Explanation Suppose that the output of given Boolean expression is Y = [ ( y ′ ∧ x ′ ∧ z ) ∨ ( y ∧ x ′ ∧ z ) ∨ ( y ′ ∧ x ∧ z ′ ) ] . In order to construct the k-map place 1 ' s in those cells corresponding to the given min terms of Boolean expression and place 0 ' s in the other cells. Examine the map for the adjacent 1 ' s and loop those 1 ' s which are not adjacent to any other 1 ' s . These are called isolated 1 ' s . Next, look for those 1 ' s which are adjacent to only one other. Loop any pair containing such a 1. Loop any octet even it contains some 1 ' s that have already been looped. Loop any quad that contains one or more 1 ' s which have not already been looped, making sure to use the minimum number of loops. Loop any pairs necessary to include any 1 ' s that have not yet been looped, making sure to use the minimum number of loops. Form the OR sum of all the terms generated by each loop. Each cell in the grid shown corresponds to a row of the truth table, with the rows of the grid corresponding to ( x ) , ( x ′ ) the columns to y ′ ∧ z ′ , y ′ ∧ z , y ∧ z , y ∧ z ′

5th Edition

ISBN: 9780134689562

Author: Dossey, John A.

Publisher: Pearson,

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Chapter 10.3, Problem 27E

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To simplify: The Boolean expression using by Karnaugh map method.

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Chapter 10.3, Problem 26E

Chapter 10.3, Problem 28E

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Discrete Mathematics

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