# 5-Variable K-Map – Karnaugh Map In Digital Electronics Tutorial Part 6

A 5-variable K-Map will have 25 = 32 cells. A function F which has maximum decimal value of 31, can be defined and simplified by a 5-variable Karnaugh Map.

#### 5-Variable K-Map

In above boolean table, from 0 to 15, A is 0 and from 16 to 31, A is 1. A 5-variable K-Map is drawn as below.

Again, as we did with 3-variable & 4-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 two squares. What we need to do is to visualize second square on first square and figure out adjacent cells. Let’s understand how to simplify 5-variables K-Map by taking couple of examples.

#### Example 1 of 5-Variable K-Map

Given function, F = Σ (1, 3, 4, 5, 11, 12, 14, 16, 20, 21, 30)

Since, the biggest number is 30, we need to have 5 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 no octet. There is one quad that is obtained by visualizing second square on first, there are 4 adjacent cells – 4,5,20 and 21. The octet is highlighted by a blue connecting line. There are 5 pairs. Similar to quad, there is one pair between two squares which is highlighted by the blue connecting line.

(4, 5, 20, 21) – B’CD’ (Since A & E are the changing variables, it is eliminated)

(12, 14) – A’BCE’ (Since D is the changing variable, it is eliminated)

(14, 30) – BCDE’ (Since A is the changing variable, it is eliminated)

(3, 11) – A’C'DE (Since B is the changing variable, it is eliminated)

(16, 20) – AB’D'E’ (Since C is the changing variable, it is eliminated)

(1, 3) – A’B'C’E (Since D is the changing variable, it is eliminated)

Thus, F = B’CD’ + A’BCE’ + BCDE’ + A’C'DE + AB’D'E’ + A’B'C’E

#### Example 2 of 5-Variable K-Map

Given function, F = Σ (0, 2, 3, 5, 7, 8, 11, 13, 17, 19, 23, 24, 29, 30)

Since, the biggest number is 30, we need to have 5 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 no octet. First we need to look for quads within each of the squares. There are none but there is a quad between two squares that is obtained by visualizing second square on first, there are 4 adjacent cells – 3, 7, 19 and 23. This quad is highlighted by blue connecting line. There are 6 pairs, out of which two are between two squares, highlighted by blue connecting line.

(3, 7, 19, 23) -  B’DE (Since A & C are the changing variables, they are eliminated)

(3, 11) – A’C'DE (Since B is the changing variables, it is eliminated)

(1, 2) – A’B'C’E’ (Since D is the changing variables, it is eliminated)

(5, 7) – A’B'CE (Since D is the changing variables, it is eliminated)

(17, 19) – AB’C'E (Since D is the changing variables, it is eliminated)

(13, 29) – BCD’E (Since A is the changing variables, it is eliminated)

(8, 24) – BC’D'E’ (Since A is the changing variables, it is eliminated)

There is 1 in cell 30, which can not be looped with any adjacent cell, hence it can not be simplified further and left as it is.

30 – ABCDE’

Thus, F = B’DE + A’C'DE + A’B'C’E’ + A’B'CE + AB’C'E + BCD’E + BC’D'E’ + ABCDE’

#### Example 3 of 5-Variable K-Map

Given function, F = Σ (0, 1, 2, 3, 8, 9, 16, 17, 20, 21, 24, 25, 28, 29, 30, 31)

Since, the biggest number is 31, we need to have 5 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 2 octets. First one is in square 2 circled in red. Another octet is between 2 squares highlighted by blue connecting lines. There are 2 quads between each of the squares.

(16, 17, 20, 21, 28, 29, 24, 25) – AD’ (Since B, C and E are changing variables, they are eliminated)

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

(0, 1, 2, 3) – A’B'C’ (Since D and E are changing variables, they are eliminated)

(28, 29, 30, 31) – ABC (Since D and E are changing variables, they are eliminated)

Thus, F = AD’ + C’D’ + A’B'C’ + ABC

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

1. why is the slot of “0″ in example no. 2 still zero? isn’t zero part of the given function?

2. Yes in example 2:
slor of 0 should be fill with 1 which further make pair with one of slot 2….:-)

3. In Example 2, (1, 2) – A’B’C’E’ is wrong. It should be (0,2)

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