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segmentation fault in simplify_indexed --> solved?
by Chris Dams 25 Jun '02

25 Jun '02

Hello everybody, I think I have solved an error in GiNaC that sometimes (only in special circumstances) causes a segmentation fault when *simplify_indexed* is called. The problem occurs when the function *diracgamma::contract_with* is sometimes called with the arguments *self* and *other* pointing to two gamma matrices of different representation label. In that case it can happen that there is a gamma5 in between and that the function *base_and_index* is called with the gamma5 in its first argument. This is wrong because base_and_index tries to look at operand number 1, which does not exist for a gamma5. This then may cause a segmentation fault. It never happens for gamma matrices of the same rep. label because the gamma5 has been permuted out of the way then. I found that the segmentation fault dissapears when inside the first *if* in *diracgamma::contract_with* I add a test that checks if rep. labels are equal (the contract-gamma-matrices algorithms there, do not seem to make much sense if rep. lables are different anyway). After my addition the code looks like if (is_a<clifford>(*other)) { unsigned char rl2 =ex_to<clifford>(*other).get_representation_label(); if(rl!=rl2) return false; ex dim = ex_to<idx>(self->op(1)).get_dim(); where the first and last line are yours and the three in the middle are mine. (BTW this is GiNaC-1.0.8). Greetings, Chris

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GiNaC 1.0.9 now available (AKA Episode I - The Indexed Menace)
by Christian Bauer 11 Jun '02

11 Jun '02

FADE IN EXT. UNIVERSITY BUILDING - EVENING It is a quiet, hot summer evening. The setting sun casts its last rays on the Physics Department building, giving the stone walls a wonderful warm golden glow. The only sounds that are heard are the raspy cries of a black swarm of ravens that circles in the sky. Two cyclists ride down the street in front of the building. CUT TO INT. UNIVERSITY BUILDING, OFFICE - CONTINUOUS There are numerous desks and computers in this small, dimly lit room in the fifth floor of the Physics Department building. The walls are lined with large cupboards, science fiction posters, GiNaC logo designs and hardcopies of 'Dilbert' cartoons. Quiet piano music fills the air. Only two people are still working at this time of day. One, RICHY K. is sitting in a corner of the room, close to the window, with his bare feet laid on top of the desk and a notebook computer in his lap. The other, CHRISTIAN B., is busy studying eBay article listings on one of two monitors on his desk in the middle of the room. RICHY K. presses a few keys on his notebook and takes a sip from a cocktail glass. Then he speaks without turning his eyes away from the screen. RICHY K. There's a tarball at the usual place now. CHRISTIAN B. Cool... CHRISTIAN B. clicks on the 'shell' icon on his computer's desktop and starts typing commands. CHRISTIAN B. So we're going to release today? RICHY K. Yup. He brings up a Mozilla window displaying the Slashdot front page on his notebook. CHRISTIAN B. Did you put the champagne in the fridge? RICHY K. Pardon? CHRISTIAN B. (louder) Did you put the champagne in the fridge? RICHY K. We don't have any champagne. CHRISTIAN B. How can we release without champagne? RICHY K. Only drinking all the time... CHRISTIAN B. types some more commands. CHRISTIAN B. Is this 1.1.0? RICHY K. Nope, 1.0.9. CHRISTIAN B. Will we have champagne when we get to 1.1.0? RICHY K. Maybe... CHRISTIAN B. slams down the 'Return' key on his keyboard and leans back. Progress messages of a compiler run start scrolling down the screen. He returns his attention to the eBay listings on the second monitor. CHRISTIAN B. They're selling an eight-CPU Origin for 180,000 here... RICHY K. Cool. Tell Hubert to buy it. CHRISTIAN B. Where is Hubert, anyway? RICHY K. I don't know... RICHY K. takes another sip from his orange-red cocktail. CHRISTIAN B. glances over the compiler output but then looks back at his browser window and types in a URL. CHRISTIAN B. Did you get yesterday's User Friendly? RICHY K. What? CHRISTIAN B. (louder) Yesterday's User Friendly. Did you understand the punch line? RICHY K. Nope. CHRISTIAN B. What does that Russian sentence mean? RICHY K. Don't know. CHRISTIAN B. You ought to know. You've been in Russia. RICHY K. They all spoke English there. CHRISTIAN B. Oh. The compiler run is still in progress. CHRISTIAN B. grabs his coffee mug, slowly stands up, and walks over to the coffee machine. After filling the mug he returns to his place. CHRISTIAN B. You're already writing the announcement? RICHY K. It's your turn this time. CHRISTIAN B. No, it's not. I wrote the announcement for 1.0.8. RICHY K. No, you didn't. I did. It's your turn now. CHRISTIAN B. It's always 'my turn'. Why is it always my turn? RICHY K. It isn't. You're only trying to sneak out of having to write it. CHRISTIAN B. I'm not. He turns back to the monitors. Seeing that the compiler run has finished successfully, he continues to enter commands. CHRISTIAN B. RPM builds fine. I'll move it to the FTP server. RICHY K. Pardon? CHRISTIAN B. (louder) I will upload the RPM to the FTP server. RICHY K. Suit yourself... CHRISTIAN B. Did you tag the CVS? RICHY K. gasps with ostentation and starts pressing keys. RICHY K. Tagged. CHRISTIAN B. Nearly... CHRISTIAN B. still hacks away at his keyboard. RICHY K. Are you writing the announcement now? CHRISTIAN B. It's your turn. Besides, I can't write announcements without champagne. RICHY K. There should still be some wine in the cupboard. CHRISTIAN B. turns around to the cupboard behind him and opens the lower door. His face brightens up when he sees several bottles of red and white wine. CHRISTIAN B. (smiling) I'll write the announcement... FADE OUT

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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; } -------------------------------------------------------------

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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

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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.

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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

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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

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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

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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

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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; }

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