use the k-map method to minimize a 4-variable (w, x, y, and z) function f f = ∑(0,2,5,7,10,13,14,15)

To minimize a four-variable Boolean function \( f(w, x, y, z) \) using the Karnaugh map (K-map) method, follow these steps. The given function is:

\[
f(w, x, y, z) = \sum(0, 2, 5, 7, 10, 13, 14, 15)
\]

### Step 1: Create a 4-variable K-map

The K-map for a 4-variable function will have a 4x4 grid, with two variables on the horizontal axis and two on the vertical axis. Let's assign the variables as follows:

- Rows: \( wy \) combination (00, 01, 11, 10)
- Columns: \( xz \) combination (00, 01, 11, 10)

The K-map looks like this:

|      | 00  | 01  | 11  | 10  |
|------|-----|-----|-----|-----|
| **00** |     |     |     |     |
| **01** |     |     |     |     |
| **11** |     |     |     |     |
| **10** |     |     |     |     |

### Step 2: Plot the minterms

Next, we fill in the K-map with 1's corresponding to the minterms \( \sum(0, 2, 5, 7, 10, 13, 14, 15) \). The binary equivalents of these minterms are:

- Minterm 0: \( (0000) \)
- Minterm 2: \( (0010) \)
- Minterm 5: \( (0101) \)
- Minterm 7: \( (0111) \)
- Minterm 10: \( (1010) \)
- Minterm 13: \( (1101) \)
- Minterm 14: \( (1110) \)
- Minterm 15: \( (1111) \)

Now, plot the 1's in the K-map at these locations:

|      | 00  | 01  | 11  | 10  |
|------|-----|-----|-----|-----|
| **00** | 1   |     |     | 1   |
| **01** |     | 1   | 1   |     |
| **11** |     |     | 1   | 1   |
| **10** |     | 1   | 1   |     |

### Step 3: Group the 1's into largest possible groups

We group adjacent 1's in powers of 2 (1, 2, 4, or 8 cells), where adjacency includes wrapping around the edges.

1. A **group of four** covering minterms \( (0, 2, 10, 14) \).
2. A **group of four** covering minterms \( (5, 7, 13, 15) \).

### Step 4: Write the simplified Boolean expression

For each group, we find the common variables:

1. **Group covering (0, 2, 10, 14):**
   - Common in rows (00, 10): \( w' \)
   - Common in columns (00, 10): \( z' \)
   - So, the term for this group is \( w'z' \).

2. **Group covering (5, 7, 13, 15):**
   - Common in rows (01, 11): \( y \)
   - Common in columns (01, 11): \( x \)
   - So, the term for this group is \( xy \).

### Step 5: Final simplified expression

The minimized Boolean expression is:

\[
f(w, x, y, z) = w'z' + xy
\]

This is the simplified expression for the given function using the K-map method.

Would you like further details or have any questions?

Here are 5 related questions:
1. How does wrapping work in a K-map?
2. What happens if a minterm is not included in any group?
3. Can K-maps be used for more than 4 variables?
4. How would the solution change if we included "don't care" terms?
5. Can we use K-maps for functions of fewer variables?

**Tip:** Always try to form the largest possible groups in the K-map to minimize the number of terms in the expression.

use the k-map method to minimize a 4-variable (w, x, y, and z) function f = ∑(0,2,5,7,10,13,14,15)

Learn how to minimize the Boolean function f(w, x, y, z) = ∑(0,2,5,7,10,13,14,15) using the Karnaugh map (K-map) method. This step-by-step guide explains how to plot minterms, group adjacent 1's, and find the simplified Boolean expression. Suitable for high school and college students.

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