After processing an infinite number of marbles, not once but twice in solving the puzzle Marbles Bonanza, you've grown rather tired of moving all these marbles around. Nevertheless, it is your duty to press on forward and try a third experiment. This time, though, you decide that you'll cut down on the amount of work by not removing any balls ever.

At the start of the minute, you put marbles 1-9 in the bag, and then add a 0 to the end of marble 1 (so that you now have duplicate marble 10s, one in the bag that you just modified, one out). Therefore you now have marbles numbered 2-10 in the bag, as in scenario B of the previous puzzle, and marble 10 outside the bag as in scenario A. 30 seconds later, you put marbles 11-19 in the bag, and add a 0 to marble 2, so that now you have two marbles numbered 20 - one in, one out. You continuously repeat this process, with each interval half as long as the one before. In general, for the nth operation, you put marbles 10n-9 to 10n-1 in the bag, and add a zero to marble n in the bag, so that it becomes marble 10n in the bag.

• How many marbles are in the bag at the end of the minute?
• What are the numbers on the marbles ?
• Is the situation inside the bag identical to either of the previous two problems after 31 seconds? 50 seconds? at the end of the minute? How about the situation outside the bag?

(In reply to re(10): No Subject by Ken Haley)

In this problem you have more than just a function.

If you define a function as f(n) = 9n, where f is the number of marbles in a bag, and n is the step you are at, then yes, their is enough information to know what happens as you aproach an infinite number of steps; there will be an infinite amount of marbles.

But, since we are talking a bout ordered sets there is more to consider. Here is a very simple example that ilustrates the mistake of asuming that a simple function can describe the result of algorithms performed on an ordered set;

Start with the simple set {1,2,3,4,5,6,7,8,9,10,11}. The alogrithm "remove even balls" leaves 6 marbles, and the algorithm "remove odd balls" leaves 5 marbles. If one does these operations in setps by adding two marbles at a time and removing one, one could fall into the temptation of describing the number of marbles after step n as f(n) = n for all steps. But this is false, more information is needed, because the entire set has to be considered as a whole.

You can also see that relabeling the balls will give different answers for the number of marbles left, but that is expected and it is not a logical inconcistency.

Posted by ajosin on 2005-06-27 21:32:25