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//Example 3.19: Reduce expression using k-map
clc; //clears the window
clear; //clears all existing variables
//Mapping the expression//
disp(' a''b'' a''b ab ab'' ')
disp('c''d'' 1 1 1 1')
disp('cd'' 1 X ')
disp('cd 1 1 1')
disp('cd'' 1 1 1')
disp(' From the map, high outputs are for 0,3,5,6,7,9,10,11,12,13,14') //given logic equation
a=[0 0 0 0;0 0 1 0;0 1 0 1;0 1 1 0;0 1 1 1;1 0 0 1;1 0 1 0 ;1 0 1 1;1 1 0 0;1 1 0 1;1 1 1 0]
disp(a)
for i=1: 11
if a(i,1)==1 then
b(i,1)='a'
else
b(i,1)='a'''
end
if a(i,2)==1 then
b(i,2)='b'
else
b(i,2)='b'''
end
if a(i,3)==1 then
b(i,3)='c'
else
b(i,3)='c'''
end
if a(i,4)==1 then
b(i,4)='d'
else
b(i,4)=' d'''
end
end
disp(' Evaluating expression from truth table and map ')
x1=strcat([ b(1,1),b(1,2),b(1,3),b(1,4) ])
x2=strcat([ b(2,1),b(2,2),b(2,3),b(2,4) ])
x3=strcat([ b(3,1),b(3,2),b(3,3),b(3,4) ])
x4=strcat([ b(4,1),b(4,2),b(4,3),b(4,4) ])
x5=strcat([ b(5,1),b(5,2),b(5,3),b(5,4) ])
x6=strcat([ b(6,1),b(6,2),b(6,3),b(6,4) ])
x7=strcat([ b(7,1),b(7,2),b(7,3),b(7,4) ])
x8=strcat([ b(8,1),b(8,2),b(8,3),b(8,4) ])
x9=strcat([ b(9,1),b(9,2),b(9,3),b(9,4) ])
x10=strcat([ b(10,1),b(10,2),b(10,3),b(10,4) ])
x11=strcat([ b(11,1),b(11,2),b(11,3),b(11,4) ])
x=[x1"+",x2"+",x3"+",x4"+",x5"+",x6"+",x7"+",x8"+",x9"+",x10"+",x11]
disp(x)
//Expression is displayed//
disp('There are 32 different minimum solutions to the given problem. some of them are')
disp('F = a''b''c''d'' + a''cd + bc''d + ab''d + abc''+ acd''')
disp('F = a''b''c''d'' + a''cd + bc''d + abd'' + bcd''+ ab''c')
disp('F = a''b''c''d'' + b''cd + a''bd + bcd''+ ab''c+ ab''d')
disp('F = a''b''c''d'' + abc'' + a''bd + bcd''+ ab''c+ ab''d')
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