* These exams test solving small linear systems of symbolic equations. */
/*
- * GiNaC Copyright (C) 1999-2000 Johannes Gutenberg University Mainz, Germany
+ * GiNaC Copyright (C) 1999-2003 Johannes Gutenberg University Mainz, Germany
*
* This program is free software; you can redistribute it and/or modify
* it under the terms of the GNU General Public License as published by
#include "exams.h"
-static unsigned exam_lsolve1(void)
+static unsigned exam_lsolve1()
{
- // A trivial example.
- unsigned result = 0;
- symbol x("x");
- ex eq, aux;
-
- eq = (3*x+5 == numeric(8));
- aux = lsolve(eq, x);
- if (aux != 1) {
- result++;
- clog << "solution of 3*x+5==8 erroneously returned "
- << aux << endl;
- }
-
- return result;
+ // A trivial example.
+ unsigned result = 0;
+ symbol x("x");
+ ex eq, aux;
+
+ eq = (3*x+5 == numeric(8));
+ aux = lsolve(eq, x);
+ if (aux != 1) {
+ ++result;
+ clog << "solution of 3*x+5==8 erroneously returned "
+ << aux << endl;
+ }
+
+ return result;
}
-static unsigned exam_lsolve2a(void)
+static unsigned exam_lsolve2a()
{
- // An example from the Maple online help.
- unsigned result = 0;
- symbol a("a"), b("b"), x("x"), y("y");
- lst eqns, vars;
- ex sol;
-
- // Create the linear system [a*x+b*y==3,x-y==b]...
- eqns.append(a*x+b*y==3).append(x-y==b);
- // ...to be solved for [x,y]...
- vars.append(x).append(y);
- // ...and solve it:
- sol = lsolve(eqns, vars);
- ex sol_x = sol.op(0).rhs(); // rhs of solution for first variable (x)
- ex sol_y = sol.op(1).rhs(); // rhs of solution for second variable (y)
-
- // It should have returned [x==(3+b^2)/(a+b),y==(3-a*b)/(a+b)]
- if (!normal(sol_x - (3+pow(b,2))/(a+b)).is_zero() ||
- !normal(sol_y - (3-a*b)/(a+b)).is_zero()) {
- result++;
- clog << "solution of the system " << eqns << " for " << vars
- << " erroneously returned " << sol << endl;
- }
-
- return result;
+ // An example from the Maple online help.
+ unsigned result = 0;
+ symbol a("a"), b("b"), x("x"), y("y");
+ lst eqns, vars;
+ ex sol;
+
+ // Create the linear system [a*x+b*y==3,x-y==b]...
+ eqns.append(a*x+b*y==3).append(x-y==b);
+ // ...to be solved for [x,y]...
+ vars.append(x).append(y);
+ // ...and solve it:
+ sol = lsolve(eqns, vars);
+ ex sol_x = sol.op(0).rhs(); // rhs of solution for first variable (x)
+ ex sol_y = sol.op(1).rhs(); // rhs of solution for second variable (y)
+
+ // It should have returned [x==(3+b^2)/(a+b),y==(3-a*b)/(a+b)]
+ if (!normal(sol_x - (3+pow(b,2))/(a+b)).is_zero() ||
+ !normal(sol_y - (3-a*b)/(a+b)).is_zero()) {
+ ++result;
+ clog << "solution of the system " << eqns << " for " << vars
+ << " erroneously returned " << sol << endl;
+ }
+
+ return result;
}
-static unsigned exam_lsolve2b(void)
+static unsigned exam_lsolve2b()
{
- // A boring example from Mathematica's online help.
- unsigned result = 0;
- symbol x("x"), y("y");
- lst eqns, vars;
- ex sol;
-
- // Create the linear system [3*x+y==7,2*x-5*y==8]...
- eqns.append(3*x+y==7).append(2*x-5*y==8);
- // ...to be solved for [x,y]...
- vars.append(x).append(y);
- // ...and solve it:
- sol = lsolve(eqns, vars);
- ex sol_x = sol.op(0).rhs(); // rhs of solution for first variable (x)
- ex sol_y = sol.op(1).rhs(); // rhs of solution for second variable (y)
-
- // It should have returned [x==43/17,y==-10/17]
- if (!(sol_x - numeric(43,17)).is_zero() ||
- !(sol_y - numeric(-10,17)).is_zero()) {
- result++;
- clog << "solution of the system " << eqns << " for " << vars
- << " erroneously returned " << sol << endl;
- }
-
- return result;
+ // A boring example from Mathematica's online help.
+ unsigned result = 0;
+ symbol x("x"), y("y");
+ lst eqns, vars;
+ ex sol;
+
+ // Create the linear system [3*x+y==7,2*x-5*y==8]...
+ eqns.append(3*x+y==7).append(2*x-5*y==8);
+ // ...to be solved for [x,y]...
+ vars.append(x).append(y);
+ // ...and solve it:
+ sol = lsolve(eqns, vars);
+ ex sol_x = sol.op(0).rhs(); // rhs of solution for first variable (x)
+ ex sol_y = sol.op(1).rhs(); // rhs of solution for second variable (y)
+
+ // It should have returned [x==43/17,y==-10/17]
+ if ((sol_x != numeric(43,17)) ||
+ (sol_y != numeric(-10,17))) {
+ ++result;
+ clog << "solution of the system " << eqns << " for " << vars
+ << " erroneously returned " << sol << endl;
+ }
+
+ return result;
}
-static unsigned exam_lsolve2c(void)
+static unsigned exam_lsolve2c()
{
- // An example from the Maple online help.
- unsigned result = 0;
- symbol x("x"), y("y");
- lst eqns, vars;
- ex sol;
-
- // Create the linear system [I*x+y==1,I*x-y==2]...
- eqns.append(I*x+y==1).append(I*x-y==2);
- // ...to be solved for [x,y]...
- vars.append(x).append(y);
- // ...and solve it:
- sol = lsolve(eqns, vars);
- ex sol_x = sol.op(0).rhs(); // rhs of solution for first variable (x)
- ex sol_y = sol.op(1).rhs(); // rhs of solution for second variable (y)
-
- // It should have returned [x==-3/2*I,y==-1/2]
- if (!(sol_x - numeric(-3,2)*I).is_zero() ||
- !(sol_y - numeric(-1,2)).is_zero()) {
- result++;
- clog << "solution of the system " << eqns << " for " << vars
- << " erroneously returned " << sol << endl;
- }
-
- return result;
+ // A more interesting example from the Maple online help.
+ unsigned result = 0;
+ symbol x("x"), y("y");
+ lst eqns, vars;
+ ex sol;
+
+ // Create the linear system [I*x+y==1,I*x-y==2]...
+ eqns.append(I*x+y==1).append(I*x-y==2);
+ // ...to be solved for [x,y]...
+ vars.append(x).append(y);
+ // ...and solve it:
+ sol = lsolve(eqns, vars);
+ ex sol_x = sol.op(0).rhs(); // rhs of solution for first variable (x)
+ ex sol_y = sol.op(1).rhs(); // rhs of solution for second variable (y)
+
+ // It should have returned [x==-3/2*I,y==-1/2]
+ if ((sol_x != numeric(-3,2)*I) ||
+ (sol_y != numeric(-1,2))) {
+ ++result;
+ clog << "solution of the system " << eqns << " for " << vars
+ << " erroneously returned " << sol << endl;
+ }
+
+ return result;
}
-unsigned exam_lsolve(void)
+static unsigned exam_lsolve2S()
{
- unsigned result = 0;
-
- cout << "examining linear solve" << flush;
- clog << "----------linear solve:" << endl;
-
- result += exam_lsolve1(); cout << '.' << flush;
- result += exam_lsolve2a(); cout << '.' << flush;
- result += exam_lsolve2b(); cout << '.' << flush;
- result += exam_lsolve2c(); cout << '.' << flush;
-
- if (!result) {
- cout << " passed " << endl;
- clog << "(no output)" << endl;
- } else {
- cout << " failed " << endl;
- }
-
- return result;
+ // A degenerate example that went wrong in GiNaC 0.6.2.
+ unsigned result = 0;
+ symbol x("x"), y("y"), t("t");
+ lst eqns, vars;
+ ex sol;
+
+ // Create the linear system [0*x+0*y==0,0*x+1*y==t]...
+ eqns.append(0*x+0*y==0).append(0*x+1*y==t);
+ // ...to be solved for [x,y]...
+ vars.append(x).append(y);
+ // ...and solve it:
+ sol = lsolve(eqns, vars);
+ ex sol_x = sol.op(0).rhs(); // rhs of solution for first variable (x)
+ ex sol_y = sol.op(1).rhs(); // rhs of solution for second variable (y)
+
+ // It should have returned [x==x,y==t]
+ if ((sol_x != x) ||
+ (sol_y != t)) {
+ ++result;
+ clog << "solution of the system " << eqns << " for " << vars
+ << " erroneously returned " << sol << endl;
+ }
+
+ return result;
+}
+
+static unsigned exam_lsolve3S()
+{
+ // A degenerate example that went wrong while trying to improve elimination
+ unsigned result = 0;
+ symbol b("b"), c("c");
+ symbol x("x"), y("y"), z("z");
+ lst eqns, vars;
+ ex sol;
+
+ // Create the linear system [y+z==b,-y+z==c] with one additional row...
+ eqns.append(ex(0)==ex(0)).append(b==z+y).append(c==z-y);
+ // ...to be solved for [x,y,z]...
+ vars.append(x).append(y).append(z);
+ // ...and solve it:
+ sol = lsolve(eqns, vars);
+ ex sol_x = sol.op(0).rhs(); // rhs of solution for first variable (x)
+ ex sol_y = sol.op(1).rhs(); // rhs of solution for second variable (y)
+ ex sol_z = sol.op(2).rhs(); // rhs of solution for third variable (z)
+
+ // It should have returned [x==x,y==t,]
+ if ((sol_x != x) ||
+ (sol_y != (b-c)/2) ||
+ (sol_z != (b+c)/2)) {
+ ++result;
+ clog << "solution of the system " << eqns << " for " << vars
+ << " erroneously returned " << sol << endl;
+ }
+
+ return result;
+}
+
+unsigned exam_lsolve()
+{
+ unsigned result = 0;
+
+ cout << "examining linear solve" << flush;
+ clog << "----------linear solve:" << endl;
+
+ result += exam_lsolve1(); cout << '.' << flush;
+ result += exam_lsolve2a(); cout << '.' << flush;
+ result += exam_lsolve2b(); cout << '.' << flush;
+ result += exam_lsolve2c(); cout << '.' << flush;
+ result += exam_lsolve2S(); cout << '.' << flush;
+ result += exam_lsolve3S(); cout << '.' << flush;
+
+ if (!result) {
+ cout << " passed " << endl;
+ clog << "(no output)" << endl;
+ } else {
+ cout << " failed " << endl;
+ }
+
+ return result;
}