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//Example 3.17: 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 ')
disp('cd'' 1 1 ')
disp('cd 1 1')
disp('cd'' 1 1')
disp(' From the map, high outputs are for 0,1,5,7,8,10,14,15') //given logic equation
a=[0 0 0 0;0 0 0 1;0 1 0 1;0 1 1 1;1 0 0 0;1 0 1 0 ;1 1 1 0;1 1 1 1]
disp(a)
for i=1: 8
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) ])
x=[x1"+",x2"+",x3"+",x4"+",x5"+",x6"+",x7"+",x8]
disp(x)
//Expression is displayed//
disp('The resulting two possible solution are')
disp('F = a''c''d'' + bc''d + acd + b''cd''')
disp('F = a''b''d'' + a''bc'' + abd + ab''c')
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