clojure - 使用 Clojure 帮助建立替换模型 [Sicp]

Question

我正在学习sicp书，我对程序的替换模型有疑问:

(defn A
   [x,y]
     (cond (= y 0) 0
           (= x 0) (* 2 y)
           (= y 1) 2
           :else (A (- x 1) (A x (- y 1)))))

此过程是练习 1.10 的一部分。 如果我使用以下参数 (A 1 10) 在 REPL 中运行该函数，结果是 1024。我决定使用替换模型验证结果，但结果是 2048。

这是我写的替代模型。有什么不对劲，但我不知道是什么。

(A 1 10)
(A (- 1 1) (A 1 (- 10 1))))
(A 0 (A 1 9)))
(A 0 (A (- 1 1) (A 1 (- 9 1)))))
(A 0 (A 0 (A 1 8))))
(A 0 (A 0 (A (- 1 1) (A 1 (- 8 1))))))
(A 0 (A 0 (A 0 (A 1 7)))))
(A 0 (A 0 (A 0 (A (- 1 1) (A 1 (- 7 1)))))))
(A 0 (A 0 (A 0 (A 0 (A 1 6)))))) 
(A 0 (A 0 (A 0 (A 0 (A (-1 1) (A 1 (- 6 1))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 5))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A (-1 1) (A 1 (- 5 1))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 4)))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A (-1 1) (A 1 (- 4 1)))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 3))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A (-1 1) (A 1 (-3 1))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 2)))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A (-1 1) (A 1 (- 2 1))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 1))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (* 2 1))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 2)))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (* 2 2)))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 4))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (* 2 4))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 8)))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (* 2 8)))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 16))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (* 2 16)))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 32))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (* 2 32))))))
(A 0 (A 0 (A 0 (A 0 (A 0 64)))))
(A 0 (A 0 (A 0 (A 0 (* 2 64)))))
(A 0 (A 0 (A 0 (A 0 128))))
(A 0 (A 0 (A 0 (* 2 128))))
(A 0 (A 0 (A 0 256)))
(A 0 (A 0 (* 2 256)))
(A 0 (A 0 512))
(A 0 (* 2 512))
(A 0 1024)
2048 ????

谁能指出我做错了什么？ 对于问题的长度，我深表歉意。

Answer 1

考虑这些行:

(A 0 (A 0 (A 0 (A 1 7)))))
(A 0 (A 0 (A 0 (A (- 1 1) (A 1 (- 7 1)))))))
(A 0 (A 0 (A 0 (A 0 (A 1 6)))))) 
(A 0 (A 0 (A 0 (A 0 (A (-1 1) (A 1 (- 6 1))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 5))))))))

剥掉多余的外层:

(A 1 7))
(A (- 1 1) (A 1 (- 7 1))))
(A 0 (A 1 6)))
(A 0 (A (-1 1) (A 1 (- 6 1)))))
(A 0 (A 0 (A 0 (A 1 5)))))

这里的某个地方你最终得到了不匹配的括号，但这并不重要。请注意，从 A 1 7 开始至A 1 6 ，单个外层A 0 _已创建，如预期。从 A 1 6 出发至A 1 5 ，您已经有了两个新层 A 0 _ 。其中每一个最终都会使结果加倍，所以这就是为什么你的答案偏离了 2 倍。