This one is a puzzle question. I’m personally not the biggest fan of these types of problems and would never ask something like this in an interview. That being said, they are kind of fun, and the following problem is no exception. Its an entertaining logic question that can give insight into a candidate’s thought process and how they approach an optimization problem.

There are 25 race horses and you want to find the fastest one. You can only race 5 horses at a time, and you can’t write down their times. How many races does it take to find the fastest **three** horses?

Here’s how I approached it. The first thing to do is figure out all the assumptions inherent in any puzzle question. The first and most crucial assumption is that the finish times for each horse remains the same across races. Without this assumption, the task is impossible, as the fastest horse could change between races: horse #1 could beat horse #2 in the first race, but lose to it in the second race. The second assumption is that there are no ties. There is a clear ranking in finish times for the horses.

Now with those out of the way, we can approach the problem. The most obvious thing to do first is to divide the 25 horses into five groups, and then race each group. This will identify the fastest horse in each heat. Now we just race the five fastest horses from each group. The winner will be the fastest out of all 25 horses, and we will have identified the fastest horse in just six races. Let’s call this winner horse #1, and assume he came from group #1. That’s the easy part. Now the question remains, how many races do we now need to find the second fastest horse? Should we just pick the second fastest horse from race #6? Should we race all the second place finishers from the five groups as well?

At this point it would be helpful to do some whiteboarding. Let’s write down some concrete race times.

Race #1:

Horse #1 1 minute

Horse #2 2 minute

Horse #3 3 minute

Horse #4 4 minute

Horse #5 5 minute

Race #2:

Horse #6 1.6 minutes

Horse #7 1.7 minutes

Horse #8 1.8 minutes

Horse #9 1.9 minutes

Horse #10 1.95 minutes

By writing out the first two races and assigning some times we can see that it is entirely possible that all the horses in the second heat are actually faster than all of the other horses in the first heat save for horse #1. Let’s write down some more race times for the other heats.

Race #3:

Horse #11 1.1 minutes

Horse #12 1.2 minutes

Horse #13 1.3 minutes

Horse #14 1.4 minutes

Horse #15 1.5 minutes

Again, we see that for the third heat, it is entirely possible that all the horses are faster than all of the other horses in the first heat save for horse #2. At this point it may seem that it is necessary to race all the second place finishers in each heat, before racing the winner of the seventh race against the second place finisher of the sixth race. But that’s not quite true.

Race #4:

Horse #16 1.01 minutes

Horse #17 1.02 minutes

Horse #18 1.03 minutes

Horse #19 1.04 minutes

Horse #20 1.05 minutes

Race #5:

Horse #21 1.001 minutes

Horse #22 1.002 minutes

Horse #23 1.003 minutes

Horse #24 1.004 minutes

Horse #25 1.005 minutes

Race #6:

Horse #1 comes in first place at 1 minute

Horse #21 comes in second at 1.001 minutes

Horse #16 comes in third at 1.01 minutes

Horse #11 comes in fourth at 1.1 minutes

Horse #6 comes in fifth at 1.6 minutes

From this we can quickly see that the second place finisher of race #6 (horse #21 from heat 5 from our whiteboard example) is faster than the first place finishers of heats 2,3, and 4. This means that he is also going to be faster than all the horses in heats 2,3, and 4. We can quickly eliminate all those horses as being the second fastest. Thus the only horses to consider for being second fastest are either the second place finisher in heat 6, or the second fastest horse from the winning heat the fastest horse was from (horse #2 from heat 1 in this case). So we race them. So far, we’ve had to do seven races.

So what about identifying the third fastest horse? Based on the same logic as above, we know that the third place finisher of race #6 (horse #16 from heat 4 from our whiteboard example) is faster than all the horses in heats 2 and 3. We do not know if this third place finisher is faster than the second or third place finisher from the winning heat (Horse #2 and #3 from heat 1), or the second place finisher from the heat that the second fastest horse came from (horse #22 from heat 5 in this case). Note for the latter case, we don’t need to consider the third place finisher from this heat, because at best it could only be the fourth fastest horse overall (we know by its placement that it must be slower than the first and second place finishers in its heat and slower than the fastest horse overall). Since we don’t care about the fourth fastest horse, we can eliminate it from consideration. So, there are only four horses to consider for third fastest, and we will need to have them compete in an eighth race to find the third fastest horse overall.

Or do we? In race #7, we raced horse #2 from heat 1 against horse #21 from heat 5. In race #8, we raced horse #2 from heat 1 again. There are only five horses we need to consider for second and third fastest, so we can combine races #7 and #8 into one. The second and third place finishers of the seventh race are the second and third fastest horses, respectively. In total, we just need seven races.

The general algorithm is:

Divide the 25 horses into groups of five.

Race each group.

Select the winner from each group and race them.

The winner from this race is the fastest horse. Let heat A denote the winning heat this horse is from, heat B denote the heat that the second place winner is from, and heat C denote the heat that the third place winner is from.

Now race horse #2 and #3 from heat A, horse #1 and #2 from heat B, and horse #1 from heat C (where #1, #2, and #3 denotes what place the horse finished for that given heat) in a final seventh race.

The first and second place winners of this race are the second and third fastest horses.