If I were to start it in Common Lisp, I'd probably do something like this. Next steps might be to make normal variable declarations automatic by scanning guard

If I were to start it in Common Lisp, I'd probably do something like this. Next steps might be to make normal variable declarations automatic by scanning guard expressions (which would prevent unwittingly mentioning a variable too early); and depending on the intended usage, maybe permit infix math via the infix-math package. (defpackage solver (:use :cl :iterate) (:shadow pi)) (in-package :solver) (defmacro defsolver (name default constraints result) (flet ((normal-var (var) `(iter (for ,var ,@(rest default)))) (custom-var (var spec) `(iter (for ,var in ,spec))) (guard (expr) `(when ,expr)) (solved () `(return-from ,name ,result))) (macrolet ((yield (form) `(append ,form (list (constraint (rest cs)))))) (labels ((constraint (cs &aux (c (first cs))) (cond ((null cs) nil) ((eq c :solved) (solved)) ((symbolp c) (yield (normal-var c))) ((eq (first c) 'for) (yield (custom-var (second c) (third c)))) (t (yield (guard c)))))) `(defun ,name () ,(constraint `(,@constraints :solved))))))) (defsolver result (:default from 1 to 26) (th o r (= (+ (* 2 (expt th 2)) (expt o 2) (expt r 2)) 1000) s z (= (+ (* (expt th 2) (expt z 2) (- s z)) (expt s 2)) 7225) g d xi (= (+ (expt (1- g) 2) (expt d 2) (expt xi 2) (- xi)) 600) (for (et i) `((,(- xi 7) ,(- xi 11)) (,(- xi 11) ,(- xi 7)) (,(+ xi 7) ,(+ xi 11)) (,(+ xi 11) ,(+ xi 7)))) (and (< 0 et) (<= et 26) (< 0 i) (<= i 26)) (for (a mu) '((4 1) (2 2) (1 4))) pi (= (* 4 g) (* a (+ a pi))) b n (= (+ (expt b 3) (expt z 3) (expt n 3) (expt o 9)) 1997) (for k '(3 4)) e (= (+ (expt (/ a k) 2) (expt (/ b n) 2) (expt (/ d xi) 2) (expt (/ e pi) 2) (expt (/ et mu) 2) (expt (/ i s) 2)) 6)) (list a b g d e z et th i k th mu n xi o pi r s)) SOLVER> (time (result)) Evaluation took: 0.010 seconds of real time 0.012000 seconds of total run time (0.012000 user, 0.000000 system) 120.00% CPU 23,160,723 processor cycles 851,968 bytes consed (4 9 19 12 15 3 1 20 5 4 20 1 9 12 2 15 14 5) Edit: formatting, simplification The b, n, k, e definitions are reordered better as in @eutectic's haskell example.