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Topic: Please help me with the following question
Replies: 78   Last Post: Mar 25, 2013 1:38 AM

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Robert Hansen

Posts: 7,615
From: Florida
Registered: 6/22/09
Re: Please help me with the following question
Posted: Mar 13, 2013 5:07 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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.

Bob Hansen

On Mar 13, 2013, at 11:02 AM, Joe Niederberger <niederberger@comcast.net> 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



Date Subject Author
2/24/13
Read Please help me with the following question
hydraflap
2/24/13
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Richard Strausz
2/25/13
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James Elander
2/27/13
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Wayne Bishop
3/1/13
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GS Chandy
3/1/13
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GS Chandy
3/5/13
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johnykeets
3/7/13
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Joe Niederberger
3/7/13
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Robert Hansen
3/7/13
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Joe Niederberger
3/7/13
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Joe Niederberger
3/7/13
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Robert Hansen
3/7/13
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Joe Niederberger
3/7/13
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Joe Niederberger
3/7/13
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Robert Hansen
3/7/13
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Robert Hansen
3/7/13
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GS Chandy
3/7/13
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Robert Hansen
3/7/13
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GS Chandy
3/7/13
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GS Chandy
3/7/13
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Robert Hansen
3/7/13
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GS Chandy
3/7/13
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GS Chandy
3/7/13
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Joe Niederberger
3/7/13
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Joe Niederberger
3/8/13
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GS Chandy
3/8/13
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GS Chandy
3/8/13
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GS Chandy
3/8/13
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GS Chandy
3/8/13
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GS Chandy
3/8/13
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Robert Hansen
3/8/13
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Joe Niederberger
3/8/13
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Robert Hansen
3/8/13
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Joe Niederberger
3/8/13
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Robert Hansen
3/8/13
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Joe Niederberger
3/8/13
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Robert Hansen
3/8/13
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Joe Niederberger
3/9/13
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Robert Hansen
3/8/13
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Joe Niederberger
3/8/13
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Joe Niederberger
3/9/13
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GS Chandy
3/9/13
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Robert Hansen
3/8/13
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GS Chandy
3/9/13
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GS Chandy
3/8/13
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GS Chandy
3/9/13
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Joe Niederberger
3/9/13
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Robert Hansen
3/10/13
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Robert Hansen
3/9/13
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GS Chandy
3/10/13
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GS Chandy
3/10/13
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GS Chandy
3/10/13
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Joe Niederberger
3/10/13
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Robert Hansen
3/10/13
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GS Chandy
3/11/13
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Robert Hansen
3/11/13
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Joe Niederberger
3/11/13
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Robert Hansen
3/11/13
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Robert Hansen
3/11/13
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Joe Niederberger
3/11/13
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Robert Hansen
3/11/13
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Joe Niederberger
3/11/13
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Robert Hansen
3/11/13
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Robert Hansen
3/12/13
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Joe Niederberger
3/12/13
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GS Chandy
3/13/13
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GS Chandy
3/13/13
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Robert Hansen
3/13/13
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Joe Niederberger
3/13/13
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Robert Hansen
3/13/13
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Joe Niederberger
3/14/13
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Robert Hansen
3/14/13
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Robert Hansen
3/15/13
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GS Chandy
3/14/13
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Joe Niederberger
3/14/13
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Robert Hansen
3/14/13
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Joe Niederberger
3/15/13
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GS Chandy
3/25/13
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