Am I missing something?

]]>(An even number of people though)

Giorgis

]]>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 â€˜oddâ€™ number of people. That along with my original pairing idea Has put me onto what I think is the right track.

â€˜Oddâ€™ 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 â€˜there is a unique closest person to squirtâ€™. 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

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

]]>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 ]]>