The Box Confusion?

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There is a box. We put into it k smaller boxes OR no ones. Then, for each of the new boxes, we put k smaller boxes into it OR no ones. That means that for each box there are only two alternatives: either it is left empty or it is filled with k new boxes. Which alternative has been realized for every given box is not known. And this procedure is repeated for an unknown (finite, of course) ammount of times.

Now there are m full boxes in the structure.

Given m and k, find the number of empty boxes!

If you don't understand the riddle question are welcome.
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too be techinical, this is not a riddle. could be the same m could equal k or even m< k
Of course, m=k, m>k and m0-1m and k are integers.

The numbers m and k are given. You are to calculate the number of empty boxes through m and k, that is, find a formula.
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Ant_222There is a box. We put into it k smaller boxes OR no ones. Then, for each of the new boxes, we put k smaller boxes into it OR no ones. That means that for each box there are only two alternatives: either it is left empty or it is filled with k new boxes. Which alternative has been realized for every given box is not known. And this procedure is repeated for an unknown (finite, of course) ammount of times. Now there are m full boxes in the structure. Given m and k, find the number of empty boxes! If you don't understand the riddle question are welcome.
Well, suppose we numerate the boxes as we add them and define by the function f(n) = {0,1} is the n-th filled with k new boxes {1} or not {0}. In general case for any function f it is not possible to give a shorthand formula.

In general, after we load all boxes all we can do is to count.

For some cases like when we know at least something about how we fill in the boxes, for example we know that in half cases we fill and in half cases we leave a box empty, we can give a statistical formula. For some special cases there is a way to find a quick way (to find a formula) to calculate the final number of empty boxes, for example, if we know that we fill boxes this way: 5 we fill 5 we do not. In these cases, we use elements of fractal math. Without that there is no closed formula apart from a pure counting.

There are some techniques that give some superior approximation for every case of f(n), they require a lot of knowledge about f(n) but they are all far from the average English forum user to understand. There is no particular difference between this last mentioned analysis and counting, actually what we have to know about f(n) is more complex than counting. Yet, this last mentioned form of analysis allows for generalization. In case of counting there is no possible generalization, no formula.

So what f(n) did you have in mind? How you wanted the box to be filled? What is the procedure used? Can we assume at least that on average in half cases we have box filled an in half we do not?
«In general, after we load all boxes all we can do is to count»

In general yes. But since I have given a certain structure build by certain rules (put 0 or k boxes in each empty box) and you know the number of filled boxes, that's not a pure general case. Use the data you have.

«In these cases, we use elements of fractal math. Without that there is no closed formula apart from a pure counting.»

Happily, no fractals needed. And yes, there is a closed formula. No probabilities are given, so the formula cannot be statistical. Olny an exact, explicit, 100% formula.

«So what f(n) did you have in mind? How you wanted the box to be filled? What is the procedure used? Can we assume at least that on average in half cases we have box filled an in half we do not?»

The procedure is described in the first post. No additional assumptions needed.

If nobody solves it in week, I'll post the answer and a solution (I know at least two exist).

Good Luck.
Ant_222«In general, after we load all boxes all we can do is to count» In general yes. But since I have given a certain structure build by certain rules (put 0 or k boxes in each empty box) and you know the number of filled boxes, that's not a pure general case. Use the data you have. «In these cases, we use elements of fractal math. Without that there is no closed formula apart from a pure counting.» Happily, no fractals needed. And yes, there is a closed formula. No probabilities are given, so the formula cannot be statistical. Olny an exact, explicit, 100% formula. «So what f(n) did you have in mind? How you wanted the box to be filled? What is the procedure used? Can we assume at least that on average in half cases we have box filled an in half we do not?» The procedure is described in the first post. No additional assumptions needed. If nobody solves it in week, I'll post the answer and a solution (I know at least two exist). Good Luck.
Then, if we take this as a trick question the answer is

mn-m+1

or

m(n-1)+1

Of course I could write a complete answer, but I'll leave room for the logic of the answer to other people.
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Of course! But, yeah, even the formula itself is a hint...

Why is it a trick question?
Ant_222Of course! But, yeah, even the formula itself is a hint... Why is it a trick question?
I'll leave to the other to think. It is interesting - though.

I don't understand do you ask me why I think it is a trick question (because you think it is not) or you ask me why it is a trick question (and you thinks that as well).

Whatever, the trick is in:

"Now there are m full boxes in the structure."

It is not said when.

Now if you think that you start with m boxes and follow the rules for each of them... you will need a PhD to solve the puzzle.
«I don't understand do you ask me why I think it is a trick question...»
The answer would depend on what you mean by "trick": a lurk or a thing that requires smart (more or less) thinking. The second is my opinion.

«Now if you think that you start with m boxes and follow the rules for each of them... you will need a PhD to solve the puzzle.»

If to change the problem this way no PhD will help... Maybe a clairvoyant, to pry into the author's mind.
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