Find a triad in a network with a C program - c

Question

I've the following problem: I have two file file1.dat and file2.dat with 9 numbers put in a column each. For exemple:
file1.dat = 1,1,1,2,2,4,4,4,7
file2.dat = 2,4,7,3,4,1,3,7,1
I've tried to write a program that should find the triad in this network, where for triads I mean a group of three numbers that start with one of this number, pass to another two that are linked toghether and return back to the original one. In this case the file1.dat describe the "nodes" from where to start, and file2.dat describe the nodes where you arrive (1->2, 1->4, 1->7, 2->3 ...). There are two triads in this little "network" and they are composed from 1,2,4 and 1,4,7. I wrote the following program:
#include <stdio.h>
#include <stdlib.h>
#define N 9
int main (void){
int A[N],B[N],i,j,l,m,k;
int x,y;
int valueA,valueB,count,middle_value,new_value;
FILE *fp,*fq;
if ((fp = fopen("file1.dat", "r")) == NULL ) {
printf("Error opening file 1\n");
exit(EXIT_FAILURE);
}
for (i = 0; i < N; i++) {
fscanf(fp,"%d", &x);
A[i] = x;
}
if ((fq = fopen("file2.dat", "r")) == NULL ) {
printf("Error opening file 2\n");
exit(EXIT_FAILURE);
}
for (i = 0; i < N; i++) {
fscanf(fq,"%d", &y);
B[i] = y;
}
the up code is used to fill the two array A and B with all the data in the two files.
for (i=0;i<N;i++){
valueA=0;
valueB=0;
valueA=A[i];
valueB=B[i];
count=0;
middle_value=0;
new_value=0;
//Start the research of the first node of the file2.dat in the file1.dat
for(k=i+1;k<N;k++){
if(A[k]==valueB){
count++;
Up here I put a counter "count" that increase to 1 let me know if I find in the first file some number that have the same value as the first attending node. Now if the counter = 1, I want the computer to memorize the valueB = A[k] to write it at the end of the file as the middle term of the triad. Then I give to "valueB" the new value that is the "node arrival" corrisponding at the start one of A[k]--> B[k];
if(count==1){
middle_value = valueB;
valueB = B[k];
Then I look forward in the file, icreasing the for from where I stop before (l = k+1 ) for the last value of the triad that have to be like A[l] --> first value of A == B[l]. I increase then the counter and when the counter reach 2 the program should print me the 3 values, otherwise it should leave counter == 1, and if it can't find the first valueB in the file1, leave the counter == 0 (that's why I put the else at the end..)
for(l=k+1;l<N;l++){
new_value=A[l];
if(new_value==valueB && valueA==B[l]){
count++;
if(count==2){
printf ("%d,%d,%d\n\n",valueA,middle_value,valueB);
}else{
count=1;
}
}
}
}else{
count=0;
}
}
}
}
fclose(fp);
fclose(fq);
return (0);
}
But it doesn't work as I want. But if, e.g., I create 2 new files like
file1.dat = 1,1,1,2,3
file2.dat = 2,4,7,3,1
where there is the triad 1,2,3, the program works (must put # define N 6 in the 3rd row).. Someone can help me?
The problem that I think to have is that the program, with the beginning files, when it associate to valueA = 1, valueB=2, then it looks again to file 1 the position of 2 (that now becomes "the middle_value") and it gives to the new valueB = 3. Now it will search the 3 in the file1 but it don't find it. So it should pass at the second '2' that is in the file1 and give to the new valueB = 4 and then everything should goes well. But it doesn't. I don't understand why

Answer 1

So, I see a number of problems with your method. Perhaps the biggest problem is that you have three loops:
for(i=0 ;i<N;i++){
for(k=i+1;k<N;k++){
for(l=k+1;l<N;l++){
And each loop begins where the other ends. But, in general, you may need to look at smaller nodes in order to find loops. For instance, if you have the following graph:
1 2 3
3 1 2
you can verify that your approach will not work. So we know right away that we need to expand the range of the loops.
Also, you have a lot of variables floating around. Sometimes (but not always), that's an indication that the program is needlessly complex. Often times, considerably complicated problems can be solved without too many intermediate variables.
Your x and y variables were obviously unnecessary because you were using them as simple intermediates, so I cut them.
You used two file pointer variables, but really only needed one. I cut out the second one.
I also moved the fclose statements closer to where you read in the files, so that they would be open for the shortest possible time, which is a polite way to read a file.
Your valueA, valueB, and count variables immediately struck me as being questionable. Especially this bit:
valueA=0;
valueB=0;
valueA=A[i];
valueB=B[i];
You give them values, and then immediately change the values.
But a deeper look at the code shows that you're using valueA and valueB to keep track of which A and B you're looking at. And you use count to keep track of how deep you are in the loops. But each of these is inherent in the structure of the code!
So I cut all that out and widened the loop range, as discussed above, resulting in this:
#include <stdio.h>
#include <stdlib.h>
#define N 9
int main () {
int A[N], B[N];
FILE *fp;
if ((fp = fopen ("file1.dat", "r")) == NULL) {
printf ("Error opening file 1\n");
exit (EXIT_FAILURE);
}
for (i = 0; i < N; i++)
fscanf (fp, "%d", &A[i]);
fclose (fp);
if ((fp = fopen ("file2.dat", "r")) == NULL) {
printf ("Error opening file 2\n");
exit (EXIT_FAILURE);
}
for (i = 0; i < N; i++)
fscanf (fp, "%d", &B[i]);
fclose (fp);
for(int i=0; i<N; i++)
for(int j=0; j<N; j++)
for(int k=0; k<N; k++)
if(B[i]==A[j] && B[j]==A[k] && B[k]==A[i])
printf("%d,%d,%d\n",A[i],A[j],A[k]);
return 0;
}
It will find each triad-loop three times, which is unfortunate, but it will find all the loops. There are slightly more complicated algorithms which would find loops once or find them more efficiently, but this algorithm most closely matches your original code, so it may be the most useful for you now.