Comments on: CT. Odd People http://mathfactor.uark.edu/2007/08/ct-odd-people/ The Math Factor Podcast Site Fri, 08 Aug 2014 12:52:06 +0000 hourly 1 https://wordpress.org/?v=4.9.25 By: strauss http://mathfactor.uark.edu/2007/08/ct-odd-people/comment-page-1/#comment-439 Fri, 06 Feb 2009 18:27:15 +0000 http://mathfactor.uark.edu/2007/08/20/ct-odd-people/#comment-439 Hi there! Yes, we have to assume that everyone has a unique person to shoot at, that there are no “ties”. This is pretty natural– in the real world, one wouldn’t expect two distances to be exactly the same!

]]> By: thebiggnome http://mathfactor.uark.edu/2007/08/ct-odd-people/comment-page-1/#comment-438 Fri, 06 Feb 2009 17:10:24 +0000 http://mathfactor.uark.edu/2007/08/20/ct-odd-people/#comment-438 Say three odd people stand on the vertices of an equilateral triangle, facing the center.  If each chooses to squirt the odd person on his or her left, no one remains dry.

Am I missing something?

]]> By: strauss http://mathfactor.uark.edu/2007/08/ct-odd-people/comment-page-1/#comment-118 Fri, 31 Aug 2007 15:01:36 +0000 http://mathfactor.uark.edu/2007/08/20/ct-odd-people/#comment-118 — but is it necessarily so, that someone must remain dry, regardless of how the folks are arranged?

]]> By: giorgis http://mathfactor.uark.edu/2007/08/ct-odd-people/comment-page-1/#comment-117 Fri, 31 Aug 2007 10:33:11 +0000 http://mathfactor.uark.edu/2007/08/20/ct-odd-people/#comment-117 If a bunch of people are on the circumference of a circle with one poor fellow at the center, as long as they a little more than r appart everybody shoots the dude at the center. You put equispaced 5 to a circle, the dude in the center gets it, and only one on the rim gets it leaving 4 out of six dry.
(An even number of people though)

Giorgis

]]> By: kraDen http://mathfactor.uark.edu/2007/08/ct-odd-people/comment-page-1/#comment-116 Mon, 27 Aug 2007 11:22:55 +0000 http://mathfactor.uark.edu/2007/08/20/ct-odd-people/#comment-116 number of people. That along with my original pairing idea Has put me onto what I think is the right track. ‘Odd’ is necessary to avoid the situation where people are in a configuration where they do pair up. So with an even number of people everyone could get shot. So if we now take another look at the puzzle. We can eliminate those who form pairs as the shoot each other. We are sure that no larger self groups of equidistant people exist by the condition that ‘there is a unique closest person to squirt’. We are now left with a group again containing an odd number of people that do not pair up (i.e. in no case will the person you shoot be shooting back). Of these remaining group there must be one who is the furthest from any one else. By definition they can not be shot as all the rest are closer to someone else. It is possible with certain layouts that multiple people may not be shot but the above shows that we a guaranteed of at least 1. Cheers Ken]]> Hi,
Though a straight line can hardly be considered random (unless we are dealing with a 1 dimensional world) I see your point. Based on my new understanding I reread the question and thought about why you had specified an ‘odd’ number of people. That along with my original pairing idea Has put me onto what I think is the right track.

‘Odd’ is necessary to avoid the situation where people are in a configuration where they do pair up. So with an even number of people everyone could get shot.

So if we now take another look at the puzzle. We can eliminate those who form pairs as the shoot each other. We are sure that no larger self groups of equidistant people exist by the condition that ‘there is a unique closest person to squirt’. We are now left with a group again containing an odd number of people that do not pair up (i.e. in no case will the person you shoot be shooting back). Of these remaining group there must be one who is the furthest from any one else. By definition they can not be shot as all the rest are closer to someone else.

It is possible with certain layouts that multiple people may not be shot but the above shows that we a guaranteed of at least 1.

Cheers
Ken

]]> By: strauss http://mathfactor.uark.edu/2007/08/ct-odd-people/comment-page-1/#comment-115 Sun, 26 Aug 2007 17:51:51 +0000 http://mathfactor.uark.edu/2007/08/20/ct-odd-people/#comment-115 This example might help:

Folks might not pair up, and there really may be more than one dry person!

]]> By: kraDen http://mathfactor.uark.edu/2007/08/ct-odd-people/comment-page-1/#comment-114 Fri, 24 Aug 2007 10:54:54 +0000 http://mathfactor.uark.edu/2007/08/20/ct-odd-people/#comment-114 Assumption that there is a unique person to shoot at counts out the case where a number of people are at the corners of a regular polygon.
With 3 people 2 are equally close and will shoot each other. 3rd person will shoot whoever is closer of the other 2 but will remain dry.
Same argument works for any odd number of people. The fact that person A is closest to person B is an if and only if condition i.e. they will shoot each other. So if we have 2N+1 people they pair up into N pairs who shoot each other leaving 1 person left who shoots whoever they are closest too but who no one shoots.
Implies he/she remains dry.
Only thing I disagree with is that your statement “there is a unique closest person to squirt” implies that there “will always be at least
person left dry” where as in fact there will always be one and only one
person left dry.
cheers
Ken
cheers
Ken

]]>