Sounds like cryptography.  I can't understand however, why the last number could be 256.  That isn't in the extended ascii, nor could it be held by a character (in computers).

You want a function that is one to one.   What this means is that:

if f(a, b) = f(c, d) then a=c and b=d

Now you don't say that this function has to use all of it's range.  That is, for every b in the result (the range) of the function, there is a y that gives you it:

For all b in Range, there exists an a in Domain such that f(a) = b.

This is called onto.

One such function that is 1-1 (one to one) but isn't onto is summation of primes:

f(a, b) = a + b, where a and b are primes.

I believe this is 1-1, but I don't think I can prove it.  However, this is fairly hard to get the original values a and b out when you are given a number such as 12.

But any answer for what i asked??is there any way???u can use and or opeartions also...thats is convetion t o binary and back all///

u think u dnt get what i mean.....we are given any two numbers between 0 and 256.From this two numbers we must obtain a 3rd number suc that afterwards we may be able to reproduce this two numbers from the single number..In wha u explained only i can be done to perfect square numbers...Guive a general one for all numbers

Also discussed here

Is there any method by doing some combination of different AND OR operations or any such operations on the two numbers and later then doing some other steps to obtain the two numbers from the single number??

TRY it its very interseting..............

Well, we have had fun with this one!

I now want to see soorejmg's solution

No soorejmg, that's not what I said.  It will _NOT_ exist.  There is _NO_ such function.

These are fairly in depth things.  I will try to explain it the best I can, but I won't be surprised if you don't understand.  These topics take quite a while to fully grasp.  If you want to look up more information for yourself, this is known as Set Theory and more specifically, the Pigeon Hole Principle.  If you have any questions, just ask.

You have sets A and B.  A is the domain, B is the range.  Also keep in mind that |A| means the number of elements in set A.  1-1 means:

For all x∈A and all y∈A, if f(x)=f(y), then x=y.

What this basically means is that every domain value goes to one range value, and no two different domain values go to the same range value.  Your function must be 1-1 to work, otherwise, you wouldn't be able to tell the first two numbers from the third.

An example of a non 1-1 function is f(x) = x^2.  (-3)^2 = 3^2, but -3 ≠ 3.  An example of a 1-1 function is f(x) = x.  Because if f(x) = f(x), x = x, which is always true for any value x.

If a function is 1-1, than the number of values in the range has to be greater or equal to the number of values in the domain.  |B| ≥ |A|

This is no problem.  But you wish your function tho be onto:

For all b∈B, there exists an a∈A such that f(a) = b.

What this says is that for every single value in the range, there is some value in the domain which goes to it.

An example of a non onto function is (again) f(x) = x^2.  There is no value in this function which goes to -1.  f(x) = -1 just isn't possible.  On the other hand, f(x) = x is onto.  You can find any real value in the range.

If a function is onto, than the number of values in the domain must be greater than or equal to the number of values in the range. |A| ≥ |B|.

Put 1-1 and onto together, and you get |A| ≥ |B| and |B| ≥ |A|.  This means that |A| = |B|.  However, for you, your domain is 256*256 and your range is 256.  256*256 ≠ 256.  Therefore, it is impossible to make a function that is 1-1 and onto.

Last edited by Ricky (2006-01-23 03:09:16)

Thanks irspow.  That is certainly one of my options.  I'm not quite sure what to do yet.  At least I have a few years to decide.

You are correct, f(x, y) is different than f(x) and f(y), but so is the explanation.

Remember, I never named the sets I was talking about in my explanation.  This is because it applies to every set and every function between those sets.

Let A = {(x, y) | x ∈ [0,255], y ∈ [0,255]}, the domain.  A is a set of ordered points, just like those in your function.  How many elements are in A?  255*255.

Let B = {z | z ∈ [0, 256]}, the range.

Now apply my previous post.

Good question, by the way.

rickyoswaldiow, that only works for 3.

f(x, y) = x + y

f(x, y) = 15, what are x and y?

Ricky wrote:

You are correct, f(x, y) is different than f(x) and f(y), but so is the explanation.

Remember, I never named the sets I was talking about in my explanation.  This is because it applies to every set and every function between those sets.

Let A = {(x, y) | x ∈ [0,255], y ∈ [0,255]}, the domain.  A is a set of ordered points, just like those in your function.  How many elements are in A?  255*255.

Let B = {z | z ∈ [0, 256]}, the range.

.

This one i understood.But not hat i asked.let A be the set above and B too.The connection between the domain and the range is through the function aplied on the domain set know?

This functions like onto  etc etc(i have studied it earlier dnt remember) comes when the FUNCTION is applied to each indiviual  member on domain.Here what is being done is the function is applied to two members of the set A.

so  how can it be?? Dnt kow if i am wrong..