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Topic: Black and White Balls (Read 1770 times) |
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alexeigor
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Black and White Balls
« on: Mar 28th, 2008, 3:42am » |
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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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| « Last Edit: Mar 28th, 2008, 6:29am by alexeigor » |
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towr
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Re: Black and Whit Balls
« Reply #1 on: Mar 28th, 2008, 4:50am » |
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Nice puzzle.
I was trying to work out a recurrence formula when suddenly I noticed, the number of white balls modulo 2 is invariant.
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| « Last Edit: Mar 28th, 2008, 4:52am by towr » |
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alexeigor
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Re: Black and Whit Balls
« Reply #2 on: Mar 28th, 2008, 6:25am » |
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Good observasion, so
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?
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| « Last Edit: Mar 28th, 2008, 6:37am by alexeigor » |
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towr
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Re: Black and White Balls
« Reply #4 on: Mar 28th, 2008, 7:10am » |
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on Mar 28th, 2008, 6:25am, alexeigor wrote:if White odd, last ball is White
if White even, last ball is Black
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Yup, that's what I got.
on Mar 28th, 2008, 6:32am, gotit wrote:| I think this problem is quite similiar to this |
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Quite similar indeed; just give all the black balls an even number, and the white balls an odd number.
When replacing two balls, number the replacement with the difference.
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| « Last Edit: Mar 28th, 2008, 7:11am by towr » |
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