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Q6E

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Discrete Mathematics and its Applications
Found in: Page 818
Discrete Mathematics and its Applications

Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

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Short Answer

Use a table to express the values of each of these Boolean functions.

\(\begin{array}{l}{\bf{a) F(x,y,z) = \bar z}}\\{\bf{b) F(x,y,z) = \bar xy + \bar yz}}\\{\bf{c) F(x,y,z) = x\bar yz + }}\overline {{\bf{(xyz)}}} \\{\bf{d) F(x,y,z) = \bar y(xz + \bar x\bar z)}}\end{array}\)

a) The table of the given Boolean function \(F(x,y,z) = \bar z\) is

\(\begin{array}{*{20}{l}}x&y&z&{\bar z}\\0&0&0&1\\0&0&1&0\\0&1&0&1\\0&1&1&0\\1&0&0&1\\1&0&1&0\\1&1&0&1\\1&1&1&0\end{array}\)

b) The table of the given Boolean function \({\bf{F(x,y,z) = \bar xy + \bar yz}}\) is

\(\begin{array}{*{20}{c}}x&y&z&{\overline x }&{\overline y }&{\overline x \cdot y}&{\overline y \cdot z}&{\overline {x \cdot } y + \overline y \cdot z}\\0&0&0&1&1&0&0&0\\0&0&1&1&1&0&1&1\\0&1&0&1&0&1&0&1\\0&1&1&1&0&1&0&1\\1&0&0&0&1&0&0&0\\1&0&1&0&1&0&1&1\\1&1&0&0&0&0&0&0\\1&1&1&0&0&0&0&0\end{array}\)

c) The table of the given Boolean function \(F(x,y,z) = x\bar yz + \overline {(xyz)} \) is

\(\begin{array}{*{20}{c}}{\bf{x}}&{\bf{y}}&{\bf{z}}&{{\bf{\bar y}}}&{{\bf{x}} \cdot {\bf{\bar y}}}&{{\bf{x}} \cdot {\bf{\bar y}} \cdot {\bf{z}}}&{{\bf{x}} \cdot {\bf{y}}}&{{\bf{x}} \cdot {\bf{y}} \cdot {\bf{z}}}&{\overline {{\bf{(x}} \cdot {\bf{y}} \cdot {\bf{z)}}} }&{{\bf{x}} \cdot {\bf{\bar y}} \cdot {\bf{z + }}\overline {{\bf{(x}} \cdot {\bf{y}} \cdot {\bf{z)}}} }\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\\{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\\{\bf{0}}&{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\\{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}\end{array}\)

d) The table of the given Boolean function \({\bf{F(x,y,z) = \bar y(xz + \bar x\bar z)}}\) is

\(\begin{array}{*{20}{c}}x&{ y}&{ z}&{ \bar x}&{ \bar y}&{ \bar z}&{ x \cdot z}&{ \bar x \cdot \bar z}&{ x \cdot z + \bar x \cdot \bar z}&{ \bar y \cdot (x \cdot z + \bar x \cdot \bar z)}\\0&0&0&1&1&1&0&1&1&1\\0&0&1&1&1&0&0&0&0&0\\0&1&0&1&0&1&0&1&1&0\\0&1&1&1&0&0&0&0&0&0\\1&0&0&0&1&1&0&0&0&0\\1&0&1&0&1&0&1&0&1&1\\1&1&0&0&0&1&0&0&0&0\\1&1&1&0&0&0&1&0&1&0\end{array}\)

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Definition

The complement of an element: \({\bf{\bar 0 = 1}}\) and \({\bf{\bar 1 = }}0\)

The Boolean sum + or \(OR\) is 1 if either term is 1.

The Boolean product \( \cdot \) or \(AND\) is 1 if both terms are 1.

Step 2: (a) Using the definition of complement

\(F(x,y,z) = \bar z\)

The function has three variables x, y and z. Each of these variables can take on the value of 0 or 1.

Therefore, the table of the given Boolean function \(F(x,y,z) = \bar z\) is

\(\begin{array}{*{20}{l}}x&y&z&{\bar z}\\0&0&0&1\\0&0&1&0\\0&1&0&1\\0&1&1&0\\1&0&0&1\\1&0&1&0\\1&1&0&1\\1&1&1&0\end{array}\)

Step 3: (b) Using Boolean product and sum

\({\bf{F(x,y,z) = \bar xy + \bar yz}}\)

The function has three variables x, y and z. Each of these variables can take on the value of 0 or 1.

Note: \(\bar xy\) represents \(\bar x \cdot y\) and \(\bar yz\) represents \(\bar y \cdot z\)

Therefore, the table of the given Boolean function \({\bf{F(x,y,z) = \bar xy + \bar yz}}\) is

\(\begin{array}{*{20}{c}}x&y&z&{\overline x }&{\overline y }&{\overline x \cdot y}&{\overline y \cdot z}&{\overline {x \cdot } y + \overline y \cdot z}\\0&0&0&1&1&0&0&0\\0&0&1&1&1&0&1&1\\0&1&0&1&0&1&0&1\\0&1&1&1&0&1&0&1\\1&0&0&0&1&0&0&0\\1&0&1&0&1&0&1&1\\1&1&0&0&0&0&0&0\\1&1&1&0&0&0&0&0\end{array}\)

Step 4: (c) Using Boolean product and sum

\(F(x,y,z) = x\bar yz + \overline {(xyz)} \)

The function has three variables x, y and z. Each of these variables can take on the value of 0 or 1.

Note: \(xyz\) represents \(x \cdot y \cdot z\) and \(x\bar y\) represents \(x \cdot \bar y\)

Therefore, the table of the given Boolean function \(F(x,y,z) = x\bar yz + \overline {(xyz)} \) is

\(\begin{array}{*{20}{c}}{\bf{x}}&{\bf{y}}&{\bf{z}}&{{\bf{\bar y}}}&{{\bf{x}} \cdot {\bf{\bar y}}}&{{\bf{x}} \cdot {\bf{\bar y}} \cdot {\bf{z}}}&{{\bf{x}} \cdot {\bf{y}}}&{{\bf{x}} \cdot {\bf{y}} \cdot {\bf{z}}}&{\overline {{\bf{(x}} \cdot {\bf{y}} \cdot {\bf{z)}}} }&{{\bf{x}} \cdot {\bf{\bar y}} \cdot {\bf{z + }}\overline {{\bf{(x}} \cdot {\bf{y}} \cdot {\bf{z)}}} }\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\\{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\\{\bf{0}}&{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\\{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}\end{array}\)

Step 5: (d) Using Boolean product and sum

\({\bf{F(x,y,z) = \bar y(xz + \bar x\bar z)}}\)

The function has three variables x, y and z. Each of these variables can take on the value of 0 or 1.

Note: \(yz\) represents \(y \cdot z\) and \(\bar y\bar z\) represents \(\bar y \cdot \bar z\)

Therefore, the table of the given Boolean function \({\bf{F(x,y,z) = \bar y(xz + \bar x\bar z)}}\) is

\(\begin{array}{*{20}{c}}x&{ y}&{ z}&{ \bar x}&{ \bar y}&{ \bar z}&{ x \cdot z}&{ \bar x \cdot \bar z}&{ x \cdot z + \bar x \cdot \bar z}&{ \bar y \cdot (x \cdot z + \bar x \cdot \bar z)}\\0&0&0&1&1&1&0&1&1&1\\0&0&1&1&1&0&0&0&0&0\\0&1&0&1&0&1&0&1&1&0\\0&1&1&1&0&0&0&0&0&0\\1&0&0&0&1&1&0&0&0&0\\1&0&1&0&1&0&1&0&1&1\\1&1&0&0&0&1&0&0&0&0\\1&1&1&0&0&0&1&0&1&0\end{array}\)

Most popular questions for Math Textbooks

Show that if \(F\) and \(G\) are Boolean functions of degree \(n\), then

\(\begin{array}{l}a)F \le F{\bf{ + }}G\\b)FG \le F\end{array}\)

Draw the Hasse diagram for the poset consisting of the set of the \({\bf{16}}\)Boolean functions of degree two (shown in Table \({\bf{3}}\) of Section \({\bf{12}}{\bf{.1}}\)) with the partial ordering \( \le \).

\({\bf{a)}}\)Draw a \(K{\bf{ - }}\)map for a function in three variables. Put a \(1\) in the cell that represents \(\bar xy\bar z\).

\({\bf{b)}}\) Which minterms are represented by cells adjacent to this cell\(?\)

\(a)\) Draw a \(K{\bf{ - }}\) map for a function in four variables. Put a \(1\) in the cell that represents \(\bar wxy\bar z\).

\(b)\) Which min terms are represented by cells adjacent to this cell\(?\)

Which rows and which columns of a \(4{\bf{ \ast }}16\) map for Boolean functions in six variables using the Gray codes \({\bf{1111}},{\bf{1110}},{\bf{1010}},{\bf{1011}},{\bf{1001}},{\bf{1000}},{\bf{0000}},{\bf{0001}},{\bf{0011}},{\bf{0010}},{\bf{0110}},{\bf{0111}},{\bf{0101}},{\bf{0100}},{\bf{1100}},{\bf{1101}}\) to label the columns and \({\bf{11}},{\bf{10}},{\bf{00}},{\bf{01}}\) to label the rows need to be considered adjacent so that cells that represent min-terms that differ in exactly one literal are considered adjacent\(?\)
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