Exercise 1.20 最大公约数算法

The process that a procedure generates is of course dependent on the rules used by the interpreter. As an example, consider the iterative gcd procedure given above. Suppose we were to interpret this procedure using normal-order evaluation, as discussed in section 1.1.5. (The normal-order- evaluation rule for if is described in exercise 1.5.) Using the substitution method (for normal order), illustrate the process generated in evaluating (gcd 206 40) and indicate the remainder operations that are actually performed. How many remainder operations are actually performed in the normal- order evaluation of (gcd 206 40)? In the applicative-order evaluation? 

 

采用Applicative-order：

gcd(206, 40)

=gcd(40, 6)

=gcd(6,4)

=gcd(4,2)

=gcd(2,0)

共4次递归，因此，采用Applicative-order，Remainder函数被调用4次。

采用Normal-order

gcd(206, 40)

=gcd( 40, R(206,40)) //这里判断R(206,40)是否为零，有一次Remainder计算

=gcd(

          R(206,40),

          R(40, R(206,40))

  )//这里判断R(40, R(206,40))是否为零，有两次Remainder计算

=gcd(

　　R(40, R(206, 40)),

      R(

            R(206,40),

            R(40, R(206, 40))

      )//判断是否为零，有4次Remainder计算

)

=gcd(

      R(

            R(206,40),

            R(40, R(206, 40))

      ),//返回结果，有4次计算

     R(

            R(40, R(206, 40)),

            R(

                R(206,40),

                R(40, R(206, 40))

            ),

     )//判断是否为零，有7次Remainder计算

)

总共有1+2+4+7=14次Remainder调用计算