逻辑门:仅使用异或门实现或门
我正在为我的考试做一些复习,去年试卷中经常提出的问题之一是如何仅使用异或门来实现或门。我从过去 1 小时开始就在尝试做,也在互联网上进行了搜索。但失败了。希望在这里得到一些帮助。
I am doing some revision for my exams and one of the questions that frequently asked in last year question paper is how to realize a OR gate using only XOR gates. I am trying to do from last 1 hour and also searched on Internet. But failed. Hope to get some help here.
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您不能仅从 XOR(或 XNOR)门导出 OR(或 AND、NOR、NAND)门,因为它们不是通用门。
要成为通用门,门需要满足以下两个条件:
You can't derive an OR (or AND, NOR, NAND) gate from only XOR (or XNOR) gates, because they are not universal gates.
To be an universal gate, a gate needs to satisfy both the following conditions:
将两条输入线仅连接到异或门的一个输入引脚,使另一输入引脚始终为假,从而产生或门。
Tie both input lines to only one input pin of the XOR gate, leaving the other input pin as always being false, thus producing an OR gate.