Commit 73e1727c by Mattias Engdegård

### Fix linear equation system solving in Calc (bug#35374)

```* lisp/calc/calcalg2.el (math-try-solve-for):
To solve Ax^n=0 where A is a nonzero constant and x the variable to
solve for, solve x^n=0 instead of solving A=0 (which obviously fails)
or something equally stupid.
* test/lisp/calc/calc-tests.el (calc-test-solve-linear-system): New.```
parent bba9757a
Pipeline #3271 passed with stage
in 36 minutes and 59 seconds
 ... ... @@ -2417,6 +2417,12 @@ ((= (length math-t1) 2) (apply 'math-solve-linear (car math-t2) math-try-solve-sign math-t1)) ((= (length math-t1) 1) ;; Constant polynomial. (if (eql (nth 2 math-t2) 1) nil ; No possible solution. ;; Root of the factor, if any. (math-try-solve-for (nth 2 math-t2) 0 nil t))) (math-solve-full (math-poly-all-roots (car math-t2) math-t1)) (calc-symbolic-mode nil) ... ...
 ... ... @@ -138,6 +138,109 @@ An existing calc stack is reused, otherwise a new one is created." (nth 1 (calcFunc-cos 1))) 0 4)))))) (ert-deftest calc-test-solve-linear-system () "Test linear system solving (bug#35374)." ;; x + y = 3 ;; 2x - 3y = -4 ;; with the unique solution x=1, y=2 (should (equal (calcFunc-solve '(vec (calcFunc-eq (+ (var x var-x) (var y var-y)) 3) (calcFunc-eq (- (* 2 (var x var-x)) (* 3 (var y var-y))) -4)) '(vec (var x var-x) (var y var-y))) '(vec (calcFunc-eq (var x var-x) 1) (calcFunc-eq (var y var-y) 2)))) ;; x + y = 1 ;; x + y = 2 ;; has no solution (should (equal (calcFunc-solve '(vec (calcFunc-eq (+ (var x var-x) (var y var-y)) 1) (calcFunc-eq (+ (var x var-x) (var y var-y)) 2)) '(vec (var x var-x) (var y var-y))) '(calcFunc-solve (vec (calcFunc-eq (+ (var x var-x) (var y var-y)) 1) (calcFunc-eq (+ (var x var-x) (var y var-y)) 2)) (vec (var x var-x) (var y var-y))))) ;; x - y = 1 ;; x + y = 1 ;; with the unique solution x=1, y=0 (should (equal (calcFunc-solve '(vec (calcFunc-eq (- (var x var-x) (var y var-y)) 1) (calcFunc-eq (+ (var x var-x) (var y var-y)) 1)) '(vec (var x var-x) (var y var-y))) '(vec (calcFunc-eq (var x var-x) 1) (calcFunc-eq (var y var-y) 0)))) ;; 2x - 3y + z = 5 ;; x + y - 2z = 0 ;; -x + 2y + 3z = -3 ;; with the unique solution x=1, y=-1, z=0 (should (equal (calcFunc-solve '(vec (calcFunc-eq (+ (- (* 2 (var x var-x)) (* 3 (var y var-y))) (var z var-z)) 5) (calcFunc-eq (- (+ (var x var-x) (var y var-y)) (* 2 (var z var-z))) 0) (calcFunc-eq (+ (- (* 2 (var y var-y)) (var x var-x)) (* 3 (var z var-z))) -3)) '(vec (var x var-x) (var y var-y) (var z var-z))) ;; The `float' forms in the result are just artefacts of Calc's ;; current solver; it should be fixed to produce exact (integral) ;; results in this case. '(vec (calcFunc-eq (var x var-x) (float 1 0)) (calcFunc-eq (var y var-y) (float -1 0)) (calcFunc-eq (var z var-z) 0)))) ;; x = y + 1 ;; x = y ;; has no solution (should (equal (calcFunc-solve '(vec (calcFunc-eq (var x var-x) (+ (var y var-y) 1)) (calcFunc-eq (var x var-x) (var y var-y))) '(vec (var x var-x) (var y var-y))) '(calcFunc-solve (vec (calcFunc-eq (var x var-x) (+ (var y var-y) 1)) (calcFunc-eq (var x var-x) (var y var-y))) (vec (var x var-x) (var y var-y))))) ;; x + y + z = 6 ;; x + y = 3 ;; x - y = 1 ;; with the unique solution x=2, y=1, z=3 (should (equal (calcFunc-solve '(vec (calcFunc-eq (+ (+ (var x var-x) (var y var-y)) (var z var-z)) 6) (calcFunc-eq (+ (var x var-x) (var y var-y)) 3) (calcFunc-eq (- (var x var-x) (var y var-y)) 1)) '(vec (var x var-x) (var y var-y) (var z var-z))) '(vec (calcFunc-eq (var x var-x) 2) (calcFunc-eq (var y var-y) 1) (calcFunc-eq (var z var-z) 3)))) ;; x = 3 ;; x + 4y^2 = 3 (ok, so this one isn't linear) ;; with the unique (double) solution x=3, y=0 (should (equal (calcFunc-solve '(vec (calcFunc-eq (var x var-x) 3) (calcFunc-eq (+ (var x var-x) (* 4 (^ (var y var-y) 2))) 3)) '(vec (var x var-x) (var y var-y))) '(vec (calcFunc-eq (var x var-x) 3) (calcFunc-eq (var y var-y) 0))))) (provide 'calc-tests) ;;; calc-tests.el ends here ... ...
Markdown is supported
0% or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!