vv

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@byrktr 

Sorry, I don't know. But usually in such problems some heuristic is used.
BTW, I have corrected a typo in cond3.

Edit.
Here is an idea.
Solve the problem in real x[i,j,k],  in the interval 0..1. Maple can do it quickly.
One obtains an "approx" solution (with some conditions violated). Maybe you can start from here.

i1:=indets([cond1])<=~1:
i0:=indets([cond1])>=~0:
i01:=i0[],i1[]:

fracsol:=Optimization:-LPSolve( add(q[i],i=1..20), {seq(cond||i, i=1..6), i01} );
sol01_approx:=map(t -> (lhs(t)=round(rhs(t))), fracsol[2]);

@byrktr 

It is always a good idea to post the mathematical presentation of the problem. Then, an experienced user could come with a simpler/alternative solution.

@student_md 

Actually, it's the absolute error. For the relative error, use

numapprox:-infnorm((y-mExact)/mExact, 0..1);

 

@denbkh 

I can easily give recurrences with "complex" sums which cannot be solved.
If you have a specific one, why don't you post it?

You have just produced a duplicate. You already asked this in your previous question and have received a partial answer.
Why don't you wait, maybe someone will come with the rest.
Note that usually such duplicates are simply deleted!

@Chrono1 

For a polyhedron the same method can be used but the formulae for  anew,bnew in doit are more complicated.
Actually even for a non-regular tetrahedron these formulae are much more complicated.

 

@Carl Love 

And it's not so?

@Carl Love 

You are right, LinearAlgebra:-Generic works with GF. I had an error when I have checked but probably I forgot a restart.
So, a conversion is not needed.

@Magma 

LinearAlgebra[Generic] seems to be incompatible with GF.
So, I made some conversions.

I have also changed your example, because it was already in a reduced form and had the rank=3.

restart;
G := GF(2, 4);
p:=2;
ex := G:-extension;
z:=indets(%)[];
poly:=convert(ex,polynom);
#  A := `~`[G:-input](Matrix(3, 6, [[1, 0, 0, 5, 0, 4], [0, 1, 0, 8, 7, 2], [0, 0, 1, 3, 0, 1]]));
A0 := <1, 0, 0, 5, 0, 4; 2, 0, 3, 10, 7, 8; 1, 0, 3, 5, 7, 4>;
A := G:-input~(A0);
AA:=G:-ConvertOut~(A);
with(LinearAlgebra[Generic]):
GG[`0`] := 0:
GG[`1`] := 1:
GG[`=`] := `=`:
GG[`+`] := ()->modp(`+`(args),p):
GG[`-`] := (a,b)->`if`(nargs=1,modp(-a,p),modp(a-b,p)):
GG[`*`] := (a,b)->modp(Rem(a*b,poly,z),p):
GG[`/`] := proc(a,b) local i;
             if b=0 then error "division by zero"; end if;
             Gcdex(b,poly,z,'i') mod p;
             Rem(a*i,poly,z) mod p;
           end proc:
ReducedRowEchelonForm[GG](AA);
G:-ConvertIn~(%);
G:-output~(%);

 

@mmcdara 

Yes, such an m exists. But unfortunately it depends on (a,b,c) which does not make the inequality very interesting.

@mmcdara 

You seem to consider the question as being about uniform distributions.
The solution was just a simple and quick way to obtain the counterexample. The "guess measure" is just a (not very useful) by-product. But I agree that a more natural (not necessarily quicker) and uniform way do generate a,b,c is  sort( ['r()' $3] ).

@Rouben Rostamian  

The problem has much in common with dynamical billiards (see wiki), so, an animation would be also nice.

@Preben Alsholm 

But waiting, we also have a measure of how far was the guess.

@mmcdara 

OK, but I met Olympiad students who know Newton's formulae. It's a good opportunity for the OP to know about them.

@mmcdara 

The standard and straightforward way to solve such problems is via Newton's identities.

s[0]:=3: s[1]:=1: s[2]:=2:    #   s[3]:=3:
for n from 3 to 30 do
s[n]:=s[n-1]+1/2*s[n-2]+1/6*s[n-3] od;
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