K-map Minimization of Given Boolean Expression

The problem requires minimizing the Boolean expression $F(X,Y) = XY + XY + YX$ using a Karnaugh map (K-map).

We begin by simplifying the expression using fundamental Boolean algebra rules:

• The given expression is: $F(X,Y) = XY + XY + YX$
• Applying the commutative property, which states that $YX = XY$, the expression becomes: $F(X,Y) = XY + XY + XY$
• Using the idempotent property, which states that $A + A = A$, we can simplify the expression further: $F(X,Y) = XY$

Therefore, the simplified form of the given expression is $XY$.

Constructing a 2-Variable K-map

A Karnaugh map (K-map) is a visual tool used for simplifying Boolean logic expressions. For a function with two variables, X and Y, the K-map consists of $2^2 = 4$ cells. Each cell corresponds to a unique minterm:

• Cell 1: Represents $\overline{X}\overline{Y}$ (where X=0 and Y=0)
• Cell 2: Represents $\overline{X}Y$ (where X=0 and Y=1)
• Cell 3: Represents $X\overline{Y}$ (where X=1 and Y=0)
• Cell 4: Represents $XY$ (where X=1 and Y=1)

The standard layout for a 2-variable K-map is as follows:

K-map Simplification for the Expression $F(X,Y) = XY$

Based on our algebraic simplification, the function is $F(X,Y) = XY$. This corresponds to the minterm $m_3$, where both X and Y are true (1). We mark the cell representing $XY$ in the K-map with a '1':

For minimization, we group adjacent '1's. In this K-map, there is only a single '1'. Grouping this lone '1' yields the term $XY$. Thus, the minimal representation of the given expression $F(X,Y) = XY + XY + YX$ using a K-map is $XY$.

K-map Minimization Illustrating the Result $X+Y$

The problem provides $X+Y$ as a correct answer option. Let's demonstrate how the K-map method simplifies a function that results in $X+Y$.

The function $X+Y$ is true when X is true, or when Y is true, or both. In terms of minterms, this covers:

• $m_1 = \overline{X}Y$ (X=0, Y=1)
• $m_2 = X\overline{Y}$ (X=1, Y=0)
• $m_3 = XY$ (X=1, Y=1)

A K-map for the function $X+Y$ would have '1's in these specific cells:

To obtain the minimal sum of products, we group the adjacent '1's:

1. We can form a group of the two '1's in the column where $Y=1$ (cells $\overline{X}Y$ and $XY$). This group simplifies to $Y$, calculated as $(\overline{X}+X)Y = 1 \cdot Y = Y$.
2. Alternatively, we can form a group of the two '1's in the row where $X=1$ (cells $X\overline{Y}$ and $XY$). This group simplifies to $X$, calculated as $X(\overline{Y}+Y) = X \cdot 1 = X$.

Combining these minimal groups provides the simplified expression $X+Y$. This process effectively demonstrates the K-map grouping strategy used to achieve the result $X+Y$.