eliminated horses even tho he never raced them from the 4th place group and 5th place Group at the beginning because the 4th and 5th already beat the 8 horses who got eliminated its obvious that horse in 3rd place faster than 10 so far counting 4th n 5th place. Horse 1st and 2nd beat everyone in their group the eliminated ones from there group never raced the horse 3rd place horse 1st and 2nd could had beaten someone faster than the horse in 3rd place now but they got eliminated so can't race horse 3rd place so we can't know for sure. All we know is 8 is left we can only pick 5 so 85% chance we need 7 try 12.5% chance we need a 6th try only.
First thought on this was 8. Race all groups (5 races), then race the winners of each (1 race) then race the winners again replacing the fastest overall with the runner up in the group repeating twice (2 races) Didn't think to remove the e and d groups like you showed and let them keep racing even though they could never be top 3
I had thought 6, but since you can theoretically get 3 horses faster than the 3rd fastest from the first race in each subsequent race, a minimum 7th race is required to resolve the final rankings (and possibly several more). My original bid was 11 - 5x5, and then 5x3 gives the fastest 9 horses. The 3rd fastest from the first of these races signposts the contenders from the remaining 4, which can be re-raced with the fastest two from the previous race in the final race.
They said minimum, meaning best scenario. So minimum is in fact 6. Run 5, then use 3rd place to run against the other 20. 3rd place beats all therefor you had your fastest 3 in only 6 turns
@@SharamanONorgannon you are assuming that the 3rd fastest from the first race is the 3rd fastest *overall.* What if horses #17 & #19 are faster? Sure - one of these is faster than the other, but you can't know whether both are faster than the 2nd fastest from the first race without a re-race. So now you have the 4 fastest horses, and don't know which of these are the 3 fastest - in fact, you don't even know which *single* horse is the fastest.
@@testname5042 They said best case. So you test the 3rd best from the first set against all the rest. In best case scenario the 3rd fastest from the first race IS the 3rd fastest overall therefor in 6 turns you can figure it out.
Horse C in 3rd place beats 4th and 5th place horses hes faster than the eliminated horses even tho he never raced them from the 4th place group and 5th place Group at the beginning because the 4th and 5th already beat the 8 horses who got eliminated its obvious that horse in 3rd place faster than 10 so far counting 4th n 5th place. Horse 1st and 2nd beat everyone in their group the eliminated ones from there group never raced the horse 3rd place horse 1st and 2nd could had beaten someone faster than the horse in 3rd place now which is the eliminated ones and couldn't race to know for sure. so we can't know for sure. All we know is 8 is left we can only pick 5 so 85% chance we need 7 try 12.5% chance we need a 6th try only.
Since I've heard this puzzle, I've always pictured it as a story around a wealthy Sicilian businessman who is interested in horse-racing and who wants to buy (real, not mechanical) horses from a horsebreeder who has 25 horses; and so he hires you to figure out and prove to him which are the three fastest. But in order to keep his potential rivals/competition in the dark about what he's exactly buying he doesn't want any records to exist/be created of how fast the horses are. So no watches or time-measuring devices are allowed, and also no recording devices, cameras, smartphones, etc. And since organizing a race costs money, he'd appreciate it if you can do the job in as few races as possible. (Or maybe he's willing to pay you a fixed amount of money that's enough for, say, 11 races, but you can keep the money that you don't spend, as long as you complete the job.) And then I read the comments and see how many viewers are willing to cheat on this job. And apparently would openly admit to it... in a job interview!
Because every horse in group D or E is by definition slower than top C which is AT BEST the third fastest(top B and top A beat it in a race). So at best every horse in D and E can be fourth fastest and can therefore be eliminated.
So if following this example the times for slowest to fastest in each group are as follows: A: 10, 9, 8, 7, 6 B: 14, 13, 12, 11, 10 C: 15, 14, 13, 12, 11 D: 16, 15, 14, 13, 12 E: 17, 16, 15, 14, 13 How is it valid to eliminate a3-a5 as being slower than b1?
You're not eliminating A3, you're eliminating A4-A5, B3-B5, C2-C5, all of D and all of E. Which will always be valid. You're not eliminating them for being slower than B1, you're eliminating them for being slower than AT LEAST 3 other horses. The final race is A3, A2, B1, B2, and C1 since those are the only horses left that aren't confirmed to be slower than at least 3 others.
I came up with 8 races. Race each set of 5, remove the bottom 3 from each race, leaving you with 10 horses. For the next 2 races, count seconds it took for the third horse to finish. The slower race removes three horses, the faster race removes 2. This leaves you with your last 5 horses, and one last race for top 3. Admittedly very flawed compared to the solution, especially the counting.
I came up with the optimal solution in a few minutes with pen and paper, but I guess it's easy to get stuck if you miss some piece of information. I thought it was easier to think backwards: first you definitely need 5 races to race each horse once. The two slowest horses from each race is definitely not part of the top 3. You are now left with five groups with three horses (15 in total). You don't want to race them all against each other, so you can rank the groups relatively to each other by racing the slowest or the fastest in each group. Since we're looking for the fastest ones, intuitively it makes more sense to rank by the fastest, so we race the fastest horse in each group in the semifinal. All the horses in the two groups represented by the last two horses in this race will all be slower than the fastest three horses in this race. We're left with three groups, with three horses in each (9 in total). Some observations at this point: - The fastest horse is identified, as the fastest horse from the semifinal - There are three ways that the three fastest horses can be arranged into the three fastest groups: the top three from the fastest group, the fastest horse plus the two fastest horses in the second fastest group, or finally the fastest horse in each of the three fastest groups. This gives us 6 candidates for the three fastest horses. Realizing that the fastest one is already identified leaves us with five horses to race the final. The two fastest horses in the final are the second and third fastest horses. Alternative solutions: - 1 race: Race 3 randomly selected horses. The other ones have never competed so their time is undefined. - 0 races: Shoot 22 horses.
I got the same result with the same reasoning, except note that when you say "there are three ways..." there are actually four; you could also have the fastest horse, the second fastest horse from that same group, and the fastest horse from the second fastest group. That doesn't change anything about the rest of your explanation though :)
I got 8 because I didn’t think to rearrange the groups based on the results of the 6th race. I was adding an additional step where all the 2nd place horses from each group raced but it was basically an inefficient way to get the same result that you’d get in the 7th race in this solution.
Had the exact same thought process, but a total eureka as soon as the video said "we split them into groups" and I went "OH WAIT WAIT WAIT IS SEE IT no point in racing the nr 4 and 5 horses from the winner race, that frees up 2 more slots!"
@@codrincx same and i got stuck on how to fit a3 in there cos what if all 3 fastest horses were in the same race, you dont need to race the fastest again freeing up another slot *d'oh* 🤦♂🤦♂🤦♂
@@onepun9583 If I'm understanding your comment right, then: The d and e group cannot imaginably matter, since after the first races, we know for sure that a1 > b1 > c1. Therefore, any horse from group e will be slower than *both* b1 and c1. The only horses that could be faster than b1 come from group a (as all horses in group c, d and e are all slower than b1), and then the only horses that could be faster than c1 come from groups a and b. Notice that in both scenarios you've written, b1 and c1 are the fastest, therefore the only ones actually relevant. If a2 or a3 (could also be a4 or a5, but we only care about top 3) isn't faster than b1, then b1 *is* the 2nd fastest horse. Similarly if b2 isn't faster than c1, then c1 *is* the 3rd fastest horse.
This is something I learned in grad school. For any top-N problem, the tournament algorithm produces the minimal time solution, same O-notation complexity, but lower constant factor due to decreased pairwise comparisons.
The general extension of this algo would have been an excellent interview question with a runtime and proof of completeness. It felt good to know I got this one right when I did it.
I destroy all but 3 horses. Now they are the fastest and need to put in the work for the other 22. Apparently the best answer to get a job in human resources.
Nice puzzle, my first instinct was 11, should have thought about it more to think about what the results of the first five races meant. I thought about racing the fastest, but glossed over that eliminating the slowest two also limited the other four that were in their original groups AND the slower horses in the other groups.
Yeah i also think its 11 by eliminate every horse on 4 and 5 spot. And i think about this just under a minute. But yeah if i had think long enough and do critical thinking i could arrive at the answer for 7
11 seems correct. The solution of this video does not seem correct. Without racing #2 of race 1 against #1 of race 2, how does one know that #2 of race 1 should be eliminated? What if all 3 fastest horses were in fact within race 1, but #2 and #3 of race 1 were inadvertently removed from subsequent comparisons against all other slower horses?
@@thangd7124 yeah 11 is also correct. The solution in this video is correct too. In step two, because we already found the fastest horse in the group of 25 (1st place of the group "a"), we know that the 3 fastest horses might be all three in group "a" (case 1). In group "b", we only consider 2nd place and 1st place horses, we remove the 3rd place horse in this group because we only need to find the 3 fastest horses (one is known in group "a"), now the 3 fastest horses might be "1st place group a, 1st place group b, 2nd place group b" (case 2). In group "c", we consider only the 1st place horse, the 3 fastest horses might be "1st place group a, 1st place group b, 1st place group c" (case 3). At 7:21, we have three cases to consider knowing that the fastest horse is 1st place group a horse. case 1: 1st place group a, 2nd place group a, 3rd place group a case 2: 1st place group a, 1st place group b, 2nd place group b case 3: 1st place group a, 1st place group b, 1st place group c The three fastest horses could be in one of the cases, but we don't know which case. What we know is the 1st place group a horse is the fastest. Therefore, we can leave out the 1st place group a horse and race the others to find the second and the third fastest horses. case 1: 2nd place group a, 3rd place group c case 2: 1st place group b, 2nd place group b case 3: 1st place group b, 1st place group c race 2nd place group a, 3rd place group c, 1st place group b, 2nd place group b, 1st place group c
I couldnt really come up with why I would need the 7th race(after 6 races and I knew I had at least one more race to figure out the top 3 but I could not really come up with how to match up 5 horses for the 7th one) and now I can say I know why. Thank you.
Great thing about algorithms is that there are pros and cons to everything. Guaranteed 7 races solution has a con that you use a lot of memory, because you need to store information about first three horses of all races until 6th race. We can have a solution that at worst uses 10 races, at best 6 and stores information about only 2 horses at a time.
My solution was to grab 5 horses, race them, put the slowest two aside and grab two more to race off against the fastest 3 so far identified, after this race, the losers will be put aside and two new horses will face off against the fastest three from the second race, etc. Not particularly quick, but since the horses won't get tired, you will eventually end up with the three fastest horses
@@fayt9121 that doesn’t work. Lets say the second fastest horse was in the same heat as the fastest horse of all 25 horses. Then the second fastest horse would never be included in your final (6th) run.
@@Macarthurmaintenance this is true. so you are saying you have to race a total of 8 times since you want the 3 fastest horses right? you do your first six to find which one is the fastest and then you race second and third place against the winners from groups 2-5. I get that. And with this hypothetical, its possible that the horses don't get tired either. Ill be honest I cant put my finger on it but something's bothering me about this.
Exactly. It's an IRL variant of sorting algorithms which only go by relation (more, less, equal) of the input data, this time a variant which takes 5 inputs, and outputs relations of every pair within those. It's less relevant IRL now, but might have been very relevant before you could time the races. Thinking about these problems becomes easier if you visualize the ranking of a horse as the # of horses it would lose against, making rank 0 the fastest. Sometimes, ties tend to mess things up, but in this case, they don't.
I’m rethinking this and hadn’t realized that the 2nd and 3rd place horse would be from other groups. I take back my comment about there being a logic flaw. Mea Culpa!
My guess was 11. This is because each race is guaranteed to tell you that the 4th and 5th horses are not in the top 3. I couldn't come up with any way to guarantee elimination of more than two horses at once. That sixth race amazingly eliminates 10 horses at once, while the others still eliminate 2. I did not imagine so many at once would be possible. By establishing a hierarchical structure, entire branches can be eliminated. Very clever.
you gotta know what your looking for and yes, copy pasting is convenient but you also wanna know what your looking for and if it fits your needs. i bet coders do it all the time but they wont be fooled be shitty codding
1:33 I was able to get a 10-race solution. Racing , and thereby sorting the 5 rows of horses, I obtain 5 columns, the last two of which can be discarded as they definitely don't have the 3 fastest horses. I sort the 3 columns, and I obtain a group of 6 horses, of which I know the 3 fastest exist. By racing two different groups of 5 of the 6 horses, you can figure out which are the fastest over all.
Not sure if it's the minimum, but the first thing that comes to mind is 11. Start with five horses racing. That's 1. Then keep racing the three fastest, replacing the two slowest horses each time for the remaining 20 horses, giving you 10 more races.
There's a faster way... It only requires 1, 2, or 7 races. 25 horses... Either you race all of the horses at once, then take the top 3 who touched the finish line... Or... You race the horses in groups of 5, and the average of horses above 50% gets to race twice. The other horses are dismissed. Then pick the top 3 fastest horses from the new average. This works better if you're dealing with 100+ contestants, and if you don't want an outlier.
I really like this solution, because the one presented in the video doesn't actually work. But, he did have a good point - you can disqualify any horses that lost to 3 or more horses. So you can get rid of columns 4 and 5, swapping out the remaining 10 horses like you said.
I also came up w/ 11 races w/ same rationale. It's simple, easy to understand, and works. I'll take software written that way over clever any day of the week. However, if it's really a question of optimization (fewest races) then 11 races is too many and this video shows the optimized solution.
@@jackbastian4072 the video method incorrectly infers that a1 beating e1 means e1/e2 etc would lose to a2. Because we don't know the times and only the orders, we don't know the margin by which a2 lost to a1 and so on. If you write out 1 through 25 starting from the bottom and moving up, this could present a situation in which the video method results in the 2nd and 3rd fastest are actually the 6th and 11th fastest overall. It's easier to show on paper but I can't attach a photo here.
My one issue with the 5x5 grid of slowest/fastest winners was I thought to my self "How do you get that after 5 races???" But then I realized that the 6th race will tell you who was the fastest and slowest of winners. Great stuff.
This part is key and I don't think it get's the attention needed in the explanation. 5 races is minimum to race each horse at least once, and the 6th race is key because not only do you get the winning horse, but you also get the ordering of the groups. This allows you to determine the horses for the last round no. 7. Without the info from race no.6 you have no way to compare the groups between themselves.
@@Productlivity_ZZ I don't get why you even need the groupings though. Wouldn't the top 5 winners already be the fastest of their groups So you only need to race the winners to get the fastest horses.
@@izzy123123123 If the three quickest horses happen to be placed in group a, the third horse of group a could be faster than the winner of any of the other groups
No race needed. I'd butcher 22 horses. Not only would the remaining 3 become the fastest by default but I could stack up on delicious horse meat and land the Google job for my out of the box problem solving.
If the Google Executives hire you, they may apply a similar method when it's downsizing time and they have to determine who their three most productive employees are.
I got stumbled at "5 races give five " Fastest" Horses from each group, and the fastest among them is the first overall. But how exactly do I proceed with the 2nd and 3rd? " And I am happy about your explanation
@@pryo2460 I'm guessing you're the kind of person that yells at everyone when your Iphone Battery stops working, or your Electric Car starts to wear down, while shouting, "It's not broken! It was working just fine yesterday!" Jokes aside, but "Mechanical" doesn't mean "indestructible, or immune to wear..." it means, "Consistent Performance on Repeated Runs." Even two Ford 150's aren't going to perform the same forever.
Zakaria Rakhrour but six makes more sense. Five races to get the top five, then one race to see the top three. It doesn't specify that you can only find one winner/loser per race.
That doesn't work, though. If a horse wins its initial race, that just means that it's fastest than the 4 horses it raced. It could still be slower than all 20 of the other horses.
+Anon E. Mous except we don't know that. it might or might not be faster then another group. we just aren't given that information. all we have is 25 horses and we need to find the shortest amount of races needed to find the fastest three .
lnoah95 I think it's implied that we're removing the element of chance. Otherwise, you could just choose 3 horses at random, without ever racing them, and there's a chance (however small) that those are the fastest 3. In that case, the answer would be 0.
The solution to the question "Find a method that always determines the three fastest horses in 7 races" isn't that difficult. But the solution to the question "Prove that this method is optimal" is not easy (if even possible); the supposed "proof" that Presh gives in the video is incorrect/incomplete.
7 million views! Thank you! Here's a Microsoft interview puzzle I think you will enjoy. A cat is hiding in one of five boxes that are lined up in a row. The boxes are numbered 1 to 5. Each night the cat hides in an adjacent box, exactly one number away. Each morning you can open a single box to try to find the cat. Can you win this game of hide and seek? What is your strategy to find the cat? What if there are n boxes? Watch the video for the solution! ruclips.net/video/yZyx9gHhRXM/видео.html
I know this is a little bit late but i got a different result and i want to know if it is right.
My result was 6,because that in the same race that we used to discover the first fastest horse we could use to discover the second and third fastest horse.
Sorry if it is hard to understand, it's because i'm not english.
@@sandrathiel4475Sorry, no, you can't put more than five horses in one race.
You provided 7 million people with an inaccurate method.
@@MrJelle18 I was saying that,in the race that we discovered the fastest one,we could say that the second and third place would be the second and third fastest horses.
@@sandrathiel4475 How do you know the 3 fastest aren't in A.
Pick any 3 horses and destroy the remaining 22 horses. These three horses will be the fastest.
@@hayn10cause they are now 100% of the horses
I think that's how they hire people to some companies in Germany. This is why German online services suck.
Reminds me of Stalin sort :D
You score high in Machiavellianism. Congrats. ♥
😂
If I race 5 horses at a time there's no way I can beat any of them, horses are fast
Depends on the race. Run them through an Army obstacle course. I'm guessing they'll get hung up on the wall climbing, rope swinging and belly crawl parts.
It also depends on the length of the race. Native American runners could outrun horses in an endurance race. People are willing to keep going when animals would quit. So, if you're in shape for a marathon, you'll probably win.
@@flatebo1 Humans have an innate advantage in an endurance race vs most animals our species evolved to hunt critters by running them to exhaustion, that is to say we are naturally built for persistence hunting.
Truer words were never spoken
@@flatebo1 If the race is in high temperatures, then humans have an advantage. In colder temperatures the advantage is slimmer or maybe non existent.
@@seraphina985 have you ever rode a horse? He is really enduring, why do you think he was used as a transport mean in the past? Even if its an endurance race he will leave a long distance with the human so he he could rest and start again, but don't worry he will not get tired before minimum 5 hours
You race 5 horses, sell the slowest one, and buy a watch.
The rest is trivial
As if the buyers are on standby.
Best answer!
Expensive watch 😂😂😂
@@alessioandreoli2145 of course, a very accurate one, with stopwatch function included 🙂
this is a question not a business....
When you showed step three I was confused at first, then after a few seconds my brain understood and it BLEW my mind, what a great video
In a Google's interview the better answer should always be: Google It!
Hahaha
U are in
Someone hire this man
Ok then the answer of Google will be
: Plse Google another job for urself 😂😂😂😂
I agree
I got 6, then realised i hadn't considered that the second fastest might be faster than another group's winner. Nice one
Same 😐😐
Same
Same! I just thought the second and third of the sixth race would also be the fastest 🤦🏽♀️
what i dont get is why we need 7 races :
we must do five races for the five groups and from that we have found the 5 fastest horses of them all
secondly we just race those five horses and the first three to finish are the fastest three horses right ?
so that a total of 6 races !
P.S : no where in the question does it state that you need to know their ranking aka whos fastest of the three ...
@@shadi3993 But if the 3 fastest horses are in the same group your method will only identify one of them.
Here is how you can really solve this problem with 6 races: Let 5 horses race and label the 3 fastest horses from that group A, B, C. Assume that C is the 3rd fastest horse of all 25 horses and proove that by letting C race against the remaining 20 horses with 5 more races (6 in total by now). In the unlikely case that C is indeed the 3rd fastest horse, you've identified the 3 fastest horses (A, B, C) with 6 races. This is'nt an efficent method because it requires more than 7 races on average but sometimes it can solve the problem with only 6 races.
Last season in F1 we had 20 drivers and it took 17 races to identify the fastest 3 drivers
This is since they are human and not mechanical horses... so their times differ from race to race.
@@you2uber530 its a joke
It took 0 races
@@you2uber530 r/whoooosh
Lmao
Yeah this was great. I was having a hard time until I saw you group them in to a,b,c,d,e based off of the rank of their race. I was getting stuck on the three fastest horses being in the same race initially but the re-grouping did the trick. The visual really helps.
Feel you, i was stuck at a minimum of 11 races , because i wasnt regrouping and i thought id always have to let the top 3 of every race continue in case the 3 fastest horses happened to be in the same race at the start.
Outcomes are dependent on competition. This is a software developer question written in a bubble. And he didn't specify time , vs, outcome.
Race six runs the winner of each group, but it takes a minimum of two races to remove all the second and third place finishers. Therefore his solution is wrong.
It's not right.
I'm still a little confused. How can you order the groups based on how fast they were if you don't know their finishing times?
@@sasquatchrosefarts What do you mean? After race 6 you know no horse behind the third place horse can win, 1 horse behind the second place horse gets a chance for third place, 2 horses behind the first place horse get a chance for second and third place, the winner of the winners definitely is first place, leaving 5 horses to be checked so we finally know the second and third place.
When you identify a horse is slower than a horse that turns out to be slower than a slow horse then you also learn something about the first horse even though you didn't let it race again. Horses can be identified to be slower after a later race. If I misunderstood can you explain why you consider the solution to be wrong?
@@esco8778 Well you order the groups based on the place their "winner horse" made at the "winners race". So you have the 5 races at the start, after that the 5 winner horses will race each other. After that you order the groups by the finishing order of the groups fastest horse in the "winners race".
So if the winner horse of group C also wins the "winners race" group C will be the "fastest group".
Google: Which 3 of those 25 horses are fastest?
Meanwhile on bing: Leave only 3 alive
LMAO
LOL
That leaves you with the slowest three. You would have to have four left to give it a possibility
@@stevenmorris3181 but they would be fastest three too, right?
@@momamilosevic2465 Accidently killed the top 22...These things happen
I am not getting smarter, i am just upgrading my memory over puzzles for when i do a job interview in like, four or five years
If they ask questions like this you do NOT want that job.
@@TheRealInscrutable : Agreed. Speaking as someone on one of Google's hiring committees, we don't ask interview questions like this. :)
Job interview questions are simpler and shorter. Tom has 3 apples and Mark has 2 apples. Calculate the distance between Earth and the Sun
@@LivingAmnesia the answer is 7 cars
@Dpg Dpg You didn't calculate it, you looked it up. Sorry you don't pass
In order to calculate it, you must take drag a line from the edge of the earth to the edge of the sun
Sell all but 3 of the horses….
Those 3 will be the “fastest” horses you have. 😎
0 races needed.
Perfect score. 💯
Tell the horses they're being sold, and the three that don't get caught are the fastest.
#OrwellWuzHere
"And then they burst out in laughter."
When you apply to Google, but it's economics dpt and not IT
This is elon musk
for people getting 6 races, we’re trying to find a method which guarantees the top 3 horses will be found. after the first 5 races, there are 15 horses which could still potentially be top 3 overall, and after the 6th race, there are still 5 in contention, including some of those 15 who didn’t win their initial race.
I do not give a crap it is six for me
@@ilikefarting edgy
@@frommelow exactly how
@@ilikefarting "I dont care if i was proven wrong i still think im right"
this is why humanity has SO many problems
@@deemcgann1695 is this math problem a 1st world problem? also it was an obvious joke
As an employer, this is the answer I want: "We need a watch so we won't have to waste time with unnecessary horse races"
And it is open to interpretation, so some will interpret it in a way so you can label the horses and get 7 races, others will not label them and get 11 races. I would think a good answer would also be that the instructions are not complete
If you are looking to hire a programmer that's definitely not the answer you're looking for. If this were a simulation it would take less resources to program this solution in than to program in a clock. Programmers have to always think abstractly to find the most practical solutions.
except you actually make more money having more races, the house always wins.
@@bro748 That's also a very good answer. Also, if I hired a programmer, the only reason I'd have them out racing horses without proper equipment, would be as a punishment. Or maybe as a reward, depending on the programmer.
@@bro748 If you are looking to hire a programmer, you want to get a solution that does not blow up in your face when you get a 26th horse and the algorithm is just not working anymore. So you should phrase this as a more general problem with N horses. And then it gets interesting whether you want to find a generalizable algorithm that performs well on both low and high numbers or something that switches in between.
Adding a clock scales nicely O(n), which is the best you can get with this, so this is actually one of the best answers. :P
I knew it with cows, that's why I couldn't solve it.
usaco?
Eduardo Torres lmao
With cats, the problem is even harder than with cows!
LOL
The hamster version was straightforward....horses don't make no sense!
I got the setup exactly the same up until the 6th round, but at the seventh round I couldn't quite get the final 2 horses. I thought c2 could be faster than a2 and b2 and got confused how I'd identify the 2nd and 3rd fastest horses with such complexity. What I missed is that c2 could infact be faster than a2 and b2, but it's irrelevant as in that case there is still a1, b1 and c1 faster.
Looking at it from the other side and eliminating all the horses that can not possibly qualify for top 3 is by far the easiest and smartest solution. I didn't come up with that.
Exactly, also this implies how subjective is categorizing letters a to e in the same group as 2 to 5 since there is no timer, there's no way to know if A2 is actually faster than E1 because E1 could've for example finished in 5 seconds, A1 7, but that doesn't mean A2 couldn't have finished in 4.9 making it actually slower than E1
I still didn’t get it. C2 could in fact be faster than a1,a2, b1, b2. So we are missing that comparison.
But you said that is irrelevant so just want to understand what I’m missing.
@@bhupendrasatpathy C2 can never be in the top 3 because it is already slower than at least 3 horses: A1, B1, and C1. It is slower than C1 because it lost the race with all the Cs. It’s slower than A1 and B1 because A1 and B1 are faster than C1 since they beat it in the race of the fastest among groups.
But all of B2 3 4 5 just means they slower than B1.
All pf A2 3 4 5 just slower than A1.
Theres no line where B2 3 4 5 faster than A2 3 4 5 because groups are ranked by heats.
Say A1 is 10sec.
B1 is 9sec.
A2 3 4 5 could all be 4sec.
B2 3 4 5 could all be 8sec.
Nm
I forgot how many horses we were going for.
This only works for spherical horses in a vacuum.
best to make them frictionless, spherical horses in a vacuum :)
@@backwashjoe7864 lmao how do they run?
And all collisions are elastic
ideal horses too, so individually they have no volume but occupy a finite space together.
They also do not attract each other so little ponies won't be made
@@backwashjoe7864 I’d argue that being frictionless and in a vacuum were redundant, but…
Instead I’m going to add that they should also be made up of non-baryonic matter so as not to be affected by virtual particles.
Frictionless non-baryonic spherical horses in a vacuum.
I think it also depends on who is keeping track and actually watching the races. Alice or Bob would be able to figure it out. Charlie is either selling popcorn or shoveling the stalls.
the answer he gave is not correckt! The problem is that you dont have a time/ watch nor a distance! If you let 5 groups run, and you take the 1st. of each group, you can not be sure if these horses are one of the fastest 3. 1st horse in group (a) maybe can run just 2Kmh and the 1st of group (c) can run 40kmh while 2nd of group (E) can run 60 kmh! . you have to have a constant to really get the 3 fastest! My logic: So you let the first 5 race, then you let the first second and third place on the track and get 2 new ones in, every race the 4,5 will be replaced. with this method you have always a constant for first, second and third place without needing a watch ! each race, 11 in total , will naturaly eliminate 4 and 5 place ! imagine this: lets say the first group are the 5 slowest of all, so the slowest up to the 5th slowest of all. in the second race the 4th and 5th place gets replaced with (in this simplyfiyed case)faster ones, and maybe one is the fastest of all and the other one the 2nd fastest of all. they will be 1st and 2nd place of the 2nd race and they will stay on this position. the 3th will be found after all other races are finished. You can switch the start situation, that the first 5er group have two of the fastest of all in it, so the first and second place will be set after the first race, and again the other races and just to find the real 3th place.---- i think its 11 races ! 😃
hahahahaha~
Goddammit Charlie! Who keeps hiring this guy?
@@crashoverwrite5196 > 1st horse in group (a) maybe can run just 2Kmh and the 1st of group (c) can run 40kmh while 2nd of group (E) can run 60 kmh!
No. If you look carefully at the solution, the groups "a, b, c, d, e" are named *after* the 6th race. the winner of group a is the 1st place of the winners, the winner of group b is the 2nd place of the winners, so by definition a_1 > b_1 > c_1 > d_1 > e_1 - and then, you have a tree from the winner, you only want the top 3 so you can eliminate any that are 4th or more from the winner, and you are left with exactly 5 horses to race and find the 2nd, 3rd overall
@@sodiboo You dont know how fast they are! so my asumtion is valid. You need a fix point for the 1st,2nd and 3rd place. my comment was based on the crux, not the solution he gave.
Animating all those letters at 1:29 must've been a bitch.
I know I will never work at these kinds of places, but to see the thought processes in these puzzles is awesome. Training yourself to look at the same real world problem in different perspectives to solve real world engineering/ self reliance problems are invaluable. Great video
Never say never 😊
@@Neil.Menezesthats an oxymoron
@@pickleism253 fine, never say never except to say the phrase 'never say never'.
This was before logic and rational thought was deemed racist. Now states like NY are outlawing 'tests' like this in the pursuit of 'equity' and we're slowly starting to fall behind other countries like China.
I was getting 9 as a first effort but then as soon as you said 7 I was able to figure out how to get 7… It’s truly wonderful how much knowing that something can be done can influence your ability to do that something
Same here
I did the same thing as well
exactly the same words
Just like our full potential as a living being. Very few people know this and it then makes it impossible to achieve it until you are taught it by someone who has achieved their full potential. Most believe it is impossible only because they don't know how.
"Knowing where the trap is-that's the first step in evading it" ― Frank Herbert, Dune
Kill 22 of the horses, then by default the remaining 3 must now be the fastest
So you're proposing mass destruction of company property during an interview?
@@brianvalenti1207 I think you mean "streamlining equestrian surpluses"
You would never get a job with that I guess ...
@@abhiramp7094 Hey, this man just solved how to prevent a food shortage. He must be hired.
@@Aniaas1 Cultural fit: 4.0
I just want to say, your question is exponentially better than the common interview question. Me sitting here with tired horses, one fell in a race....
there's nothing exponential about this. Exponentially doesn't mean "a lot more".
@@fmobus i bet it gets girls at the club reeeal hot when you tell them that
@@fmobus you do understand what an exponential equation looks like right? Calling something exponentially better is in comparison to calling it linearly better, where something exponentially better reaches its limit faster than something that reaches its limit linearly. So “exponentially better” means “a lot better” and “exponentially more” means “a lot more”…
@@fmobusAre you the type of kid who tries to correct a teacher?
It’s a classic quant question
If I was doing a job interview at Google and they would ask me this question , explaining I dont have a watch , I'd be like "Smartphone DOH , you know howmany stopwatch apps there are ? Are you sure you are from Google , geez"
I realized pretty quickly that 6 races isn't enough, at which point I decided it's to complex and might require like 20 races or something.
Glad to see such an elegant solution, thank you!
So close
That was my exact chain of thoughts
I got 11 races lol
Why isn’t it 6? When you run the 6th race full of the winner horses, the 1st 2nd and 3rd places would be the fastest horses, right?
@@galacticlava1475
2nd place horse in 6th race might be slower than second place from the race the first place horse raced in.
for all we know, that specific race might have consisted of the 5 fastest horses in the herd.
thats why we take the 2nd and 3rd horse from that group, because even if 2nd place horse in race 6 is faster than 2nd place horse in race with the overall fastest horse, it may still be slower than 3rd place horse in that race.
you consider the 2nd place horse in the race producing the 2nd fastest in the 6th race as u want to find the top 3 horses in speed, as 2nd place horse in 6th race might be faster than 2nd place horse in race that produced the fastest horse.
Lastly, u include no3 from 6th race as all these extra horses included can still be slower than it, and if so, u still have ur fastest 3.
Lazy programmer to Google: `Just run 11 races and leave optimization for problems with 25 million horses.'.
I don´t know if you are even allowed to label the horses, since it is not said you can. And if you can´t label the horses, 11 horses is the right answer, because every race exactly 2 horses can´t go on for further testing. It all depends on the interpretation of the questions.
yeah i think 11 races is the common answer
@@JonathanMandrake thats the "you aren't good enough" answer
That's sometimes a good approach - when the optimization is a complex process. But in this case the optimization is simple enough that there's no reason not to use it on small numbers of horses too.
@@a0flj0 Sure, but without knowing this answer beforehand, who is actually going to think about this? Google should want to see that you know how to solve a problem in front of you, not necessarily do it in the best and most accurate way possible.
In real life you'll have resources to assist you in researching what the best way to do something is. You're not just given an issue and told to fix it immediately without any outside help.
The mathematically optimal answer shown in the video is satisfying to discover! The use of this method seems to be dependent on the specific numbers of total horses and horses per race presented in the problem, though...
While trying to solve the problem, I came up with an algorithm which has the advantage of generalizing smoothly to any variation of the problem with M horses in the pool, N horses allowed per race and K spots at the top of the field, where K ≤ (N+1)/2. Here it is:
Start: Have a group of untested horses, and a "leaderboard" with K spots. All M horses start in the untested group.
Place N untested horses in the lineup for the next race. Run [Race Type A].
[Race Type A]
Race the horses in the group; place the top K on the leaderboard, in order, and throw out the (N - K) horses at the bottom.
Now add the horse in Kth place on the leaderboard, and (N - 1) untested horses, to the racing group. Run [Race Type B].
[Race Type B]
Run the race with the current group. The next step depends on the outcome of the race:
If the horse currently in Kth place on the leaderboard wins the race, eliminate all the other horses in the race and set up the next race with the Kth horse and (N - 1) new horses.
No change to the leaderboard. Run [Race Type B].
If the Kth horse comes in 2nd or worse, eliminate the Kth horse and all horses in the race who were slower. If K or more horses beat the Kth horse, add the top K finishers to the next race;
otherwise, add all horses who placed above the Kth horse.
Add the other K-1 horses on the leaderboard to the race. Fill the remaining spots in the race, if any, with more untested horses. Run [Race Type A].
When no more untested horses remain, the leaderboard contains the K fastest horses.
I ran a Monte Carlo simulation of this algorithm with 10,000 trials, and for the values given in the problem (M=25, N=5, K=3), the expected number of races is just under 8. Not bad!
This is interesting!
What does a simulation find in the situation (M=25, N=5, K=2)?
By the way, the approach of the video's method isn't necessarily the optimal one though. For example, in the case of (M=4, N=2, K=2), the video's approach would suggest that 4 races is optimal. However, it's relatively easy to find a method that takes 3 races in 1/3 of all possible scenarios, and 4 races in the other 2/3 of scenarios, thereby resulting in the optimal method that requires 11/3 = 3.6666... races on average; which is more efficient than 4 races. And maybe 7 races is optimal for the case finding the Top 3 out of 25 horses, but the video failed to give a solid proof of it (the "proof of minimality" in the video is flawed).
LOL, your generalized algorithm actually results in the optimal method for the (M=4, N=2, K=2) case! :-)
@@yurenchu when I run your first scenario, I get right around 7.5!
Google doesn't actually use these sorts of logic puzzles in interviews. At least they haven't for the last 20 years or so. They ask algorithmic/coding questions, system design questions, questions that check technical knowledge and the like. There are lots of videos posted by Google about how to prep for a Google interview. But tl;dr, re-read Introduction to Algorithms (AKA the mobile book).
This _is_ an algorithmic question. To be more specific, a sorting question.
So something we talked about in economics the other day is how if you take too much time to solve a problem in the most efficient way it can be more costly than just doing the inefficient way that instantly comes to mind. The answer I came up with was 11. I know it’s not even close to 7 but if I was able to have a system ready in 10 seconds that works I’m still proud of that answer
Edit: very happy with the conclusion of the video though. Super cool thought process!!
But then again, if you spent a bit longer thinking more efficient solution you would save 4 races.
That's 20 races you would have saved those mechanical horses from doing. If you do this more than once the maintenance costs are going to pass your thinking time.
well, this makes sense to me, but i think that if u create the best solutions u will be faster to create the best solutions next time and everytime, and maybe just at the start is harder, maybe
Depends on the context. If it's a one time thing then sure but if you have to compare horses thousands of times then you have to use the better answer
I think he is wrong, he would not be hired by Google. (About the answer in the video)
I think he assumes that all horses finish at equal time intervals. The fastest horse, without any timing available, doesnt say anything aboit the slowest horse.
The horse that finished 2nd to the fastest of all horses (the fastest group) could still be slower than the horse that finished 2nd to the horse that finished #5 in the 'race of winners'
So without also determining which are the slowest horses this is just assumption.
I'd say one need at least 8 races therefore.
11 is making sure anyway. Cross reference and you can make the whole top25.
@@Coen80 With 7 races you always identify the three fastest horses
2 million views! Thank you!
Thank you!!
Can you please change your theme to a black board
Thankyou
What if the 5 horses in group "a" were actually faster than all other horses ... How you will solve it without a watch
I guessed 7 does that count
@Moh Taw,
- "What if the 5 horses in group "a" were actually faster than all other horses ... How you will solve it without a watch"
That's exactly why horses a2 and a3 also participate in the seventh race (along with b1, b2 and c1), which they will win in your scenario.
As a project manager: “Build a stopwatch then we can time the horses”.
Doesn't work in scenarios where a stopwatch can't exist, e.g. in distributed systems. Ref Logical Clocks, from the 1970s.
@@davidhawley1132 ok thanks David. I’ve noted your assumptions. I’ll speak to the client in relation to your concerns of legacy constraints and find out if these are going to be problematic... I have liased with them and it turns out that we live in 2021 and we’re dealing with a startup. It’s also apparent that there are already timing tools available on the market, so I’ve conducted build vs buy analysis and it makes sense to procure the stop watch instead. $3 from Amazon. Since finding the fastest horse is going to be repetitive it also means we can make huge time savings in efficiency across the product life cycle and there is less risk involving the horses. So I’m grateful for your input as this led to a cost saving of tens of thousands. I’m giving you a pay rise and I’ll write a case study based on your input for lessons learned in the next knowledge sharing workshop so we can all benefit from your unrivalled wisdom.
Lol obviously this was not a serious solution dude.
@@lukeh7854 Code bootcamp grads would try to pull the move you suggest. My Comp Sci degree was in the Math dept, and we studied problems like this.
@@davidhawley1132 We have "stopwatches" that work just fine in distributed simulations. It's a very difficult problem with an embarrassingly simple solution. (But, I'm also talking software - getting circuits in time without a common clock is something I'm not going to mess with, unless performance is not a concern.) (And who wants to work on something where performance is not a concern?) :D
@@xnadave It's a classic problem with a good-enough solution. But from the comments, I bet few posting here even know there is a problem.
I had to answer this question to get accepted into university. I got the right answer by making a mistake that led to the same answer. I got accepted 🗿
I'm the interviewer, I'm now wishing I'd never asked this question after hearing 27 variations of how to solve it. It's late Friday and I want to finish work. . . . . . .First one to tell me, 'Sell 22 of them, keep the cash!' Gets the job! I don't care, I'm retiring next week.
I said shoot 22 horses but this is so much better!
Hey its been a few weeks. How is the retirement going?
@@apollyon1 So did I!
What has happened about retiring next week? Are you so busy making the papers ready that you forgot you had announced your retirement publicly on RUclips?
@@apollyon1 ingenious :)
I get it, but jumped to the initial conclusion of 6 races. I’ll say that I knew I was wrong and that I was missing something.
Good lesson here, I’m glad I watched.
i did the same thing
i also end with 6 races
I was about to, but then I thought "what if all 3 fastest horses is in the same group? This can't be the solution." I couldn't figure out the right answer though.
Mäx-Tick absolutely not
Sorry, Mäx-Tick, but someone with a background in math posted a solid proof in this comment section 8 months ago, showing that it's impossible to design a method that is guaranteed to identify the three fastest horses every time in just 6 races.
I paused and played the video only after solving the problem, now most important question is how much time does Google give in interview? 😁
2 minutes only
@@writer_gupta_ji wow, that is really hard
@@writer_gupta_ji*(edit this method is flawed as pointed out by others below me)*
I solved it in less then 2 minutes, tho I solved for 6-7 (ill explain that later) using a different method which is race 5 horses then race first place against 4 new repeat 5 times keeping track of the second place horses at the end of the sixth race if the fastest horse of race 5 isn't first or second then the top 3 are the 3 fastest else race 2nd place horse of race 5 against 3rd place of race 6 and first place of race 7 is 3rd fastest. This is a better solution because there is an 84.43% chance u will only need 6 races. *>>> AGAIN THIS METHOD IS FLAWED SO PLZ STOP COMMENTING THAT IM WRONG, I AM WELL AWARE.
i even solve this question only reading at thumbnail and then open video 😅
@@nothingnothing1799 Maybe I am missing something, but how does your method work? Imagine if your very first race had all three fastest horses in it, and you basically eliminate the 2nd and 3rd fastest ones at the very beginning, and never return to them in your procedure above, how do you get to the top three?
0.
Destroy 22 of the houses. The last 3 are now the fastest. (You can also sell them to be more humane but in a hypothetical scenario where you HAVE to know which ones are the fastest)
This is the smallest amount of races required to determine the 3 faster horses.
There are ways to learn this in 0 races. Consider disassembling one horse, reverse engineering its software to find where the horses movement rate is located, then hack the other 24 horses and learn their rates without the horse moving.
The question is "The three fastest horses" not "Rank the horses by the fastest" Another way is to destroy all but three of the horses, and then you automatically know the three fastest horses.
Technically correct, the best kind of correct.
For machines... Yeap.
"...destroy all but three of the horses."
*hold up*
@Heberth R. Vc aqui? Pensei que só assistia o Tadeu
but how do you choose which one to destroy you don't know their speed relative to other horses and hence you cannot put a threshold
Anyone that can use the logic this guy does to answer an interview question like this is probably too good for the job, anyway.
I dont really think so maybe he is or maybe he is not
It's a programming problem, ie why Google would ask it to a potential employee. It doesn't really matter what horse is the 1st, 2nd, and 3rd fastest, they're trying to get you to come up with a problem-solving technique to identify what those horses are. Programming is all about developing these techniques, the actual numbers/results don't matter so long as the techniques are correct.
@@digiquo8143 Agreed. There are all kinds of videos on you tube (including several other of the 25 horses problem) that pose intriguing puzzles that challenge your brain to think, find possible solutions, look at possibilities, and come up with sound techniques that will work in any case.
Someone else here mentioned six races, and when I worked through his logic, I found out that he was in fact correct, but it only worked in 2 possible cases, both of which were statistically insignificant. The correct technique, as shown in this video, will hold up rigorously for all cases.
@@digiquo8143 results matter if there are many results, i.e. many applicants for 1 job
@@nobodyknows3180 nah man
Shoot 22 of the horses. By definition, you will be left with the 3 fastest.
That's a tad psychotic, but damned if it isn't the sort of off-the-wall answer Google would hire you for.
It was 'RACE the horses', not 'ERASE' !
You had one job......
@@DreadX10 race horses in sets of 5 and kill the ones not 1st place, 5 races.
Race remaining horses and kill 5th and 4th place. 6 races.
You are now certain that the 3 horses you picked are the fastest of the original 25.
Got to laugh at this one.
@@ozzygilliam9194 ...but two or three of the fastest horses might (by chance) be in your group of 5, so then you would have shot one or two of the fastest horses.
you could race them in groups of 5 and the two slowest are retired (see Blade Runner).
I thought it would be 6 because the 3 winners of the race between the 5 fastest horses would be the 3 fastest horses🤦
Exactly
@@MontagoDK Right but what if more than one of the top 3 was in a group together?
Horse C in 3rd place beats 4th and 5th place horses hes faster than the eliminated horses even tho he never raced them from the 4th place group and 5th place Group at the beginning because the 4th and 5th already beat the 8 horses who got eliminated its obvious that horse in 3rd place faster than 10 so far counting 4th n 5th place. Horse 1st and 2nd beat everyone in their group the eliminated ones from there group never raced the horse 3rd place horse 1st and 2nd could had beaten someone faster than the horse in 3rd place now which is the eliminated ones and couldn't race to know for sure. so we can't know for sure. All we know is 8 is left we can only pick 5 so 85% chance we need 7 try 12.5% chance we need a 6th try only.
I was wondering how you could find the 2nd and 3rd fastest overall.
The other method I came up with was more of a slider and would require 11 races, which works but isn't minimal.
(Line up the horses, race the first 5, keep the top 3, add the next two horses, race and repeat)
That was my first thought too
I would race 5 groups of 5 and aswell remove to two slowest in each. Leaves 15, repeat, leaves 9, then 5 and of those it's the fastest three.
11 races
(5 x 5horses, 3x5, 2x 5, and last race with 5)
@@aramisortsbottcher8201 Wrote out a very similar thing about a day after you did. I just continued to take the top 3, even when it left we with 9 instead of 10. When I got to 6, I ran one race with 5 and left one behind. Then ran it again with the last horse eliminated and the spare 6th plugged in. For a small touch of extra certainty.
@@robertdavis5693 I first had this idea too, but when the point with nine horses was arrived, I raced the first five and after that the next four, but filled the fifth slot with one of the previous race. That way I could sort out two of five instead of one of four horses. This left me with 5 instead of 6 horses, so they all fit into one race :D
Came up with the exact method
The smartest person seems to be the person who comes up with the question with the answer in mind. Honestly takes a lot of brain power to come up with a problem and solution a lot of people cant solve
@@Naughty_Squad Which is why mathematics and linguistics are bad bedfellows.
@@Naughty_Squad I disagree, the problem itself is very simplistic and makes sense what is being asked. All you need is logic.
@@commandercaptain4664 Strange bad fellows
After watching this video, horse neither looks or sounds like a word anymore
wow whar a relief I'm not the only one!
Buffalo buffalo buffalo buffalo... Oh wait, that’s something else.
Vsauce! Michael here. What you experienced is what we call "Jamais Vu". It's what you experience when a word has been spoken so many times that it loses its meaning. Thanks for watching, this has been Michael... or am I?
yeah its called semantic satiation
I relate to this on an existential level with other words.
you can do it just in 6 races, 5+4+4+4+4+4=25
first race, 5 random
second race, best of first race and 4 more
do it in 6 races, you must mark where the horses are when the winner of the first race reaches the finish line.
I came up with 8 races and proved it using pennies as horses. I didn't consider the 7th race could determine both the 2nd and 3rd fastest horses by eliminating the entire field of horses that only had 1 winner (race 1 -5). Thanks for posting.
Even i got 8 as my answer
I got 8 as well, but then when I saw the actual solution was 7 I immediately noticed we repeat the 7th and 8th race with the bottom 2 horses that have absolutely no hope of ever winning a top spot, and so worked out the best solution from there. So I worked out how to do it in 7 after seeing the solution was 7, but before seeing how to do it.
@@MrSupdup for some reason i kept looking for the top 5 fastest horses haha
needed them place bet odds i guess
I also got 8. It is easier to find a solution that is not optimal, but that is fine in computer science. After you found a solution that works, you should seek to find a better solution, if necessary. 👍
I got 6, but I was wrong. I assumed the first 3 winners of the 6th race were the 3 fastest, without considering that they hadn't raced yet against those of higher level
It would have a lot of pressure since it’s asked during an interview
And that's when you pull out your horses
I suppose that's the point in the interview more than any other thing.
How about n horses? And we are only allowed m horses in each race. Is there a general formula for this?
Great question! Thumbs up, so Presh can notice it.
and we need to find the top x horse
I'm not sure about all combinations of numbers of horses etc, but the method in the video works for ¼(k-1)²(k+2)² horses where you can race ½(k-1)(k+2) horses in each race.
I think it's more adequately written as (½(k-1)(k+2))² horses and ½(k-1)(k+2) horses per race.
That makes it more obvious the former is the square of the latter.
+codebeard Is k the number of horses we're trying to find, or is it just a coincidence that k=3 for 25 horses? If k is the number of horse we're trying to find, the minimum number of races is ½(k-1)(k+2) + 2.
2:28 couldn't you skip the last race? If you race the top 5 horses you immediately know the top three since, like you described earlier, we get a list of the order that horses finished, not just the winning horse only.
Never mind I think i see why now
Yes its 6
@@MbtaVideoClipsplease do explain why its 7
@@30GBofnegativeluckso basically if you think about it the fastest horse in group a could be (for example) 25 mph and the second fastest in group a could be 20mph but it doesn’t get send to the fastest race bc it wasn’t the fastest in its group and we can say the fastest group e horse was only 19 mph but the fastest in its group
Something i think could be explained as i didnt initially think of it and other ppl dont seem to have noticed is the fastest horse of a given race could still be slower than the slowest horse in another race. Just depends on luck of the initial groups
I was thinking about this too, but with this method, it won't matter. Even if the horses are arranged strictly by performance (the slowest in group A is faster than the fastest in group B), it will just mean that the winners of race 7 (2nd and 3rd fastest overall) will all come from group A
Better to put forward with timings .... I mean minutes and seconds
There is another problem in every race the horses are randomly asigned and you don't have the time so by that definition in one of the groups you can still have the 1 fastes hourse and 2nd fastes from all 25 hourses, but by this method of elimination you have already decided that the secound fastes hourse is no longer competing and the 1st fastes hourse is compeeting to the once left from other groups.
Exactly. In some competitions, the best of the best go straight to the finals, but it does not mean the newcomers aren't stronger.
@@mitsuriKoi which is why the whole mechanical horse with a pre-designated time that always stays the same made this make a LOT more sense for me!!
Better solution : sell all the horses and buy a motorcycle if you need speed.
Edit : dont forget to buy a watch if you don't have one
What u gonaa do with speed and time when u don't have brain
@@BabyIshii i have a brain which can atleast understand a joke which u couldn't. 🤦
@@E7visuals ohhh my bad...I didn't know ur sense of humor died as well
that's what how you solve the problem after you get the job. *because Google $$$$$
I'm genuinely curious what percentage of people still have watches. They seem kind of redundant now TBH. Edit - and by have I mean "actually wear and use". Buried in the couch or the back of a junk drawer isn't what I'm talking about.
The answer is ZERO races! Just kill any 22 horses (the stated rules do not preclude this option) The remainder will always be the 3 fastest horses.
Adam Mangler But you still wouldn't know which of them is the fastest
Of course you would - (of the survivors anyway!)
*_:0)_*
Adam Mangler you just killed your top 3 fastest horses and spared the slowest of em all... you never know :p
forest wow He's saying that if they're are only 3 horses left then they have to be the fastest three horses because all the other horses are dead.
Adam Mangler...you get the top prize for thinking outside the box...also for being a leader among the ratards :-)
Shoot 22 horses, sell them for glue or Tesco spaghetti bolognaise then you have the fastest 3 horses
I understood how you got the three fastest horses in seven races (in a few minutes after I finished watching the video). I drew the diagrams you drew, and I began eliminating horses.
Its not correckt! The problem is that you dont have a time/ watch nor a distance! If you let 5 groups run, and you take the 1st. of each group, you can not be sure if these horses are one of the fastest 3. 1st horse in group (a) maybe can run just 2Kmh and the 1st of group (c) can run 40kmh while 2nd of group (E) can run 60 kmh! . you have to have a constant to really get the 3 fastest! My logic: So you let the first 5 race, then you let the first second and third place on the track and get 2 new ones in, every race the 4,5 will be replaced. with this method you have always a constant for first, second and third place without needing a watch ! each race, 11 in total , will naturaly eliminate 4 and 5 place ! imagine this: lets say the first group are the 5 slowest of all, so the slowest up to the 5th slowest of all. in the second race the 4th and 5th place gets replaced with (in this simplyfiyed case)faster ones, and maybe one is the fastest of all and the other one the 2nd fastest of all. they will be 1st and 2nd place of the 2nd race and they will stay on this position. the 3th will be found after all other races are finished. You can switch the start situation, that the first 5er group have two of the fastest of all in it, so the first and second place will be set after the first race, and again the other races and just to find the real 3th place.---- i think its 11 races ! 😃
@@crashoverwrite5196 Yeah. My first thought was to do what you described, but I abandoned that quickly because I knew it would require more races than necessary. I figured it out quickly enough. The answer is 7. Just pay attention to what is said in the video.
Now stop eliminating horses before it's too late!
Have you considered career in CS and interview with Google?
@@crashoverwrite5196 What about if you ran the first 6 races as described in the video but for the seventh, ran all 2nd place finishing horses from the first 5 races against one another and for the 8th (and final race) run the top two horses from both of the 6th and 7th races?
My first idea was taking the 2 slowest off each time as well. Ended up with 11 races however I appreciate it can be done in less.
But upon further contemplation I figured if I'm at Google I'd just pick 3 random horses. The truth is what we say it is.
Same idea here. A way with less races may possibly exist, but designing and verifying its veracity likely takes longer than the few races that it might save us, so we better simply get those horses to run.
One could ask the interviewer, "is the question here really about the best algorithm, or is it about figuring out which are the three fastest horses in the most efficient way?"
yes man. you got it with 11.
The problem is that you dont have a time/ watch nor a distance! If you let 5 groups run, and you take the 1st. of each group, you can not be sure if these horses are one of the fastest 3. 1st horse in group (a) maybe can run just 2Kmh and the 1st of group (c) can run 40kmh while 2nd of group (E) can run 60 kmh! . you have to have a constant to really get the 3 fastest! My logic: So you let the first 5 race, then you let the first second and third place on the track and get 2 new ones in, every race the 4,5 will be replaced. with this method you have always a constant for first, second and third place without needing a watch ! each race, 11 in total , will naturaly eliminate 4 and 5 place ! imagine this: lets say the first group are the 5 slowest of all, so the slowest up to the 5th slowest of all. in the second race the 4th and 5th place gets replaced with (in this simplyfiyed case)faster ones, and maybe one is the fastest of all and the other one the 2nd fastest of all. they will be 1st and 2nd place of the 2nd race and they will stay on this position. the 3th will be found after all other races are finished. You can switch the start situation, that the first 5er group have two of the fastest of all in it, so the first and second place will be set after the first race, and again the other races and just to find the real 3th place.---- i think its 11 races ! 😃
I started out saying 10, then realized I could eliminate the 4th and 5th place horses from the first 5 races. Just as I was patting myself on the back and saying the answer was 8 races, the rest of the video played. Guess I'm not as smart as I thought I was.
@@johns9652 6 races, because reality isn't so comfortable; you have the winners run, and the winner of winners is the fastest with 2nd & 3rd place marked behind it.
If UFC, or other events operated like y'all are suggesting, then all competition sports should be eliminated, and all champions should return their trophies.
I used that approach initially. 25 horses to start, 5 races, two slowest eliminated each race, 15 horses left. 3 races, two slowest eliminated each race, 9 horses left. 2 races, but with nine horses, you'll have one group of five and one group of four, and you need to keep the fastest three from each group, so eliminate the two slowest from the group of five, but only one from the group of four. This creates somewhat of a dilemma, as you have already run 10 races and you still have 6 horses.
Run a race of five, with the sixth sitting out. Knock out the two slowest from that group, leaving the three fastest and with the one that sat out, that's four remaining. That was 11 races so far
The final race will show who is 1-2-3 out of the original 25. But that was 12 races.
break 22 horses, the ones left are now the three fastest horses
yes, i will simply run 5 death races, and kill 22 horses out of the 25
@@storm0fnova No, he means kill all 22 before even racing. His answer is 0 races, because it's a foregone conclusion that the 3 survivors would be fastest.
Shoot one of them and tell the rest to get back to work. You'll have 24 horses going just as fast as they possibly can!
Another Captain Kirk type solution! :)
HAHA
Actually it's 6 races. Your original question says; "What is the minimum number of races needed so you CAN identify the fastest 3 horses?" You didn't say "MUST".
Therefore; (H3 = 3rd finishing horse from R1)
R1: H1-5
R2: H3 + H6-H9
R3: H3 + H10-H13
R4: H3 + H14-H17
R5: H3 + H18-H21
R6: H3 + H22-H25
- H3 was faster than Horses 4-21, we now have our 3 fastest horses.
gg
LOL
goat
The word "can" implies an ability to perform a task with a (near) 100% likelihood of success.
If I'd say "I _can_ multiply two two-digit numbers from the top of my head within 3 seconds", then my claim implies that I can (pretty much) _always_ perform that multiplication within 3 seconds, and not just in the unlikely "lucky" scenario that one of the two-digit numbers happens to be the number 10.
Likewise, "number of races needed so you can identify the fastest three horses" implies a number of races in which you'd generally manage to successfully identify the fastest three horses. (The word "minimum" then asks that from all numbers of races in which you'd generally succeed to identify the fastest three horses, which number is the minimum number? So, for example, suppose you master three different methods that successfully identify the fastest three horses: the first method generally requires 9 races, the second method generally requires 8 races, and the third method generally requires 7 races; then the minimum number of races in which you can identify the three fastest horses, is 7 races.)
bruh then you can do it in 0 races. just pick any three, you'll be right sometimes
You can actually do it with 0 races if you just guess
I made it seven, but it was a bit of a guess. I got as far as racing five groups of five and then racing the winners, giving six races. After that I figured one more race should do it, but more by intuition than reasoning. The step that eluded me was that any horse beaten by three others could be eliminated, although that seems obvious, now you mention it. I think I would have got there eventually, but it's easy to say that and my working would most likely have involved some extra unnecessary steps. Thanks for the clarification 🙂
The most important information of this question is the fact that they are mechanical horses.
Damn I feel like I got pretty close. I immediately had 2 ideas. The first method was to just keep the top 3 and add another 2 new horses until complete, which is 11 races. The other idea was to use a bracket and I figured out to get the fastest overall requires 6 races, but for second and third place I just did the same thing and compared the second fastest from the winner's group to the second fastest from the initial groups which gave me 8. Intuitively I could feel this was wrong though, and if I had actually wrote any of this down instead of trying to solve in my head I probably would have done a lot better, but even still I'm not sure I would have seen that last step. What I've learned here is knowing when and how to use grids can be extremely useful, and that I should note things down.
Google watching these videos to figure out which questions to leave out of their next interview:
I was originally going to ask why 6 wouldnt work, but the visuals really helped me find that out. Great video!
@@jasonbernard5468 it’s not a solution if it doesn’t work in all cases
@@jordantyler148 yea my ideaonly works in one case anyhow otherwise it needs 8. I was wrong when i commented before
@@jordantyler148 actually Im not sure, imight have been right, but not the way i thought. You say it isnt a solution unless it works in all cases, but what if i had a method that gives the answer in 6 rounds if and only if all top 3 horses are in the first group, and gives it in 7 rounds if not? Im not sure if I have such a method, but if I did it would be a better answer than the video, would it not?
@@jasonbernard5468 It would be interesting if you had a method that sometimes worked in 6 rounds, but occasionally needed an additional round, but only if you knew if it was done after 6 rounds, not if you do the 6 rounds and you dont know if its right.
@@jordantyler148 There is a method that works for all cases and in best case is only 6 races, but worst case 11. It does not involve doing 5 races with all horses like this. You do an initial race which gives you an initial #1, #2, #3. You take you 3rd place and race it against 4 new horses. If any are faster than current #3 then you race the fastest 3 of those against your current #1 and #2, giving you a new set of finishers. And continue with whatever the current #3 is. Best case, they are in order and you do 1 initial race plus 5 additionals where none are faster that that first #3, so 6 total races. Worst case they are in reverse order. and you will have to re-race after each of the 5 races, so 11 max. The question is to find the minimum, not the most optimal solution. So this would be a "better" answer.
Lazy solution: run 11 races and eliminate the 2 slowest each time. 22 eliminated, all that's left are top three
This is the correct answer because there is no fact that the 2nd, 3rd, and 4th heats are any faster then the first three winners.
This is what I came up with
I get a paper and label the horses so and I based my time on the fastest one in each group so then I can tell which ones are the fastest Ones ( I will race the fastest of each group to see which one is the absolute fastest and based the speed on that one
My best answer would be run a horse race gambling game while doing the laziest solution done everyday making sure to keep horses always healthy, do it for 6 months to a year, you get money and you get the best horses trained after months of running 🥳.
Since the fastest horse now wont necessarily be the consistently fastest horse, you need reliable statistics!
That's actually the correct answer, because it's said you don't have a watch, so you could never compare the times of each group's fastest horses, 2nd, 3rd, etc (one race at a time, remember?) The 2nd place from group A may be faster than group B's fastest horse.
The guy who made this video may be good at math, but he surely lacks text interpretation...
I was started to write a long comment throughly explaining why I disagreed with your answer. I read through the comments before responding to see if others came up with the same conclusion. Upon reading those comments, others had realized what I had, and explained how they were confused and why your answer made sense. I see where I was wrong, after reading the comments.
If the 4th, and 5th place horse from the the 6th race were the fastest in their group, that entire group is eliminated. Even the 3rd place horse would be the very minimum possibility. Excellent mind puzzle, made me think thank you.
I was starting to do the same - only when I scanned through comments I was only surprised that everyone except me was in agreement. If the horses are divided randomly into even groups of 5 - how do we know that all of the top 3 are not in the same group and therefore 2nd and 3rd place are not automically incorrectly eliminated?
No it actually is not. On placing yes, but with actual times it can be very different. For instance A1 can run 1min but A2 needed 5min and the whole group E can still lie between them
Horse C in 3rd place beats 4th and 5th place horses hes faster than the eliminated horses even tho he never raced them from the 4th place group and 5th place Group at the beginning because the 4th and 5th already beat the 8 horses who got eliminated its obvious that horse in 3rd place faster than 10 so far counting 4th n 5th place. Horse 1st and 2nd beat everyone in their group the eliminated ones from there group never raced the horse 3rd place horse 1st and 2nd could had beaten someone faster than the horse in 3rd place now but they got eliminated so can't race horse 3rd place so we can't know for sure. All we know is 8 is left we can only pick 5 so 85% chance we need 7 try 12.5% chance we need a 6th try only.
my dumbass would keep the top 3 in the first race and then replace the last too until every horse was in a race
personally I thought it was 6, 25/5 for the 5 initial races and then 1 with the winners, surely first, second and third place meet all those requirements, 3 fastest horses, I hadn't thought about the fact that 2nd place in group A could be faster than 3rd place Group C in race 6, interesting, something to think about
Its 5 races...his answer its too complicated...and wrong...find my answer,to see the simpl logic...
Horse C in 3rd place beats 4th and 5th place horses hes faster than the eliminated horses even tho he never raced them from the 4th place group and 5th place Group at the beginning because the 4th and 5th already beat the 8 horses who got eliminated its obvious that horse in 3rd place faster than 10 so far counting 4th n 5th place. Horse 1st and 2nd beat everyone in their group the eliminated ones from there group never raced the horse 3rd place horse 1st and 2nd could had beaten someone faster than the horse in 3rd place now but they got eliminated so can't race horse 3rd place so we can't know for sure. All we know is 8 is left we can only pick 5 so 85% chance we need 7 try 12.5% chance we need a 6th try only.
First thought on this was 8. Race all groups (5 races), then race the winners of each (1 race) then race the winners again replacing the fastest overall with the runner up in the group repeating twice (2 races) Didn't think to remove the e and d groups like you showed and let them keep racing even though they could never be top 3
I had thought 6, but since you can theoretically get 3 horses faster than the 3rd fastest from the first race in each subsequent race, a minimum 7th race is required to resolve the final rankings (and possibly several more).
My original bid was 11 - 5x5, and then 5x3 gives the fastest 9 horses. The 3rd fastest from the first of these races signposts the contenders from the remaining 4, which can be re-raced with the fastest two from the previous race in the final race.
They said minimum, meaning best scenario. So minimum is in fact 6. Run 5, then use 3rd place to run against the other 20. 3rd place beats all therefor you had your fastest 3 in only 6 turns
@@SharamanONorgannon you are assuming that the 3rd fastest from the first race is the 3rd fastest *overall.*
What if horses #17 & #19 are faster?
Sure - one of these is faster than the other, but you can't know whether both are faster than the 2nd fastest from the first race without a re-race. So now you have the 4 fastest horses, and don't know which of these are the 3 fastest - in fact, you don't even know which *single* horse is the fastest.
@@testname5042 They said best case. So you test the 3rd best from the first set against all the rest. In best case scenario the 3rd fastest from the first race IS the 3rd fastest overall therefor in 6 turns you can figure it out.
Horse C in 3rd place beats 4th and 5th place horses hes faster than the eliminated horses even tho he never raced them from the 4th place group and 5th place Group at the beginning because the 4th and 5th already beat the 8 horses who got eliminated its obvious that horse in 3rd place faster than 10 so far counting 4th n 5th place. Horse 1st and 2nd beat everyone in their group the eliminated ones from there group never raced the horse 3rd place horse 1st and 2nd could had beaten someone faster than the horse in 3rd place now which is the eliminated ones and couldn't race to know for sure. so we can't know for sure. All we know is 8 is left we can only pick 5 so 85% chance we need 7 try 12.5% chance we need a 6th try only.
“You do not have a watch”
*pulls out phone*
Since I've heard this puzzle, I've always pictured it as a story around a wealthy Sicilian businessman who is interested in horse-racing and who wants to buy (real, not mechanical) horses from a horsebreeder who has 25 horses; and so he hires you to figure out and prove to him which are the three fastest. But in order to keep his potential rivals/competition in the dark about what he's exactly buying he doesn't want any records to exist/be created of how fast the horses are. So no watches or time-measuring devices are allowed, and also no recording devices, cameras, smartphones, etc.
And since organizing a race costs money, he'd appreciate it if you can do the job in as few races as possible. (Or maybe he's willing to pay you a fixed amount of money that's enough for, say, 11 races, but you can keep the money that you don't spend, as long as you complete the job.)
And then I read the comments and see how many viewers are willing to cheat on this job. And apparently would openly admit to it... in a job interview!
To the people saying 6 is the answer
What if the 3 fastest horses are placed in the same group by chance
Do you get why there are 7 race needed now?
If so why removing group e and d would be correct since you dont have a watxh
Simon REEVES after racing the 5 fastest horses from each group you can automatically eliminate these 2
Because every horse in group D or E is by definition slower than top C which is AT BEST the third fastest(top B and top A beat it in a race). So at best every horse in D and E can be fourth fastest and can therefore be eliminated.
So if following this example the times for slowest to fastest in each group are as follows:
A: 10, 9, 8, 7, 6
B: 14, 13, 12, 11, 10
C: 15, 14, 13, 12, 11
D: 16, 15, 14, 13, 12
E: 17, 16, 15, 14, 13
How is it valid to eliminate a3-a5 as being slower than b1?
You're not eliminating A3, you're eliminating A4-A5, B3-B5, C2-C5, all of D and all of E. Which will always be valid. You're not eliminating them for being slower than B1, you're eliminating them for being slower than AT LEAST 3 other horses. The final race is A3, A2, B1, B2, and C1 since those are the only horses left that aren't confirmed to be slower than at least 3 others.
Oh wow! I actually got this right. I almost never get any of your puzzles right so I’m beyond thrilled! Love your channel!
Meanwhile in Yandex interview: give all the horses a shot of vodka!! It'll be a tie.
That, or we get a Bolshevik Revolution led by Joseph Stallion wherein World War 3 kicks off against the Emu Nation.
I came up with 8 races. Race each set of 5, remove the bottom 3 from each race, leaving you with 10 horses. For the next 2 races, count seconds it took for the third horse to finish. The slower race removes three horses, the faster race removes 2. This leaves you with your last 5 horses, and one last race for top 3.
Admittedly very flawed compared to the solution, especially the counting.
I came up with the optimal solution in a few minutes with pen and paper, but I guess it's easy to get stuck if you miss some piece of information.
I thought it was easier to think backwards: first you definitely need 5 races to race each horse once. The two slowest horses from each race is definitely not part of the top 3.
You are now left with five groups with three horses (15 in total). You don't want to race them all against each other, so you can rank the groups relatively to each other by racing the slowest or the fastest in each group. Since we're looking for the fastest ones, intuitively it makes more sense to rank by the fastest, so we race the fastest horse in each group in the semifinal.
All the horses in the two groups represented by the last two horses in this race will all be slower than the fastest three horses in this race. We're left with three groups, with three horses in each (9 in total).
Some observations at this point:
- The fastest horse is identified, as the fastest horse from the semifinal
- There are three ways that the three fastest horses can be arranged into the three fastest groups: the top three from the fastest group, the fastest horse plus the two fastest horses in the second fastest group, or finally the fastest horse in each of the three fastest groups.
This gives us 6 candidates for the three fastest horses. Realizing that the fastest one is already identified leaves us with five horses to race the final.
The two fastest horses in the final are the second and third fastest horses.
Alternative solutions:
- 1 race: Race 3 randomly selected horses. The other ones have never competed so their time is undefined.
- 0 races: Shoot 22 horses.
Professor Layton ass solutions
thats dark men
I got 8 races in 10 seconds and no paper...
@@thomgizziz Same
I got the same result with the same reasoning, except note that when you say "there are three ways..." there are actually four; you could also have the fastest horse, the second fastest horse from that same group, and the fastest horse from the second fastest group. That doesn't change anything about the rest of your explanation though :)
I got 8 because I didn’t think to rearrange the groups based on the results of the 6th race. I was adding an additional step where all the 2nd place horses from each group raced but it was basically an inefficient way to get the same result that you’d get in the 7th race in this solution.
Had the exact same thought process, but a total eureka as soon as the video said "we split them into groups" and I went "OH WAIT WAIT WAIT IS SEE IT no point in racing the nr 4 and 5 horses from the winner race, that frees up 2 more slots!"
@@codrincx same and i got stuck on how to fit a3 in there cos what if all 3 fastest horses were in the same race, you dont need to race the fastest again freeing up another slot *d'oh* 🤦♂🤦♂🤦♂
me neither!
@@codrincx i have a question: at this part 7:40, what if e2 < d2 < b2 < c2 < c1 < b1? And what if b2 < c2 < d2 < e2 < e1 < d1 < c1 < b1?
@@onepun9583 If I'm understanding your comment right, then:
The d and e group cannot imaginably matter, since after the first races, we know for sure that a1 > b1 > c1. Therefore, any horse from group e will be slower than *both* b1 and c1. The only horses that could be faster than b1 come from group a (as all horses in group c, d and e are all slower than b1), and then the only horses that could be faster than c1 come from groups a and b.
Notice that in both scenarios you've written, b1 and c1 are the fastest, therefore the only ones actually relevant. If a2 or a3 (could also be a4 or a5, but we only care about top 3) isn't faster than b1, then b1 *is* the 2nd fastest horse. Similarly if b2 isn't faster than c1, then c1 *is* the 3rd fastest horse.
This is something I learned in grad school. For any top-N problem, the tournament algorithm produces the minimal time solution, same O-notation complexity, but lower constant factor due to decreased pairwise comparisons.
The general extension of this algo would have been an excellent interview question with a runtime and proof of completeness. It felt good to know I got this one right when I did it.
I destroy all but 3 horses.
Now they are the fastest and need to put in the work for the other 22.
Apparently the best answer to get a job in human resources.
Nice puzzle, my first instinct was 11, should have thought about it more to think about what the results of the first five races meant. I thought about racing the fastest, but glossed over that eliminating the slowest two also limited the other four that were in their original groups AND the slower horses in the other groups.
11?💀
Yeah i also think its 11 by eliminate every horse on 4 and 5 spot. And i think about this just under a minute. But yeah if i had think long enough and do critical thinking i could arrive at the answer for 7
11 seems correct. The solution of this video does not seem correct. Without racing #2 of race 1 against #1 of race 2, how does one know that #2 of race 1 should be eliminated? What if all 3 fastest horses were in fact within race 1, but #2 and #3 of race 1 were inadvertently removed from subsequent comparisons against all other slower horses?
@@thangd7124 a2 and a3 are competing in the final race. Rewatch the video.
@@thangd7124 yeah 11 is also correct. The solution in this video is correct too.
In step two, because we already found the fastest horse in the group of 25 (1st place of the group "a"), we know that the 3 fastest horses might be all three in group "a" (case 1).
In group "b", we only consider 2nd place and 1st place horses, we remove the 3rd place horse in this group because we only need to find the 3 fastest horses (one is known in group "a"), now the 3 fastest horses might be "1st place group a, 1st place group b, 2nd place group b" (case 2).
In group "c", we consider only the 1st place horse, the 3 fastest horses might be "1st place group a, 1st place group b, 1st place group c" (case 3).
At 7:21, we have three cases to consider knowing that the fastest horse is 1st place group a horse.
case 1: 1st place group a, 2nd place group a, 3rd place group a
case 2: 1st place group a, 1st place group b, 2nd place group b
case 3: 1st place group a, 1st place group b, 1st place group c
The three fastest horses could be in one of the cases, but we don't know which case. What we know is the 1st place group a horse is the fastest. Therefore, we can leave out the 1st place group a horse and race the others to find the second and the third fastest horses.
case 1: 2nd place group a, 3rd place group c
case 2: 1st place group b, 2nd place group b
case 3: 1st place group b, 1st place group c
race 2nd place group a, 3rd place group c, 1st place group b, 2nd place group b, 1st place group c
I couldnt really come up with why I would need the 7th race(after 6 races and I knew I had at least one more race to figure out the top 3 but I could not really come up with how to match up 5 horses for the 7th one) and now I can say I know why. Thank you.
In the original question without explaining they're mechanical, I would have wondered if the horses got tired and slower running multiple races lol
They never mentioned the time scope. For all we know the same horse may get some rest time between races
Great thing about algorithms is that there are pros and cons to everything. Guaranteed 7 races solution has a con that you use a lot of memory, because you need to store information about first three horses of all races until 6th race. We can have a solution that at worst uses 10 races, at best 6 and stores information about only 2 horses at a time.
My solution was to grab 5 horses, race them, put the slowest two aside and grab two more to race off against the fastest 3 so far identified, after this race, the losers will be put aside and two new horses will face off against the fastest three from the second race, etc.
Not particularly quick, but since the horses won't get tired, you will eventually end up with the three fastest horses
that's the solution i got. it's only 4 races more than needed, so I'm still proud of myself for coming up with that solution.
@@haydenhuss8758 you just do sets of five and then race the winners of each set for a total of 6 races.
this works because each horse will be equally as tired since each horse will have had one race.
@@fayt9121 that doesn’t work. Lets say the second fastest horse was in the same heat as the fastest horse of all 25 horses. Then the second fastest horse would never be included in your final (6th) run.
@@Macarthurmaintenance this is true. so you are saying you have to race a total of 8 times since you want the 3 fastest horses right? you do your first six to find which one is the fastest and then you race second and third place against the winners from groups 2-5. I get that. And with this hypothetical, its possible that the horses don't get tired either. Ill be honest I cant put my finger on it but something's bothering me about this.
If you ever end up with 25 horses and you forget their speed, not that you ever will but still, well, now you know...
It's a logical problem with practical labeling. It's not that difficult, really.
Exactly. It's an IRL variant of sorting algorithms which only go by relation (more, less, equal) of the input data, this time a variant which takes 5 inputs, and outputs relations of every pair within those. It's less relevant IRL now, but might have been very relevant before you could time the races.
Thinking about these problems becomes easier if you visualize the ranking of a horse as the # of horses it would lose against, making rank 0 the fastest. Sometimes, ties tend to mess things up, but in this case, they don't.
I’m rethinking this and hadn’t realized that the 2nd and 3rd place horse would be from other groups. I take back my comment about there being a logic flaw. Mea Culpa!
6 races
@@babarshahzad7967 not enough.
@@babarshahzad7967 Explain one by one what you think these 6 races would be
0:35 "mechanical horses"?
Damn, I guess it's true that we learn something new every day.
My guess was 11. This is because each race is guaranteed to tell you that the 4th and 5th horses are not in the top 3. I couldn't come up with any way to guarantee elimination of more than two horses at once. That sixth race amazingly eliminates 10 horses at once, while the others still eliminate 2. I did not imagine so many at once would be possible. By establishing a hierarchical structure, entire branches can be eliminated. Very clever.
11 is the brute force answer. Anything less is optimization
Simple. Pet rock wins
Why go thru all this interview struggle when u gonna copy & paste the code from stackoverflow anyway?
Maybe they want to know if you're one of those guys who wastes company's time by writing stuff on stackoverflow? :-P
yuri renner good one 😂
you gotta know what your looking for and yes, copy pasting is convenient but you also wanna know what your looking for and if it fits your needs. i bet coders do it all the time but they wont be fooled be shitty codding
LOL
what company do u work for haha
1:33 I was able to get a 10-race solution. Racing , and thereby sorting the 5 rows of horses, I obtain 5 columns, the last two of which can be discarded as they definitely don't have the 3 fastest horses.
I sort the 3 columns, and I obtain a group of 6 horses, of which I know the 3 fastest exist. By racing two different groups of 5 of the 6 horses, you can figure out which are the fastest over all.
Not sure if it's the minimum, but the first thing that comes to mind is 11. Start with five horses racing. That's 1. Then keep racing the three fastest, replacing the two slowest horses each time for the remaining 20 horses, giving you 10 more races.
There's a faster way... It only requires 1, 2, or 7 races.
25 horses... Either you race all of the horses at once, then take the top 3 who touched the finish line... Or... You race the horses in groups of 5, and the average of horses above 50% gets to race twice. The other horses are dismissed. Then pick the top 3 fastest horses from the new average.
This works better if you're dealing with 100+ contestants, and if you don't want an outlier.
I really like this solution, because the one presented in the video doesn't actually work. But, he did have a good point - you can disqualify any horses that lost to 3 or more horses. So you can get rid of columns 4 and 5, swapping out the remaining 10 horses like you said.
@@jeremystarczewski4748 I might be missing something... why doesn't the method work?
I also came up w/ 11 races w/ same rationale. It's simple, easy to understand, and works. I'll take software written that way over clever any day of the week. However, if it's really a question of optimization (fewest races) then 11 races is too many and this video shows the optimized solution.
@@jackbastian4072 the video method incorrectly infers that a1 beating e1 means e1/e2 etc would lose to a2. Because we don't know the times and only the orders, we don't know the margin by which a2 lost to a1 and so on. If you write out 1 through 25 starting from the bottom and moving up, this could present a situation in which the video method results in the 2nd and 3rd fastest are actually the 6th and 11th fastest overall. It's easier to show on paper but I can't attach a photo here.
My one issue with the 5x5 grid of slowest/fastest winners was I thought to my self "How do you get that after 5 races???" But then I realized that the 6th race will tell you who was the fastest and slowest of winners. Great stuff.
This part is key and I don't think it get's the attention needed in the explanation. 5 races is minimum to race each horse at least once, and the 6th race is key because not only do you get the winning horse, but you also get the ordering of the groups. This allows you to determine the horses for the last round no. 7. Without the info from race no.6 you have no way to compare the groups between themselves.
@@Productlivity_ZZ I don't get why you even need the groupings though.
Wouldn't the top 5 winners already be the fastest of their groups
So you only need to race the winners to get the fastest horses.
Never mind I think I see now
@@izzy123123123 If the three quickest horses happen to be placed in group a, the third horse of group a could be faster than the winner of any of the other groups
@@siepie830 yeah it was more like representing their group but because they lost in their final race their group is eliminated
Horse # Six: "I am not a number, I am a free horse!"
The new Horse # 2: HAHAHAHAHAHAHAHAHA!
@@cobbler88 Spot on with the "new" :)
@@Walter-Montalvo And I think one of them returned a second time.
Unexpected prisoner
They want infor neigh tion
Honestly thought it was 6 races.
5 horses, 5 races, fastest of each race in a final race, top 3 found, 6 races.
I had a racehorse once that was amazingly strong. It always took every other horse in the race to beat him.
So he lost?
@@solome6478 He last.
I made it in 11 races.
I raced 5 random non-eliminated horses and eliminated bottom-2 in each race.
That would be 100% accurate
@@sudhirkumar-zx9zg It's not the correct answer because the question asked about fewest number of races, and 11 is not it.
You can get the answer in 6 races
Yes but what if the bottom 2 horses in race one were faster than every other remaining horse
@@DanielRodriguez-bu8du for 6 races you need luck, but not 7 races.
No race needed.
I'd butcher 22 horses. Not only would the remaining 3 become the fastest by default but I could stack up on delicious horse meat and land the Google job for my out of the box problem solving.
If the Google Executives hire you, they may apply a similar method when it's downsizing time and they have to determine who their three most productive employees are.
yuri renner
The circle of life
@__ Drake,
Ah, of(f) course, The Lion King -- lions eat horses. ;-)
funny:)
Good luck eating mechanical horses haha
I got stumbled at "5 races give five " Fastest" Horses from each group, and the fastest among them is the first overall. But how exactly do I proceed with the 2nd and 3rd? " And I am happy about your explanation
Me to interviewer: "are the horses mechanical or do they get tired and slow from racing over and over?"
He said they are mechanical.
@@Guppypants Mechanical horses can wear down as well...
@@killerdinamo08 nope they won't u keep charging them after each race
@@pryo2460 I'm guessing you're the kind of person that yells at everyone when your Iphone Battery stops working, or your Electric Car starts to wear down, while shouting, "It's not broken! It was working just fine yesterday!"
Jokes aside, but "Mechanical" doesn't mean "indestructible, or immune to wear..." it means, "Consistent Performance on Repeated Runs." Even two Ford 150's aren't going to perform the same forever.
ok FUNNY
The number of people saying 6 after watching the video is too damn high
Welp, I would have said 8 otherwise
Zakaria Rakhrour but six makes more sense. Five races to get the top five, then one race to see the top three. It doesn't specify that you can only find one winner/loser per race.
That doesn't work, though. If a horse wins its initial race, that just means that it's fastest than the 4 horses it raced. It could still be slower than all 20 of the other horses.
+Anon E. Mous except we don't know that. it might or might not be faster then another group. we just aren't given that information. all we have is 25 horses and we need to find the shortest amount of races needed to find the fastest three .
lnoah95 I think it's implied that we're removing the element of chance. Otherwise, you could just choose 3 horses at random, without ever racing them, and there's a chance (however small) that those are the fastest 3. In that case, the answer would be 0.
I answered by myself in the same manner. But the presentation seems tougher than the solution
That's true
I did too. But I got 8 races since it seems possible c_2 could be faster than all the others in the 7th race.
The solution to the question "Find a method that always determines the three fastest horses in 7 races" isn't that difficult. But the solution to the question "Prove that this method is optimal" is not easy (if even possible); the supposed "proof" that Presh gives in the video is incorrect/incomplete.
@@yurenchu good point,
You can easily prove that the answer is at least 6 and then prove that 6 is actually bot enough. But he did not do it.
Its 11 - sorry folks
Kill all but 3 horses. You now have the 3 fastest horses.