Hidden numbers?

[Pumices]

I am really stumped on this one, I have been working on it for nearly 2 weeks now and I am beginning to wonder if it is even possible. So I though I might throw it out there because there are a whole lot of people that are alot smarter than me that might be able to figure it out. So lets say you have A,B,C, and D all integers 0 - 1000. Now I am trying to find an equation that will allow me to take A,B,C and D and come out with A,Z, and D. But the hard part is I have to be able to take A, Z, and D and come out with the original B and C. Any help at all would be greatly appreciated. Thanks

[Pumices]

I forgot to mention that z also has to be a integer 0-1000 thanks again

[ydoaPs]

is [math]0{\leq}A<B<C<D<Z{\leq}1000[/math] or [math]0{\leq}A{\leq}B{\leq}C{\leq}D{\leq}Z{\leq}1000[/math] or is it a different order or is there no order?

[Pumices]

There is no order, A,B,C,D, and Z can be any given number 0-1000

[Daecon]

B and C need to be prime numbers for it to work, I imagine.

 

Multiply them together to get Z.

[timo]

I don´t really get what you´re doing but maybe this helps:

You seem to map (B,C) -> Z and want the inverse Z -> (B,C). As the number of combinations (B,C) |{(B,C)}| is about 1 million while |{Z}| = 1001 the map (B, C) -> cannot be injective. Therefore it cannot be bijective.

In the case that Z is not limited to an integer in the range of 0-999 (1000 elements is easier to talk about and the extension to 1001 is not too hard) but to 1 milion elements, a bijection can be made rather easily by: (B, C) -> Z = 1000*B + C with the reverse Z -> (B, C) = (Z div 1000, Z mod 1000).

[Pumices]

So basically it is impossible. So in theory for any given set of numbers n long where all numbers in the set fall in the same range. That set of numbers can not be converted to a set of numbers n-1 long with all numbers in the set having the same range as the numbers in the first set and then be converted back to created the original sequence. Hum... I don't like impossibilities

[Pumices]

Ok let me throw this out to you guys. What if those numbers 0-999(to make it easier) were able to represent any given set of numbers (ex. 1 = 1200, 7 = 9000 etc.) and you could have any number of fixed points for instance you could have A, B, C, D, E, and F and come out with A, Z, D, E, and F or something along that line. Pretty much anything goes. The only stipulation is that all the numbers have to be in the same range. Is this possible in anyway no matter how extreme or if you have any ideas how if could possibly be done please share. Thanks again for all the help.

[timo]

To create a bijection between two sets, both sets must have the same number of elements. If the sets A, B, C, D, E, F and Z all have the same finite number of elements, then no bijection {A}x{B}x{C}x{D}x{E}x{F} <-> {A}x{Z}x{D}x{E}x{F} can exists.

[ElijahJones]

To create a bijection between two sets, both sets must have the same number of elements. If the sets A, B, C, D, E, F and Z all have the same finite number of elements, then no bijection {A}x{B}x{C}x{D}x{E}x{F} <-> {A}x{Z}x{D}x{E}x{F} can exists.

 

Yeah its easy you cannot have a bijection between sets of different sizes. That doe snot mean you cannot find maps that have useful structure though. On pumices problem let me see what I get form his original question.

 

Ok Pumice you want

 

[MATH]f_1(A,B,C,D)=g_1(A,Z,D)[/MATH]

[MATH]f_2(A, Z,D)=g_2(B,C)[/MATH]

 

I think you are generalizing your problem too much and by restricting you numbers to 0-1000 are actually making the problem harder. Just let everything be integers any integers what soever or if you have a fetish say let them be any natural numbers.

 

If you would give me more details about your specific problem or possible insight I can saymore but here is somethin gthat satisfies your requirements from the first post.

 

[MATH]A+B+C+D=A+Z+D[/MATH]

 

This satisfies both of your conditions because you can substract A and D from both sides to get B+C=Z. Or multiplicatively you could go

 

[MATH]ABCD=AZD[/MATH]

 

Again the same thing is implied now just divide both sides by AD to get BC=Z.

 

What was your reason for asking this question if I can ask?

[Pumices]

Hey, thanks for all the replies and sorry i haven't been around for a while but the question came about when me and my friend were talking about data compression. Basically if the above problem were in someway true it would be possible to take the bits of a file numbers 0-255 and compress them but then be able to do it again. Basically you would be able to compress a given file to nearly any size you want.

[BigMoosie]

It is impossible to compress completely random bits, compression algorithms take advantage of common patterns and such.

[cosine]

Another way (at least the way I'm looking at it) to show why you can't do this, is because your going from something with 4 dimensions (4 variables) to 3 dimensions (3 variables) which you can do, but then trying to go backwards. I would just like to let you realize this intuitively by imagining projecting every point on a plane onto a line. This shouldn't be too hard, but what about projecting every point on a line onto a plane? The projection could only make a line. To make more you would be inserting new information.

 

Hope this helps.