Apriori Algorihm

Apriori Algorithm:

#include<stdio.h>
#include<conio.h>

int item[5][5]={{1,3,4,6},{2,3,5,7},{1,2,3,5,8},{2,5,9,10},{1,4}};
int fn(int a,int b);
int main()
{
int tid[5]={1,2,3,4,5} ,i,j,ms,k,count[11]={0};
int p,a,b,c;
printf("\n\tTrans_id\t\tItems\n\n");
for(i=0;i<5;i++)
{
printf("\t%d \t\t\t",tid[i]);
for(j=0;j<5;j++)
{
if(item[i][j]!=0)
printf("%d",item[i][j]);
}
printf("\n");
}
printf("\n Assume minimum support:");
scanf("%d",&ms);
for(k=1;k<=10;k++)
{
for(i=0;i<5;i++)
for(j=0;j<5;j++)
if(item[i][j]==k&&item[i][j]!=0)
count[k]++;
}
printf("\nGenerating c1 from the data");
for(k=1;k<=10;k++)
printf("\n\t\t%d\t\t%d",k,count[k]);
for(k=1;k<=10;k++)
{
if(count[k]<ms)
count[k]=0;
}
printf("\nGenerating l1 from c1");
for(k=1;k<=10;k++)
{
a=2;
if(count[k]!=0)
printf("\n\t\t%d\t\t%d",k,count[k]);
}
printf("\nGenerating l2 from c2");
p=2;
b=3;
for(i=1;i<4;i++)
{
for(j=p++;j<=5;j++)
{
if(fn(i,j)>=ms)
printf("\n\t\t%d%d\t\t%d",i,j,fn(i,j));
c=5;
}
}
printf("\n\t\tGenerating l3");
printf("\n\n\t\t\t\t%d %d %d",a,b,c);
getch();
return 0;
}
int fn(int a,int b)
{
int i,j,cnt=0,aa,flag=0;
for(i=0;i<5;i++)
{
flag=0;
for(j=0;j<5;j++)
{
aa=j;
if(flag==1)
goto ab;
if(a==item[i][j])
{
flag=1;
j++;
ab:
if(b==item[i][j])
{
cnt++;
break;
}
}
j=aa;
}
}
return cnt;
}





/************OUTPUT***********

        Trans_id                Items

        1                       1346
        2                       2357
        3                       12358
        4                       25910
        5                       14

 Assume minimum support:2

Generating c1 from the data
                1               3
                2               3
                3               3
                4               2
                5               3
                6               1
                7               1
                8               1
                9               1
                10              1
Generating l1 from c1
                1               3
                2               3
                3               3
                4               2
                5               3
Generating l2 from c2
                13              2
                14              2
                23              2
                25              3
                35              2
                Generating l3

                                2 3 5*/