帮助替换 model [Sicp]，使用 Clojure

Question

我正在研究 sicp 书，我对一个程序的替换 model 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)))))

这个过程是练习 1.10 的一部分。 如果我使用以下参数（A 1 10）在 REPL 中运行 function，结果为 1024。我决定使用 Substitution Model 验证结果，但结果为 2048。

这是我写的替换 model。 有问题，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？ 我很抱歉这个问题的长度。

Answer 1

考虑以下几行：

(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))))))))

剥去多余的外层：

(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)))))

在这里的某个地方，您最终得到了不匹配的括号，但这并不重要。 请注意，在从A 1 7到A 1 6的过程中，按预期创建了A 0 _的外层。 从A 1 6到A 1 5 ，你有两个新的A 0 _层。 这些中的每一个最终都会使结果翻倍，这就是为什么您的答案会相差 2 倍。