UPDATE: Turns out there was a bug in the parser that generated the trees. More in Final edit.
Let T be a binary tree such that every internal node has exactly two children. For this tree, we want to code a function that for every node v in T finds the number of nodes in the subtree defined by v.
Example
Input
Desired output
With red I indicate the numbers that we want to compute. The nodes of the tree will be stored in an array, let us call it TreeArray by following the preorder layout.
For the example above, TreeArray will contain the following objects:
10, 11, 0, 12, 13, 2, 7, 3, 14, 1, 15, 16, 4, 8, 17, 18, 5, 9, 6
A node of the tree is described by the following struct:
struct tree_node{
long long int id; //id of the node, randomly generated
int numChildren; //number of children, it is 2 but for the leafs it's 0
int size; //size of the subtree rooted at the current node,
// what we want to compute
int pos; //position in TreeArray where the node is stored
int lpos; //position of the left child
int rpos; //position of the right child
tree_node(){
id = -1;
size = 1;
pos = lpos = rpos = -1;
numChildren = 0;
}
};
The function to compute all the size values is the following:
void testCache(int cur){
if(treeArray[cur].numChildren == 0){
treeArray[cur].size = 1;
return;
}
testCache(treeArray[cur].lpos);
testCache(treeArray[cur].rpos);
treeArray[cur].size = treeArray[treeArray[cur].lpos].size +
treeArray[treeArray[cur].rpos].size + 1;
}
I would like to understand why this function is faster when T looks like this (almost like a left going chain):
and slower when T looks like this (almost like a right going chain):
The following experiments were run on Intel(R) Core(TM) i5-3470 CPU @ 3.20GHz with 8 GB of RAM, L1 cache 256 KB, L2 cache 1 MB, L3 cache 6 MB.
Each dot in the graphs is the result of the following for loop (the parameters are defined by the axis):
for (int i = 0; i < 100; i++) {
testCache(0);
}
n corresponds to the total number of nodes, and time is measured in seconds. As we can see it is clear that as n grows the function is much faster when the tree looks like a left going chain, even though the number of nodes is exactly the same in both cases.
Now let us try to find where the bottleneck is. I used the PAPI library to count interesting hardware counters.
The first counter is the instructions, how many instructions do we actually spend? Is there a difference when the trees look different?
The difference is not significant. It looks like for large inputs the left going chain requires fewer instructions, but the difference is so small, so I think it is safe to assume that they both require the same number of instructions.
Seeing that we have stored the tree in a nice pre order layout inside treeArray it makes sense to see what is happening in cache. Unfortunately for L1 cache my computer does not provide any counters, but I have for L2 and L3.
Let's look at the accesses to L2 cache. The accesses to L2 cache happen when we get a miss in L1 cache, so that is an indirect counter for L1 misses as well.
As we can see the right going tree requires fewer L1 misses, so it seems that it uses the cache efficiently.
Same for L2 misses, the right going tree seems to be more efficient. Still nothing to indicate why the right going trees are so slower. Let's look at L3.
In L3 things explode for the right going trees. So the problem seems to be in L3 cache. Unfortunately I could not explain the reason behind this behavior. Why do things get messed up in L3 cache for the right going trees?
Here is the entire code together with the experiment:
#include <iostream>
#include <fstream>
#define BILLION 1000000000LL
using namespace std;
/*
*
* Timing functions
*
*/
timespec startT, endT;
void startTimer(){
clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &startT);
}
double endTimer(){
clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &endT);
return endT.tv_sec * BILLION + endT.tv_nsec - (startT.tv_sec * BILLION + startT.tv_nsec);
}
/*
*
* tree node
*
*/
//this struct is used for creating the first tree after reading it from the external file, for this we need left and child pointers
struct tree_node_temp{
long long int id; //id of the node, randomly generated
int numChildren; //number of children, it is 2 but for the leafs it's 0
int size; //size of the subtree rooted at the current node
tree_node_temp *leftChild;
tree_node_temp *rightChild;
tree_node_temp(){
id = -1;
size = 1;
leftChild = nullptr;
rightChild = nullptr;
numChildren = 0;
}
};
struct tree_node{
long long int id; //id of the node, randomly generated
int numChildren; //number of children, it is 2 but for the leafs it's 0
int size; //size of the subtree rooted at the current node
int pos; //position in TreeArray where the node is stored
int lpos; //position of the left child
int rpos; //position of the right child
tree_node(){
id = -1;
pos = lpos = rpos = -1;
numChildren = 0;
}
};
/*
*
* Tree parser. The input is a file containing the tree in the newick format.
*
*/
string treeNewickStr; //string storing the newick format of a tree that we read from a file
int treeCurSTRindex; //index to the current position we are in while reading the newick string
int treeNumLeafs; //number of leafs in current tree
tree_node ** treeArrayReferences; //stack of references to free node objects
tree_node *treeArray; //array of node objects
int treeStackReferencesTop; //the top index to the references stack
int curpos; //used to find pos,lpos and rpos when creating the pre order layout tree
//helper function for readNewick
tree_node_temp* readNewickHelper() {
int i;
if(treeCurSTRindex == treeNewickStr.size())
return nullptr;
tree_node_temp * leftChild;
tree_node_temp * rightChild;
if(treeNewickStr[treeCurSTRindex] == '('){
//create a left child
treeCurSTRindex++;
leftChild = readNewickHelper();
}
if(treeNewickStr[treeCurSTRindex] == ','){
//create a right child
treeCurSTRindex++;
rightChild = readNewickHelper();
}
if(treeNewickStr[treeCurSTRindex] == ')' || treeNewickStr[treeCurSTRindex] == ';'){
treeCurSTRindex++;
tree_node_temp * cur = new tree_node_temp();
cur->numChildren = 2;
cur->leftChild = leftChild;
cur->rightChild = rightChild;
cur->size = 1 + leftChild->size + rightChild->size;
return cur;
}
//we are about to read a label, keep reading until we read a "," ")" or "(" (we assume that the newick string has the right format)
i = 0;
char treeLabel[20]; //buffer used for the label
while(treeNewickStr[treeCurSTRindex]!=',' && treeNewickStr[treeCurSTRindex]!='(' && treeNewickStr[treeCurSTRindex]!=')'){
treeLabel[i] = treeNewickStr[treeCurSTRindex];
treeCurSTRindex++;
i++;
}
treeLabel[i] = '\0';
tree_node_temp * cur = new tree_node_temp();
cur->numChildren = 0;
cur->id = atoi(treeLabel)-1;
treeNumLeafs++;
return cur;
}
//create the pre order tree, curRoot in the first call points to the root of the first tree that was given to us by the parser
void treeInit(tree_node_temp * curRoot){
tree_node * curFinalRoot = treeArrayReferences[curpos];
curFinalRoot->pos = curpos;
if(curRoot->numChildren == 0) {
curFinalRoot->id = curRoot->id;
return;
}
//add left child
tree_node * cnode = treeArrayReferences[treeStackReferencesTop];
curFinalRoot->lpos = curpos + 1;
curpos = curpos + 1;
treeStackReferencesTop++;
cnode->id = curRoot->leftChild->id;
treeInit(curRoot->leftChild);
//add right child
curFinalRoot->rpos = curpos + 1;
curpos = curpos + 1;
cnode = treeArrayReferences[treeStackReferencesTop];
treeStackReferencesTop++;
cnode->id = curRoot->rightChild->id;
treeInit(curRoot->rightChild);
curFinalRoot->id = curRoot->id;
curFinalRoot->numChildren = 2;
curFinalRoot->size = curRoot->size;
}
//the ids of the leafs are deteremined by the newick file, for the internal nodes we just incrementally give the id determined by the dfs traversal
void updateInternalNodeIDs(int cur){
tree_node* curNode = treeArrayReferences[cur];
if(curNode->numChildren == 0){
return;
}
curNode->id = treeNumLeafs++;
updateInternalNodeIDs(curNode->lpos);
updateInternalNodeIDs(curNode->rpos);
}
//frees the memory of the first tree generated by the parser
void treeFreeMemory(tree_node_temp* cur){
if(cur->numChildren == 0){
delete cur;
return;
}
treeFreeMemory(cur->leftChild);
treeFreeMemory(cur->rightChild);
delete cur;
}
//reads the tree stored in "file" under the newick format and creates it in the main memory. The output (what the function returns) is a pointer to the root of the tree.
//this tree is scattered anywhere in the memory.
tree_node* readNewick(string& file){
treeCurSTRindex = -1;
treeNewickStr = "";
treeNumLeafs = 0;
ifstream treeFin;
treeFin.open(file, ios_base::in);
//read the newick format of the tree and store it in a string
treeFin>>treeNewickStr;
//initialize index for reading the string
treeCurSTRindex = 0;
//create the tree in main memory
tree_node_temp* root = readNewickHelper();
//store the tree in an array following the pre order layout
treeArray = new tree_node[root->size];
treeArrayReferences = new tree_node*[root->size];
int i;
for(i=0;i<root->size;i++)
treeArrayReferences[i] = &treeArray[i];
treeStackReferencesTop = 0;
tree_node* finalRoot = treeArrayReferences[treeStackReferencesTop];
curpos = treeStackReferencesTop;
treeStackReferencesTop++;
finalRoot->id = root->id;
treeInit(root);
//update the internal node ids (the leaf ids are defined by the ids stored in the newick string)
updateInternalNodeIDs(0);
//close the file
treeFin.close();
//free the memory of initial tree
treeFreeMemory(root);
//return the pre order tree
return finalRoot;
}
/*
*
*
* DOT FORMAT OUTPUT --- BEGIN
*
*
*/
void treeBstPrintDotAux(tree_node* node, ofstream& treeFout) {
if(node->numChildren == 0) return;
treeFout<<" "<<node->id<<" -> "<<treeArrayReferences[node->lpos]->id<<";\n";
treeBstPrintDotAux(treeArrayReferences[node->lpos], treeFout);
treeFout<<" "<<node->id<<" -> "<<treeArrayReferences[node->rpos]->id<<";\n";
treeBstPrintDotAux(treeArrayReferences[node->rpos], treeFout);
}
void treePrintDotHelper(tree_node* cur, ofstream& treeFout){
treeFout<<"digraph BST {\n";
treeFout<<" node [fontname=\"Arial\"];\n";
if(cur == nullptr){
treeFout<<"\n";
}
else if(cur->numChildren == 0){
treeFout<<" "<<cur->id<<";\n";
}
else{
treeBstPrintDotAux(cur, treeFout);
}
treeFout<<"}\n";
}
void treePrintDot(string& file, tree_node* root){
ofstream treeFout;
treeFout.open(file, ios_base::out);
treePrintDotHelper(root, treeFout);
treeFout.close();
}
/*
*
*
* DOT FORMAT OUTPUT --- END
*
*
*/
/*
* experiments
*
*/
tree_node* T;
int n;
void testCache(int cur){
if(treeArray[cur].numChildren == 0){
treeArray[cur].size = 1;
return;
}
testCache(treeArray[cur].lpos);
testCache(treeArray[cur].rpos);
treeArray[cur].size = treeArray[treeArray[cur].lpos].size + treeArray[treeArray[cur].rpos].size + 1;
}
int main(int argc, char* argv[]){
string Tnewick = argv[1];
T = readNewick(Tnewick);
n = T->size;
double tt;
startTimer();
for (int i = 0; i < 100; i++) {
testCache(0);
}
tt = endTimer();
cout << tt / BILLION << '\t' << T->size;
cout<<endl;
return 0;
}
Compile by typing g++ -O3 -std=c++11 file.cpp
Run by typing ./executable tree.txt. In tree.txt we store the tree in the newick format.
Here is a left going tree with 10^5 leafs
Here is a right going tree with 10^5 leafs
The running times I get: ~0.07 seconds for the left going trees ~0.12 seconds for the right going trees
I apologize for the long post but given how narrow the problem seems to be, I couldn't find a better way to describe it.
Thank you in advance!
EDIT:
This is a follow up edit after MrSmith42's answer. I understand that locality plays a very big role, but I am not sure I understand that this is the case here.
For the two example trees above let us see how we access the memory over time.
For the left going tree:
For the right going tree:
To me it seems like in both cases we have locally access patterns.
EDIT:
Here is a plot about the number of conditional branches:
Here is a plot about the number of branch mispredictions:
Here is a left going tree with 10^6 leafs
Here is a right going tree with 10^6 leafs
FINAL EDIT:
I would like to apologize for wasting everyone's time, the parser I was using had a parameter for how "left" or "right" going I would like to make my tree look like. That was a floating number, it had to be close to 0 to make it left going and close to 1 to make it right going. However to make it look like a chain it had to be very small, like 0.000000001 or 0.999999999. For small inputs the tree looked like a chain even for values like 0.0001. I thought this number was small enough and that it would also give a chain for larger trees, however as I will show isn't the case. If you use numbers like 0.000000001 the parser stops working due to floating point problems.
vadikrobot's answer showed that we have locality issues. Inspired by his experiment I decided to generalize the access pattern diagram above to see how it behaves not just in the example trees, but in any trees.
I modified vadikrobot's code to look like this:
void testCache(int cur, FILE *f) {
if(treeArray[cur].numChildren == 0){
fprintf(f, "%d\t", tim++);
fprintf (f, "%d\n", cur);
treeArray[cur].size = 1;
return;
}
fprintf(f, "%d\t", tim++);
fprintf (f, "%d\n", cur);
testCache(treeArray[cur].lpos, f);
fprintf(f, "%d\t", tim++);
fprintf (f, "%d\n", cur);
testCache(treeArray[cur].rpos, f);
fprintf(f, "%d\t", tim++);
fprintf (f, "%d\n", cur);
fprintf(f, "%d\t", tim++);
fprintf (f, "%d\n", treeArray[cur].lpos);
fprintf(f, "%d\t", tim++);
fprintf (f, "%d\n", treeArray[cur].rpos);
treeArray[cur].size = treeArray[treeArray[cur].lpos].size +
treeArray[treeArray[cur].rpos].size + 1;
}
Access patterns generated by the wrong parser
Let's look at a left tree with 10 leafs.
Looks very nice, as predicted in the diagrams above (I only forgot in the above diagrams the fact that when we find the size of a node, we also access the size parameter of that node, cur in the source code above).
Let's look at a left tree with 100 leafs.
Looks as expected. What about 1000 leafs?
This is definitely not expected. There is a small triangle in the top right corner. And the reason for that is because the tree doesn't look like a left going chain, there is a small subtree hanging out somewhere in the end. The problem becomes even larger when the leafs are 10^4.
Let's look at what happens with right trees. When the leafs are 10:
Looks nice, what about 100 leafs?
Looks good too. This is why I questioned the locality of right trees, to me both seemed at least theory local. Now if you try increasing the size something interesting happens:
For 1000 leafs:
For 10^4 leafs things get even messier:
Access patterns generated by the correct parser
Instead of using that general parser I created one for this specific question:
#include <iostream>
#include <fstream>
using namespace std;
int main(int argc, char* argv[]){
if(argc!=4){
cout<<"type ./executable n{number of leafs} type{l:left going, r:right going} outputFile"<<endl;
return 0;
}
int i;
int n = atoi(argv[1]);
if(n <= 2){cout<<"leafs must be at least 3"<<endl; return 0;}
char c = argv[2][0];
ofstream fout;
fout.open(argv[3], ios_base::out);
if(c == 'r'){
for(i=0;i<n-1;i++){
fout<<"("<<i<<",";
}
fout<<i;
for(i=0;i<n;i++){
fout<<")";
}
fout<<";"<<endl;
}
else{
for(i=0;i<n-1;i++){
fout<<"(";
}
fout<<1<<","<<n<<")";
for(i=n-1;i>1;i--){
fout<<","<<i<<")";
}
fout<<";"<<endl;
}
fout.close();
return 0;
}
Now the access patterns look as expected.
For left trees with 10^4 leafs:
in the black part we go from a low place to a high place, but the distance between the previous low and the current low is small, same for previous high and current high. Hence the cache must be smart enough to hold two blocks, one for the low places and one for the high places, giving a small amount of cache misses.
For right trees with 10^4 leafs:
The original experiments again. This time I could only try up to 10^5 leafs, because as Mysticial noticed, we will get a stack overflow because of the height of the trees, which wasn't the case in the previous experiments since the height was smaller than the one expected.
Time wise they seem to perform the same, however cache and branch wise not. The right trees beat the left trees in branch predictions, the left trees beat the right trees in cache.
Maybe my PAPI usage has been wrong, output from perf:
left trees:
right trees:
I might have messed something up again, and I apologize for this. I included my attempt here just in case anyone wants to continue investigating.
