All computer languages can perform numerical computations. Lisp can also manipulate formulas symbolically. We will represent equations in Lisp notation; for example, the equation f = m * a is written (= f (* m a)). We will assume that all operations are unary or binary.
(vars '()) => NIL or () (vars 'a) => (a) (vars 'pi) => NIL or () (vars '(sqrt x)) => (x) (vars '(+ a a)) => (a) (vars '(= a b)) => (a b) (vars '(+ a 2)) => (a)
solve will return nil if it fails, or a formula with v on the lhs.
(solve '(= f (* m a)) 'f) => (= f (* m a))
(solve '(= (* m a) f) 'f) => (= f (* m a))
For example, given the original equation (= x (- y z)), we could (a) add z to both sides to give (= (+ x z) y), or (b) subtract y from both sides and negate to give (= (- y x) z) (these two operations can be combined as a single transformation).
(solve '(= x (- y z)) 'y) (solve '(= (- y x) z) 'y) ; first try: nil (solve '(= (+ x z) y) 'y) ; second try: succeeds => (= y (+ x z))
Hint: first write the solve function so that it handles only the base cases and the + operator. Do (trace solve) so you can observe how it works on more complex expressions.
You should handle the operators + - * / sqrt exp log and (expt x 2). The operator - can be either unary (having only one argument, i.e. minus) or binary (having two arguments, i.e. difference), which must be treated differently. We will assume that all other operators will have a fixed number of arguments (either one or two).
Demonstrate that you can solve the following equations for any of their variables. These are defined in the file formulas.lsp .
(= s (* 0.5 (* a (expt t 2)))) (= s (+ s0 (* v t))) (= a (/ f m)) (= v (* a t)) (= f (/ (* m v) t)) (= f (/ (* m (expt v 2)) r)) (= h (- h0 (* 4.94 (expt t 2)))) (= c (sqrt (+ (expt a 2) (expt b 2)))) (= v (* v0 (- 1 (exp (/ (- t) (* r c))))))