K-Map MCQ Quiz - Objective Question with Answer for K-Map - Download Free PDF

The correct answer is C' + D'

Key PointsTo simplify the Boolean function F(A, B, C, D) using the Karnaugh Map (K-map) method, we first need to construct a 4-variable K-map and then fill in the values based on the given minterms (0, 1, 2, 4, 5, 6, 8, 9, 10, 12, 13, 14). The minterms can be represented in binary form to determine their positions on the K-map.

The K-map for F(A, B, C, D) would look like this:

simplified expression =C’ + D’

Hence the correct answer is C’ + D’

Simplified Boolean Function in POS Form

Problem Statement: The problem requires simplifying the given Boolean function f(A, B, C, D), which is expressed in terms of minterms (Σm) as:

f(A, B, C, D) = Σ(0, 2, 4, 5, 6, 7, 8, 10, 13, 15)

The task is to express the given Boolean function in its Product of Sums (POS) form and determine the correct simplified expression among the given options. The correct answer is Option 4.

Steps to Solve:

To simplify the Boolean function in POS form, follow these steps:

Step 1: Identify the Minterms and Maxterms

The given function is expressed in minterms as Σ(0, 2, 4, 5, 6, 7, 8, 10, 13, 15). To convert this into POS form, we first determine the maxterms, which are the complements of the minterms. The maxterms are the indices not included in the minterm list:

Minterms: 0, 2, 4, 5, 6, 7, 8, 10, 13, 15

Maxterms: 1, 3, 9, 11, 12, 14

Step 2: Write the Maxterms in Standard POS Form

Each maxterm corresponds to the complement of the minterm. For a 4-variable function, the maxterms are expressed using the variables A, B, C, D in their complemented or uncomplemented forms:

• Minterm 1 → Maxterm: A + B' + C' + D'
• Minterm 3 → Maxterm: A + B' + C + D'
• Minterm 9 → Maxterm: A' + B + C' + D
• Minterm 11 → Maxterm: A + B' + C + D
• Minterm 12 → Maxterm: A' + B' + C + D'
• Minterm 14 → Maxterm: A + B' + C' + D

The POS form is the product (AND operation) of these maxterms.

Step 3: Simplify the POS Expression Using Boolean Algebra

Using Boolean algebra, the given function can be simplified step by step. After simplification, the Boolean function in POS form is:

f(A, B, C, D) = (B + D')(A' + B' + D)

Step 4: Match the Simplified Expression with the Given Options

The simplified POS expression matches Option 4. Hence, the correct answer is:

Option 4: (B + D')(A' + B' + D)

F (P, Q, R, S) = ∑ 0, 2, 5, 7, 8, 10, 13, 15

Don’t care min terms are 2, 7, 8, 13

By plotting the K-map, the minimal SOP (sum of products) can be found.

Explanation –

While putting the terms to k-map following things happen,

• 3^rd and 4^th columns are swapped
• 3^rd and 4^th rows.
• term 2 is going to (0, 3) column instead of (0, 2)
• 8 is going to (3, 0) instead of (2,0)

Solving, the above K-map, we get Q̅S̅ + QS

Concept:

We are given a Boolean function in the nested form: \( f(f(xy, y), z) \). To simplify this, we need to understand how the function \( f \) behaves and use Boolean algebra rules.

Step 1: Assume definition of function

Let us assume \( f(a, b) = a + b \), a common definition used in Boolean simplification unless otherwise specified.

So,

\( f(xy, y) = xy + y \)

Using the identity: \( xy + y = y \) (Since \( y(x + 1) = y \))

Step 2: Substitute into outer function

Now substitute into the outer function:

\( f(y, z) = y + z \)

Hence, the simplified expression is:

\( y + z \)

Compare with Options:

• Option 1: \( \bar{x} + z \) ❌
• Option 2: \( xyz \) ❌
• Option 3: \( xy̅ + z \) ❌
• Option 4: None of the options ✅

Final Answer: ✅ None of the options

The correct answer is : A̅ B̅ + C

Explanation:

To find the minimum product of sums (POS) for the given Karnaugh map (K-map), we'll follow these steps:

1. Analyze the K-map:

• Identify groups of 1s and determine their corresponding sum terms.
• Identify groups of 0s and determine their corresponding product terms.
• We'll use the don't-care conditions (marked as 'x') to simplify the expression.

Given: \(\rm Y=f(ABCD)=\Sigma(0,1,2,6,7,10,14)+\Sigma d(3,8,11,15)\)

Grouping the given equation we will get the boolean function as: Y = A̅ B̅ + C

The correct answer is option 1.

Concept:

The given K-Map is,

F = a'b'c+a'bc'+ab'c'+abc

F= a'(b'c+bc')+a (b'c'+bc)

F=a'(b⊕c)+a(b⊙c)

F=a'(b⊕c)+a(b⊕c)'

F=a⊕b⊕c

Hence the correct answer is Exclusive OR.

The correct answer is option 4

K-maps

F(A, B, C, D) = ∑ (0, 1, 2, 3, 6, 12, 13, 14, 15)

Two K-Maps can be constructed from the given boolean function

The expression for K-Map 1 is AB + A'B' + A'CD'

The expression for K-Map 2 is AB +A'B' + BCD'

Concept:

Draw the K- map, convert the K-map into a SOP (sum of product) or POS (product of sum) form. While reducing the K-map in these forms, a don’t care will be needed only when with the use of don’t cares we can reduce the term size.

Diagram: K – Map

From the K-map simplification:

F(a, b, c, d) = a̅.c

F(a, b, c, d) = \(\overline {\left( {a + \bar c} \right)}\)

Diagram:

Therefore, only one NOR gate is needed to implement the minimized function

Concept:

We can simplify the given boolean function with the help of K-Map.

The K-map is a systematic way of simplifying Boolean expressions. With the help of the K-map method, we can find the simplest POS and SOP expression, which is known as the minimum expression.

Just like the truth table, a K-map contains all the possible values of input variables and their corresponding output values.

The K-map method is used for expressions containing 2, 3, 4, and 5 variables.

Calculation:

Given ;  F (x, y, z) = Σ(0, 2, 4, 5, 6) 

3 variable K-map:

The grouping of cells has shown below

The expression obtained from the K-Map → F = z' + xy'

Additional Information Note 1 − If outputs are not defined for some combination of inputs, then those output values will be represented with the don’t care symbol ‘x’. That means, we can consider them as either ‘0’ or ‘1’.

Note 2 − If don’t care terms are also present, then place doesn’t care ‘x’ in the respective cells of the K-map. Consider only the don’t care ‘x’ that are helpful for grouping the maximum number of adjacent ones. In those cases, treat the don’t care value as ‘1’.

F(P, Q, R, S) = PQ + P̅QR + P̅QR̅S

F(P, Q, R, S) = PQ(R + R̅)(S + S̅) + P̅QR(S + S̅) + P̅QR̅S  

F(P, Q, R, S) = PQRS + PQRS̅ + PQR̅S + PQR̅S̅ + P̅QRS + P̅QRS̅ + P̅QR̅S

F(P, Q, R, S) = ∑ (15 + 14 + 13 + 12 + 7 + 6 + 5)  

K-Map:

F(P, Q, R, S) = PQ + QR + QS

Concept:

The K-map is a graphical method that provides a systematic method for simplifying and manipulating the Boolean expressions or to convert a truth table to its corresponding logic circuit in a simple, orderly process.

In an 'n' variable K map, there are 2n cells

For 4 variables there will be 24 = 16 cells as shown:

Calculations:

F(a, b, c) = ∑(0, 2, 4, 5, 6)

F = c' + ab'

Karnaugh map (K-map):

• The Karnaugh map (K-map) is a method of simplifying Boolean algebra expressions.
• The Karnaugh map reduces the need for extensive calculations.
• Karnaugh map can be explained as An array that contains 2k number of cells, where k is the number of variables in the Boolean expression that is to be reduced or optimized.

For 4 variables there will be 24 = 16 cells as shown:

Number of cells in 2 variable k-map = 22 = 4

Number of cells in 3 variable k-map = 23 = 8

Number of cells in 4 variable k-map = 24 = 16

Number of cells in 5 variable k-map = 2⁵ = 32

The “Don’t Care” conditions allow us to replace the empty cell of a k-map and form a grouping of the variables which is larger than that of original groups. 

While forming groups of cells, we can consider a “Don’t Care” cell as 1 or 0 or we can also ignore that cell. 

Let us assume that A, B, C, and D are the 4 variables.

K- map for given function:

f(A, B, C, D) = BD + B̅ D̅

Implementing the above function by using XOR gates with complements of variables:

\(f\left( {A,\;B,\;C,\;D} \right) = \overline {\overline {B + \bar D} + \overline {\bar B + D} }\)

\(f\left( {A,\;B,\;C,\;D} \right) = \left( {B + \bar D} \right).\left( {\bar B + D} \right)\)

f(A, B, C, D) = BD + B̅ D̅

Explanation:-

As we already know from 'k' variables, (2^k) numbers can be formed. Thus the number of possible functions with 'k' variables as input such that there are 'm' minterms are,

2^kC_m or C(2^k, m)

Where (2^k ) is a possible number from 'k' variables and 'm' are the desired number of minterm for which the number of the function needs to be calculated. 

For example,

Counting the number of Boolean functions possible with two variables such that there are exactly two minterms.

Calculation:-

As we already know from two variables (a and b) four numbers (0, 1, 2, 3) can be formed i.e, in binary digits 00, 01, 10, 11 and possible Min terms are a'b', 

a’b, ab’, ab respectively which gives ‘1’ as the output for respective binary digits as input.

Thus the number of possible functions with two variables as input such that there are exactly two minterms are,

⁴C₂ = 4! / 2!(4 - 2)! = 24 / (2 × 2) = 24/4 = 6

Where ‘4’ is the possible number from two variables and ‘2’ is the desired number of minterms for which the number of the function needs to be calculated.

Explanation :

Given:

f(w, 0, 0, z)= 1

f(1, x, 1, z)= x+z

f(w, 1, y, z)= wz+y

Only thing makes this Question complicated is how you fill the K map cells.

1. f(w, 0, 0, z) = 1 ; means whenever x = 0 and y = 0 (total 4 cells) then put 1 in k-map cell.

2. f(1, x, 1, z) = x+z ; means whenever w=1 and y=1 (total 4 cells) then evaluate the expression “x+z” as per each cell (only those where w=1 and y=1) and fill the result in that particular cell.

3. f(w, 1, y, z) = wz+y ; means whenever x=1( total 8 cells) then evaluate the expression “wz+y” as per each cell (only those where x=1) and fill the result in that particular cell.

1. In the Below K map first we mark x = 0 and y = 0

  1001, 0000 , 0001 , 1000 these 4 cells will be marked as 1 .

  (wx’y’z , w’x’y’z’ , w’x’y’z , wx’y’z’) likewise we can fill the case where w=1 and y=1.

2. Now x=1( total 8 cells)

                     0100,     0101,           0110,          0111,          1100,        1101,        1110,        1111

                     w’xy’z’,      w’xy’z,      w’xyz’,        w’xyz,        wxy’z’,      wxy’z,        wxyz’,       wxyz

Likewise we can fill the K map as given in Image.

So after minimum sum of product expression will be x’y’ + xy + wz ; so total 6 literals. Answer will be 6.