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

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* 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.
This commit is contained in:
Mattias Engdegård 2019-09-22 15:03:02 +02:00
parent bba9757a1f
commit 73e1727c40
2 changed files with 109 additions and 0 deletions
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6
lisp/calc/calcalg2.el
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@ -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)

103
test/lisp/calc/calc-tests.el
View file

@ -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