May 2002 - GiNaC-list

More Spam...
by Richard B. Kreckel 28 Oct '02

28 Oct '02

Dear folks, We apologize for the #@&*%! Spam you received through this list. Please bear with us while a solution (likely SpamAssassin) will be deployed. Meanwhile, since it hasn't been on our lists yet, I am pleased to spam you with some shameless ad about the Workshop on Open Source Computer Algebra Tuesday 21st - Thursday 23rd of May 2002 Lyon, France <http://www.lapcs.univ-lyon1.fr/~nthiery/CalculFormelLibre/workshop.html> Hurry up, in case you are interested. ¡Hasta la vista! -richy. -- Richard B. Kreckel <Richard.Kreckel(a)Uni-Mainz.DE> <http://wwwthep.physik.uni-mainz.de/~kreckel/>

2 1

small inconsistency?
by Ben Sapp 27 May '02

27 May '02

Hi, Below I provide a small example program which I think illustrates a small inconsistency in GiNaC. The program prints out x==4 x==4 and then exits. I think that if the "==" operator were handled consistently with other operators it should print out y==4 x==4 If I change the "==" operator to "+" it behaves as I expect. Cheers, Ben. ------------------------------------------------------------- #include <ginac/ginac.h> using namespace GiNaC; int main(int argc, char *argv[]) { symbol x("x"), y("y"); // change the "==" to a "+" and it works as expected ex b = (x == 4); cout << b.subs(x == y) << std::endl; cout << b << std::endl; return 0; } -------------------------------------------------------------

2 1

Note: compilation problem / CLN / "missing libraries"
by bullet.train 27 May '02

27 May '02

Dear List, I have had a long chase trying to compile GiNaC- The problem has been trying to compile the CLN tests and libraries. The symptoms are that the CLN test break and GiNaC fails to compile- compiler g++ --version 2.96 (nasty RH version) under linux the problem is caused by the following line in cln-config libdirs=-L${exec_prefix}/lib If this is present then the compiler flags see -L/usr/local/etc/etc and then the compiler cannot find any of the usual libraries and so breaks this test and doesn't make. Commenting out this line with '#' fixes the problem. This is probably not a ginac problem- so this is for information only- SA

2 1

relational creation
by Ben Sapp 20 May '02

20 May '02

Hi, I noticed in relational.cpp there is a line like this: (Should be one line but my mailer brakes it up) ----------------------- new relational(lh.eval(level-1),rh.eval(level-1),o))->setflag(status_flags::dynallocated | status_flags::evaluated); ---------------------- The question I have is if I am creating relationals using the "new" operator do I need to call the "->setflag(status_flags::dynallocated)" function? Thanks, Ben.

2 2

expressions inside a user-defined class.
by Juan Jose Arribas 17 May '02

17 May '02

Hi,: I am trying to introduce symbolic functions in a numeric fem code. To do that I want to store expressions (functions) inside a class. I read outside the class a string with the function (example, sin(t) ), then I create an object for store it. I use the constructor to pass the string. In the constructor I create the expression, and I intend to use it later (with the valor function), but in the constructor I always get a segmentation fault when I call the ex constructor. It must be something obvious but I am unable to find it. I have tried to debug with gdb and put a try--catch group but I have not found the explanation of the segmentation fault error. The code I am using to test it is as follows: ========================================================================= #include<iostream.h> #include<ginac/ginac.h> #include<ginac/lst.h> using namespace std; using namespace GiNaC; class fext { public: string funcionstr; ex funcion ; symbol tiempo ; lst l ; public: fext (string &); // Constructor double valor(double); }; fext::fext (string & funcionc) { funcionstr = string(funcionc) ; tiempo = symbol("t") ; funcion = ex(funcionstr,lst(tiempo)) ; } double fext::valor(double tt) { return( ex_to<numeric>(funcion.subs(tiempo==tt)).to_double() ) ; } int main() { string st ; cin >> st ; fext fexx = fext(st) ; cout << "\n input t-value " ; double valorcillo ; cin >> valorcillo ; cout << "\n fexx value" << fexx.valor(valorcillo) ; } ================================================================= I am using debian-linux (woody) with the following versions: ginac-config --version 1.0.8 g++ -v Reading specs from /usr/lib/gcc-lib/i386-linux/2.95.4/specs gcc version 2.95.4 20011006 (Debian prerelease) To compile it I use g++ prusimple.cc -o prusimple -lginac -lcln Thanks in advance for any idea. Juan Jose Arribas Mechanics Department Civil Engineers School Politechnic University Madrid Spain

3 3

suggestion for the manual
by Chris Dams 13 May '02

13 May '02

Hello, I think the manual should mention that one can use the complex numer I, by simply typing I. Perhaps it could be included in the list of constants on page 15. Greetings, Chris Dams

2 1

print_latex
by Alessandro Zaccagnini 10 May '02

10 May '02

I have been using print_latex and I found that the following piece of code #include <iostream> #include <ginac/ginac.h> using namespace std; using namespace GiNaC; int main() { GiNaC::ex q; q = 2 * pow(6, numeric(1,3)); q.print(print_latex(std::cout)); } produces the output 2 6^{{(\frac{1}{3})}} which latex then typesets in the wrong way, since the space is lost. Putting a \cdot in place of the blank seems to fix the problem. Best wishes. Alessandro Zaccagnini Universita' di Parma

2 1

function has
by Tatiana Zolo 09 May '02

09 May '02

Hi, the behavior of function 'has' in ginsh seems inexplicable: ginsh - GiNaC Interactive Shell (GiNaC V1.0.8) __, _______ Copyright (C) 1999-2002 Johannes Gutenberg University Mainz, (__) * | Germany. This is free software with ABSOLUTELY NO WARRANTY. ._) i N a C | You are welcome to redistribute it under certain conditions. <-------------' For details type `warranty;'. Type ?? for a list of help topics. > has(b*x*x^a, $0*$1*$1^$2); 1 > has(c*x*x^a, $0*$1*$1^$2); 1 > has(d*x*x^a, $0*$1*$1^$2); 1 > has(e*x*x^a, $0*$1*$1^$2); 0 > has(f*x*x^a, $0*$1*$1^$2); 1 > has(g*x*x^a, $0*$1*$1^$2); 1 > has(h*x*x^a, $0*$1*$1^$2); 1 > has(i*x*x^a, $0*$1*$1^$2); 0 > has(l*x*x^a, $0*$1*$1^$2); 1 > exit; Then I cosed and opened again ginsh and ... Type ?? for a list of help topics. > has(e*x*x^a, $0*$1*$1^$2); 1 > has(f*x*x^a, $0*$1*$1^$2); 1 > has(g*x*x^a, $0*$1*$1^$2); 1 > has(h*x*x^a, $0*$1*$1^$2); 0 > has(i*x*x^a, $0*$1*$1^$2); 1 > has(l*x*x^a, $0*$1*$1^$2); 1 > has(m*x*x^a, $0*$1*$1^$2); 1 > has(n*x*x^a, $0*$1*$1^$2); 0 > has(o*x*x^a, $0*$1*$1^$2); 1 > exit; [zolo@sandbox zolo]$ ginsh ginsh - GiNaC Interactive Shell (GiNaC V1.0.8) __, _______ Copyright (C) 1999-2002 Johannes Gutenberg University Mainz, (__) * | Germany. This is free software with ABSOLUTELY NO WARRANTY. ._) i N a C | You are welcome to redistribute it under certain conditions. <-------------' For details type `warranty;'. Type ?? for a list of help topics. > has(h*x*x^a, $0*$1*$1^$2); 1 > has(i*x*x^a, $0*$1*$1^$2); 1 > has(l*x*x^a, $0*$1*$1^$2); 1 > has(m*x*x^a, $0*$1*$1^$2); 0 > has(n*x*x^a, $0*$1*$1^$2); 1 > Furthermore, with the number it says always false, is it right? > has(2*x*x^a, $0*$1*$1^$2); 0 > has((-2)*x*x^a, $0*$1*$1^$2); 0 > has(sqrt(2)*x*x^a, $0*$1*$1^$2); 0 Thank you very much. Best regards Tatiana Zolo

2 1

method 'subs()'
by Tatiana Zolo 08 May '02

08 May '02

Hi all, I don't understand the behavior of method 'subs()' in the following case: > subs(a^x*b^x, $1^$2*$3^$2==($1*$3)^$2); (a*b)^x > subs(a^5*b^5, $1^$2*$3^$2==($1*$3)^$2); b^5*a^5 I also check it with a little program, that I attach, and the output is Expression 1: b^x*a^x List of match: {$1==b,$2==x,$3==a} Expression 1 after subs: (b*a)^x Expression 2: b^5*a^5 List of match: {$1==b,$2==5,$3==a} Expression 2 after subs: b^5*a^5 Why is different the behavior of subs() when it is applied on b^x*a^x or on b^5*a^5? Best regards Tatiana Zolo #include <iostream> #include <ginac/ginac.h> using namespace std; using namespace GiNaC; typedef GiNaC::ex GExpr; typedef GiNaC::symbol GSymbol; typedef GiNaC::numeric GNumber; typedef GiNaC::lst GList; int main() { GSymbol x("x"); GSymbol a("a"); GSymbol b("b"); GList l_1, l_2; GExpr p = pow(a, x) * pow(b, x); cout << endl << "Expression 1: " << p << endl; match(p, pow(wild(1), wild(2)) * pow(wild(3), wild(2)), l_1); cout << "List of match: " << l_1 << endl; p = p.subs(pow(wild(1), wild(2)) * pow(wild(3), wild(2)) == pow(wild(1)*wild(3), wild(2))); cout << "Expression 1 after subs: " << p << endl << endl; GExpr q = pow(a, 5) * pow(b, 5); cout << "Expression 2: " << q << endl; match(q, pow(wild(1), wild(2)) * pow(wild(3), wild(2)), l_2); cout << "List of match: " << l_2 << endl; q = q.subs(pow(wild(1), wild(2)) * pow(wild(3), wild(2)) == pow(wild(1)*wild(3), wild(2))); cout << "Expression 2 after subs: " << q << endl << endl; return 0; }

2 1