Le=C2=A0lun. 20 juil. 2020 =C3=A0=C2=A010:36, Hans Hagen &l= t;j.hagen@xs4all.nl> a =C3=A9cr= it=C2=A0:

On 7/2= 0/2020 7:56 AM, Jeong Dal wrote:
> Dear Fabrice,
>
> You may split Binom(n,k) function into two functions as following:
>
=C2=A0> see original mail
=C2=A0>
> Dalyoung
Best stay in a protected namespace ...

\startluacode

=C2=A0 =C2=A0 =C2=A0local function fact (n)
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0if n <=3D 0 then
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0return 1
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0else
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0return n * fact(n-1)
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0end
=C2=A0 =C2=A0 =C2=A0end

=C2=A0 =C2=A0 =C2=A0local function ncr(n,r)
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0return fact(n)/(fact(r)*fact(n-r))
=C2=A0 =C2=A0 =C2=A0end

=C2=A0 =C2=A0 =C2=A0userdata.P =3D {
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0fact =3D fact,
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0ncr=C2=A0 =3D ncr,
=C2=A0 =C2=A0 =C2=A0}

=C2=A0 =C2=A0 =C2=A0function MP.pascal_ncr(n, r)
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0mp.print(ncr(n,r))
=C2=A0 =C2=A0 =C2=A0end

\stopluacode

Watch the last definition. This permits

=C2=A0 =C2=A0 =C2=A0 =C2=A0% tt :=3D lua("mp.print(userdata.P.ncr(&quo= t; & decimal n & "," &
decimal r & " ))");

replaced by

=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0tt :=3D lua.MP.pascal_ncr(n,r);

which looks nicer.

\startbuffer[pt1]
=C2=A0 =C2=A0 =C2=A0numeric n, r, s, u, dx, dy, tt;
=C2=A0 =C2=A0 =C2=A0path p, q;
=C2=A0 =C2=A0 =C2=A0pair A, B, start, now;
=C2=A0 =C2=A0 =C2=A0u :=3D 1.8cm;
=C2=A0 =C2=A0 =C2=A0A :=3D dir(210)*u;
=C2=A0 =C2=A0 =C2=A0B :=3D dir(-30)*u;
=C2=A0 =C2=A0 =C2=A0dy :=3D sind(30)*u;
=C2=A0 =C2=A0 =C2=A0dx :=3D 2*cosd(30)*u;
=C2=A0 =C2=A0 =C2=A0for n=3D0 upto 4:
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0start :=3D n*dir(210)*u;
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0for r=3D0 upto n:
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0s :=3D n-r;
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0% tt :=3D lua("mp.print(userd= ata.P.ncr(" & decimal n & "," &
decimal r & " ))");
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0tt :=3D lua.MP.pascal_ncr(n,r); =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0now :=3D start+r*right*dx;<= br> =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0dotlabel.top(textext("= $\displaystyle {" & decimal n &
"\choose" & decimal r & "} =3D "& decimal t= t & "$"),now);
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0draw (now+A) -- now -- (now= +B);
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0endfor;
=C2=A0 =C2=A0 =C2=A0endfor;
\stopbuffer

Now, in context lmtx we can have a different kind of abstraction. We can do this:

=C2=A0 =C2=A0 =C2=A0function MP.pascal_ncr_x()
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0mp.print(ncr(mp.scan.pair()))
=C2=A0 =C2=A0 =C2=A0end

and then use:

=C2=A0 =C2=A0 =C2=A0tt :=3D runscript("MP.pascal_ncr_x()") (n,r) = ;

Of course one can decide to pick to two numerics instead, like

=C2=A0 =C2=A0 =C2=A0tt :=3D runscript("MP.pascal_ncr_x()") n r ;<= br>
but i leave that as exercise.

=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0% tt :=3D runscript mp_pascal_ncr = (n,r) ;
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0tt :=3D pascal_ncr (n,r) ;<= br>
However, we still have the rather verbose runscript here, so we go
further, we register pascal as script:

\startluacode
=C2=A0 =C2=A0 =C2=A0metapost.registerscript("pascal_ncr",MP.pasca= l_ncr_x)
\stopluacode

And then define an alias at the metafun end:

\startMPextensions
=C2=A0 =C2=A0 =C2=A0newinternal mp_pascal_ncr ; mp_pascal_ncr :=3D scriptin= dex "pascal_ncr" ;

=C2=A0 =C2=A0 =C2=A0def pascal_ncr =3D
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0runscript mp_pascal_ncr
=C2=A0 =C2=A0 =C2=A0enddef ;
\stopMPextensions

The internal permits this:

=C2=A0 =C2=A0 tt :=3D runscript mp_pascal_ncr (n,r) ;

while the additional def permits

=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0tt :=3D pascal_ncr (n,r) ;<= br>
Now watch out, because we define pascal_ncr here, something
lua.MP.pascal_ncr(n,r) won't work because the last part gets expanded <= br> because that is what mp does (i'll probably cook something for that som= e
day).

Now, to come back to

=C2=A0 =C2=A0 "I couldn=E2=80=99t wikify it at that time because I don= =E2=80=99t know
=C2=A0 =C2=A0 =C2=A0how to.=C2=A0 I=E2=80=99ll do it soon."

looks like you suddenly have an additional challenge,

Hans

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