形式が不十分だったので、この質問を書き直しています。
(define (reduce f)
((lambda (value) (if (equal? value f) f (reduce value)))
(let r ((f f) (g ()))
(cond ((not (pair? f))
(if (null? g) f (if (eq? f (car g)) (cadr g) (r f (caddr g)))))
((and (pair? (car f)) (= 2 (length f)) (eq? 'lambda (caar f)))
(r (caddar f) (list (cadar f) (r (cadr f) g) g)))
((and (not (null? g)) (= 3 (length f)) (eq? 'lambda (car f)))
(cons 'lambda (r (cdr f) (list (cadr f) (gensym (cadr f)) g))))
(else (map (lambda (x) (r x g)) f))))))
; (reduce '((lambda x x) 3)) ==> 3
; (reduce '((lambda x (x x)) (lambda x (lambda y (x y)))))
; ==> (lambda #[promise 2] (lambda #[promise 3] (#[promise 2] #[promise 3])))
; Comments: f is the form to be evaluated, and g is the local assignment
; function; g has the structure (variable value g2), where g2 contains
; the rest of the assignments. The named let function r executes one
; pass through a form. The arguments to r are a form f, and an
; assignment function g. Line 2: continue to process the form until
; there are no more conversions left. Line 4 (substitution): If f is
; atomic [or if it is a promise], check to see if matches any variable
; in g and if so replace it with the new value. Line 6 (beta
; reduction): if f has the form ((lambda variable body) argument), it is
; a lambda form being applied to an argument, so perform lambda
; conversion. Remember to evaluate the argument too! Line 8 (alpha
; reduction): if f has the form (lambda variable body), replace the
; variable and its free occurences in the body with a unique object to
; prevent accidental variable collision. [In this implementation a
; unique object is constructed by building a promise. Note that the
; identity of the original variable can be recovered if you ever care by
; forcing the promise.] Line 10: recurse down the subparts of f.
ラムダ式でラムダ削減を行う上記のコードがあります(これが私が欲しいものです)。ここでの私の問題は、誰かがこの実装を書き直すのを手伝ってくれることです(私はSchemeの経験がないので)。これにより、本体からアルファ変換を行う部分を抽出し、それを別の関数に入れます。ベータ還元も。関数 reduce は再帰的であるため、新しく作成された 2 つの関数は single-step である必要があります。つまり、1 つの境界付き変数のみを変換し、1 つの式のみを還元します。