jsguy jsguy - 2 months ago 7
C++ Question

Why is DFS slower in one tree and faster in the other?

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

enter image description here

Desired output

enter image description here

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):

enter image description here

and slower when
T
looks like this (almost like a right going chain):

enter image description here

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);
}


enter image description here

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?

enter image description here

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.

enter image description here

As we can see the right going tree requires fewer L1 misses, so it seems that it uses the cache efficiently.

enter image description here

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.

enter image description here

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:

enter image description here

For the right going tree:

enter image description here

To me it seems like in both cases we have locally access patterns.

EDIT:

Here is a plot about the number of conditional branches:

enter image description here

Here is a plot about the number of branch mispredictions:

enter image description here

Here is a left going tree with 10^6 leafs

Here is a right going tree with 10^6 leafs

Answer

UPDATE:

I plot the number of accessed element in array in time

void testCache(int cur, FILE *f) {
   if(treeArray[cur].numChildren == 0){
       fprintf (f, "%d\n", cur);
       treeArray[cur].size = 1;
       return;
   }

   fprintf (f, "%d\n", cur);
   testCache(treeArray[cur].lpos, f);
   fprintf (f, "%d\n", cur);
   testCache(treeArray[cur].rpos, f);

   fprintf (f, "%d\n", treeArray[cur].lpos);
   fprintf (f, "%d\n", treeArray[cur].rpos);
   treeArray[cur].size = treeArray[treeArray[cur].lpos].size + treeArray[treeArray[cur].rpos].size + 1;
}

as a result I plot 999990 element of resulted text file: enter image description here

You can see that for the left tree all elements are locally accessed, but for the right one there exist non-uniformity in accessing.

OLD:

I tried to calculate number of memory reads using valgrind. for right one

valgrind --tool=callgrind --cache-sim ./a.out right
==11493== I   refs:      427,444,674
==11493== I1  misses:          2,288
==11493== LLi misses:          2,068
==11493== I1  miss rate:        0.00%
==11493== LLi miss rate:        0.00%
==11493== 
==11493== D   refs:      213,159,341  (144,095,416 rd + 69,063,925 wr)
==11493== D1  misses:     15,401,346  ( 12,737,497 rd +  2,663,849 wr)
==11493== LLd misses:        329,337  (      7,935 rd +    321,402 wr)
==11493== D1  miss rate:         7.2% (        8.8%   +        3.9%  )
==11493== LLd miss rate:         0.2% (        0.0%   +        0.5%  )
==11493== 
==11493== LL refs:        15,403,634  ( 12,739,785 rd +  2,663,849 wr)
==11493== LL misses:         331,405  (     10,003 rd +    321,402 wr)
==11493== LL miss rate:          0.1% (        0.0%   +        0.5%  )

and for left one

valgrind --tool=callgrind --cache-sim=yes ./a.out left

==11496== I   refs:      418,204,722
==11496== I1  misses:          2,327
==11496== LLi misses:          2,099
==11496== I1  miss rate:        0.00%
==11496== LLi miss rate:        0.00%
==11496== 
==11496== D   refs:      204,114,971  (135,076,947 rd + 69,038,024 wr)
==11496== D1  misses:     19,470,268  ( 12,661,123 rd +  6,809,145 wr)
==11496== LLd misses:        306,948  (      7,935 rd +    299,013 wr)
==11496== D1  miss rate:         9.5% (        9.4%   +        9.9%  )
==11496== LLd miss rate:         0.2% (        0.0%   +        0.4%  )
==11496== 
==11496== LL refs:        19,472,595  ( 12,663,450 rd +  6,809,145 wr)
==11496== LL misses:         309,047  (     10,034 rd +    299,013 wr)
==11496== LL miss rate:          0.0% (        0.0%   +        0.4%  )

As you can see number of memory read 'rd' in 'right' case bigger that in left