@sursumCorda The last example shown on that Mathematica help page is a permutation. Indeed, it is the example that you used in your first version of this Question, and it was what I coded my Answer for.

Maple has a command apply, which handles the vast majority of cases. It is quite rare that one one needs to resort to `?()` in top-level code, although one may need to overload it in an object module. The same applies to index (used twice in my Answer), which is an easier-to-use version of `?[]`. 

@sursumCorda If you could specify an algorithm for this (rather than just an example), I'd be happy to encode it in Maple as efficiently as possible. 

@sursumCorda Your example rules contain a=b. How is that rule represented in the output 

[[g, d, c], [e, a], [f, d, c], [b]]

I could do it with a two-line procedure if your rules were a permutation (or bijection), but they aren't in your example. And given that they aren't, I don't understand what you mean by "cycles". Your rules don't even represent a function, let alone a bijection.

@mmcdara K is the SI unit of temperature; kg is the SI unit of mass.

Just making sure I have this right: You want to solve a system of only two equations but 2014 unknowns, with no other restrictions?

Maple has a package, DeepLearning, that may be able to help.

If you already have an algorithm in mind, if you describe it precisely, it should be possible to implement in Maple. You may also find the ImageTools package useful.

I'm not sure if this helps: Maple can do both integrals, without restrictions on m, n, a, or b. The case m=n should be treated separately, but Maple has no trouble with that integral either.

@Vortex I wasn't trying to "clarify" anything or to criticize your terminology. There's nothing incorrect about calling it an integro-functional equation. 

I haven't investigated this in detail, but at first glance my guess is that the two equations are redundant, i.e., they say essentially the same thing.

Anyway, I totally agree with mmcdara's Answer.

Maple has the ability to numerically (and only numerically) solve "delay-differential" equations (DDEs). The "functional" aspect of your integro-functional equation is entirely of the "delay" type. I don't know whether there's any possibility of converting your equation to a DDE, but it's worth considering. 

Note that p is a bound variable (in the sense of mathematical logic), so it may be better to treat q as the equation's independent variable.

Even if the equation cannot be converted to a DDE, it may be worth considering using a custom algorithm that imitates DDE numeric techniques, for example, a very fine mesh of points in the interval p..p+q.

@sharmamanjeet391 Please attach a worksheet, as you did above, showing your failed attempt to implement the solution that I described.

@sharmamanjeet391 None of your systems contain any dependence on t. It just seems to be a placeholder. Thus they are actually ODEs not PDEs. Remove all occurences of t and use dsolve. I'm not saying that this will work, but it's worth trying.

And the method I described was not meant to be a workaround for this particular bug; it's just basic mathematical advice applicable even for algebraic equations. 

@Sphericalmoments Here is a procedure to create random multipartite tournaments having any number of partite sets. As you suggested, it was easy to do this by generalizing my line of code that you highlighted.

RandomMultipartiteTournament:= (N::seq(posint))->
local i:= 0, j, r:= rand(0..1), V:= combinat:-choose([seq]([$i+1..(i+= j)], j= N), 2), v;
    GraphTheory:-Graph({seq}(seq(seq(`if`(r()=0, [i,j], [j,i]), i= v[1]), j= v[2]), v= V))
:
G:= RandomMultipartiteTournament(3,4,5,6);
    G := Graph 4: a directed unweighted graph with 18 vertices and 119 arc(s)

#Verify correct number of arcs three ways:
3*(4+5+6)+4*(5+6)+5*6;
                              119
add(mul~(combinat:-choose([3,4,5,6], 2)));
                              119 
N:= [$3..6]: add(N[k]*add(N[k+1..]), k= 1..nops(N));
                              119

#Verify number of partite parts:
GraphTheory:-ChromaticNumber(GraphTheory:-UnderlyingGraph(G));
                               4

Ordinarily, I'd write you a lengthy explanation of every syntactic element of my command that you asked about, but I have a limited ability to type at the moment (the pain extends down to the fingertips of my right hand). So, can you tell me specifically which parts you don't understand? 

@Sphericalmoments Although this solution is not immediately obvious to me (especially considering my current condition), I suspect that the maximum count that you want can be determined for arbitrary (n x m) bipartite tournaments by simple combinatorics. Perhaps your reason for exploring the problem iteratively is to get ideas for such a formula.