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Simplification of Sample Boolean Expressions
Dr. Chuck Cartledge
Wednesday 13th^ June, 2018
Contents
1 Introduction 1
2 Fall 2017 example 1 2.1 Boolean identities................................... 1 2.2 Truth table....................................... 2 2.3 Karnaugh map..................................... 2 2.4 2-Input NAND gates................................. 2
3 Summer 2018 example 3 3.1 Boolean identities................................... 3 3.2 Karnaugh map..................................... 3
4 Conclusion 4
List of Tables
1 Boolean identities to simplify F(x, y, z)= (xy + z)(yz) + z.............. 5 2 Truth table for F(x, y, z)= (xy + z)(yz) + z....................... 6 3 Truth table for F(x, y, z)= (xy + z)(yz) + z and F(x, y, z)= z + xy.......... 6 4 Karnaugh map of F(x, y, z)= xyz + xyz + xyz + z.................... 6 5 Boolean identities to simplify F(w, x, y, z)= wxyz + xyz + xyz + wxyz........ 7 6 Karnaugh map of F(w, x, y, z)= wxyz + xyz + xyz + wxyz............... 7
List of Figures
1 F(x, y, z)= z + xy as 2-input NAND gates....................... 3
i
1 Introduction
Tidewater Community College (TCC) Computer Science (CSC) 205 examines the hierarchical structure of computer architecture. The course:
- focuses on multi-level machine organization.
- uses a simple assembler language to complete programming projects.
- and, includes processors, instruction, execution, addressing techniques, data representation and digital logic.
Lecture 6 of the Fall 2017 semester presented the problem of implementing the following Boolean expression in hardware:
F(x, y, z)= (xy + z)(yz) + z
using NAND gates. Class discussions during lecture 7 of the Summer 2018 presentted a problem of simplifying the following Boolean expression using Boolean identities and a Karnaugh map.
F(w, x, y, z)= wxyz + xyz + xyz + wxyz
Included in this document is a full solution to the problem using Boolean identities, truth tables, and Karnaugh maps.
2 Fall 2017 example
Different approaches will be used to simplify the Boolean expression:
F(x, y, z)= (xy + z)(yz) + z
2.1 Boolean identities
Boolean identities are used to simplify:
F(x, y, z)= (xy + z)(yz) + z
to:
F(x, y, z)= z + xy
(see Table 1).
Figure 1: F(x, y, z)= z + xy as 2-input NAND gates.
3 Summer 2018 example
Different approaches will be used to simplify the Boolean expression:
F(w, x, y, z)= wxyz + xyz + xyz + wxyz
3.1 Boolean identities
Boolean identities are used to simplify:
F(w, x, y, z)= wxyz + xyz + xyz + wxyz
to:
F(w, x, y, z)= yz + wxz + wxyz
(see Table 5).
3.2 Karnaugh map
The original Boolean expression is used to populate a Karnaugh map (Kmap) to “read” the simpli- fied Boolean expression directly (see Table 6). The original expression is “reworked” to make it easier to find the original terms in the Kmap.
F(w, x, y, z)= wxyz + xyz + xyz + wxyz F(w, x, y, z)= A + B +C + D
4 Conclusion
The original 2017 Boolean expression:
F(x, y, z)= (xy + z)(yz) + z
Was reduced to:
F(x, y, z)= z + xy
using Boolean identities, validated using truth tables, and again simplified using sum of products (SOP) and a Karnaugh map. Ultimately, the simplified expression was “made real” using 2-input NAND gates. The 2018 Boolean expression:
F(w, x, y, z)= wxyz + xyz + xyz + wxyz
Was reduced to:
F(w, x, y, z)= yz + wxz + wxyz
using Boolean identities and a Karnaugh map (Kmap).
Table 2: Truth table for F(x, y, z)= (xy + z)(yz) + z. Inputs Int. ops Ans. x y z xy +z (yz) F(x, y, z) 0 0 0 0 0 1 0 0 0 1 0 1 1 1 0 1 0 0 0 1 0 0 1 1 0 1 0 1 1 0 0 0 0 1 0 1 0 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1
Table 3: Truth table for F(x, y, z)= (xy + z)(yz) + z and F(x, y, z)= z + xy. Inputs Int. ops Ans. Simp. x y z xy +z (yz) F(x, y, z) F(x, y, z) 0 0 0 0 0 1 0 0 0 0 1 0 1 1 1 1 0 1 0 0 0 1 0 0 0 1 1 0 1 0 1 1 1 0 0 0 0 1 0 0 1 0 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 0 1 1
Table 4: Karnaugh map of F(x, y, z)= xyz + xyz + xyz + z. Simplifying to F(x, y, z)= z + xy. yz
x
AD
BD C
D
D
Table 5: Boolean identities to simplify F(w, x, y, z)= wxyz + xyz + xyz + wxyz
Expression Explanation
wxyz + xyz + xyz + wxyz Original
wxyz + xyz(w + w) + xyz(w + w) + wxyz Inverse
wxyz + wxyz + wxyz + wxyz + wxyz + wxyz Distributive
wxyz + wxyz + wxyz + wxyz + wxyz + wxyz + wxyz Idompotent
wxyz + yz(wx + wx + wx + wx) + wxyz + wxyz Distributive
wxyz + yz(x(w + w) + x(w + w)) + wxz(y + y) Distributive
wxyz + yz(x( 1 ) + x( 1 )) + wxz( 1 ) Inverse
wxyz + yz(x + x) + wxz Inverse
wxyz + yz( 1 ) + wxz Identity
wxyz + yz + wxz Identity
yz + wxz + wxyz Final
Table 6: Karnaugh map of F(w, x, y, z)= wxyz + xyz + xyz + wxyz. Original expression is simplified to:F(w, x, y, z)= yz + wxz + wxyz yz
wx
A B
B
C
C
D
