@asa12 An equation of the form H(x) + H(y) = H(x+y) is called a functional equation. As far as I know, Maple has no facility whatsoever for solving functional equations. I also have not heard of any other software for it.

You erased your original Question after it was Answered. That is a major no-no. Major. Please put back the original Question and especially its title (or something close to them), and ask the above as a separate and new Question. I won't answer any of your questions until you take care of this.

Even if they're not ASCII, they're still printable characters; so, does it really matter?

If it does matter, then group them into groups of 7 bits instead of 8.

If you can find the algorithm for it, I'll code it in Maple.

Update: I've found an algorithm, and I'm working on it.

Update 2: I've finished coding the algorithm, but before I post it, please provide some two-dimensional data for me the test it on.

@peter2108 In the same place that there is the button to Vote Up the Answer, there is also a button that looks like a trophy cup. You can use that button to select an Answer as the Best Answer to your Question. I think that using that button doesn't require any reputation points.

@marc sancandi In the absence of any statement to the contrary, I think that it's reasonable to assume that the OP wanted a uniform distribution over all 21 pairs (x,y) such that 1 <= x < y <= 7. I don't see how you can say a priori that one distribution is "right" and the other "wrong"---it all depends on what is desired. 

@torabi Yes, Maple has no numeric solver for elliptic PDEs. You're out of luck.

@John Fredsted I think that it's the best option so far (for a small integer like 7). I didn't hold out any hope of the combinat:-randcomb being efficient.

Is f3(x) a known function? Can h3 be given a value before solving? On a quick read through, I don't see anything that makes this system nonlinear.

@Carl Love Here's a novel approach, and it's a short one-liner.

GenXY:= ()-> combinat:-randcomb(7, 2)[];

This generates a uniform distribution.

@Kitonum The distribution that you're sampling has the same problem as John's. See the histograms that I posted in a Reply to his Answer.

@John Fredsted 

The difference in the distributions being sampled is profound, as the following histograms show. Figuring out why this difference occurs might be a good problem for a junior- or senior-level probability course.

Download two_distributions.mw

@yongling93 In that case, you want to store your bits in a Matrix.  Then all your shifting (what you call "crossover") can be done with surprisingly simple indexing commands. Like this:

B:= Matrix([[0,0,1,1,0,0,1], [0,0,1,1,1,0,0],[0,1,0,1,0,1,0], [1,0,0,1,1,1,0]]):
Bnew:= copy(B):
Bnew[[4,1,2,3], 3..5]:= B[.., 3..5]:
Bnew;

B;

I converted this from a Question to a Post because the question asks for open-ended discussion and opinions rather than definite answers.

@Carl Love Please let me know the outcome of this assignment. I particularly want to know if my guess about the professor trying to stump you with a impossible n was correct.