I have gone through Google and Stack Overflow search, but nowhere I was able to find a clear and straightforward explanation for how to calculate time complexity.
What do I know already?
Say for code as simple as the one below:
char h = 'y'; // This will be executed 1 time
int abc = 0; // This will be executed 1 time
Say for a loop like the one below:
for (int i = 0; i < N; i++) {
Console.Write('Hello, World!!');
}
int i=0; This will be executed only once.
The time is actually calculated to i=0 and not the declaration.
i < N; This will be executed N+1 times
i++ This will be executed N times
So the number of operations required by this loop are {1+(N+1)+N} = 2N+2. (But this still may be wrong, as I am not confident about my understanding.)
OK, so these small basic calculations I think I know, but in most cases I have seen the time complexity as O(N), O(n^2), O(log n), O(n!), and many others.
How to find time complexity of an algorithm
You add up how many machine instructions it will execute as a function of the size of its input, and then simplify the expression to the largest (when N is very large) term and can include any simplifying constant factor.
For example, lets see how we simplify 2N + 2 machine instructions to describe this as just O(N).
Why do we remove the two 2s ?
We are interested in the performance of the algorithm as N becomes large.
Consider the two terms 2N and 2.
What is the relative influence of these two terms as N becomes large? Suppose N is a million.
Then the first term is 2 million and the second term is only 2.
For this reason, we drop all but the largest terms for large N.
So, now we have gone from 2N + 2 to 2N.
Traditionally, we are only interested in performance up to constant factors.
This means that we don't really care if there is some constant multiple of difference in performance when N is large. The unit of 2N is not well-defined in the first place anyway. So we can multiply or divide by a constant factor to get to the simplest expression.
So 2N becomes just N.
This is an excellent article: Time complexity of algorithm
The below answer is copied from above (in case the excellent link goes bust)
The most common metric for calculating time complexity is Big O notation. This removes all constant factors so that the running time can be estimated in relation to N as N approaches infinity. In general you can think of it like this:
statement;
Is constant. The running time of the statement will not change in relation to N.
for ( i = 0; i < N; i++ )
statement;
Is linear. The running time of the loop is directly proportional to N. When N doubles, so does the running time.
for ( i = 0; i < N; i++ ) {
for ( j = 0; j < N; j++ )
statement;
}
Is quadratic. The running time of the two loops is proportional to the square of N. When N doubles, the running time increases by N * N.
while ( low <= high ) {
mid = ( low + high ) / 2;
if ( target < list[mid] )
high = mid - 1;
else if ( target > list[mid] )
low = mid + 1;
else break;
}
Is logarithmic. The running time of the algorithm is proportional to the number of times N can be divided by 2. This is because the algorithm divides the working area in half with each iteration.
void quicksort (int list[], int left, int right)
{
int pivot = partition (list, left, right);
quicksort(list, left, pivot - 1);
quicksort(list, pivot + 1, right);
}
Is N * log (N). The running time consists of N loops (iterative or recursive) that are logarithmic, thus the algorithm is a combination of linear and logarithmic.
In general, doing something with every item in one dimension is linear, doing something with every item in two dimensions is quadratic, and dividing the working area in half is logarithmic. There are other Big O measures such as cubic, exponential, and square root, but they're not nearly as common. Big O notation is described as O ( <type> ) where <type> is the measure. The quicksort algorithm would be described as O (N * log(N )).
Note that none of this has taken into account best, average, and worst case measures. Each would have its own Big O notation. Also note that this is a VERY simplistic explanation. Big O is the most common, but it's also more complex that I've shown. There are also other notations such as big omega, little o, and big theta. You probably won't encounter them outside of an algorithm analysis course. ;)
Taken from here - Introduction to Time Complexity of an Algorithm
1. Introduction
In computer science, the time complexity of an algorithm quantifies the amount of time taken by an algorithm to run as a function of the length of the string representing the input.
2. Big O notation
The time complexity of an algorithm is commonly expressed using big O notation, which excludes coefficients and lower order terms. When expressed this way, the time complexity is said to be described asymptotically, i.e., as the input size goes to infinity.
For example, if the time required by an algorithm on all inputs of size n is at most 5n3 + 3n, the asymptotic time complexity is O(n3). More on that later.
A few more examples:
1 = O(n)
n = O(n2)
log(n) = O(n)
2 n + 1 = O(n)
3. O(1) constant time:
An algorithm is said to run in constant time if it requires the same amount of time regardless of the input size.
Examples:
array: accessing any element
fixed-size stack: push and pop methods
fixed-size queue: enqueue and dequeue methods
4. O(n) linear time
An algorithm is said to run in linear time if its time execution is directly proportional to the input size, i.e. time grows linearly as input size increases.
Consider the following examples. Below I am linearly searching for an element, and this has a time complexity of O(n).
int find = 66;
var numbers = new int[] { 33, 435, 36, 37, 43, 45, 66, 656, 2232 };
for (int i = 0; i < numbers.Length - 1; i++)
{
if(find == numbers[i])
{
return;
}
}
More Examples:
Array: Linear Search, Traversing, Find minimum etc
ArrayList: contains method
Queue: contains method
5. O(log n) logarithmic time:
An algorithm is said to run in logarithmic time if its time execution is proportional to the logarithm of the input size.
Example: Binary Search
Recall the "twenty questions" game - the task is to guess the value of a hidden number in an interval. Each time you make a guess, you are told whether your guess is too high or too low. Twenty questions game implies a strategy that uses your guess number to halve the interval size. This is an example of the general problem-solving method known as binary search.
6. O(n2) quadratic time
An algorithm is said to run in quadratic time if its time execution is proportional to the square of the input size.
Examples:
Bubble Sort
Selection Sort
Insertion Sort
7. Some useful links
Big-O Misconceptions
Determining The Complexity Of Algorithm
Big O Cheat Sheet
Several examples of loop.
O(n) time complexity of a loop is considered as O(n) if the loop variables is incremented / decremented by a constant amount. For example following functions have O(n) time complexity.
// Here c is a positive integer constant
for (int i = 1; i <= n; i += c) {
// some O(1) expressions
}
for (int i = n; i > 0; i -= c) {
// some O(1) expressions
}
O(nc) time complexity of nested loops is equal to the number of times the innermost statement is executed. For example, the following sample loops have O(n2) time complexity
for (int i = 1; i <=n; i += c) {
for (int j = 1; j <=n; j += c) {
// some O(1) expressions
}
}
for (int i = n; i > 0; i += c) {
for (int j = i+1; j <=n; j += c) {
// some O(1) expressions
}
For example, selection sort and insertion sort have O(n2) time complexity.
O(log n) time complexity of a loop is considered as O(log n) if the loop variables is divided / multiplied by a constant amount.
for (int i = 1; i <=n; i *= c) {
// some O(1) expressions
}
for (int i = n; i > 0; i /= c) {
// some O(1) expressions
}
For example, [binary search][3] has _O(log n)_ time complexity.
O(log log n) time complexity of a loop is considered as O(log log n) if the loop variables is reduced / increased exponentially by a constant amount.
// Here c is a constant greater than 1
for (int i = 2; i <=n; i = pow(i, c)) {
// some O(1) expressions
}
//Here fun is sqrt or cuberoot or any other constant root
for (int i = n; i > 0; i = fun(i)) {
// some O(1) expressions
}
One example of time complexity analysis
int fun(int n)
{
for (int i = 1; i <= n; i++)
{
for (int j = 1; j < n; j += i)
{
// Some O(1) task
}
}
}
Analysis:
For i = 1, the inner loop is executed n times.
For i = 2, the inner loop is executed approximately n/2 times.
For i = 3, the inner loop is executed approximately n/3 times.
For i = 4, the inner loop is executed approximately n/4 times.
…………………………………………………….
For i = n, the inner loop is executed approximately n/n times.
So the total time complexity of the above algorithm is (n + n/2 + n/3 + … + n/n), which becomes n * (1/1 + 1/2 + 1/3 + … + 1/n)
The important thing about series (1/1 + 1/2 + 1/3 + … + 1/n) is around to O(log n). So the time complexity of the above code is O(n·log n).
References:
1
2
3
Time complexity with examples
1 - Basic operations (arithmetic, comparisons, accessing array’s elements, assignment): The running time is always constant O(1)
Example:
read(x) // O(1)
a = 10; // O(1)
a = 1,000,000,000,000,000,000 // O(1)
2 - If then else statement: Only taking the maximum running time from two or more possible statements.
Example:
age = read(x) // (1+1) = 2
if age < 17 then begin // 1
status = "Not allowed!"; // 1
end else begin
status = "Welcome! Please come in"; // 1
visitors = visitors + 1; // 1+1 = 2
end;
So, the complexity of the above pseudo code is T(n) = 2 + 1 + max(1, 1+2) = 6. Thus, its big oh is still constant T(n) = O(1).
3 - Looping (for, while, repeat): Running time for this statement is the number of loops multiplied by the number of operations inside that looping.
Example:
total = 0; // 1
for i = 1 to n do begin // (1+1)*n = 2n
total = total + i; // (1+1)*n = 2n
end;
writeln(total); // 1
So, its complexity is T(n) = 1+4n+1 = 4n + 2. Thus, T(n) = O(n).
4 - Nested loop (looping inside looping): Since there is at least one looping inside the main looping, running time of this statement used O(n^2) or O(n^3).
Example:
for i = 1 to n do begin // (1+1)*n = 2n
for j = 1 to n do begin // (1+1)n*n = 2n^2
x = x + 1; // (1+1)n*n = 2n^2
print(x); // (n*n) = n^2
end;
end;
Common running time
There are some common running times when analyzing an algorithm:
O(1) – Constant time
Constant time means the running time is constant, it’s not affected by the input size.
O(n) – Linear time
When an algorithm accepts n input size, it would perform n operations as well.
O(log n) – Logarithmic time
Algorithm that has running time O(log n) is slight faster than O(n). Commonly, algorithm divides the problem into sub problems with the same size. Example: binary search algorithm, binary conversion algorithm.
O(n log n) – Linearithmic time
This running time is often found in "divide & conquer algorithms" which divide the problem into sub problems recursively and then merge them in n time. Example: Merge Sort algorithm.
O(n2) – Quadratic time
Look Bubble Sort algorithm!
O(n3) – Cubic time
It has the same principle with O(n2).
O(2n) – Exponential time
It is very slow as input get larger, if n = 1,000,000, T(n) would be 21,000,000. Brute Force algorithm has this running time.
O(n!) – Factorial time
The slowest!!! Example: Travelling salesman problem (TSP)
It is taken from this article. It is very well explained and you should give it a read.
When you're analyzing code, you have to analyse it line by line, counting every operation/recognizing time complexity. In the end, you have to sum it to get whole picture.
For example, you can have one simple loop with linear complexity, but later in that same program you can have a triple loop that has cubic complexity, so your program will have cubic complexity. Function order of growth comes into play right here.
Let's look at what are possibilities for time complexity of an algorithm, you can see order of growth I mentioned above:
Constant time has an order of growth 1, for example: a = b + c.
Logarithmic time has an order of growth log N. It usually occurs when you're dividing something in half (binary search, trees, and even loops), or multiplying something in same way.
Linear. The order of growth is N, for example
int p = 0;
for (int i = 1; i < N; i++)
p = p + 2;
Linearithmic. The order of growth is n·log N. It usually occurs in divide-and-conquer algorithms.
Cubic. The order of growth is N3. A classic example is a triple loop where you check all triplets:
int x = 0;
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
for (int k = 0; k < N; k++)
x = x + 2
Exponential. The order of growth is 2N. It usually occurs when you do exhaustive search, for example, check subsets of some set.
Loosely speaking, time complexity is a way of summarising how the number of operations or run-time of an algorithm grows as the input size increases.
Like most things in life, a cocktail party can help us understand.
O(N)
When you arrive at the party, you have to shake everyone's hand (do an operation on every item). As the number of attendees N increases, the time/work it will take you to shake everyone's hand increases as O(N).
Why O(N) and not cN?
There's variation in the amount of time it takes to shake hands with people. You could average this out and capture it in a constant c. But the fundamental operation here --- shaking hands with everyone --- would always be proportional to O(N), no matter what c was. When debating whether we should go to a cocktail party, we're often more interested in the fact that we'll have to meet everyone than in the minute details of what those meetings look like.
O(N^2)
The host of the cocktail party wants you to play a silly game where everyone meets everyone else. Therefore, you must meet N-1 other people and, because the next person has already met you, they must meet N-2 people, and so on. The sum of this series is x^2/2+x/2. As the number of attendees grows, the x^2 term gets big fast, so we just drop everything else.
O(N^3)
You have to meet everyone else and, during each meeting, you must talk about everyone else in the room.
O(1)
The host wants to announce something. They ding a wineglass and speak loudly. Everyone hears them. It turns out it doesn't matter how many attendees there are, this operation always takes the same amount of time.
O(log N)
The host has laid everyone out at the table in alphabetical order. Where is Dan? You reason that he must be somewhere between Adam and Mandy (certainly not between Mandy and Zach!). Given that, is he between George and Mandy? No. He must be between Adam and Fred, and between Cindy and Fred. And so on... we can efficiently locate Dan by looking at half the set and then half of that set. Ultimately, we look at O(log_2 N) individuals.
O(N log N)
You could find where to sit down at the table using the algorithm above. If a large number of people came to the table, one at a time, and all did this, that would take O(N log N) time. This turns out to be how long it takes to sort any collection of items when they must be compared.
Best/Worst Case
You arrive at the party and need to find Inigo - how long will it take? It depends on when you arrive. If everyone is milling around you've hit the worst-case: it will take O(N) time. However, if everyone is sitting down at the table, it will take only O(log N) time. Or maybe you can leverage the host's wineglass-shouting power and it will take only O(1) time.
Assuming the host is unavailable, we can say that the Inigo-finding algorithm has a lower-bound of O(log N) and an upper-bound of O(N), depending on the state of the party when you arrive.
Space & Communication
The same ideas can be applied to understanding how algorithms use space or communication.
Knuth has written a nice paper about the former entitled "The Complexity of Songs".
Theorem 2: There exist arbitrarily long songs of complexity O(1).
PROOF: (due to Casey and the Sunshine Band). Consider the songs Sk defined by (15), but with
V_k = 'That's the way,' U 'I like it, ' U
U = 'uh huh,' 'uh huh'
for all k.
For the mathematically-minded people: The master theorem is another useful thing to know when studying complexity.
O(n) is big O notation used for writing time complexity of an algorithm. When you add up the number of executions in an algorithm, you'll get an expression in result like 2N+2. In this expression, N is the dominating term (the term having largest effect on expression if its value increases or decreases). Now O(N) is the time complexity while N is dominating term.
Example
For i = 1 to n;
j = 0;
while(j <= n);
j = j + 1;
Here the total number of executions for the inner loop are n+1 and the total number of executions for the outer loop are n(n+1)/2, so the total number of executions for the whole algorithm are n + 1 + n(n+1/2) = (n2 + 3n)/2.
Here n^2 is the dominating term so the time complexity for this algorithm is O(n2).
Other answers concentrate on the big-O-notation and practical examples. I want to answer the question by emphasizing the theoretical view. The explanation below is necessarily lacking in details; an excellent source to learn computational complexity theory is Introduction to the Theory of Computation by Michael Sipser.
Turing Machines
The most widespread model to investigate any question about computation is a Turing machine. A Turing machine has a one dimensional tape consisting of symbols which is used as a memory device. It has a tapehead which is used to write and read from the tape. It has a transition table determining the machine's behaviour, which is a fixed hardware component that is decided when the machine is created. A Turing machine works at discrete time steps doing the following:
It reads the symbol under the tapehead.
Depending on the symbol and its internal state, which can only take finitely many values, it reads three values s, σ, and X from its transition table, where s is an internal state, σ is a symbol, and X is either Right or Left.
It changes its internal state to s.
It changes the symbol it has read to σ.
It moves the tapehead one step according to the direction in X.
Turing machines are powerful models of computation. They can do everything that your digital computer can do. They were introduced before the advent of digital modern computers by the father of theoretical computer science and mathematician: Alan Turing.
Time Complexity
It is hard to define the time complexity of a single problem like "Does white have a winning strategy in chess?" because there is a machine which runs for a single step giving the correct answer: Either the machine which says directly 'No' or directly 'Yes'. To make it work we instead define the time complexity of a family of problems L each of which has a size, usually the length of the problem description. Then we take a Turing machine M which correctly solves every problem in that family. When M is given a problem of this family of size n, it solves it in finitely many steps. Let us call f(n) the longest possible time it takes M to solve problems of size n. Then we say that the time complexity of L is O(f(n)), which means that there is a Turing machine which will solve an instance of it of size n in at most C.f(n) time where C is a constant independent of n.
Isn't it dependent on the machines? Can digital computers do it faster?
Yes! Some problems can be solved faster by other models of computation, for example two tape Turing machines solve some problems faster than those with a single tape. This is why theoreticians prefer to use robust complexity classes such as NL, P, NP, PSPACE, EXPTIME, etc. For example, P is the class of decision problems whose time complexity is O(p(n)) where p is a polynomial. The class P do not change even if you add ten thousand tapes to your Turing machine, or use other types of theoretical models such as random access machines.
A Difference in Theory and Practice
It is usually assumed that the time complexity of integer addition is O(1). This assumption makes sense in practice because computers use a fixed number of bits to store numbers for many applications. There is no reason to assume such a thing in theory, so time complexity of addition is O(k) where k is the number of bits needed to express the integer.
Finding The Time Complexity of a Class of Problems
The straightforward way to show the time complexity of a problem is O(f(n)) is to construct a Turing machine which solves it in O(f(n)) time. Creating Turing machines for complex problems is not trivial; one needs some familiarity with them. A transition table for a Turing machine is rarely given, and it is described in high level. It becomes easier to see how long it will take a machine to halt as one gets themselves familiar with them.
Showing that a problem is not O(f(n)) time complexity is another story... Even though there are some results like the time hierarchy theorem, there are many open problems here. For example whether problems in NP are in P, i.e. solvable in polynomial time, is one of the seven millennium prize problems in mathematics, whose solver will be awarded 1 million dollars.
I'm trying to come up with a good way to evaluate the following function
double foo(std::vector<double> const& x, double c = 0.95)
{
auto N = x.size(); // Small power of 2 such as 512 or 1024
double sum = 0;
for (auto i = 0; i != N; ++i) {
sum += (x[i] * pow(c, double(i)/N));
}
return sum;
}
My two main concerns with this naive implementation are performance and accuracy. So I suspect that the most trivial improvement would be to reverse the loop order: for (auto i = N-1; i != -1; --i) (The -1 wraps around, this is OK). This improves accuracy by adding smaller terms first.
While this is good for accuracy, it keeps the performance problem of pow. Numerically, pow(c, double(i)/N) is pow(c, (i-1)/N) * pow(c, 1/N). And the latter is a constant. So in theory we can replace pow with repeated multiplication. While good for performance, this hurts accuracy - errors will accumulate.
I suspect that there's a significantly better algorithm hiding in here. For instance, the fact that N is a power of two means that there is a middle term x[N/2] that's multiplied with sqrt(c). That hints at a recursive solution.
On a somewhat related numerical observation, this looks like a signal multiplication with an exponential, so I naturally think : "FFT, trivial convolution=shift, IFFT", but that seems to offer no real benefit in terms of accuracy or performance.
So, is this a well-known problem with known solutions?
The task is a polynomial evaluation. The method for a single evaluation with the least operation count is the Horner scheme. In general a low operation count will reduce the accumulation of floating point noise.
As the example value c=0.95 is close to 1, any root will be still closer to 1 and thus lose accuracy. Avoid that by computing the difference to 1 directly, z=1-c^(1/n), via
z = -expm1(log(c)/N).
Now you have to evaluate the polynomial
sum of x[i] * (1-z)^i
which can be done by careful modification of the Horner scheme. Instead of
for(i=N; i-->0; ) {
res = res*(1-z)+x[i]
}
use
for(i=N; i-->0; ) {
res = (res+x[i])-res*z
}
which is mathematically equivalent but has the loss of digits in 1-z happening as late as possible without using more involved method like doubly accurate addition.
In tests those two methods contrary to the intent gave almost the same results, a substantial improvement could be observed by separating the result into its value at c=1, z=0 and a multiple of z as in
double res0 = 0, resz=0;
int i;
for(i=N; i-->0; ) {
/* res0+z*resz = (res0+z*resz)*(1-z)+x[i]; */
resz = resz - res0 -z*resz;
res0 = res0 + x[i];
}
The test case that showed this improvement was for the coefficient sequence of
f(u) = (1-u/N)^(N-2)*(1-u)
where for N=1000 the evaluations result in
c z=1-c^(1/N) f(1-z) diff for 1st proc diff for 3rd proc
0.950000 0.000051291978909 0.000018898570629 1.33289104579937e-17 4.43845264361253e-19
0.951000 0.000050239954368 0.000018510931892 1.23765066121009e-16 -9.24959978401696e-19
0.952000 0.000049189034371 0.000018123700958 1.67678642238461e-17 -5.38712954453735e-19
0.953000 0.000048139216599 0.000017736876972 -2.86635949350855e-17 -2.37169225231204e-19
...
0.994000 0.000006018054217 0.000002217256601 1.31645860662263e-17 1.15619997300212e-19
0.995000 0.000005012529261 0.000001846785028 -4.15668713370839e-17 -3.5363625547867e-20
0.996000 0.000004008013365 0.000001476685973 8.48811716443534e-17 8.470329472543e-22
0.997000 0.000003004504507 0.000001106958687 1.44711343873661e-17 -2.92226366802734e-20
0.998000 0.000002002000667 0.000000737602425 5.6734266807093e-18 -6.56450534122083e-21
0.999000 0.000001000499833 0.000000368616443 -3.72557383333555e-17 1.47701370177469e-20
Yves' answer inspired me.
It seems that the best approach is to not calculate pow(c, 1.0/N) directly, but indirectly:
cc[0]=c; cc[1]=sqrt(cc[0]), cc[2]=sqrt(cc[1]),... cc[logN] = sqrt(cc[logN-1])
Or in binary,
cc[0]=c, cc[1]=c^0.1, cc[2]=c^0.01, cc[3]=c^0.001, ....
Now if we need x[0b100100] * c^0.100100, we can calculate that as x[0b100100]* c^0.1 * c^0.0001. I don't need to precalculate a table of size N, as geza suggested. A table of size log(N) is probably sufficient, and it can be created by repeatedly taking square roots.
[edit]
As pointed out in a comment thread on another answer, pairwise summation is very effective in keeping errors under control. And it happens to combine extremely nicely with this answer.
We start by observing that we sum
x[0] * c^0.0000000
x[1] * c^0.0000001
x[2] * c^0.0000010
x[3] * c^0.0000011
...
So, we run log(N) iterations. In iteration 1, we add the N/2 pairs x[i]+x[i+1]*c^0.000001 and store the result in x[i/2]. In iteration 2, we add the pairs x[i]+x[i+1]*c^0.000010, etcetera. The chief difference with normal pairwise summation is that this is a multiply-and-add in each step.
We see now that in each iteration, we're using the same multiplier pow(c, 2^i/N), which means we only need to calculate log(N) multipliers. It's also quite cache-efficient, as we're doing only contiguous memory access. It also allows for easy SIMD parallelization, especially when you have FMA instructions.
If N is a power of 2, you can replace the evaluations of the powers by geometric means, using
a^(i+j)/2 = √(a^i.a^j)
and recursively subdivide from c^N/N.c^0/N. With preorder recursion, you can make sure to accumulate by increasing weights.
Anyway, the speedup of sqrt vs. pow might be marginal.
You can also stop recursion at a certain level and continue linearly, with mere products.
You could mix repeated multiplication by pow(c, 1./N) with some explicit pow calls. I.e. every 16th iteration or so do a real pow and otherwise move forward with the multiply. This should yield large performance benefits at negligible accuracy cost.
Depending on how much c varies, you might even be able to precompute and replace all pow calls with a lookup, or just the ones needed in the above method (= smaller lookup table = better caching).
I don't have any experience in openMP , so I want to kow how to do the following:
for (int i = 1; i <= NumImages; i++) {
//call a function
myfunction(...);
for (int k = 0 ; k < SumNumber k++) {
for (int l = 0; l < ElNum ; l++) {
//do 2 summing up calculations inside a while loop
}//end k loop
}//end i loop
Now , I have 40 cores in my disposal.
NumImages will be from 50 to 150 ,more usual 150.
SumNumber will be around 200.
ElNum will be around 5000.
So , the best dealing with this is assigning every thread to a function call and also execute in parallel the l loop?
And if yes , it will be like:
#pragma omp parallel for num_threads(40)
for (int i = 1; i <= NumImages; i++) {
myfunction(...);
for (int k = 0 ; k < SumNumber k++) {
#pragma omp for
for (int l = 0; l < ElNum ; l++) {
And the above means (for NumImages = 150) that myfunction will be executed 40 times in parallel and also l loop and then ,when l loop and k loop finishes , the next 40 threads will call again the function and the next 40 , so 3*40 = 120 and then the next 30?
Generally the best way is the way that splits the work evenly, to maintain efficiency (no cores are waiting). E.g. in your case probably static scheduling is not a good idea, because 40 does not divide 150 evenly, for the last iteration you would loose 25% of computing power. So it might turn out, that it would be better to put parallel clause before second loop. It all the depends on the mode you choose, and how really work is distributed within loops. E.g., If myfunction does 99% then its a bad idea, if 99% of work is within 2 inner loops it might be good.
Not really. There are 3 scheduling modes. But none of them works in a way, that it blocks other threads. There is a pool of tasks (iterations) that is distributed among the threads. Scheduling mode describes the strategy of assigning tasks to threads. When one thread finishes, it just gets next task, no waiting. The strategies are described in more detail here: http://en.wikipedia.org/wiki/OpenMP#Scheduling_clauses (I am not sure if balant-copy paste from wiki is a good idea, so I'll leave a link. It's a good material.)
Maybe what is not written there is that the modes overhead are presented in order of the amount of overhead they introduce. static is fastest, then dynamic, then guided. My advice when to use which would be, this is not the exact best, but good rule of thumb IMO:
static if you know will be divided evenly among the threads and take the same amount of time
dynamic if you know the tasks will not be divided evenly or their execution times are not even
guided for rather long tasks that you pretty much cannot tell anything
If your tasks are rather small you can see an overhead even for static scheduling (E.g. why my OpenMP C++ code is slower than a serial code?), but I think in your case dynamic should be fine and best choice.
I have
cilk_for (int i = 0; i < 100; i++)
x = fib(35);
the above takes 6.151 seconds
and
for (int i = 0; i < 100; i++)
x = cilk_spawn fib(35);
takes 5.703 seconds
The fib(x) is the horrible recursive Fibonacci number function. If I dial down the fib function cilk_for does better than cilk_spawn, but it seems to me that regardless of the time it takes to do fib(x) cilk_for should do better than cilk_spawn.
What don't I understand?
Per comments, the issue was a missing cilk_sync. I'll expand on that to point out exactly how the ratio of time can be predicted with surprising accuracy.
On a system with P hardware threads (typically 8 on a i7) for/cilk_spawn code will execute as follows:
The initial thread will execute the iteration for i=0, and leave a continuation that is stolen by some other thread.
Each thief will steal an iteration and leave a continuation for the next iteration.
When each thief finishes an iteration, it goes back to step 2, unless there are no more iterations to steal.
Thus the threads will execute the loop hand-over-hand, and the loop exits at a point where P-1 threads are still working on iterations. So the loop can be expected to finish after evaluating only (100-P-1) iterations.
So for 8 hardware threads, the for/cilk_spawn with missing cilk_sync should take about 93/100 of the time for the cilk_for, quite close to the observed ratio of about 5.703/6.151 = 0.927.
In contrast, in a "child steal" system such as TBB or PPL task_group, the loop will race to completion, generating 100 tasks, and then keep going until a call to task_group::wait. In that case, forgetting the synchronization would have led to a much more dramatic ratio of times.
My application does some operations on matrices of large size.
I recently came accross the concept of cache & the performance effect it can have through this answer.
I would like to know what would be the best algorithm which is cache friendly for my case.
Algorithm 1:
for(int i = 0; i < size; i++)
{
for(int j = i + 1; j < size; j++)
{
c[i][j] -= K * c[j][j];//K is a constant double variable
}//c is a 2 dimensional array of double variables
}
Algorithm 2:
double *A = new double[size];
for(int n = 0; n < size; n++)
A[n] = c[n][n];
for(int i = 0; i < size; i++)
{
for(int j = i + 1; j < size; j++)
{
c[i][j] -= K * A[j];
}
}
The size of my array is more than 1000x1000.
Benchmarking on my laptop shows Algorithm 2 is better than 1 for size 5000x5000.
Please note that I have multi threaded my application such that a set of rows is operated by a thread.
For example: For array of size 1000x1000.
thread1 -> row 0 to row 249
thread2 -> row 250 to row 499
thread3 -> row 500 to row 749
thread4 -> row 750 to row 999
If your benchmarks show significant improvement for the second case, then it most likely is the better choice. But of course, to know for "an average CPU", we'd have to know that for a large number of CPU's that can be called average - there is no other way. And it will really depend on the definition of Average CPU. Are we talking "any x86 (AMD + Intel) CPU" or "Any random CPU that we can find in anything from a watch to the latest super-fast creation in the x86 range"?
The "copy the data in c[n][n]" method helps because it gets its own address, and doesn't get thrown out of the (L1) cache when the code walks its way over the larger matrix [and all the data you need for the multiplication is "close together". If you walk c[j][j], every j steps will jump sizeof(double) * (size * j + 1) bytes per iteration, so if size is anything more than 4, the next item needed wont be in the same cache-line, so another memory read is needed to get that data.
In other words, for anything that has a decent size cache (bigger than size * sizeof(double)), it's a definite benefit. Even with smaller cache, it's quite likely SOME benefit, but the chances are higher that the cached copy will be thrown out by some part of c[i][j].
In summary, the second algorithm is very likely better for nearly all options.
Algorithm2 benefits from what's called "spatial locality", moving the diagonal into a single dimension array makes it reside in memory in consecutive addresses, and thereby:
Enjoys the benefit of fetching multiple useful elements per a single cache line (presumably 64byte, depending on your CPU), better utilizing cache and memory BW (whereas c[n][n] would also fetch a lot of useless data since it's in the same lines).
Enjoys the benefits of a HW stream prefetchers (assuming such exist in your CPU), that aggressively run ahead of your code along the page and brings the data in advance to the lower cache levels, improving the memory latency.
It should be pointed that moving the data to A doesn't necessarily improve cacheability since A would still compete against a lot of data constantly coming from c and thrashing the cache. However, since it's used over and over, there's a high chance that a good LRU algorithm would make it stay in the cache anyway. You could help that by using streaming memory operations for array c. It should be noted that these are very volatile performance tools, and may on some scenarios lead to perf reduction if not used correctly.
Another potential benefit could come from mixing SW prefetches slightly ahead of reaching every new array line.