How to synchronize CUDA threads when they are in the same loop and we need to synchronize them to execute only a limited part - cuda

Question

I have written some code, and now I want to implement this on CUDA GPU but I'm new to synchronization. Below I'm presenting the code and I want to that LOOP1 to be executed by all threads (hence I want to this portion to take advantage of CUDA and the remaining portion (the portion other from the LOOP1) is to be executed by only a single thread.
do{
point_set = master_Q[(*num_mas) - 1].q;
List* temp = point_set;
List* pa = point_set;
if(master_Q[num_mas[0] - 1].max)
max_level = (int) (ceilf(il2 * log(master_Q[num_mas[0] - 1].max)));
*num_mas = (*num_mas) - 1;
while(point_set){
List* insert_ele = temp;
while(temp){
insert_ele = temp;
if((insert_ele->dist[insert_ele->dist_index-1] <= pow(2, max_level-1)) || (top_level == max_level)){
if(point_set == temp){
point_set = temp->next;
pa = temp->next;
}
else{
pa->next = temp->next;
}
temp = NULL;
List* new_point_set = point_set;
float maximum_dist = 0;
if(parent->p_index != insert_ele->point_index){
List* tmp = new_point_set;
float *b = &(data[(insert_ele->point_index)*point_len]);
**LOOP 1:** while(tmp){
float *c = &(data[(tmp->point_index)*point_len]);
float sum = 0.;
for(int j = 0; j < point_len; j+=2){
float d1 = b[j] - c[j];
float d2 = b[j+1] - c[j+1];
d1 *= d1;
d2 *= d2;
sum = sum + d1 + d2;
}
tmp->dist[tmp->dist_index] = sqrt(sum);
if(maximum_dist < tmp->dist[tmp->dist_index])
maximum_dist = tmp->dist[tmp->dist_index];
tmp->dist_index = tmp->dist_index+1;
tmp = tmp->next;
}
max_distance = maximum_dist;
}
while(new_point_set || insert_ele){
List* far, *par, *tmp, *tmp_new;
far = NULL;
tmp = new_point_set;
tmp_new = NULL;
float level_dist = pow(2, max_level-1);
float maxdist = 0, maxp = 0;
while(tmp){
if(tmp->dist[(tmp->dist_index)-1] > level_dist){
if(maxdist < tmp->dist[tmp->dist_index-1])
maxdist = tmp->dist[tmp->dist_index-1];
if(tmp == new_point_set){
new_point_set = tmp->next;
par = tmp->next;
}
else{
par->next = tmp->next;
}
if(far == NULL){
far = tmp;
tmp_new = far;
}
else{
tmp_new->next = tmp;
tmp_new = tmp;
}
if(parent->p_index != insert_ele->point_index)
tmp->dist_index = tmp->dist_index - 1;
tmp = tmp->next;
tmp_new->next = NULL;
}
else{
par = tmp;
if(maxp < tmp->dist[(tmp->dist_index)-1])
maxp = tmp->dist[(tmp->dist_index)-1];
tmp = tmp->next;
}
}
if(0 == maxp){
tmp = new_point_set;
aloc_mem[*tree_index].p_index = insert_ele->point_index;
aloc_mem[*tree_index].no_child = 0;
aloc_mem[*tree_index].level = max_level--;
parent->children_index[parent->no_child++] = *tree_index;
parent = &(aloc_mem[*tree_index]);
tree_index[0] = tree_index[0]+1;
while(tmp){
aloc_mem[*tree_index].p_index = tmp->point_index;
aloc_mem[(*tree_index)].no_child = 0;
aloc_mem[(*tree_index)].level = master_Q[(*cur_count_Q)-1].level;
parent->children_index[parent->no_child] = *tree_index;
parent->no_child = parent->no_child + 1;
(*tree_index)++;
tmp = tmp->next;
}
cur_count_Q[0] = cur_count_Q[0]-1;
new_point_set = NULL;
}
master_Q[*num_mas].q = far;
master_Q[*num_mas].parent = parent;
master_Q[*num_mas].valid = true;
master_Q[*num_mas].max = maxdist;
master_Q[*num_mas].level = max_level;
num_mas[0] = num_mas[0]+1;
if(0 != maxp){
aloc_mem[*tree_index].p_index = insert_ele->point_index;
aloc_mem[*tree_index].no_child = 0;
aloc_mem[*tree_index].level = max_level;
parent->children_index[parent->no_child++] = *tree_index;
parent = &(aloc_mem[*tree_index]);
tree_index[0] = tree_index[0]+1;
if(maxp){
int new_level = ((int) (ceilf(il2 * log(maxp)))) +1;
if (new_level < (max_level-1))
max_level = new_level;
else
max_level--;
}
else
max_level--;
}
if( 0 == maxp )
insert_ele = NULL;
}
}
else{
if(NULL == temp->next){
master_Q[*num_mas].q = point_set;
master_Q[*num_mas].parent = parent;
master_Q[*num_mas].valid = true;
master_Q[*num_mas].level = max_level;
num_mas[0] = num_mas[0]+1;
}
pa = temp;
temp = temp->next;
}
}
if((*num_mas) > 1){
List *temp2 = master_Q[(*num_mas)-1].q;
while(temp2){
List* temp3 = master_Q[(*num_mas)-2].q;
master_Q[(*num_mas)-2].q = temp2;
if((master_Q[(*num_mas)-1].parent)->p_index != (master_Q[(*num_mas)-2].parent)->p_index){
temp2->dist_index = temp2->dist_index - 1;
}
temp2 = temp2->next;
master_Q[(*num_mas)-2].q->next = temp3;
}
num_mas[0] = num_mas[0]-1;
}
point_set = master_Q[(*num_mas)-1].q;
temp = point_set;
pa = point_set;
parent = master_Q[(*num_mas)-1].parent;
max_level = master_Q[(*num_mas)-1].level;
if(master_Q[(*num_mas)-1].max)
if( max_level > ((int) (ceilf(il2 * log(master_Q[(*num_mas)-1].max)))) +1)
max_level = ((int) (ceilf(il2 * log(master_Q[(*num_mas)-1].max)))) +1;
num_mas[0] = num_mas[0]-1;
}
}while(*num_mas > 0);

Answer 1

If you want a single kernel to execute parts of its code on only a single thread within a block, use an if statement to execute a portion of code for only a single threadIdx and then barrier synchronize. Perhaps you should take a stab at writing a kernel and posting that for people to look at.