Friday Puzzler -- Pinewood derby races

Re: Friday Puzzler -- Pinewood derby races

John Veerkamp

assuming( not realistic) that all cars run at the same speed each heat it runs. I can do it in 6 heats. Maybe another solution but I cannot think of one.

Re: Friday Puzzler -- Pinewood derby races

Alex Y

Good point on the unrealistic but necessary assumption that each car will run at the same speed each time it is run. Sorry for omitting that.

Would be interested in seeing your 6-race solution. The best I have seen is seven races, but I've never seen any demonstration that seven is the least, and have had the feeling that it may be possible to do better.

Re: Friday Puzzler -- Pinewood derby races

Don Orr

I agree with the 6 races.

5 races of 5 cars each =5 winners.

5 winners race and top 3 go on to regionals.

What do I win ???

Re: Friday Puzzler -- Pinewood derby races

Henry Higginbotham

But what of the possibility that the three fastest of the 25 are all in the same first-round race?

On the other hand, the best I've come up with is 12 races--kind of a brute force solution. Maybe some coffee will help.

Re: Friday Puzzler -- Pinewood derby races

Larry Barrett

I'm with you, Henry. You have to keep the top three from each race because any three could be the fastest of the original 25. So the first round of 5x5 eliminates the slowest two from each race = 10; 15 left. The second round of 3x5 eliminates 6 more, 9 left. The third round could be one race of 5 and one race of 4; this will eliminate 3 more; 6 left. The fourth round would be one race with 5 in it, and one car standing by (draw straws). The three winners on the one race plus the standby are in the last race, and the top three are the winners.

Total races = 5+3+2+1+1=12.

Re: Friday Puzzler -- Pinewood derby races

Larry Barrett

Here may be a way to reduce the races a little. Lets identify the first round races as A, B, C, D, and E, and the cars in each race as A1, A2, ... A5, etc.

Run the first 5x5 races and eliminate the slowest two from each race; ltes say for convenience that the slowest cars in each of the first round were cars 4 and 5 in each race. So after the first 5 races, 15 cars remain: A1, A2, A3, B1, B2, ... , E1, E2, E3. Again for convenience lets say that they finished in this same order: A1 won the first race, A2 finished 2nd, A3 finished 3rd, etc.

For the next race, select the fastest car from each of the first 5 races: A1, B1, ... E1. After this race, lets say that they finished in this order: C1, B1, A1, D1, and E1. So D1 and E1 are eliminated, and also D2, D3, E2, and E3 are eliminated since they were slower than D1 and E1 in the first round. Also, we can eliminate A2 and A3 since A1 finished in 3rd place in this race and A2 and A3 were slower in the first round, and we can eliminate B3 since we know that C1, B1, and B2 are faster.

So after the 6th race the remaining cars are C1, B1, A1, and C2, C3, and B2. B3.

This s as far as I can go for the time being.

Almost there! :-)

Re: Almost there! :-)

Larry Barrett

So after the 6th race we know that C1 is the fastest car. We also know that that B1 beat A1 and we know the results of earlier races. To determine the 2nd and 3rd cars we race one more time with C2, C3, B1, B2, A1.

Based on earlier races, the only possible 1st and 2nd places in this race are:

B1, B2,

B1, A1

B1, C2

C2, B1

C2, C3

So C1, combined with one of these outcomes, are the three fastest cars.

Of course, in the actual pinewood derby, the kids will insist on many more races just to see their cars run. It is a lot of fun for all.

Correct!