Someone said that it could be done in 5 races, just pick the top 3 times. Unfortunately, this problem isn't stated as a mathematical problem.
Assuming that we do not know the time and assuming that "top 3" means that they have beaten everyone else directly or indirectly...
Step 1 - 5 races. These are the groups and they are in order 1st, 2nd, 3rd, ...
Step 2 - 1 race of 1st place winners of Step 1. Winner is #1. The rider that was 2nd in his group now replaces him.
Step 3 - 1 race of remaining 1st place and the replacement. Winner is #2. The rider that was behind him in the original group replaces him.
Step 4 - Same as step 3, Winner is #3.
That would be 8 races.
In each race the original racers are advancing up one as a member in their original group wins. Basically a merge sort with 5 groups.
On Mar 13, 2013, at 11:02 AM, Joe Niederberger <firstname.lastname@example.org> wrote:
> Getting back to the original problem... > > Johnykeets says: >> and finally Race 7:2nd and 3rd cyclists from the category of winner of Race 6, 1st and 2nd cyclists from the category of runners-up of Race 6, 3rd of Race 6, > Other wise, we may also try this If we have two best players in one team we get win...ners in 7 races. After sixth race we get topper who initially belonged to say race 1, and then rest 4 competes with the second position holder of the race 1. so we get top three. Please let me know about the more easy methods in solving this problem. > > I too think you can do it with 7 races also but I cannot follow your description of the setup for race 7. > > Anyone care to prove that it cannot be done in 6? > > Cheers > Joe N