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+
+//Example 6-19//
+//reduce expression using k-map by finding POS//
+clc
+//clears the window//
+clear
+//clears all existing variables//
+//Mapping the expression//
+disp('       C''D'' C''D CD CD'' ')
+disp('A''B''   0     0    0  1 ')
+disp('AB''     1     0    0  0 ')
+disp('AB       1     1    0  0 ')
+disp('AB''     1     1    1  0 ')
+disp(' From the map, low outputs for 0,1,3,5,6,7,10,14,15')
+//given logic equation//
+c=[0 0 0 0;0 0 0 1;0 0 1 1;0 1 0 1;0 1 1 0;0 1 1 1;1 0 1 0;1 1 1 0;1 1 1 1]
+disp(c)
+for i=1: 9
+    if c(i,1)==1 then
+       d(i,1)='A'''
+   else
+       d(i,1)='A'
+    end
+    if c(i,2)==1 then
+       d(i,2)='B'''
+   else
+       d(i,2)='B'
+    end
+     if c(i,3)==1 then
+       d(i,3)='C'''
+   else
+       d(i,3)='C'
+    end
+    if c(i,4)==1 then
+       d(i,4)='D'''
+   else
+       d(i,4)=' D '
+    end
+end
+disp(' evaluating expression from truth table and map')
+l=strcat([ d(1,1),d(1,2),d(1,3),d(1,4)])
+m=strcat([ d(2,1),d(2,2),d(2,3),d(2,4)])
+n=strcat([ d(3,1),d(3,2),d(3,3),d(3,4)])
+o=strcat([ d(4,1),d(4,2),d(4,3),d(4,4)])
+x=strcat([l"+",m"+",n"+",o ])
+disp('the sum of product expression is:')
+disp(x)
+disp('Reading the SOP form')
+disp(' A''B''CD''+BC''D''+AB''D+AC'' ')
+//Expression is displayed//
+disp('now reducing expression using boolean algebra from POS')
+disp(' (A+B+C)(A''+C''+D)(A+D'')(B''+C'') ')
+disp('POS has 14 inputs,SOP has 16 inputs')
+disp('Therefore,POS form is less expensive')