Karnaugh Maps - Digital Logic Simplification PDF

CHAPTER 4 WAN NOR HAFIDZAH BINTI WAN MOHD  LESSON OUTCOMES At the end of this lesson, the students should be able to: Construct the Karnaugh Map based on number of input variables (2-input, 3-input and 4-input variables). Mapping the value into the Karnaugh Map....

CHAPTER 4 WAN NOR HAFIDZAH BINTI WAN MOHD  LESSON OUTCOMES At the end of this lesson, the students should be able to: Construct the Karnaugh Map based on number of input variables (2-input, 3-input and 4-input variables). Mapping the value into the Karnaugh Map. Simplify the truth table and Boolean expression using Karnaugh Map. Understand the concept of ‘don’t care’ term on Karnaugh Map.  KARNAUGH MAPS One of the easiest ways to simplify logic circuits is to use Karnaugh Map method. Karnaugh map is a graphical representation of the output for a given logic expression. It contains the same information as a truth table.  KARNAUGH MAPS The K-Map contains a cell for each input combination. A logic expression or truth table with n input variables has 2n cells on the K-Map. A 2 variable K-Map has 22 = 4 cells. AB=00 corresponds to cell 𝐴𝐴. 𝐵𝐵 or cell 0. AB=01 corresponds to cell 𝐴𝐴. 𝐵𝐵 or cell 1. AB=10 corresponds to cell 𝐴𝐴. 𝐵𝐵 or cell 2. AB=11 corresponds to cell 𝐴𝐴. 𝐵𝐵 or cell 3.  KARNAUGH MAPS The K-Map cells are labeled so that horizontally & vertically adjacent cells differ only in one variable. For example: Cell 𝐴𝐴. 𝐵𝐵 (0) and cell 𝐴𝐴. 𝐵𝐵 (2): differ only in one variable, A. Cell 𝐴𝐴. 𝐵𝐵 (0) and cell 𝐴𝐴. 𝐵𝐵 (1): differ only in one variable, B.  KARNAUGH MAPS A 3 variable K-Map has 23 = 8 cells. The labeling of a 3 variable K-Map is not in counting order. They are labeled as (0 1), (2 3), (6 7) and (4 5), so that horizontally and vertically adjacent cells differ only in one variable.  l KARNAUGH MAPS For example: Cell (0) and cell (1): only C is different Cell (2) and cell (3): only C is different Cell (6) and cell (7): only C is different Cell (4) and cell (5): only B is different Cell (0) and cell (4) are considered adjacent: only A is different – the left and right side of the map are connected as it rolled into a cylinder.  KARNAUGH MAPS A 4 variable K-Map has 24=16 cells. The cells are labeled in the order of: (0 1 3 2), (4 5 7 6), (12 13 15 14) and (8 9 11 10). For example, cell 0 is considered adjacent to cell 1, 4 8 and 2. Cell 1 is adjacent to cell 0, 3, 5 and 9. SIMPLIFICATION USING 2 VARIABLE K-MAP The steps in simplifying a logic expression using a K-Map: 1) Write a sum of product expression from the truth table. 2) Plot a 1 on the K-Map for each product term, or plot a 1 on the K-Map for each output Y = 1 3) Draw loops around adjacent cells containing two 1’s on the K-Map. The loops may overlap. SIMPLIFICATION USING 2 VARIABLE K-MAP 4) Repeat step 2 to 3. 5) Each loop produces a simplified product term. The minimum product term for 2 variable map is as follows: a) 1 cell group yield 2 variable product term b) 2 cell group yield 1 variable product term c) 4 cell group yield of 1 for the expression 6) Logically OR the simplified product term. SIMPLIFICATION USING 2 VARIABLE K-MAP Example 1: Given an OR gate truth table: The logic expression in the form of sum of product is 𝑌𝑌 = 𝐴𝐴. 𝐵𝐵 + 𝐴𝐴. 𝐵𝐵 + 𝐴𝐴. 𝐵𝐵 which can also be written as 𝑌𝑌 = ∑(1,2,3). Plot 1 on the K-Map (cell 1, 2 and 3). Loop the adjacent cells. There are 2 pairs of 1’s that can be loop: cell (1 3) and cell (2 3). Each loop contains two 1’s. SIMPLIFICATION USING 2 VARIABLE K-MAP Each loop produces a simplified product term. The loop for cell (1 3) produces a product term B. This can easily be proven using Boolean algebra: 𝐴𝐴. 𝐵𝐵 + 𝐴𝐴. 𝐵𝐵 = 𝐴𝐴 𝐵𝐵 + 𝐵𝐵 = 𝐴𝐴. The loop for cell (2 3) produces a product term A, that can proven using Boolean algebra: 𝐴𝐴. 𝐵𝐵 + 𝐴𝐴. 𝐵𝐵 = 𝐵𝐵 𝐴𝐴 + 𝐴𝐴 = 𝐵𝐵. Logically OR the simplified product term. The simplified Boolean expression is 𝑌𝑌 = 𝐴𝐴 + 𝐵𝐵. SIMPLIFICATION USING 2 VARIABLE K-MAP Example 2: Shown below is a NAND gate truth table & K-Map: The loop for cell (0 1) produces 𝐴𝐴. The loop for cell (0 2) produces 𝐵𝐵. The simplified expression is 𝑌𝑌 = 𝐴𝐴 + 𝐵𝐵 which is equivalent to 𝐴𝐴. 𝐵𝐵. SIMPLIFICATION USING 2 VARIABLE K-MAP Example 3: Shown below is an XOR gate truth table & K-Map. Cell 1 is not considered adjacent to cell 2, because both variables are different - not possible to form a loop. We cannot simplify the expression. So the expression is, 𝑌𝑌 = 𝐴𝐴. 𝐵𝐵 + 𝐴𝐴. 𝐵𝐵 which is equivalent to 𝑌𝑌 = 𝐴𝐴 ⊕ 𝐵𝐵. SIMPLIFICATION USING 3 VARIABLE K-MAP A 3 variable K-Map allows loop containing two, four and eight 1’s. For example, a loop containing two 1’s are: (0 1), (0 2) and (0 4). A loop containing four 1’s are: (0 1 2 3), (2 3 6 7), (6 7 4 5), (0 1 4 5), (0 2 6 4) and (1 3 7 5). SIMPLIFICATION USING 3 VARIABLE K-MAP Steps in simplifying a 3 variable K-Map: 1) Draw loop around adjacent cells containing four 1s. A loop containing many 1’s will produce a simpler product term. So, find out if there are other possible loops containing four 1’s. The loops may overlap. 2) After finishing loops containing four 1’s, draw loops containing two 1’s. 3) Each loop produces a simplified product term. The minimum product term for 3 variable map is as follows: a) 1 cell group yield 3 variable product term b) 2 cell group yield 2 variable product term c) 4 cell group yield 1 variable product term d) 8 cell group yield of 1 for the expression 4) Logically OR the simplified product term. SIMPLIFICATION USING 3 VARIABLE K-MAP Example 1: Write the simplified expression from the truth table shown below. SIMPLIFICATION USING 3 VARIABLE K-MAP The loop for cell (0 1) produces 𝐴𝐴. 𝐵𝐵. This can be shown algebraically 𝐴𝐴. 𝐵𝐵. 𝐶𝐶 + 𝐴𝐴. 𝐵𝐵. 𝐶𝐶 = 𝐴𝐴. 𝐵𝐵 𝐶𝐶 + 𝐶𝐶 = 𝐴𝐴. 𝐵𝐵. The loop for cell (0 2) produces 𝐴𝐴. 𝐶𝐶. This can be shown algebraically 𝐴𝐴. 𝐵𝐵. 𝐶𝐶 + 𝐴𝐴. 𝐵𝐵. 𝐶𝐶 = 𝐴𝐴. 𝐶𝐶 𝐵𝐵 + 𝐵𝐵 = 𝐴𝐴. 𝐶𝐶. The simplified expression is 𝑌𝑌 = 𝐴𝐴. 𝐵𝐵 + 𝐴𝐴. 𝐶𝐶. We can prove this using Boolean algebra: 𝑌𝑌 = 𝐴𝐴. 𝐵𝐵. 𝐶𝐶 + 𝐴𝐴. 𝐵𝐵. 𝐶𝐶 + 𝐴𝐴. 𝐵𝐵. 𝐶𝐶 = 𝐴𝐴. 𝐵𝐵 𝐶𝐶 + 𝐶𝐶 + 𝐴𝐴. 𝐵𝐵. 𝐶𝐶 = 𝐴𝐴. 𝐵𝐵 + 𝐴𝐴. 𝐵𝐵. 𝐶𝐶 = 𝐴𝐴 𝐵𝐵 + 𝐵𝐵. 𝐶𝐶 = 𝐴𝐴 𝐵𝐵 + 𝐵𝐵)(𝐵𝐵 + 𝐶𝐶 = 𝐴𝐴 𝐵𝐵 + 𝐶𝐶 = 𝐴𝐴. 𝐵𝐵 + 𝐴𝐴. 𝐶𝐶 SIMPLIFICATION USING 3 VARIABLE K-MAP Example 2: Write the simplified expression from the truth table shown below. SIMPLIFICATION USING 3 VARIABLE K-MAP Draw loop around cells containing four 1’s: cell (0 1 2 3). Next a loop containing two 1’s: cell (2 6). Loop (0 1 2 3) produces the term 𝐴𝐴. This can be proved as follows: 𝐴𝐴. 𝐵𝐵. 𝐶𝐶 + 𝐴𝐴. 𝐵𝐵. 𝐶𝐶 + 𝐴𝐴. 𝐵𝐵. 𝐶𝐶 + 𝐴𝐴. 𝐵𝐵. 𝐶𝐶 = 𝐴𝐴. 𝐵𝐵 𝐶𝐶 + 𝐶𝐶 + 𝐴𝐴. 𝐵𝐵 𝐶𝐶 + 𝐶𝐶 = 𝐴𝐴. 𝐵𝐵 + 𝐴𝐴. 𝐵𝐵 = 𝐴𝐴 𝐵𝐵 + 𝐵𝐵 = 𝐴𝐴 Loop (2 6) produces the term 𝐵𝐵. 𝐶𝐶. The simplified expression is 𝑌𝑌 = 𝐴𝐴 + 𝐵𝐵. 𝐶𝐶. SIMPLIFICATION USING 3 VARIABLE K-MAP Example 3: Write the simplified expression from the truth table shown below. In this example, there are two loops containing four 1’s. SIMPLIFICATION USING 3 VARIABLE K-MAP Loop (0 2 6 4) produces 𝐶𝐶. Loop (6 7 4 5) produces 𝐴𝐴. The simplified expression is 𝑌𝑌 = 𝐴𝐴 + 𝐶𝐶. We can’t form 3 loops containing two 1’s as shown below, because it will produce an unsimplified expression: 𝑌𝑌 = 𝐴𝐴. 𝐶𝐶 + 𝐴𝐴. 𝐶𝐶 + 𝐴𝐴. 𝐶𝐶 SIMPLIFICATION USING 3 VARIABLE K-MAP This unsimplified expression can be further simplified as follows: 𝑌𝑌 = 𝐴𝐴. 𝐶𝐶 + 𝐴𝐴. 𝐶𝐶 + 𝐴𝐴. 𝐶𝐶 = 𝐶𝐶(𝐴𝐴 + 𝐴𝐴) + 𝐴𝐴. 𝐶𝐶 = 𝐶𝐶 + 𝐴𝐴. 𝐶𝐶 = (𝐶𝐶 + 𝐴𝐴)(𝐶𝐶 + 𝐶𝐶) = 𝐴𝐴 + 𝐶𝐶. SIMPLIFICATION USING 3 VARIABLE K-MAP Example 4: The cells (0 4) on the K-map shown below can be connected forming a cylinder. The loop for cell (0 4) produces 𝑌𝑌 = 𝐵𝐵. 𝐶𝐶. SIMPLIFICATION USING 3 VARIABLE K-MAP The cells (0 1) and cell (4 5) below, form a loop containing four 1’s produces the term 𝑌𝑌 = 𝐵𝐵. SIMPLIFICATION USING 4 VARIABLE K-MAP A 4 variable K-Map allows loop containing two, four, eight and sixteen 1’s. For example: Loops containing two 1’s are: (0 1), (0 4), (0 2) and (0 8). Loops containing four 1’s are: (0 4 12 8), (0 1 3 2), (0 1 8 9) and (0 2 8 10). Loops containing eight1’s are: (0 1 3 2 4 5 7 6) and (0 1 3 2 8 9 11 10). SIMPLIFICATION USING 4 VARIABLE K-MAP Steps in simplifying a 4 variable K-Map: 1) Draw loop around adjacent cells containing eight 1s. A loop containing many 1’s will produce a simpler product term. So, find out if there are other possible loops containing eight 1’s. The loops may overlap. 2) After finishing loops containing eight 1’s, draw loops containing four 1’s. 3) Next draw loops containing two 1’s. SIMPLIFICATION USING 4 VARIABLE K-MAP 4) Each loop produces a simplified product term. The minimum product term for 4 variable map is as follows: a) 1 cell group yield 4 variable product term b) 2 cell group yield 3 variable product term c) 4 cell group yield 2 variable product term d) 8 cell group yield 1 variable product term e) 16 cell group yield of 1 for the expression 5) Logically OR the simplified product term. SIMPLIFICATION USING 4 VARIABLE K-MAP Example 1: Write the simplified expression from the truth table shown below. SIMPLIFICATION USING 4 VARIABLE K- MAP Solution: Loop (4 5 7 6) produces 𝐴𝐴. 𝐵𝐵. Loop (5 7 13 15) produces 𝐵𝐵. 𝐷𝐷. Loop (3 7) produces 𝐴𝐴. 𝐶𝐶. 𝐷𝐷. The simplified expression is 𝑌𝑌 = 𝐴𝐴. 𝐵𝐵 + 𝐵𝐵. 𝐷𝐷 + 𝐴𝐴. 𝐶𝐶. 𝐷𝐷. SIMPLIFICATION USING 4 VARIABLE K- MAP Example 2: Write the simplified expression from K-Map shown below: SIMPLIFICATION USING 4 VARIABLE K-MAP Solution: The loop containing eight 1’s produces the term 𝐵𝐵. The loop containing four 1’s produces the term 𝐶𝐶. 𝐷𝐷. The simplified expression is 𝑌𝑌 = 𝐵𝐵 + 𝐶𝐶. 𝐷𝐷. SIMPLIFICATION USING 4 VARIABLE K-MAP Example 3: Write the simplified expression from the K-Map shown below: 1) Connecting cell (1) and cell (9) forming a loop containing two 1’s produces the term 𝐵𝐵. 𝐶𝐶. 𝐷𝐷. SIMPLIFICATION USING 4 VARIABLE K-MAP 2) Connecting cell (4 12) and cell (6 14) forming a loop containing four 1’s produces the term 𝐵𝐵. 𝐷𝐷. SIMPLIFICATION USING 4 VARIABLE K-MAP 3) Connecting cells at the four edges forming a loop containing four 1’s (0 2 8 10) produces the term 𝐵𝐵. 𝐷𝐷.  EXERCISES 1. Simplify the following truth tables using Karnaugh map: a) A B C Y b) A B C D Y 0 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 1 1 0 0 1 0 1 0 1 0 1 0 0 1 1 0 0 1 1 1 0 1 0 0 1 1 0 0 0 0 1 0 1 1 0 1 1 0 1 1 0 1 0 0 1 1 1 0 1 1 0 1 1 0 0 0 1 1 1 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 0 1 1 1 1 0 SOLUTION: a) 𝐴𝐴. 𝐵𝐵 𝐴𝐴. 𝐵𝐵 𝐴𝐴. 𝐵𝐵 𝐴𝐴. 𝐵𝐵 𝐶𝐶 1 1 1 𝐶𝐶 1 1 𝑌𝑌 = 𝐴𝐴 + 𝐵𝐵. 𝐶𝐶 b) 𝐴𝐴. 𝐵𝐵 𝐴𝐴. 𝐵𝐵 𝐴𝐴. 𝐵𝐵 𝐴𝐴. 𝐵𝐵 𝐶𝐶. 𝐷𝐷 1 1 1 1 𝐶𝐶. 𝐷𝐷 1 1 𝐶𝐶. 𝐷𝐷 1 𝐶𝐶. 𝐷𝐷 1 1 1 𝑌𝑌 = 𝐶𝐶. 𝐷𝐷 + 𝐵𝐵. 𝐶𝐶 + 𝐴𝐴. 𝐷𝐷 + 𝐴𝐴. 𝐵𝐵. 𝐶𝐶  SIMPLIFYING BOOLEAN EXPRESSION USING KARNAUGH MAPS Simplifying Boolean expression using K-Map is easier and will usually produce a simplified expression compared to using Boolean algebra. Given a sum of product expression, we can map each product term on the K-Map.  SIMPLIFYING BOOLEAN EXPRESSION USING KARNAUGH MAPS Example 1: Suppose we a given a Boolean expression 𝑌𝑌 = 𝐴𝐴. 𝐵𝐵 + 𝐶𝐶 + 𝐴𝐴. 𝐵𝐵. 𝐶𝐶 instead of a truth table. We can fill in the K-Map by taking each of product terms and placing 1’s in the corresponding cells.  SIMPLIFYING BOOLEAN EXPRESSION USING KARNAUGH MAPS The product term 𝐴𝐴. 𝐵𝐵 will occupy cell (2 3). 𝐶𝐶 will occupy cell (1 3 7 5). 𝐴𝐴. 𝐵𝐵. 𝐶𝐶 will occupy cell (6).  SIMPLIFYING BOOLEAN EXPRESSION USING KARNAUGH MAPS Now the K-Map can be looped for simplification. The simplified expression is 𝑌𝑌 = 𝐵𝐵 + 𝐶𝐶. SIMPLIFYING BOOLEAN EXPRESSION USING KARNAUGH MAPS Example 2: Simplify the expression 𝑌𝑌 = 𝐵𝐵 𝐶𝐶𝐶𝐶 + 𝐶𝐶 + 𝐶𝐶. 𝐷𝐷(𝐴𝐴 + 𝐵𝐵 + 𝐴𝐴𝐴𝐴) using Karnaugh Map. Solution: 1) Expand the expression in the form of Sum of Product: 𝑌𝑌 = 𝐵𝐵 𝐶𝐶𝐶𝐶 + 𝐶𝐶 + 𝐶𝐶. 𝐷𝐷(𝐴𝐴 + 𝐵𝐵 + 𝐴𝐴𝐴𝐴) = 𝐵𝐵. 𝐶𝐶. 𝐷𝐷 + 𝐵𝐵. 𝐶𝐶 + 𝐶𝐶. 𝐷𝐷(𝐴𝐴. 𝐵𝐵 + 𝐴𝐴𝐴𝐴) = 𝐵𝐵. 𝐶𝐶. 𝐷𝐷 + 𝐵𝐵. 𝐶𝐶 + 𝐴𝐴. 𝐵𝐵. 𝐶𝐶. 𝐷𝐷 + 𝐴𝐴. 𝐵𝐵. 𝐶𝐶. 𝐷𝐷 SIMPLIFYING BOOLEAN EXPRESSION USING KARNAUGH MAPS 2) Map each product term on the K-Map. SIMPLIFYING BOOLEAN EXPRESSION USING KARNAUGH MAPS 3) Looping the K-Map will produce a simplified expression 𝑌𝑌 = 𝐴𝐴. 𝐵𝐵 + 𝐵𝐵. 𝐷𝐷 + 𝐵𝐵. 𝐶𝐶 + 𝐴𝐴. 𝐵𝐵. 𝐶𝐶. 𝐷𝐷  EXERCISES 1. Simplify the following equations using Karnaugh map: a) 𝑌𝑌 = 𝐴𝐴. 𝐵𝐵. 𝐶𝐶 + 𝐴𝐴. 𝐵𝐵. 𝐶𝐶 + 𝐵𝐵. 𝐶𝐶 + 𝐴𝐴. 𝐵𝐵 + 𝐴𝐴. 𝐵𝐵 b) 𝑌𝑌 = 𝐴𝐴. 𝐵𝐵. 𝐶𝐶. 𝐷𝐷 + 𝐴𝐴. 𝐵𝐵. 𝐷𝐷 + 𝐴𝐴. 𝐶𝐶. 𝐷𝐷 + 𝐴𝐴. 𝐶𝐶. 𝐷𝐷 + 𝐴𝐴. 𝐵𝐵 So lu t io n : a) 𝐴𝐴. 𝐵𝐵 𝐴𝐴. 𝐵𝐵 𝐴𝐴. 𝐵𝐵 𝐴𝐴. 𝐵𝐵 𝐶𝐶 1 1 1 𝐶𝐶 1 1 1 𝑌𝑌 = 𝐴𝐴 + 𝐵𝐵 𝐴𝐴. 𝐵𝐵 𝐴𝐴. 𝐵𝐵 𝐴𝐴. 𝐵𝐵 𝐴𝐴. 𝐵𝐵 b) 1 𝐶𝐶. 𝐷𝐷 𝐶𝐶. 𝐷𝐷 1 1 1 1 𝐶𝐶. 𝐷𝐷 1 1 𝐶𝐶. 𝐷𝐷 𝑌𝑌 = 𝐴𝐴. 𝐵𝐵. 𝐶𝐶 + 𝐶𝐶. 𝐷𝐷 + 𝐴𝐴. 𝐷𝐷  DESIGNING COMBINATIONAL LOGIC CIRCUIT In chapter 3, we have discussed the three steps in designing a combinational logic circuits: 1) Construct the truth table. 2) Write the simplified expression. 3) Draw the logic circuit. In this chapter, we will use Karnaugh Map to simplify the expression.  DESIGNING COMBINATIONAL LOGIC CIRCUIT Example 1: The diagram shown above is a system to control three valves: Pwater, Px and Py that will control the flow of water, liquid X and liquid Y into a tank. DESIGNING COMBINATIONAL LOGIC CIRCUIT During the initial state, the tank is empty. When the system starts, valve Pwater is open to allow the flow of water. As the water level reaches level A, the next valve Px is open to allow the flow of liquid X into the tank. When the mixture of water + liquid X reaches level B, valve Py is open to allow the flow of liquid Y. When it reaches level C, all valves are closed and a green light is switch on signaling a successful operation. The tank uses three float switches A, B and C. If any of the switches fail to function, all valves are closed and a red light is switch on. DESIGNING COMBINATIONAL LOGIC CIRCUIT The input of the system are three float switches A,B and C. The output are there valves Pwater, Px, Py and a green light Lh and a red light Lm. The logic levels to represent the input and output: Input Output Float off = 0 Valve opened = 1 Float on = 1 Valve closed = 0 Light on = 1 Light off = 0 DESIGNING COMBINATIONAL LOGIC CIRCUIT The system will operate as follows: Input Output Remark ABC = 000 Pwater=1, Px=0, Py=0, Lh=0, Lm=0 Start. Water. ABC = 100 Pwater=1, Px=1, Py=0, Lh=0, Lm=0 Level A. Water + X. ABC = 110 Pwater=1, Px=1, Py=1, Lh=0, Lm=0 Level B. Water + X + Y. ABC = 111 Pwater=0, Px=0, Py=0, Lh=1, Lm=0 Level C. Close all valves. Green light on. Success Others Pwater=0, Px=0, Py=0, Lh=0, Lm=1 Closed all valves. Red light on. System failure DESIGNING COMBINATIONAL LOGIC CIRCUIT The truth table: DESIGNING COMBINATIONAL LOGIC CIRCUIT Simply the expression using K-Map or Boolean algebra. DESIGNING COMBINATIONAL LOGIC CIRCUIT Draw the logic circuit:  DON’T CARE TERM ON KARNAUGH MAP Some logic circuits may be designed so that there are certain input conditions for which there are no specific output levels, usually because these input conditions will never occur. There will be certain combinations of input, where we ‘don’t care’ whether the output is 1 or 0.  DON’T CARE TERM ON KARNAUGH MAP Example: Let us consider a logic circuit that will detect an odd number from a BCD code. BCD code are numbers from 0 to 9. Six numbers are not used by the code: (1010,1011,1100,1101,1110 and 1111) –don’t care terms. Logic levels: Input Output Odd numbers: 1,3,5,7,9 Odd numbers = 1 Even numbers: 0,2,4,6,8 Even numbers = 0 Not used numbers: Not used numbers = X 10,11,12,13,14,15  DON’T CARE TERM ON KARNAUGH MAP The truth table is as follows: There are 6 don’t care terms which is labeled as X. DON’T CARE TERM ON KARNAUGH MAP  An X on the K-Map means that the cell can be a 1 or a 0.  Cells containing 1’s can be group together with cell containing X’s to form a bigger loop.  In this example the X’s in cell (11,13,15) is looped together around the adjacent cell’s containing 1’s.  Including X’s in a loop helps to further simplify the expression.  The X’s in cell (10, 12 and 14) are ignored.  The simplified expression is 𝑌𝑌 = 𝐷𝐷. THANK YOU