TEST(DFS, general){
TestingVisitor visitor;
/* on small graphs */
for (int i = 1; i <= 5; ++i){
visitor.setGraphSize(i);
for (auto &graph : getAllGraphs(i)){
runDFS(graph.get(), &visitor);
EXPECT_TRUE(visitor.isGood());
visitor.reset();
}
}
/* on big graphs */
int n = 100;
visitor.setGraphSize(n);
for (int i = 0; i <= 100; ++i){
runDFS(getRandomGraph(n, 0.01*i).get(), &visitor);
EXPECT_TRUE(visitor.isGood());
visitor.reset();
}
}
int main(int argc, char **argv) {
int i, j, k;
FILE* fptr;
fptr = fopen(argv[1], "r");
fscanf(fptr,"%d",&n);
Node = calloc(n, sizeof(node)); //declare node array
decolour(); //set all colours to 2 (no colour)
matrix = createMatrix(n);
//scan file
for(i = 0; i < n * n; i++) {
fscanf(fptr, "%d", &matrix[i]);
}
fclose(fptr);
makeSymmetric(matrix, n);
//assign neighbours
for(j = 0; j < n; j++) {
Node[j].nbr = calloc(n, sizeof(int));
for(k = 0; k < n; k++) {
Node[j].nbr[k] = matrix[k + n*j];
}
}
printMatrix(matrix, n);
//running DFS now beginning from every vertex
for(i=0; i<n; i++){
counter = 1; //set visited counter to 1
decolour(); //decolour all
Node[i].colour = 0;
runDFS(i, -1);
}
if(cyclic == 1) printf("Cyclic\n");
else printf("Not Cyclic\n");
if(counter != n){
printf("Not Connected\n");
printf("Not Tree\n");
}
else{
printf("Connected\n");
if(cyclic == 0) printf("Tree\n");
else printf("Not Tree\n");
}
if(bipartite == 1) printf("Bipartite\n");
else printf("Not Bipartite\n");
free(matrix);
free(Node);
return 0;
}
void runDFS(int i, int c){ //i = current node to check, c = from where the pointer has come
int k;
for(k = 0; k < n; k++){ //run the loop for every neighbour
if(k != c){ // skip itself
if(Node[i].nbr[k] == 1){ //if there is an edge then go ahead
if(Node[k].colour == 2){ //if uncoloured, colour it with the opposite parity
if(Node[i].colour == 0) Node[k].colour = 1;
else Node[k].colour = 0;
runDFS(k, i); //since it was uncoloured, we now shift our pointer to the next one
counter++; //since we have discovered a new (amd unconnected vertex, counter increases
}
else{
cyclic = 1; //if the next vertex is already coloured, it is also visited, hence cyclic graph
if(Node[k].colour == Node[i].colour){
bipartite = 0; //if the neighbour is of the same colour, two colouring is not possible and so it is not bipartite
}
}
}
}
}
}