- 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;
-
- if (!result) {
- cout << " passed " << endl;
- clog << "(no output)" << endl;
- } else {
- cout << " failed " << endl;
- }
-
- return result;
+ // 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;