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C program for karnaugh map

Last response: in Applications
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November 5, 2012 11:40:00 AM

please explain me this karnaugh map program


#include<stdio.h>
#include<conio.h>
void main()
{

int a[2000][10],b[2000][10],p[2000][10],q[2000][10],r[2000][10],s[2000][10],t[2000][10],u[2000][10],w[2000][10],o[2000][10],x[2000][10],y[6000][5],z[2000][10],c=1,c1=4,c2=1,c3=1,c4=1,c5=1,e=0,f=1,g=1,h=0,i,j,k,l=1,m,v;
clrscr();
printf("instruction");
printf("\nEnter 0 for not selecting and 1 for selecting\n");
for(i=0;i<=15;i++)
{
printf("value for cell %d=",i);
scanf("%d",&v);
if(v==1)
{
a[c][5]=i;
m=i;
while(c1!=0)

{
a[c][c1]=m%2;
m=m/2;
c1--;
}
c++;
c1=4;
}
}
for(i=1;i<c;i++)
{
for(j=1;j<c;j++)
{
for(k=1;k<=4;k++)
{
if(a[k]+a[j][k]==1) //a[11]+1[11],a[12][12],...a[23][23]//
e++; //e=1//
}
if(e==1)
{
for(k=1;k<=4;k++)
{
if(a[k]+a[j][k]!=1)
b[f]=a[k]; //b[11]=a[11]
else
b[f]=3; //b[11]=3//
g++; //g=1//
}
b[f][5]=a[5];
b[f][6]=a[j][5];
g=1;
f++;
h++;
}
e=0;
}
if(h==0)
{
for(k=1;k<=4;k++)
x[c2][k]=a[k];
c2++;
}
h=0;
}
c=0;
for(i=1;i<f;i++)
{
for(j=1;j<i;j++)
{
for(k=1;k<=4;k++)
{
if(b[k]==b[j][k])
c++;
}
if(c!=4)
e++;
c=0;
}
if(e==i-1)
{
for(k=1;k<=6;k++)
u[h+1][k]=b[k];
h++;
}
e=0;
}
for(i=1;i<=h;i++)
{
if(u[7]!=4)
{
for(j=1;j<=h;j++)
{
if(u[j][7]!=4)
{
if((u[6]==u[j][5])||(u[6]==u[j][5])||(u[5]==u[j][5])||(u[5]==u[j][6]))
{
for(k=1;k<=h;k++)
{
if(u[j][6]==u[k][5]||u[j][6]==u[k][6]||u[j][5]==u[k][5]||u[j][5]==u[k][6])
{
for(e=1;e<=h;e++)
{
if(i!=j&&i!=k&&i!=e&&j!=k&&j!=e&&k!=e)
{
if(u[k][6]!=u[e][5]&&u[k][6]!=u[e][6]&&u[k][6]!=u[e][6]&&u[k][6]!=u[e][5])
u[j][7]=4;
}
}
}
}
}
}
}
}
}
printf("\n");
f=h+1;
e=0;h=0;g=1;l=1;c=1;
for(i=1;i<f;i++)
{
for(j=1;j<f;j++)
{
for(k=1;k<=4;k++)
{
if(u[k]+u[j][k]==1)
e++;
if((u[k]+u[j][k]==3)||(u[k]+u[j][k]==4))
e=2;
}
if(e==1)
{
for(k=1;k<=4;k++)
{
if(u[k]+u[j][k]!=1&&u[k]+u[j][k]!=6)
z[l]=u[k];
else
z[l]=3;
g++;
}
g=1;
l++;
h++;
}
e=0;
}
if(h==0)
{
for(k=1;k<=7;k++)
w[c3][k]=u[k];
c3++;
}
h=0;
}
c=0;
for(i=1;i<l;i++)
{
for(j=1;j<i;j++)
{
for(k=1;k<=4;k++)
{
if(z[k]==z[j][k])
c++;
}
if(c!=4)
e++;
c=0;
}
if(e==i-1)
{
for(k=1;k<=4;k++)
t[h+1][k]=z[k];
h++;
}
e=0;
}
l=h+1;
e=0;h=0;g=1;m=1;c=1;c1=4;
for(i=1;i<l;i++)
{
for(j=i+1;j<l;j++)
{
for(k=1;k<=4;k++)
{
if(t[k]+t[j][k]==1)
e++;
if(t[k]+t[j][k]==3||t[k]+t[j][k]==4)
e=2;
}
if(e==1)
{
for(k=1;k<=4;k++)
{
if(t[k]+t[j][k]!=1&&t[k]+t[j][k]!=6)
y[m]=t[k];
else
y[m]=3;
g++;
}
g=1;
m++;
h++;
}
e=0;
}
if(h==0)
{
for(k=1;k<=4;k++)
r[c4][k]=t[k];
c4++;
}
h=0;
}
c=0;h=0;
for(i=1;i<m;i++)
{
for(j=1;j<i;j++)
{
for(k=1;k<=4;k++)
{
if(y[k]==y[j][k])
c++;
}
if(c!=4)
e++;
c=0;
}
if(e==i-1)
{
for(k=1;k<=4;k++)
s[h+1][k]=y[k];
h++;
}
e=0;
}
f=h+1;
e=0;h=0;g=1;l=1;c=1;
for(i=1;i<f;i++)
{
for(j=1;j<f;j++)
{
for(k=1;k<=4;k++)
{
if(s[k]+s[j][k]==1)
e++;
if((s[k]+s[j][k]==3)||(s[k]+s[j][k]==4))
e=2;
}
if(e==1)
{
for(k=1;k<=4;k++)
{
if(s[k]+s[j][k]!=1&&s[k]+s[j][k]!=6)
q[l]=s[k];
else
q[l]=3;
g++;
}
g=1;
l++;
h++;
}
e=0;
}
if(h==0)
{
for(k=1;k<=4;k++)
p[c5][k]=s[k];
c5++;
}
h=0;
}
c=0;
for(i=1;i<l;i++)
{
for(j=1;j<i;j++)
{
for(k=1;k<=4;k++)
{
if(q[k]==q[j][k])
c++;
}
if(c!=4)
e++;
c=0;
}
if(e==i-1)
{
for(k=1;k<=4;k++)
o[h+1][k]=q[k];
h++;
}
e=0;
}
for(i=1;i<=h;i++)
{
printf("1");
}
printf("\n");
for(i=1;i<c2;i++)
{
if(x[1]==0)
printf("A'");
if(x[1]==1)
printf("A");
if(x[2]==0)
printf("B'");
if(x[2]==1)
printf("B");
if(x[3]==0)
printf("C'");
if(x[3]==1)
printf("C");
if(x[4]==0)
printf("D'");
if(x[4]==1)
printf("D");
printf("+");
}
c=0;
for(i=1;i<c3;i++)
{
if(w[7]!=4)
{
if(w[1]==0)
printf("A'");
if(w[1]==1)
printf("A");
if(w[2]==0)
printf("B'");
if(w[2]==1)
printf("B");
if(w[3]==0)
printf("C'");
if(w[3]==1)
printf("C");
if(w[4]==0)
printf("D'");
if(w[4]==1)
printf("D");
printf("+");
}
}
printf("\n");
c=0;
for(i=1;i<c4;i++)
{
for(j=1;j<=h;j++)
{
if(((o[j][1]-r[1])==2||(o[j][1]-r[1])==3||(o[j][1]-r[1])==0)&&((o[j][2]-r[2])==2||(o[j][2]-r[2])==3||(o[j][2]-r[2])==0)&&((o[j][3]-r[3])==2||(o[j][3]-r[3])==0||(o[j][3]-r[3])==3)&&((o[j][4]-r[4])==2||(o[j][4]-r[4])==0||(o[j][4]-r[4])==3))
c++;
}
for(j=1;j<c5;j++)
{
if(((p[j][1]-r[1])==2||(p[j][1]-r[1])==3||(p[j][1]-r[1])==0)&&((p[j][2]-r[2])==2||(p[j][2]-r[2])==3||(p[j][2]-r[2])==0)&&((p[j][3]-r[3])==2||(p[j][3]-r[3])==0||(p[j][3]-r[3])==3)&&((p[j][4]-r[4])==2||(p[j][4]-r[4])==0||(p[j][4]-r[4])==3))
c++;
}
if(c==0)
{
if(r[1]==0)
printf("A'");
if(r[1]==1)
printf("A");
if(r[2]==0)
printf("B'");
if(r[2]==1)
printf("B");
if(r[3]==0)
printf("C'");
if(r[3]==1)
printf("C");
if(r[4]==0)
printf("D'");
if(r[4]==1)
printf("D");
printf("+");
}
c=0;
}
printf("\n");
for(i=1;i<c5;i++)
{
if(p[1]==0)
printf("A'");
if(p[1]==1)
printf("A");
if(p[2]==0)
printf("B'");
if(p[2]==1)
printf("B");
if(p[3]==0)
printf("C'");
if(p[3]==1)
printf("C");
if(p[4]==0)
printf("D'");
if(p[4]==1)
printf("D");
printf("+");
}
getch ();
}

More about : program karnaugh map

a b L Programming
November 5, 2012 12:34:37 PM

Ah, another homework assignment. You need to do better than simply posting code.

Is this your own work, or did you simply cut/paste from a book? (my bet is that it's a cut/paste job)
Does it compile?
Does it run?
What is the output supposed to look like?
What specifically do you want to know about this code?
m
0
l
a b L Programming
November 7, 2012 9:05:01 AM

siddharth_21 said:
please explain me this karnaugh map program


You have got to be kidding. That code is more unreadable than most minified JavaScript. If you got that from a textbook (although it looks worse than the rubbish in most textbooks too) please let me know which one so I can go and burn a few copies of it in protest.
m
0
l
a b L Programming
November 7, 2012 9:57:18 AM

So, what's the problem?
m
0
l
!