Bijection from (Z_n)^3 to (Z_n)^2

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Dynotec
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Bijection from (Z_n)^3 to (Z_n)^2

Postby Dynotec » Sun May 29, 2011 3:43 pm UTC

I'm looking for a nice bijection from {1,...,n}^3 to {1,...,n}^2. I'm hoping to use it for a hack with an unfortunate statistical modeling program that doesn't handle vectors very well beyond two dimensions.

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The Sleeping Tyrant
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Re: Bijection from (Z_n)^3 to (Z_n)^2

Postby The Sleeping Tyrant » Sun May 29, 2011 4:00 pm UTC

I'm afraid that whenever n is bigger than 1 there isn't any bijection between them, since the first contains n^3 elements while the second has only n^2 elements.

Dynotec
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Re: Bijection from (Z_n)^3 to (Z_n)^2

Postby Dynotec » Sun May 29, 2011 4:33 pm UTC

The Sleeping Tyrant wrote:I'm afraid that whenever n is bigger than 1 there isn't any bijection between them, since the first contains n^3 elements while the second has only n^2 elements.


Yes, of course. I feel dumb having asked the question. What about from {1,...,n}^3->{1,...n}x{1,...,n^2}?

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jestingrabbit
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Re: Bijection from (Z_n)^3 to (Z_n)^2

Postby jestingrabbit » Sun May 29, 2011 4:35 pm UTC

There's always the natural bijection, (a,b,c) -> (a, (b,c)). There are also a myriad other maps. What sort of thing are you looking for?

edit: misread the problem, saw the ^2 as outside the }. Gmal's is right.
Last edited by jestingrabbit on Sun May 29, 2011 7:39 pm UTC, edited 1 time in total.
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Dynotec
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Re: Bijection from (Z_n)^3 to (Z_n)^2

Postby Dynotec » Sun May 29, 2011 5:06 pm UTC

jestingrabbit wrote:There's always the natural bijection, (a,b,c) -> (a, (b,c)). There are also a myriad other maps. What sort of thing are you looking for?


A simple easily computable and invertible function that takes a three-tuple of integers and returns a two-tuple of integers.

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Re: Bijection from (Z_n)^3 to (Z_n)^2

Postby gmalivuk » Sun May 29, 2011 5:10 pm UTC

Dynotec wrote:What about from {1,...,n}^3->{1,...n}x{1,...,n^2}?
(a, b, c) -> (a, (b-1)n + c)
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