Re: REALLY OT: News Flash
Hans du Plooy wrote:
On Fri, 2007-02-16 at 14:55 -0600, Mike McCarty wrote:
Ask me about the machine and the ping pong balls some time.
Tell us?
I have three boxes, two of them empty, the other has a countably
infinite number of ping pong balls. The box with the balls on it
is labelled "A", the other two boxes are labelled "B" and "C".
Each ball has a number on it, i.e., 1, 2, 3, 4, ...
I have a little robotic machine, an arm. It can pick up two
ping pong balls from box A and move them into box B, and then
pick up one ball from box B and move it to box C. One of the interesting
things about this arm is that it gets faster the more balls
it has moved. It "learns" how to move faster. The first ball
movement takes one second to move, including going from A to B,
and B to C. The second ball movement takes 1/2 second. The n-th
ball movement takes 1/(2^n) seconds to complete.
I program the machine so that it always picks up the two lowest numbered
balls from box A and moves it to box B, then picks up the highest
numbered ball from box B and moves it to box C. So, the operation goes
like this...
From A move 1 and 2 to B. From B move 2 to C. (1 sec) oo, 1, 1
From A move 3 and 4 to B. From B move 4 to C. (1/2 sec) oo, 2, 2
From A move 5 and 6 to B. From B move 6 to C. (1/4 sec) oo, 3, 3
...
At the right I give the number of balls in each of the boxes A, B, C.
At each step, the number of balls in A is infinite, and the number of
balls in B and C are equal, and increasing. After two seconds,
we see that box A has no balls in it, box B an infinite number of
balls in it (numbered 1, 3, 5, 7, ...) and box C has an infinite
number of balls in it (numbered 2, 4, 6, ...). What happens with
box A is perhaps a little surprising, but not too, and what happens
with boxes B and C makes sense, as at each step we put one ball
into each box.
Now I reprogram the machine so that it does the same exact same thing,
but this time it moves the lowest numbered ball from B to C. It is
still placing one ball into each of B and C on each step. But let's
examine this more closely...
From A move 1 and 2 to B. From B move 1 to C. (1 sec) oo, 1, 1
From A move 3 and 4 to B. From B move 2 to C. (1/2 sec) oo, 2, 2
From A move 5 and 6 to B. From B move 3 to C. (1/4 sec) oo, 3, 3
Seems just as before. However after two seconds have elapsed we
find that box A is empty, as before, but box B is also empty,
and box C has all the balls in it. How can this be, in the face
of the fact that at each step we increase the number of balls
in B and in C by one ball, just as before? How can what number
is written on the ball we pick up affect the result?
NB: I know the answers, so don't instruct me. I figured this out
in about 1970 when I first encountered this. Also, if you know
the answer, I suggest you hold back, and let people think about it.
Mike
--
p="p=%c%s%c;main(){printf(p,34,p,34);}";main(){printf(p,34,p,34);}
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