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 19 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 210 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 1119 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 10n9 to 10n1 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 20050623 22:59:12 