您可以使用布尔代数定律最小化一个不关心条件的布尔表达式吗? [英] Can you minimize a Boolean expression having don't care conditions using Boolean algebraic laws?

问题描述

我必须最小化具有无关条件的SOP形式的布尔表达式.我可以通过k-map来做到这一点，但是我只能使用布尔代数定律来做到这一点.

问Sigma(0,2,3,5,6,7,8,9)+ d(10,11,12,13,14,15)

解决方案

可以使用布尔法则将给定的最小项列表和无关项列表转换为一个较短的隐含列表.

Karnaugh-Veitch 地图:

从地图上或通过检查给定的16个可能的术语中的14个，可以很清楚地看到只有两个表达式为假的术语:1和2.

因此，表达式可以写为

!(X0 & !X1 & !X2 & !X3) & !(!X0 & X1 & ! X2 & !X3)

应用 De Morgan定理:

(!X0 + X1 + X2 + X3) & (X0 + !X1 + X2 + X3)

两个因素共享(X2 + X3)，因此可以将其排除在外:

(!X0 + X1) & (X0 + X1) + X2 + X3

这导致结果:

!X0 & !X1 + X0 & X1 + X2 + X3

我们在一开始就隐式地将所有无关项设置为true.这是一个很好的猜测，但是如果没有反复试验就无法分析得出.

I have to minimise a boolean expression in SOP form having don't care conditions. I can do this by k-map but can I do this only using Boolean algebraic laws.

Q. Sigma(0,2,3,5,6,7,8,9) + d(10,11,12,13,14,15)

解决方案

The given lists of minterms and don't care terms can be transformed to one shorter list of implicants using Boolean laws.

The Karnaugh-Veitch map:

From the map or by inspection of the given 14 of 16 possible terms, it is clear that there are only two terms where the expression is false: 1 and 2.

Therefore, the expression can be written as

!(X0 & !X1 & !X2 & !X3) & !(!X0 & X1 & ! X2 & !X3)

Applying De Morgan's theorem:

(!X0 + X1 + X2 + X3) & (X0 + !X1 + X2 + X3)

Both factors share (X2 + X3) which can thus be factored out:

(!X0 + X1) & (X0 + X1) + X2 + X3

This leads to the result:

!X0 & !X1 + X0 & X1 + X2 + X3

We implicitely set all don't care terms to trueat the beginning. This was a good guess, but cannot be analytically derived without try-and-error.

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