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A Counting Problem

Blogs > blankspace
Post a Reply
blankspace
Profile Blog Joined June 2010
United States292 Posts
April 10 2011 12:45 GMT
#1
Draw n points on the circumference on a circle. Then connect all lines between all points on the circle. What is the maximum number of regions that the circle may be divided into?

Example: For the case n=3, you have a triangle inscribed in the center of the circle. The number of regions is thus 4. For the case n=4, you have a square with diagonals drawn, so that gets you 8.

I will tell you though that the answer is not 2^(n-1), but it is not complicated.

Hello friends
Oracle
Profile Blog Joined May 2007
Canada411 Posts
Last Edited: 2011-04-10 13:47:40
April 10 2011 13:15 GMT
#2
I think the answer is the (amount of faces of an n-complete graph) + n

I'll get back to you once i figure out the first part


EDIT: I dont think this is possible with my knowledge of graph theory, since # faces is ill-defined for non-planar graphs, and everything K_5 and above is nonplanar... and im stuck
Av4st
Profile Joined September 2008
Canada92 Posts
Last Edited: 2011-04-10 13:53:09
April 10 2011 13:42 GMT
#3
oops -- within the circle.. hmm
Oracle
Profile Blog Joined May 2007
Canada411 Posts
April 10 2011 13:45 GMT
#4
On April 10 2011 22:42 getty wrote:
+ Show Spoiler +
1 + 2^(n-1)


for n = 5 you have

[image loading]

which has 11 faces. Plus the 5 imaginary ones on the outside, so 16 total.

2^4 + 1 = 17
blankspace
Profile Blog Joined June 2010
United States292 Posts
April 10 2011 13:48 GMT
#5
I don't know if there is a good graph theory solution, but my solution for this does not use any. You need to find a clever way to count the regions.

Also note that the number of regions for n=7 is 57 so looking for some sort of 2^something formula is not going to work.
Hello friends
Marradron
Profile Blog Joined January 2009
Netherlands1586 Posts
Last Edited: 2011-04-10 14:00:13
April 10 2011 13:56 GMT
#6
http://mathworld.wolfram.com/CompleteGraph.html

n(n-1) /2 + n

Gotta check if its right

nvm these are edges or something not correct.
Oracle
Profile Blog Joined May 2007
Canada411 Posts
April 10 2011 14:04 GMT
#7
# VERTICES | # EDGES | # FACES
3 | 3 | 1 + 3 = 4
4 | 6 | 4 + 4 = 8
5 | 10 | 11 + 5 = 16
6 | 15 | 24 + 6 = 30
7 | 21 | 50 + 7 = 57

For anyone else attempting a graph-theory approach
Complete
Profile Joined October 2009
United States1864 Posts
Last Edited: 2011-04-10 15:23:17
April 10 2011 15:20 GMT
#8
n/m..
Complete
Profile Joined October 2009
United States1864 Posts
Last Edited: 2011-04-10 15:44:34
April 10 2011 15:43 GMT
#9
+ Show Spoiler +
(x^4-6x^3+23x^2-18x+24)/24
blankspace
Profile Blog Joined June 2010
United States292 Posts
April 10 2011 15:50 GMT
#10
@complete
+ Show Spoiler +
I think that is the correct answer, though you should realize that you can express it in a more elegant form (namely as the sum of binomial coefficients. What is your reasoning/how did you get the answer? Did you assume it was a polynomial and interpolated?
Hello friends
n.DieJokes
Profile Blog Joined November 2008
United States3443 Posts
April 10 2011 16:17 GMT
#11
It looks to me like he graphed the first couple of n's and then used software to fit the data but thats just a guess
MyLove + Your Love= Supa Love
LaLuSh
Profile Blog Joined April 2003
Sweden2358 Posts
Last Edited: 2011-04-10 18:01:54
April 10 2011 17:54 GMT
#12
+ Show Spoiler +
C(n, 4) + C(n, 2) + 1

Number of lines drawn between n points can be expressed as a series (n-1)+(n-2)+...(n-(n-1))

Or more practical as binomial coefficient: C(n,2). Expressing number of possibilities to connect 2 points in an n point circle by a line.

C(n,4) gives the number of possible intersects between any 4 of the n points in the circle.

Adding them together and the extra region in the center (+1), gives you the total.

+ Show Spoiler +
Watched the spoilers for help....
blankspace
Profile Blog Joined June 2010
United States292 Posts
April 10 2011 18:22 GMT
#13
@LaluSh

+ Show Spoiler +
That is the correct answer Can you justify why your answer is correct though? As in why should counting these intersections give you the number of regions?
Hello friends
JodoYodo
Profile Blog Joined May 2009
Canada1772 Posts
Last Edited: 2011-04-10 18:31:40
April 10 2011 18:31 GMT
#14
+ Show Spoiler +
Every time a line intersects another line, you bisect two regions into four regions
Dance dance dance 'till we run this town!
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