计算机程序的构造和解释习题3.29

计算机程序的构造和解释习题3.29

we know that A or B is equivalent to not ((not A) and (not B)) from Using De-Morgan’s rules. the diagram is:




so, the code is:

 ;;;Exercise 3.29
 (define (or-gate in1 in2 out)
   (let ((a (make-wire))
         (b (make-wire))
         (c (make-wire)))
     (inverter in1 a)
     (inverter in2 b)
     (add-gate a b c)
     (inverter c out)
     'ok))

From the diagram, we can get that :

the delay time of or-gate = the delay time of and-gate + the delay time of inverter * 2。