A 3-variable K-Map will have **2 ^{3} = 8 cells**. As told in last post, number of variables are decided by the biggest decimal number in a given function. A function F which has maximum decimal value of 7, can be defined and simplified by a

**3-variable Karnaugh Map**.

#### Boolean Table For 3 Variables

#### 3-Variable K-Map

Carefully note how cells are numbered in above diagram. First cells is denoted by 0, second by 1 and then third by 3 and not by 2. This is because, A’BC (ANDing of first row A’ and third column BC) corresponds to decimal number 3 in the boolean table. Similarly, second row, third column ABC is denoted by 7 and not by 6.

Now, let’s understand how to simplify 3-variables K-Map by taking couple of examples.

#### Example of 3-Variable K-Map

Given function, **F = Σ (1, 2, 3, 4, 5, 6)**

Since the biggest number in this function is 6, it can be defined by 3 variables.

Let’s draw K-Map for this function by writing 1 in cells that are present in function and 0 in rest of the cells. You don’t necessarily need to write 0s but it is okay to have them.

We need to apply rules for simplifying K-Map that we read in last tutorial. So, first we need to look for an octet i.e. 8 adjacent 1′s. There is none, so we should now look for a quad i.e. 4 adjacent 1′s. Again, there is none, so we will look for pairs. There are 3 pairs circled in red.

(1,3) – A’C (Since B is the changing variable between these two cells, it is eliminated)

(2,6) – BC’ (Since A is the changing variable, it is eliminated)

(4, 5) – AB’ (Since C is the changing variable, it is eliminated)

Thus,** F = A’C + BC’ + AB’**

We hope that you will find it easy when you solve next 3-variable k-map. Since, 3-variable K-Map is simplest of all to solve, we have put more focus on 4-variable and 5-variable K-Maps and you will find couple of more examples in next part of this tutorial.

Read the full series at Part 1, Part 2, Part 3, Part 4, Part 5, Part 6 and Part 7.