A 6-variable K-Map will have 2^{6} = 64 cells. A function F which has maximum decimal value of 63, can be defined and simplified by a **6-variable Karnaugh Map**.

#### Boolean Table For 6-Variables Karnaugh Map

Boolean table for 6 variables is quite big, so we have shown only values, where there is a noticeable change in values which will help us to draw the K-Map.

A = 0 for decimal values 0 to 31 and A = 1 for 31 to 63.

B = 0 for decimal values 0 to 15 and 32 to 47. B = 1 for decimal values 16 to 31 and 48 to 63.

#### 6-Variable K-Map

A 6-variable K-Map is drawn as below:

Again, as we did with 3-variable, 4-variable and 5-variable K-Map, carefully note the numbering of each cell. Now, we have two squares and we can loop octets, quads and pairs between these four squares. What we need to do is to visualize each of these squares one on another and figure out adjacent cells. Let’s understand how to simplify 6-variables K-Map by taking couple of examples.

#### Example 1 of 6-Variable K-Map

Given function, **F = Σ (0, 2, 4, 8, 10, 13, 15, 16, 18, 20, 23, 24, 26, 32, 34, 40, 41, 42, 45, 47, 48, 50, 56, 57, 58, 60, 61)**

Since, the biggest number is 61, we need to have 6 variables to define this function.

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.

Applying rules of simplifying K-Map, there is one loop which has 16 1′s – containing 1′s at all the corners of all 4 squares. We obtain it by visualizing all the 4 squares over one another but only in horizontal or vertical direction (not diagonal) and figuring out adjacent cells. All the 1′s in corners are circled in green.

There are 4 pairs, one in fourth square at bottom-right and other 3 are between the squares and are highlighted by blue connecting line.

(0, 2, 8, 10, 16, 18, 24, 26, 32, 34, 40, 42, 48, 50, 56, 58) – D’F’ (A, B, C and E are changing variables, so they are eliminated)

(41, 45, 57, 61) – ACE’F (B & D are changing variables, so they are eliminated)

(13, 15, 45, 47) – B’CDF (A & E are changing variables, so they are eliminated)

(0, 4, 16, 20) – A’C'E’F’ (B & D are changing variables, so they are eliminated)

(56, 57, 60, 61) – ABCE’ (D and F are changing variables, so they are eliminated)

There is 1 in cell 23, which can not be looped with any adjacent cell, hence it can not be simplified further and left as it is.

23 = A’BC’DEF

Thus, **F = D’F’ + ACE’F + B’CDF + A’C'E’F’ + ABCE’ + A’BC’DEF**

#### Example 2 of 6-Variable K-Map

Given function, **F = Σ (0, 1, 2, 3, 4, 5, 8, 9, 12, 13, 16, 17, 18, 19, 24, 25, 36, 37, 38, 39, 52, 53, 60, 61)
**

Since, the biggest number is 61, we need to have 6 variables to define this function.

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.

Applying rules of simplifying K-Map, there are 3 octets. First one is in in first square on top-left. Second and third octets are obtained by over-lapping square 1 (top-left) and 2 (top-right). There are two quads, one in third square (bottom-left) and second in fourth square (bottom-right).

(0, 1, 4, 5, 8, 9, 12, 13) – A’B'E’ (C, D and F are changing variables, they are eliminated)

(0, 1, 2, 3, 16, 17, 18, 19) – A’C'D’ (B, E and F are changing variables, they are eliminated)

(0, 1, 8, 9, 16, 17, 24, 25) – A’D'E’ (B, C and F are changing variables, they are eliminated)

(36, 37, 38, 39) – AB’C'D (E and F are changing variables, they are eliminated)

(52, 53, 60, 61) – ABCE’ (D and F are changing variables, they are eliminated)

Thus, **F = A’B'E’ + A’C'D’ + A’D'E’ + AB’C'D + ABCE’**

This is the last part of the series. Read the full series at Part 1, Part 2, Part 3, Part 4, Part 5, Part 6 and Part 7. We hope that you found this tutorial helpful. Do let us know your feedback, suggestions and queries in the comments section below

nice one…my doubts have been cleared …thanks

Hi I have a correction here

(52, 53, 60, 61) – ABCE’ (D and F are changing variables, they are eliminated) from Example 2 of 6-Variable K-Map

it should be ABDE’ and not ABCE’ ^^

that’s all GodBless

I agree with Jolina. ABDE’ is the correct answer for (52, 53, 60, 61) on example 2 NOT ABCE’. By the way, great tutorial after all. Thank you!

what if the problem is like f=(15,63) for the 6 variable?

CDEF will be the answer of (15 ,63)

GOOD TUTOR NICE ONE TO ALL. AT LEST I HAVE BEEN ABLE TO LEARN SOMETHING NEW

good examples and explanation