- GiNaC-list - ginac.de

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

GiNaC 1.0.8 now available
by Christian Bauer 31 Mar '02

31 Mar '02

Hello! Following our recent trend of Sunday evening releases we are proud to announce the immediate availability of GiNaC 1.0.8 which you can get from ftp://ftpthep.physik.uni-mainz.de/pub/GiNaC Changes include, but are not limited to - Improvements in memory usage of the expand() methods. Hm, wait... that's the only change!? <searches CVS logs> Aha, sorting of vector<expair> is a little faster, and the namespace fix for is_ex_the_function() is in. Alas, long gone now are the days of 0.9.x where changelog pages filled screen after screen... Bye, Christian -- / Coding on PowerPC and proud of it \/ http://www.uni-mainz.de/~bauec002/

1 0

Adding another infix operator?(programing)
by Mark Stankus 21 Mar '02

21 Mar '02

Hello, I am new to ginac. I have a computer algebra program which uses the infix operator "**" for a noncommutative multiplication (for example, a**b where a and b are symbols, but are thought of as representing matrices). (See http://math.ucsd.edu/~ncalg for more.) Is it possible/easy/hard to add a "**" infix operator to ginac? If so, where do I start looking in the code to do this. I would like to link my code up with ginac in the future. Mark Stankus mstankus(a)calpoly.edu

2 1

And GiNaC-1.0.7 shall be made legion...
by Richard B. Kreckel 18 Mar '02

18 Mar '02

Its numbers shall be increased a thousand thousand fold. The din of a million keyboards like unto a great storm shall cover the earth, and the followers of Mammon shall tremble. adapted from The Book of Mozilla, 3:31 (Red Letter Edition)

1 0

Behavior of the function 'has'
by Tatiana Zolo 13 Mar '02

13 Mar '02

I do not understand the behavior of the function 'has' in some case. I show you the problem with ginsh: > has(2^x*3^x, $0^x); 1 > has(2^x*3^x*4^x, $0^x); 1 > has(2^x*3^x, $0^x*$1^x); 1 > has(2^x*3^x*4^x, $0^x*$1^x); 0 > has(2^x*3^x*4^x, $0^x*$1^x*$2^x); 1 I would expect that also has(2^x*3^x*4^x, $0^x*$1^x); answers 'true' because I see $0^x*$1^x as a subexpression of 2^x*3^x*4^x. Same thing without to use wildcard. > has(2^x*3^x*4^x, 2^x); 1 > has(2^x*3^x*4^x, 2^x*3^x); 0 > has(2^x*3^x*4^x, 2^x*3^x*4^x); 1 Regards Tatiana Zolo

2 1

canonical ordering
by Garnet Kin-Lic Chan 11 Mar '02

11 Mar '02

Hi, I have been extending GiNaC in various ways, but something I haven't yet fathomed is the canonical ordering that automatically happens. For example, I derived a class from indexed, whose compare_same_type simply calls the inherited::compare_same_type, yet the ordering of products in my class was different from that of the usual indexed objects. What determines and triggers the canonical ordering? Perhaps this could also be explained somewhere in the reference. sincerely, Garnet

2 1