Clifford base names WAS: Re: [GiNaC-list] How can I get a GiNaC::ex's name

[Sheplyakov Alexei]

Hello!

On Thu, 2007-05-24 at 14:50 -0400, Warren Weckesser wrote:
> This works for me:
> 
> string s = ex_to<symbol>(E).get_name();
> 
 
Firstly, I'd like to note that GiNaC expressions do not have any
"names" at all. Only symbols have "names" which serve merely as
a printing representation. Two different symbols may have the same
name, as in

symbol x("x"), y("x"); // valid code

or 

namespace A {
	symbol x("x");
}

namespace B {
	symbol x("x");
}

So your code will work if (and only if) E is a symbol. And it is good
idea to check if E is indeed symbol, since ex_to<T> functions are unsafe
(as documented in the manual), e.g.

string s = "dunno";
if (is_a<symbol>(E))
	s = ex_to<symbol>(E).get_name();


On Fri, May 25, 2007 at 10:48:44AM +0200, Javier Ros Ganuza wrote:

> I have a similar problem, if I define the clifford units this way,
> 
>  varidx nu(symbol("nu", "\\nu"), 3);
>  ex     basis1 = clifford_unit(mu, diag_matrix(lst(1, 1, 1)),0);
> 
>  ex e00 = basis1.subs(mu == 0),
>     e01 = basis1.subs(mu == 1),
>     e02 = basis1.subs(mu == 2);
> 
> Then they are ploted
> 
> cout << e00 << endl;
> 
> Like this
> 
> e~0
> I want to rename using symbol.set_name(string), but
> when I do
> 
> cout << ex_to<symbol>(e00).get_name() << endl;

This should segfault. e00 is not a symbol, it is element of Clifford
algebra. So ex_to<symbol> is meaningless here...

> Anything gets plotted.

Well, some sunny day it may even print something. Probably you are
lucky :)

> So I suspect that e00 is not a symbol an then,

It is definitely not a symbol. It is an element of Clifford algebra.

> It woul be interesting to have this posibility, tough elements of
> diferent basis are right now plotted in the same way.

I don't think renaming symbols will help here.
 
> Basis1 e00, plots as e~0
> 
> Basis2 e10, plots as e~0

> Nevertheless if plotting in latex style, units are plotted in a
> different way.

Indeed. And this is 
$ cat test.cpp
#include <iostream>
#include <ginac/ginac.h>
using namespace std;
using namespace GiNaC;

int main(int argc, char** argv) {
  varidx nu(symbol("nu", "\\nu"), 3);
  ex basis1 = clifford_unit(nu, diag_matrix(lst(1, 1, 1)), 0);
	ex basis2 = clifford_unit(nu, diag_matrix(lst(3, 1, 2)), 1);
  ex e10 = basis1.subs(nu == 0),
     e20 = basis2.subs(nu == 1);
	cout << latex << 
		"e10 = " << e10 << endl <<
		"e20 = " << e20 << endl << dflt <<
		"e10 = " << e10 << endl <<
		"e20 = " << e20 << endl;
	return 0;
}

$ g++ test.cpp; ./a.out
e10 = \clifford[0]{e}^{{0} }
e20 = \clifford[0]{e}^{{1} }
e10 = e~0
e20 = e~1

> How can I deal with this?.

May be make both formats print the same (well, almost) thing, like
this patch does:

diff --git a/ginac/clifford.cpp b/ginac/clifford.cpp
index c7ce8d3..56ae54f 100644
--- a/ginac/clifford.cpp
+++ b/ginac/clifford.cpp
@@ -203,6 +203,7 @@ void clifford::do_print_dflt(const print_dflt & c, unsigned level) const
                seq[0].print(c, precedence());
                c.s << "\\";
        } else
+               c.s << "e[" << int(representation_label) << "]";
                this->print_dispatch<inherited>(c, level);
 }
 
Best regards,
 Alexei

-- 
All science is either physics or stamp collecting.

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