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:

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1. ;;;Exercise 3.29  
2. (define (or-gate in1 in2 out)  
3.   (let ((a (make-wire))  
4.         (b (make-wire))  
5.         (c (make-wire)))  
6.     (inverter in1 a)  
7.     (inverter in2 b)  
8.     (add-gate a b c)  
9.     (inverter c out)  
10.     '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。