int sum = 0;
for(int i = 1; i < n; i++) {
for(int j = 1; j < i * i; j++) {
if(j % i == 0) {
for(int k = 0; k < j; k++) {
sum++;
}
}
}
}
I don't understand how when j = i, 2i, 3i... the last for loop runs n times. I guess I just don't understand how we came to that conclusion based on the if statement.
Edit: I know how to compute the complexity for all the loops except for why the last loop executes i times based on the mod operator... I just don't see how it's i. Basically, why can't j % i go up to i * i rather than i?
Let's label the loops A, B and C:
int sum = 0;
// loop A
for(int i = 1; i < n; i++) {
// loop B
for(int j = 1; j < i * i; j++) {
if(j % i == 0) {
// loop C
for(int k = 0; k < j; k++) {
sum++;
}
}
}
}
Loop A iterates O(n) times.
Loop B iterates O(i2) times per iteration of A. For each of these iterations:
j % i == 0 is evaluated, which takes O(1) time.
On 1/i of these iterations, loop C iterates j times, doing O(1) work per iteration. Since j is O(i2) on average, and this is only done for 1/i iterations of loop B, the average cost is O(i2 / i) = O(i).
Multiplying all of this together, we get O(n × i2 × (1 + i)) = O(n × i3). Since i is on average O(n), this is O(n4).
The tricky part of this is saying that the if condition is only true 1/i of the time:
Basically, why can't j % i go up to i * i rather than i?
In fact, j does go up to j < i * i, not just up to j < i. But the condition j % i == 0 is true if and only if j is a multiple of i.
The multiples of i within the range are i, 2*i, 3*i, ..., (i-1) * i. There are i - 1 of these, so loop C is reached i - 1 times despite loop B iterating i * i - 1 times.
The first loop consumes n iterations.
The second loop consumes n*n iterations. Imagine the case when i=n, then j=n*n.
The third loop consumes n iterations because it's executed only i times, where i is bounded to n in the worst case.
Thus, the code complexity is O(n×n×n×n).
I hope this helps you understand.
All the other answers are correct, I just want to amend the following.
I wanted to see, if the reduction of executions of the inner k-loop was sufficient to reduce the actual complexity below O(n⁴). So I wrote the following:
for (int n = 1; n < 363; ++n) {
int sum = 0;
for(int i = 1; i < n; ++i) {
for(int j = 1; j < i * i; ++j) {
if(j % i == 0) {
for(int k = 0; k < j; ++k) {
sum++;
}
}
}
}
long cubic = (long) Math.pow(n, 3);
long hypCubic = (long) Math.pow(n, 4);
double relative = (double) (sum / (double) hypCubic);
System.out.println("n = " + n + ": iterations = " + sum +
", n³ = " + cubic + ", n⁴ = " + hypCubic + ", rel = " + relative);
}
After executing this, it becomes obvious, that the complexity is in fact n⁴. The last lines of output look like this:
n = 356: iterations = 1989000035, n³ = 45118016, n⁴ = 16062013696, rel = 0.12383254507467704
n = 357: iterations = 2011495675, n³ = 45499293, n⁴ = 16243247601, rel = 0.12383580700180696
n = 358: iterations = 2034181597, n³ = 45882712, n⁴ = 16426010896, rel = 0.12383905075183874
n = 359: iterations = 2057058871, n³ = 46268279, n⁴ = 16610312161, rel = 0.12384227647628734
n = 360: iterations = 2080128570, n³ = 46656000, n⁴ = 16796160000, rel = 0.12384548432498857
n = 361: iterations = 2103391770, n³ = 47045881, n⁴ = 16983563041, rel = 0.12384867444612208
n = 362: iterations = 2126849550, n³ = 47437928, n⁴ = 17172529936, rel = 0.1238518469862343
What this shows is, that the actual relative difference between actual n⁴ and the complexity of this code segment is a factor asymptotic towards a value around 0.124... (actually 0.125). While it does not give us the exact value, we can deduce, the following:
Time complexity is n⁴/8 ~ f(n) where f is your function/method.
The wikipedia-page on Big O notation states in the tables of 'Family of Bachmann–Landau notations' that the ~ defines the limit of the two operand sides is equal. Or:
f is equal to g asymptotically
(I chose 363 as excluded upper bound, because n = 362 is the last value for which we get a sensible result. After that, we exceed the long-space and the relative value becomes negative.)
User kaya3 figured out the following:
The asymptotic constant is exactly 1/8 = 0.125, by the way; here's the exact formula via Wolfram Alpha.
Remove if and modulo without changing the complexity
Here's the original method:
public static long f(int n) {
int sum = 0;
for (int i = 1; i < n; i++) {
for (int j = 1; j < i * i; j++) {
if (j % i == 0) {
for (int k = 0; k < j; k++) {
sum++;
}
}
}
}
return sum;
}
If you're confused by the if and modulo, you can just refactor them away, with j jumping directly from i to 2*i to 3*i ... :
public static long f2(int n) {
int sum = 0;
for (int i = 1; i < n; i++) {
for (int j = i; j < i * i; j = j + i) {
for (int k = 0; k < j; k++) {
sum++;
}
}
}
return sum;
}
To make it even easier to calculate the complexity, you can introduce an intermediary j2 variable, so that every loop variable is incremented by 1 at each iteration:
public static long f3(int n) {
int sum = 0;
for (int i = 1; i < n; i++) {
for (int j2 = 1; j2 < i; j2++) {
int j = j2 * i;
for (int k = 0; k < j; k++) {
sum++;
}
}
}
return sum;
}
You can use debugging or old-school System.out.println in order to check that i, j, k triplet is always the same in each method.
Closed form expression
As mentioned by others, you can use the fact that the sum of the first n integers is equal to n * (n+1) / 2 (see triangular numbers). If you use this simplification for every loop, you get :
public static long f4(int n) {
return (n - 1) * n * (n - 2) * (3 * n - 1) / 24;
}
It is obviously not the same complexity as the original code but it does return the same values.
If you google the first terms, you can notice that 0 0 0 2 11 35 85 175 322 546 870 1320 1925 2717 3731 appear in "Stirling numbers of the first kind: s(n+2, n).", with two 0s added at the beginning. It means that sum is the Stirling number of the first kind s(n, n-2).
Let's have a look at the first two loops.
The first one is simple, it's looping from 1 to n. The second one is more interesting. It goes from 1 to i squared. Let's see some examples:
e.g. n = 4
i = 1
j loops from 1 to 1^2
i = 2
j loops from 1 to 2^2
i = 3
j loops from 1 to 3^2
In total, the i and j loops combined have 1^2 + 2^2 + 3^2.
There is a formula for the sum of first n squares, n * (n+1) * (2n + 1) / 6, which is roughly O(n^3).
You have one last k loop which loops from 0 to j if and only if j % i == 0. Since j goes from 1 to i^2, j % i == 0 is true for i times. Since the i loop iterates over n, you have one extra O(n).
So you have O(n^3) from i and j loops and another O(n) from k loop for a grand total of O(n^4)
I understand that the innermost for loop is Θ(logn)
and the two outermost for loops is Θ(n^2) because it's an arithmetic sum. The if-statement is my main problem. Does anyone know how to solve this?
int tally=0;
for (int i = 1; i < n; i ++)
{
for (int j = i; j < n; j ++)
{
if (j % i == 0)
{
for (int k = 1; k < n; k *= 2)
{
tally++;
}
}
}
}
Edit:
Now I noticed loop order: i before j.
In this case for given i value j varies from i to n and there are (n/i) successful if-conditions.
So program will call then most inner loop
n/1 +n/2+n/3+..+n/n
times. This is sum of harmonic series, it converges to n*ln(n)
So inner loop will be executed n*log^2(n) times.
As you wrote, two outermost loops provide O(n^2) complexity, so overall complexity is O(n^2 + n*log^2(n)), the first term overrides the second one, loop, and finally overall complexity is quadratic.
int tally=0;
for (int i = 1; i < n; i ++)
{
// N TIMES
for (int j = i; j < n; j ++)
{
//N*N/2 TIMES
if (j % i == 0)
{
//NlogN TIMES
for (int k = 1; k < n; k *= 2)
{
//N*logN*logN
tally++;
}
}
}
}
Old answer (wrong)
This complexity is linked with sum of sigma0(n) function (number of divisors) and represented as sequence A006218 (Dirichlet Divisor problem)
We can see that approximation for sum of divisors for values up to n is
n * ( log(n) + 2*gamma - 1 ) + O(sqrt(n))
so average number of successful if-conditions for loop counter j is ~log(j)
I need to analyze the O-Notation complexity of the following code. I think I have some idea, but need some help. Do my O-Notations for each loop correct?
int sum = 0;
for (int i = 1; i <= n*2; i++ ) //N operations, ignore the multiplication by 2
sum++;
O(N)
int sum = 0;
for ( int i = 1; i <= n; i++) //N operations
for ( int j = n; j > 0; j /= 2) //N operations, ignore division by constant?
sum++;
O(N^2)
int sum = 0;
for ( int i = 1; i <= n; i++) //N ops
for ( int j = i; j <= n; j += 2) // N ops
sum++;
O(N^2)
int sum = 0;
for ( int i = 1; i <= n * n; i++) // N * N ops
for ( int j = 1; j < i; j++ ) // N opts
sum++;
O(N^3)
You iterate through 2*N elements, so its O(2N) = O(N)
You iterate through N elements and in each iteration, you go through logN elements, so its O(N*logN) - this time its different from 1., because you divide the counter j, not the limiting variable N, so , that means, if N is 20, you get this:
first iteration: j = 20
second iteration: j = 10
third iteration: j = 5
fourth iteration j = 2
fifth iteration j = 1
so its log(20) = 4.something, and that means 5 iterations...
So, it depends where is calculation perform, either in limiting factor or iteration calculation...
This is O(N^2), because you have for loop with n iterations, and in each iteration you have n/2 iterations, so its O(N * N / 2) = O(N^2)
In first loop you have N*N iterations. In each iteration you have to go to i, which depends on n, so you have:
i=1 => no iteration in inner for loop
i=2 => 1 iteration
i=3 => 2 iterations
i=4 => 3 iterations
...
i=n*n => n*n - 1 iterations
Sum of it is:
0+1+2+3+...+n*n-1 = 2+3+...n*n = 1+2+3+...+m-1(where m is n*n) = m*(m+1)/2 - m
So its: O(M^2) = O(N^4)
I have seen that in some cases the complexity of nested loops is O(n^2), but I was wondering in which cases we can have the following complexities of nested loops:
O(n)
O(log n) I have seen somewhere a case like this, but I do not recall the exact example.
I mean is there any kind of formulae or trick to calculate the complexity of nested loops? Sometimes when I apply summation formulas I do not get the right answer.
Some examples would be great, thanks.
Here is an example for you where the time complexity is O(n), but you have a double loop:
int cnt = 0;
for (int i = N; i > 0; i /= 2) {
for (int j = 0; j < i; j++) {
cnt += 1;
}
}
You can prove the complexity in the following way:
The first iteration, the j loop runs N times. The second iteration, the j loop runs N / 2 times. i-th iteration, the j loop runs N / 2^i times.
So in total: N * ( 1 + 1/2 + 1/4 + 1/8 + … ) < 2 * N = O(N)
It would be tempting to say that something like this runs in O(log(n)):
int cnt = 0;
for (int i = 1; i < N; i *= 2) {
for (int j = 1; j < i; j*= 2) {
cnt += 1;
}
}
But I believe that this runs in O(log^2(N)) which is polylogarithmic
i, j, N, sum is all integer type. N is input.
( Code1 )
i = N;
while(i > 1)
{
i = i / 2;
for (j = 0; j < 1000000; j++)
{
sum = sum + j;
}
}
( Code2 )
sum = 0;
d = 1;
d = d << (N-1);
for (i = 0; i < d; i++)
{
for (j = 0; j < 1000000; j++)
{
sum = sum + i;
}
}
How to calculate step count and time complexity for a Code1, Code2?
to calculate the time complexity try to understand what takes how much time, and by what n are you calculating.
if we say the addition ("+") takes O(1) steps then we can check how many time it is done in means of N.
the first code is dividing i in 2 each step, meaning it is doing log(N) steps. so the time complexity is
O(log(N) * 1000000)= O(log(N))
the second code is going form 0 to 2 in the power of n-1 so the complexity is:
O(s^(N-1) * 1000000)= O(2^(N-1))
but this is just a theory, because d has a max of 2^32/2^64 or other number, so it might not be O(2^(N-1)) in practice