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(2): No Subject by Ken Haley)

Answer to first paragraph,

I admit that you make a compelling physical argument. However, the person waching the balls can never tell you what happened when the minute was reached; You can see the end result, and he can go on telling you what happened in each step, but the bridge between the two can never be crossed.

The only way one can describe what happened when the minute was reached and infinite operations where applied to the marbles is by thiking of infinite sets, where the numbering of the sets is crucial, thus the difference between A and B.

What happened in each step, where the number of marbles is growing in both cases, is irrelevant. The single most important thing happens at the end of the mininute, when infinite operations are performend, and all the earlier partial results are obliterated.

Physically, there is only one important instant for the system; the one minute mark. What happened before does not matter.


Answer to second paragraph;

You are changing the operations perfored on the infinite sets, so it is no suprise that you get different results. Your first set of operations makes,

{1,2,3,4,...} -> {{1,3,5,7,...} , {1,2,3,4,...} }

Notice the arrow, not the equality. These sets are not equal. You have perfored the infinte number of operations you describe. If we represent the operatiors collectively by SOxRE (Separate Odd x Renumber Even).

SOxRE {1,2,3,4,...} =  {{1,3,5,7,...} , {1,2,3,4,..} }

Now it is legitimate to use the equal sign. Calling A and B the operations associated with the firt problem,

A{1,2,3,4,...} = {1,2,3,4,5,6,7,8,9,11,12,13,14,15,16,17,18,19,20,21,...}

B{1,2,3,4,...} = {}

Also, as you so well put it,

BxS0xRE{1,2,3,4,...} = {1,3,5,7,...}

But there are no contradicions here. There is no reason why BxS0xRE and B should be equal (you just proved that they are not!). It looks like that for some reason you are thinking that  S0xRE is the identity operator, something that isn't event  true  for finite sets. 

What is true is  that the  Number operator N (the operator that is defined only for finite sets and returns the the number of elements in a set) is invariant under S0xRE. That is,

NxSOxRE = N

So for the operator N, SOxRE is behaving like the identity operator. But there is no reason to postulate or to expect that it has to behave the same way way for A or B!.

Posted by ajosin on 2005-06-23 22:59:12