Tearing up a piece of paper and putting it back together

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Today on tearing up a piece of paper before throwing it in the bin i thought to myself how difficult would it be to put those pieces back together to read what was on it.

I'm talking about central tears every time (then piling the pieces back together and doing another central tear etc).

Just thought about it for a bit and i was actually quite surprised it's something like the factorial of (2^n) times 4 (if you take each piece as identical and rectangular say).
That's pretty amazing.
So if you tear a piece of paper up in this way say 8 times the number of ways of putting it back together is factorial of 256 times 4. The difficulty in retrieving the information is obviously related to this (not sure how exactly- would be interesting to know the maths of this btw).

Ye might check my maths there but i think that's right.
Just thought it was interesting.

ray giraffe

Looks like it should be [latex](2^n)!4^n[/latex]

Edit: Oops it should be[latex] \displaystyle (2^n)! 4^{2^n} [/latex]

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ray giraffe wrote: »

Looks like it should be [latex](2^n)!4^n[/latex]

Hi
Why is it 4^n can you tell me.
The only reason i thought it was 4 is because each remaining torn piece of paper can be oriented in 4 ways (or 2 for just one-sided text). Where does the ^n come in.

It's incredible that by just tearing a piece of paper a few times you can create that much disorder. Never realised it would be so much.

ray giraffe

take everything wrote: »

Hi
Why is it 4^n can you tell me.
The only reason i thought it was 4 is because each remaining torn piece of paper can be oriented in 4 ways (or 2 for just one-sided text). Where does the ^n come in.

It's incredible that by just tearing a piece of paper a few times you can create that much disorder. Never realised it would be so much.

Oops this should be better:

[latex] \displaystyle (2^n)! 4^{2^n} [/latex]

Each of the 2^n pieces can be in 4 orientations

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ray giraffe wrote: »

Oops this should be better:

[latex] \displaystyle (2^n)! 4^{2^n} [/latex]

Each of the 2^n pieces can be in 4 orientations

Sorry I can't get my head around this.
Can you explain 2^n as the exponent for 4.
Apologies for not getting it.

ray giraffe

take everything wrote: »

Sorry I can't get my head around this.
Can you explain 2^n as the exponent for 4.
Apologies for not getting it.

Each of the 2^n pieces can be in 4 orientations
You have a sequence of (2^n) choices, each time with 4 options.

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ray giraffe wrote: »

Each of the 2^n pieces can be in 4 orientations
You have a sequence of (2^n) choices, each time with 4 options.

OK.
So after one tear you have two pieces.
Just taking the pieces on their own, you have 4 ways to orientate each one. And 2 (or (2^1)!)ways to arrange them.
So total ways of putting them back together is (2^1)!*4.

After 2 tears you have 4 pieces. Again you still have 4 ways to orientate each individual piece. And now 4! (or (2^2)!) ways to arrange them.
So total ways of putting them back together is (2^2)!*4.

So generalising to n tears, you have (2^n) ways to arrange them, but you still have 4 ways to orientate each piece don't you?
Isn't the 4 ways of orientating each individual piece staying constant here.

What am i missing.
Sorry.

Edit:
Sorry i think i see it at last.
4 is multiplied by itself (2^n)! times.

Edit 2:
OK i still haven't got it.
Am i some way on the right track with that.

ray giraffe

take everything wrote: »

OK.
So after one tear you have two pieces.
Just taking the pieces on their own, you have 4 ways to orientate each one.

4 ways to orientate the first piece (A,B,C,D), 4 ways to orientate the second(A,B,C,D).

So there are 4x4 =16 ways to do the job of orientating both:

AA
AB
AC
AD
BA
BB
BC
BD
CA
CB
CC
CD
DA
DB
DC
DD