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If a certain number of people, n, are in a circle and every other person is eliminated until one remains standing, what is the equation that gives the relation between n and the wining position, p?

difficulty advanced
Guest Jun 1, 2015

Best Answer 

 #3
avatar+26750 
+5

I get a similar(ish) expression to Bertie, except that I count from the first person not to be eliminated on the first rotation round the circle.

 

$$p=2(n-2^{floor(\log_2{n})})+1$$

 

This generates the following positions, p, for n from 1 to 16

 positions

.

Alan  Jun 3, 2015
 #1
avatar+889 
+5

As yet I don't know how to prove this, but a formula that seems to work is

$$\displaystyle p=2\left(n-2^{\bmod\left(\frac{(\log(n)-0.001)}{\log(2)}\right)\right)-1.$$

The logs are to base 10 and mod (I hope, otherwise I need to find another operator,) delivers the integer part of the expression within the brackets.

p is measured from the first person to be eliminated, being the number of positions further on around the circle.

For example n = 6 will return a value of p = 3. So, from the first person eliminated count to the third position further round the circle.

Bertie  Jun 3, 2015
 #2
avatar+92781 
0

Thanks Bertie,

The penny just dropped, at least I understand what it being asked now.     

Melody  Jun 3, 2015
 #3
avatar+26750 
+5
Best Answer

I get a similar(ish) expression to Bertie, except that I count from the first person not to be eliminated on the first rotation round the circle.

 

$$p=2(n-2^{floor(\log_2{n})})+1$$

 

This generates the following positions, p, for n from 1 to 16

 positions

.

Alan  Jun 3, 2015

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