基本布尔最小化
我正在尝试简化以下布尔代数片段，以便我可以构建电路：
A'.B'.C.D + A'.B.C.D' + A'.B.C.D + A.B'.C'.D + A.B'.C.D + A.B.C'.D + A.B.C.D' + A.B.C.D
到目前为止，我已经得到了：
(C.D) + (B.C) + (A.C'.D)
这是正确的吗？
我想获得尽可能好的最小化。
到目前为止我所经历的步骤是：
A'.B'.C.D + A'.B.C.D' + A'.B.C.D + A+B'+C'+D + A.B'+C+D + A.B.C'.D + A.B.C.D' + A.B.C.D
= A.A'(B'.C.D) + A.A'(B.C.D') + A.A'(B.C.D) + B.B'(A.C'.D)
= (B.C.D) + (B'.C.D) + (B.C.D) + (B.C.D') + (A.C'.D)
= (C.D) + (B.C) + (A.C'.D)
我还能做更多吗？
I am trying to simplify the following piece of boolean algebra so I can construct the circuit :
A'.B'.C.D + A'.B.C.D' + A'.B.C.D + A.B'.C'.D + A.B'.C.D + A.B.C'.D + A.B.C.D' + A.B.C.D
So far I have gotten it to :
(C.D) + (B.C) + (A.C'.D)
Is this correct?
I want to get the best possible minimization.
The steps I have went through so far are :
A'.B'.C.D + A'.B.C.D' + A'.B.C.D + A+B'+C'+D + A.B'+C+D + A.B.C'.D + A.B.C.D' + A.B.C.D
= A.A'(B'.C.D) + A.A'(B.C.D') + A.A'(B.C.D) + B.B'(A.C'.D)
= (B.C.D) + (B'.C.D) + (B.C.D) + (B.C.D') + (A.C'.D)
= (C.D) + (B.C) + (A.C'.D)
Can I do any more?
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评论（4）
假设你的等式实际上是：
我刚刚通过 逻辑星期五 运行了这个，并将其分解为：
所以你可能想检查一下您的简化工作和/或检查您是否给出了正确的方程。
不过我怀疑上面的原始方程可能有拼写错误，也许应该是：
？
在这种情况下，逻辑星期五将其简化为：
Assuming your equation is actually:
I just ran this through Logic Friday and it factored it down to:
So you might want to check your simplification work and/or check that you've given the correct equation.
However I suspect there may be typos in the original equation above, and perhaps it should be:
?
In which case Logic Friday simplifies it to:
我能看到你唯一可以做的就是在左边两项中分配“C”：
或者你可以分配“D”：
对评论的回应：分配律描述如下： http://www.ee.surrey.ac.uk/Projects/Labview/布尔代数/。请参阅标题“T3”下的信息
The only thing I can see that you could possibly do is distribute the "C" in the left two terms:
Or you could distribute the "D":
Response to Comment: The distributive law is described here: http://www.ee.surrey.ac.uk/Projects/Labview/boolalgebra/. See the information under heading "T3"
这是另一个解决方案（通过暴力找到）：(
a+c).(b+d).(c+d)
Here's another solution (found by brute force):
(a+c).(b+d).(c+d)
为了简化布尔表达式，请使用卡诺图。我认为如果我们减少变量的数量，这是非常有用的。但如果我们有更多变量，那么我们可以遵循方法，因为这种方法不太可取。
for simplifying boolean expressions use karnaugh maps. i think it is very much useful if we less number of variables. but if we have more variables then we can follow methods because this method is not that preferable.