Karnaugh tables ; number of groups Hey guys i was just wondering why is the number of all groups in a Karnaugh table equal to $3^n$ ; for example :
In a $2×2$ table with $2$ variables we have $4$ groups of one and $4$ groups of two and $1$ group of four which is equal to $9$.
Or in a $2×4$ table with $3$ variables we have $8$ groups of one and $12$ groups of two and $6$ groups of four and $1$ group of eight which is equal to $27$.
 A: For readers not familiar with these Karnaugh tables, see here.
First of all, congratulations for this non-evident observation.
There is simple proof based on a certain coding. I understand what you mean by a group (which is the advantage of thinking geometrically with Karnaugh tables), but I will switch to the more tractable equivalent boolean algebra expressions.
Let us understand the situation in the case of 3 variables (corresponding to your $2 \times 4$ table).
In this case, the monomials are :
$$A,B,C,\bar{A},\bar{B},\bar{C},A\bar{B},\bar{A}B,..., ABC,A\bar{B}C,....$$
(please note that variables are listed in the alphabetic order: $A$, then $B$ then $C$)
Let us consider any boolean monomial in the above list.
We are going to code it by a list of three numbers $-1,0,1$ in the following way.
When a variable:

*

*is present in it (without a bar on it), code it +1,


*is present in it as negated (i.e., with a bar), code it -1,


*is absent, code it $0$.
In this way, to any monomial having any number of variables, we can associate a specific code with three numbers.
Examples: $A\bar{B}C$ is coded $(1,-1,1)$, $A\bar{C}$ is coded $(1,0,-1)$, $C$ is coded $(0,0,1)$ etc...
resulting in a $3^3$ count, and more generaly a $3^n$ count if there are $n$ variables.
Remarks:

*

*In fact, we implicitly take into account the "void" monomial. Otherwise the count would be $3^n-1$...


*One could have given a (more complicated) proof by distinguishing the different cases (as you did) and using relationship $\displaystyle \sum_{i=0}^n {n\choose i} 2^i = (1+2)^n=3^n$.


*I discovered almost the same question here.
