Digital-electronics 简明教程
4 Variable K-Map in Digital Electronics
已经开发出几种技术，可以将一个复杂的布尔表达式简化为最简单的形式。 K-Map 或 Karnaugh Map 就是一种这样的最小化或简化技术。
Several techniques have been developed to simplify a complex Boolean expression into its simplest form. K-Map or Karnaugh Map is one of such minimization or simplification techniques.
K-Map 或 Karnaugh 图是一个由相邻单元格的排列组成的图或图表。其中，K-Map 的每个单元格表示变量的特定组合，采用求和或乘积形式。K-map 可用于简化涉及任意数量变量的布尔函数。但是，对于涉及五个或更多变量的问题，使用 K-map 简化布尔函数变得繁琐。因此，在实际操作中，K-map 限制在六个变量。
The K-Map or Karnaugh Map is a graph or chart which composed of an arrangement of adjacent cells. Where, each cell of the K-Map represents a particular combination of variables in either sum or product form. The K-map can be used to simplify Boolean functions involving any number of variables. But, the simplification of a Boolean function using K-map becomes tedious for problems involving five or more variables. Therefore, in actual practice, the K-map is limited to six variables.
卡诺图中单元格的数量取决于给定布尔函数中变量的数量。卡诺图将有 2n 个单元格或方块，其中 n 是布尔表达式中变量的数量。因此，对于一个二元变量函数，卡诺图将有 22 = 4 个单元格；对于一个三元变量布尔函数，卡诺图将有 23 = 8 个单元格；对于一个四元变量布尔函数，卡诺图将有 24 = 16 个单元格，以此类推。
The number of cells in a K-map depends upon the number of variables in the given Boolean function. A K-map will have 2n cells or squares, where n is the number of variables in the Boolean expression. Therefore, for a two variable function, the K-map will have 22 = 4 cells, for a three variable Boolean function, the K-map will have 23 = 8 cells, and for four variable Boolean function, the K-map will have 24 = 16 cells, and so on.
在本文中，我们将讨论四变量 K-Map，并将使用其在 4 个变量中简化布尔函数。
Here, we will discuss four variable K-Map and will use it to simplify Boolean functions in 4 variables.
Four Variable K-Map
四变量 K-map 用于简化包含 4 个变量的复杂布尔表达式。众所周知，一个四变量布尔表达式可以有 24 = 16 个可能的变量组合。
A four variable K-map is used to simplify a complex Boolean expression in 4 variables. As we know, a four variable Boolean expression can have 24 = 16 possible combinations of variables.
例如，在 SOP（乘积之和）形式中，
For example, in SOP (Sum of Products) Form,
\mathrm{f(A,B,C,D) \: = \: \bar{A}\bar{B}\bar{C}\bar{D} \: + \: \bar{A}\bar{B}\bar{C}D \: + \: \bar{A}\bar{B}C\bar{D} \: + \: \dots \: + \: ABCD}
该表达式的最小项表示如下：
The minterm representation of this expression is as follows,
\mathrm{f(A,B,C,D) \: = \: m_{0} \: + \: m_{1} \: + \: m_{2} \: + \: \dotso \: + \: m_{15}}
在 POS（求和之积）形式中
In POS (Product of Sums) Form,
\mathrm{f(A,B,C,D) \: = \: (A \: + \: B \: + \: C \: + \: D)(A \: + \: B \: + \: C \: + \: \overline{D} )(A \: + \: B \: + \: \overline{C} \: + \: D) \: \dotso \: (\overline{A} \: + \: \overline{B} \: + \: \overline{C} \: + \: \overline{D})}
该表达式的最大项表示为：
The maxterm representation of this expression is as,
\mathrm{f(A,B,C,D) \: = \: M_{0}\cdot M_{1}\cdot M_{2} \: \dotso \: M_{15}}
一个 4 变量 K-map 有 16 个单元格，其中每个单元格要么表示函数的最小项，要么表示函数的最大项。四变量布尔表达式的 SOP（乘积之和）形式和 POS（求和之积）形式如图 1 所示。
A 4 variable K-map has 16 cells, where each cell represents either a minterm or a maxterm of the function. The SOP (Sum of products) form and POS (Product of Sums) form of a four variable Boolean expression is shown in Figure 1.
在此处，列和行的二进制数表示法为格雷码。这被称为 adjacent ordering 。在这些 K-map 中，地图左侧的二进制数字表示沿任何行变量 A 和 B 的条件，并且 K-map 顶部的二进制数字表示沿任何列变量 C 和 D 的条件。单元格右下角的十进制数字表示最小项或最大项名称。
Here, the binary number designations of the columns and rows are in the Gray code. This is known as adjacent ordering. In these K-maps, the binary numbers along the left side of the map indicate the conditions of variables A and B along any row, and the binary numbers along the top of the K-map indicate the conditions of the variables C and D along any column. The decimal numbers in the bottom right corners of the cells indicate the minterm or maxterm designation.
现在，让我们考虑一个示例来说明利用 4 变量 K-map 简化布尔函数。
Now, let us consider an example to illustrate the utilization of the 4 variable K-map for simplification of a Boolean function.
Example 1
使用 4 变量 K-map 简化以下布尔表达式。
Simplify the following Boolean expression using the 4-variable K-map.
\mathrm{f \lgroup A,B,C,D \rgroup \: = \: \sum m \lgroup 2,3,6,7,8,10,13,15 \rgroup}
Solution
给定布尔函数的 SOP K-map 表示如图 2 所示。
The SOP K-map representation of the given Boolean function is shown in Figure 2.
Explanation
函数的简化按照以下步骤完成 −
The simplification of the function is done as per the following steps −
K 映射中没有孤立的 1。
There are no isolated 1s in the K-Map.
最小项 m2 可以与 m3、m6 和 m7 形成一个 4-大方块。形成它并将其读取为−
The minterm m2 can form a 4-square with m3, m6, and m7. Make it and read it as −
\mathrm{\bar{A}C}()
最小项 m8 可以与 m10 形成一个 2-大方块。形成它并将其读取为 −
The minterm m8 can form a 2-square with m10. Make it and read it as −
\mathrm{AB\bar{D}}()
最小项 m13 可以与 m15 形成一个 2-大方块。形成它并将其读取为 −
The minterm m13 can form a 2-square with m15. Make it and read it as −
\mathrm{(A\bar{B}D)}
以 SOP 形式编写所有乘积。
Write all the product in SOP form.
因此，简化的 SOP 表达式为，
So the simplified SOP expression is,
\mathrm{f(A,B,C,D) \: = \: \bar{A}C \: + \: A\bar{B}D \: + \: AB\bar{D}}
Example 2
使用 4 变量 K 映射最小化以下布尔表达式。
Minimize the following Boolean expression using the 4-variable K-map.
\mathrm{f(A,B,C,D) \: = \: \prod \: M(4,6,11,14,15)}
Solution
给定布尔函数的 POS K-map 表示形式如图 3 所示。
The POS K-map representation of the given Boolean function is shown in Figure 3.
Explanation
按以下步骤完成给定函数的最小化 −
In minimization of the given function is done as per the following step −
K 图中没有孤立的零。
There are no isolated zeros in the K-map.
最大项 M4 可以与 M6 形成一个 2-大方块。形成它并将其读取为 −
The maxterm M4 can form a 2-squate with M6. Make it and read it as −
\mathrm{(A \: + \: \bar{B} \: + \: D)}
最大项 M11 可以与 M15 形成一个 2-大方块。形成它并将其读取为 −
The maxterm M11 can form a 2-square with M15. Make it and read it as −
\mathrm{(\bar{A} \: + \: \bar{C} \: + \: \bar{D})}
现在，仅剩下最大项 M14。M14 可以与 M6 或 M15 形成 2-大方块。如果我们用 M15 形成它，则将其读取为 −
Now, only maxterm M14 is left. M14 can form 2-square with M6 or M15. If we make it with M15, then read it as −
\mathrm{(\bar{A} \: + \: \bar{B} \: + \: \bar{C})}
最后，以 POS 形式写出所有和式项。
Finally, write all the sum terms in POS form.
因此，给定布尔函数的化简 POS 表达式为，
So, the reduced POS expression of the given Boolean function is,
\mathrm{f(A,B,C,D) \: = \: (A \: + \: \bar{B} \: + \: D)(\bar{A} \: + \: \bar{C} \: + \: \bar{D})(\bar{A} \: + \: \bar{B} \: + \: \bar{C})}
Numerical Problems on K-Map
Q1. 使用 4 变量 K 映射简化以下布尔函数
Q1. Reduce the following Boolean function using 4-variable K-map
\mathrm{f \lgroup A,B,C,D \rgroup \: = \: \sum m \lgroup 0,1,2,5,8,10,11,13,14,15 \rgroup }
Q2. 使用 4 变量 K 映射化简以下布尔表达式。
Q2. Minimize the following Boolean expression using 4-variable K-Map.
\mathrm{f \lgroup A,B,C,D \rgroup \: = \: \prod \: M \lgroup 0,2,8,10,11,13,15 \rgroup}
Conclusion
这一切都与 4 变量卡诺图及其应用于将布尔表达式最小化为其极小形式有关。从上述讨论中，我们可以得出结论，4 变量卡诺图是包含 4 个变量的布尔表达式的图形表示，并且以标准 SOP 或 POS 形式表示。这用于将 4 变量布尔表达式的标准 SOP 或 POS 形式或最小项形式或最大项形式转换为其最小 SOP 或 POS 形式。
This is all about the 4 variable K-map and its application to minimize a Boolean expression into its minimal form. From the above discussion, we can conclude that the 4-variable K-map is a graphical representation of a Boolean expression involving 4 variables and is represented in standard SOP or POS form. This is used to convert a standard SOP or POS form, or minterm form or maxterm form of a 4-variable Boolean expression into its minimal SOP or POS form.