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Title: Black and White Balls Post by alexeigor on Mar 28th, 2008, 3:42am Consider an urn filled with a number of balls each of which is either black or white. There are also enough balls outside the urn to play the following game. We want to reduce the number of balls in the urn to one by repeating the following process as often as necessary. Take any two balls out of the urn. If both have the same colour, throw them away, but put another black ball into the urn; if thy have different collours the return the white one to the urn and throw the black away. Each execution of the above process reduces the number f balls in the urn by one; when only one ball isleft the game is over. What, if anything, can be said about the colour of the final ball in the urn in relation to the original number of black and white balls? |
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Title: Re: Black and Whit Balls Post by towr on Mar 28th, 2008, 4:50am Nice puzzle. I was trying to work out a recurrence formula when suddenly I noticed, [hide]the number of white balls modulo 2 is invariant[/hide]. |
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Title: Re: Black and Whit Balls Post by alexeigor on Mar 28th, 2008, 6:25am Good observasion, so [hide] Transitions: w,b == w, b-1; w,b == w-2, b+1; Invariant: w == w mod 2; if White odd, last ball is White if White even, last ball is Black I'm right? [/hide] |
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Title: Re: Black and White Balls Post by gotit on Mar 28th, 2008, 6:32am I think this problem is quite similiar to this http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_easy;action=display;num=1194039906;start=12#12 |
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Title: Re: Black and White Balls Post by towr on Mar 28th, 2008, 7:10am on 03/28/08 at 06:25:25, alexeigor wrote:
on 03/28/08 at 06:32:02, gotit wrote:
When replacing two balls, number the replacement with the difference. |
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