# A question with infinity [Part 2]

If you haven't seen my first post regarding infinity you can find it here: A question with infinity Thanks for all the constructive comments on my first post, and creative answers to my questions.

In this question I will show my work in finding an equation for an infinite dimensional array of boxes with one person in each all being moved into just one row of boxes so that each person has their own room. I'm asking for help checking my logic as well as making my series of equations into one expression. Step 3 of part 5 and the solution for part 5 are still works in progress.

If you do not understand what I am talking about please read my first question, it will explain what everything means.

First I would like to thank Andreas Blass for helping me find a more efficient formula.

The idea starts with a formula used to give people in a 2 dimensional infinite series of boxes (having l and h) directions to their new rooms all at the same height, none of them sharing a room.

2^r * c - 1 is the formula used to determine someones new room. This formula works because everyone on the first floor is made to take every odd room, or every other room, leaving every even room, or every other room open. This leaves an infinite number of rooms open because infinity divided by 2 still equals infinity. The floor above moves into every forth room, leaving every forth room or 1/4 of the rooms open. Again this leaves an infinite number of rooms open because infinity/4 = infinity. The people on floor 3 (h = 3) take every 8th room... I think you can see the pattern by now.

Now you are all caught up lets move on to

Part 4

You now have an inf. number of inf. of these 2d box arrays stacked on top of each other. Giving you a cube with inf. dimensions. With the same goal as before, fit all of the people into just one line of boxes.

Each box has a label like so

[4][62][27] represented by [l][w][h] | l = length | w = width | h = height


Step 1

I looked at this problem and saw it as a bunch of 2d arrays that needed to be smushed into 1d arrays that formed a 2d array that would then be smushed into a 1d array. So 2^(l * w - 1) = h and 2^(h) * l - 1 = new room number

Solution

I then simply made the two equations into one by plugging in for h.
2^(2^(1 * w -1))*l - 1

Part 5

Now you have an inf. dimensional array of boxes with people in them that need to be moved into a inf. string of boxes. Just thinking about 4 dimensional shapes hurts my brain... inf. dimensional = nightmare

The boxes now have an inf. number of tags like so [39][29][88]...
represented by [l]...[n-1][n-1][n] | l = length and is the first tage | n = the last tag


Step 1

I picture this as a series of 2d arrays being pushed into 1d arrays forever and ever. This means that using the last tag (n) and the first tag and the equation you can find the persons spot in the 1d array that forms the the 2d array that gets smushed into a 1d array... ectra This is represented by the following equation
2^(l * n - 1) = n - 1 | two to the power of tag one times the last tag minus one equals the second to last tag

Step 2

Then using the knowledge from step 1 I was able to generate some rules, like my name I make the rules in a "java" styled format :) (the rules allow any person not in an inf. dimensioned shape to figure out how what room they need to go to)
d = number of dimensions
n = last data tag
While (d > 2)
d = d-1 && n = 2^(n*l-1)
Then
2^(n*l-1)*l-1 = new room number

Step 3

I am still working on step three. Step three is putting these rules into one equation.

Solution

I am still need to think of a way to solve this problem, using the rules if n = infinity then I would not be able to tell people what rooms to go to because there is an inf. number of tags to go through.

Part 6 You now have an in number of these inf dimensional boxes......... JK I think its time for this to end, though I could continue with these until infinity :O.

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What is the question here? –  ronno May 12 '13 at 4:09
Read the second paragraph –  java May 12 '13 at 4:15