I've been trying to understand Big-O notation. Earlier today, I was given a function to practice with and told that it has a O(n^5). I've tried calculating it on my own but don't know if I've calculated T(n) correctly.
Here are my two questions:
1) Did I calculate T(n) correctly and if not then what did I do wrong?
2) Why do we only concern ourselves with the variable to the highest power?
1 sum = 0; //1 = 1
2 for( i=0; i < n; i++) //1 + n + 2(n-1) = 1+n+2n-2 = 3n-1
3 for (j=0; j < i*i; j++) //n + n*n + 2n(n-1))= n+ n^2 + 2n^2-2n = 3n^2 -n
4 for (k=0; k < j; k++) //n + n*n + 4n(n-1))= n + n*n +4n*n-4n = 5n^2 -3n
5 sum++;
6 k++;
7 j++;
8 i++;
// so now that I have simplified everything I multiplied the equations on lines 2-4 and added line 1
// T(n) = 1 + (3n-1)(3n^2-n)(5n^2 -3n) = 45n^5 -57n^4 +23n^3 -3n^2 + 1
Innermost loop runs j times.
Second loop runs for j = 0 to i^2 -> sum of integers.
Outer loop runs to n -> sum of squares and 4th powers of integers.
We only take the highest power because as n approaches infinity, the highest power of n (or order) will always dominate, irrespective of its coefficient.
Related
What is the time complexity and tilde for the loop below?
for (int i = N/2; i < N; i++) {
for (int j = i; j < N; j++) {
doSomething(i, j);
}
}
I think that it runs N/2 + (N/2 + 1) + (N/2 + 2) + ... + (N-1) times, but how do I get it's time complexity and tilde?
For example - if N = 100, the loop will run 50 + 51 + 52 + 53 + ... + 99 times.
I am assuming doSomething(i, j); is not iterating all the elements between i and j; if this is the case, the complexity of this algorithm is O(N^2).
The outer loop for (int i = N/2; i < N; i++) { will execute O(N) times, cause N/2 is actually constant value.
The inner loop in worst case will execute N times (or N - i times) too, this will also merge with previous O(N).
Therefore, overall time complexity will be O(N^2) in worst case scenario.
The inner loop is executed:
N/2-1 times for i = N/2,
N/2-2 times for i = N/2+1
....
1 time for i = N-2
therefore the total time for the inner loop is :
(N/2-1) + (N/2-2) + .... (N/2-k) where k = N/2 - 1
= N/2*k - (1 + 2 + ... + k)
= N/2*(N/2-1) - (N/2-1)(N/2)/2
= N/2(N/2 - 1 - N/4 + 1/2)
= N/2(N/4 - 1/2)
= N^2/8 - N/4
Hence the order of growth of the code is of N^2
If you consider tilde notation which is defined as :
"∼g(n) to represent any quantity that, when divided by f(n), approaches 1 as n grows" from here, you can see that ~g(n) = ~N^2/8 because as N grows (N^2/8)/(N^2/8-N/4) approaches 1.
what is the complexity of the second for loop? would it be n-i? from my understanding a the first for loop will go n times, but the index in the second for loop is set to i instead.
//where n is the number elements in an array
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
// Some Constant time task
}
}
In all, the inner loop iterates sum(1..n) times, which is n * (n + 1) / 2, which is O(n2)
If you try to visualise this as a matrix where lines represents i and each columns represents j you'll see that this forms a triangle with the sides n
Example with n being 4
0 1 2 3
1 2 3
2 3
3
The inner loop has (on average) complexity n/2 which is O(n).
The total complexity is n*(n+1)/2 or O(n^2)
The number of steps this takes is a Triangle Number. Here's a bit of code I put together in LINQpad (yeah, sorry about answering in C#, but hopefully this is still readable):
void Main()
{
long k = 0;
// Whatever you want
const int n = 13;
for (int i = 0; i < n; i++)
{
for (int j = i; j < n; j++)
{
k++;
}
}
k.Dump();
triangleNumber(n).Dump();
(((n * n) + n) / 2).Dump();
}
int triangleNumber(int number)
{
if (number == 0) return 0;
else return number + triangleNumber(number - 1);
}
All 3 print statements (.Dump() in LINQpad) produce the same answer (91 for the value of n I selected, but again you can choose whatever you want).
As others indicated, this is O(n^2). (You can also see this Q&A for more details on that).
We can see that the total iteration of the loop is n*(n+1)/2. I am assuming that you are clear with that from the above explanations.
Now let's find the asymptotic time complexity in an easy logical way.
Big Oh, comes to play when the value of n is a large number, in such cases we need not consider the dividing by 2 ( 2 is a constant) because (large number / 2) is also a large number.
This leaves us with n*(n+1).
As explained above, since n is a large number, (n+1) can be approximated to (n).
thus leaving us with (n*n).
hence the time complexity O(n^2).
Question 1
for (i = 0; i < n; i++) {
for (j = 0; j < i * i ; j++){
}
}
Answer: O(n^3)
At first glance, O(n^3) made sense to me, but I remember a previous problem I did:
Question 2
for (int i = n; i > 0; i /= 2) {
for (int j = 0; j < i; j++) {
//statement
}
}
Answer: O(n)
For Question 2, the outer loop is O(log n) and the inner loop is O(2n / log n) resulting in O(n). The inner loop is O(2n / log n) because - see explanation here: Big O of Nested Loop (int j = 0; j < i; j++)
Why we don't do Question 1 like Question 2 since in Question 1, j also depends on i which means we should really be taking the average of how many iterations will occur in the inner loop (as we do in Question 2).
My answer would be: O(n) for the outer loop and O(n^2 / n) for the inner loop which results in O(n^2) for Question 1.
Your answer is wrong. The code is Θ(n³).
To see that note that the inner loop takes i² steps which is at most n² but for half of the outer loop iterations is at least (n/2)² = n²/4.
Therefore the number of total inner iterations is at most n * n² = n³ but at least n/2 * n²/4 = n³/8.
Your consideration is wrong in that the inner loop takes on average proportional to n² many iterations, not n² / n.
What your inner for loop is doing, in combination with the outer for loop, is calculating the sum of i^2. If you write it out you are adding the following terms:
1 + 4 + 9 + 16 + ...
The result of that is (2n^3+3n^2+n)/6. If you want to calculate the average of the number of iterations of the inner for loop, you divide it by n as this is the number of iterations of the outer for loop. So you get (2n^2+3n+1)/6, in terms of Big O notation this will be O(n^2). And having that gives you... nothing. You have not gain any new information as you already knew the complexity of the inner for loop is O(n^2). Having O(n^2) running n times gives you O(n^3) of total complexity, that you already knew...
So, you can calculate the average number of iterations of the inner for loop, but you will not gain any new information. There were no cuts in the number of iteration steps as there were in your previous question (the i /= 2 stuff).
void fun(int n, int k)
{
for (int i=1; i<=n; i++)
{
int p = pow(i, k);
for (int j=1; j<=p; j++)
{
// Some O(1) work
}
}
}
Time complexity of above function can be written as 1k + 2k + 3k + … n1k.
In your case k = 2
Sum = 12 + 22 + 32 + ... n12.
= n(n+1)(2n+1)/6
= n3/3 + n2/2 + n/6
I have been trying to figure out how the get to the solution of this for loop but I cant seem to understand why the answer is what it is. I get stuck on the inner for loop, can someone help me break it down step by step and to get to the answer of the second for loop, line 2. THIS IS FOR REVIEW NOT HOMEWORK. I am just trying to understand the concept better.
1: for (int i = 0; i < n; i++) { // Runs n+1 times
2: for (int j = n; j >= i; j--) { // Runs (n+2)+(n+1)+ n(n+1)/2 - 3 times
3: cout << i << “,” << j <<endl; // Runs (n+1) + n(n+1)/2 - 1 times
4: }
5: }
I know the second line is simplified to n(n+1)/2 +2n, but I dont understand how to even get the (n+2)+(n+1)+ n(n+1)/2 - 3 first.
1: for (int i = 0; i < n; i++) { // Runs n+1 times
This line does not run for n+1 times. Since it is i < n It runs for n times.
2: for (int j = n; j >= i; j--) {
This line runs for n - i + 1 times. Because it uses >= for comparison.
So if we write down the executions for cout , we get something like this:
1: n+1
2: n
3: n-1
...
...
n: 1
So what we need to do is simply add numbers up to n+1
which is (n+1)(n+2)/2
hope this helps
for (int i = 0; i < n; i++) {
for (int j = n; j >= i; j--) {
cout << "something" << endl;
}
}
Lets see for the values of i, how many times the inner loop is run.
i=0 --> n
i=1 --> n-1
i=2 --> n-2
i=3 --> n-3
...
...
i=n-1 --> 1
Let us increase each element to n to have an upper bound
1 + 2 + 3 +....+ n-1 + n < [n + n + n + .... + n] = O(n^2)
Lets throw half of the first elements, and the other half decrease it to n/2. For a lower bound.
1 + 2 + 3 +....+ n-1 + n > [n/2 + n/2 + ... + n/2] = (n/2)*(n/2) = (n^2)/4 = Ω(n^2)
Lower and Upper bounds are Ω(n), O(n), we can deduce it is actually ϴ(n).
We could have done this immediately by noticing it is actually an arithmetic sequence.
1 + 2 + 3 + ... + n = (1+n)*(n-1)/2 = ϴ(n^2)
I thought the time complexity of the following code is O(log N), but the answer says it's O(N). I wonder why:
int count = 0;
for (int i = N; i > 0; i /= 2) {
for (int j = 0; j < i; j++) {
count += 1;
}
}
For the inners for-loop, it runs for this many times:
N + N/2 + N/4 ...
it seems to be logN to me. Please help me understand why here. Thanks
1, 1/2, 1/4, 1/8... 1/2 ** n is a geometric sequence with a = 1, r = 1/2 (a is the first term, and r is the common ratio).
Its sum can be calculated using the following formula:
In this case, the limit of the sum is 2, so:
n + n/2 + n/4 ... = n(1 + 1/2 + 1/4...) -> n * 2
Thus the complicity is O(N)
Proceeding step by step, based on the code fragment, we obtain: