So I've been trying to get a handle on Big Oh calculations. I feel I have the basics down but am stumped on what seems a really easy calculation. So if the calculation below has a big oh of O(n log n) (I really hope I've at least got that right) what does changing the order of the loops do to the complexity? Thanks so much in advance for your time.
int ONLogN(int N) //O(n log n)
{
int iIterations = 0;
for (int i = 0; i < N; ++i)
{
++iIterations;
for (int j = 1; j < N + 1; j *= 2)
++iIterations;
}
return iIterations;
}
int WhatBigOhIsThis(int N) //???
{
int iIterations = 0;
for (int j = 1; j < N + 1; j *= 2)
{
++iIterations;
for (int i = 0; i < N; ++i)
++iIterations;
}
return iIterations;
}
The index variables on the two loops are independent, hence the resulting complexity is necessarily the same.
You're still looping for the same number of iterations. Changing the order of the loops would have no effect on complexity
Related
I was asked to find the big-O and Big-Omega if I know that function f has O(log(n)), Ω(1) and function g has O(n), Ω((log(n))^2)
for (int i = n; i >= 0; i/=2)
if (f(i) <= g(i))
for (int j = f(i)*f(i); j < n; j++)
f(j);
The big problem that I have is that I don't know how to incorporate the complexity of the funstions in the calculation. I mean I know how to calculate the complexity of loops that looks like this:
for(int i =0 ; i< n*2; i++) {
....
}
or like this
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
}
}
Thank you in advance.
This is what I've tried:
for (int i = n; i >= 0; i/=2)// this is aproximatly O(log(n))
if (f(i) <= g(i))// because O(n) < O(log(n)) this line is O(n)
for (int j = f(i)*f(i); j < n; j++)// O(n*log(n))
f(j);// O(log(n))
So by my calculation I get O(log(n)*n *n *log(n)*log(n))=O(n^2*log^3(n))
This is tricky question, because the loop execution depends on values, returned by functions f and g. However remember, that you need to estimate the worst case - so you need to assume two things:
f(i) <= g(i) is always true, so the internal loop always executes
the internal loop starts from 0, because it's a minimal value, which you can get as a result of squaring the f(i) value
So, your piece of code becomes much simpler:
for (int i = n; i >= 0; i/=2)
{
f(i);
g(i);
f(i);
f(i);
for (int j = 0; j < n; j++)
f(j);
}
I think you can take over from here.
I am studying Data Structures, but I am a bit confused about my homework assignment. I am trying to figure out how to count the precise Big-Oh values of the following:
for (int i = 0; i < n; i++)
{
for (int j = i; j < n; j++)
{
for (int k = j; k < n; k++)
{
sum += j;
}
}
}
I know how to count the Big-Oh's in terms of magnitude, but I don't know how to count it precisely. An example answer is in the form: 10N + 3n^2 + 2 (IE)
Ok, I am trying to understand the concept of Big O. I have a function I am suppose to find the Big O and I am not quite "getting it" this is an example in the book of one that is like my homework.. I know the answer is O(nk) but can someone please break this down in simplistic terms so I might better understand.
int selectkth(int a[], int k, int n)
{
int i, j, mini, tmp;
for (i=0; i < k; i++)
{
mini = i;
for (j = i+1; j < n; j++)
{
if (a[j] < a[mini])
mini = k;
tmp = a[i];
a[i] = a[mini];
a[mini] = tmp;
}
}
return a[k-1];
}
When calculating the bigO try to think of the worst time complexity, and pay attention to loops. Here we have two loops:
// Below line is run k times
for (i=0; i < k; i++)
// Worst case scenario, loop below will run n times.
for (j = i+1; j < n; j++)
bigO would be these two values multiplied togehter = k*n
Also check out this post: What is a plain English explanation of "Big O" notation?
I have been searching through the forums on big O notation and learned quite a bit. My problem is pretty specific and I think a unique case will better help me understand big O, I am ignoring constants.
To my understanding if a loop goes through all elements than it's O(n).
for(int i = 0; i < n; i++)
{
}
If a loop goes through all of n, inside another loop that goes through all of n, it's multiplied n * n = n^2
for(int i = 0; i < n; i++)
{
for(int j = 0; j < n; j++)
{
}
}
Lastly if a loop is followed by another loop that goes through all elements it is n + n = 2n
for(int j = 0; j < n; j++)
{
}
for(int k = 0; k < n; k++)
{
}
My question directly proceeds these lines of code
for(int i = 0; i < n; i++)
{
for(int j = 0; j < n; j++)
{
}
for(int k = 0; k < n; k++)
{
}
for(int l = 0; l < n; l++)
{
for(int m = 0; m < n; m++)
{
}
}
}
So based on the rules above I am calculating big O to be n * (n + n + n * n), which is n^3 + 2n^2. So would that make my big O(n^3) or would my big O be O(n^3 +2n^2). Am I going about this all wrong? Or am I somewhere close in the ballpark? Mainly I'm trying to figure out if these loops would be less than O(n^4). Thanks in advance.
The big-O notation is used to characterize the asymptotic behavior of an algorithm depending on some value n that indicates the data volume, but independent of any constant, e.g. processor speed.
In your example, n^3 grown faster than 2n^2, i.e., for large n, 2n^2 can be neglected compared to n^3. The asymptotic behavior of your nested loops thus have order O(n^3).
How would you show that this entire sequence of loops is O(n)? Or is it O(n)? At first glance, just looking at the double loops one might think it's O(n^2) but I don't think it is...
int i = 0;
int arr[N];
int idx = 0;
for (i = 0; i < 2; i++)
{
for (j = 0; j < N/2; j++)
{
idx = (i * N/2) + j;
foo(arr[idx]);
}
}
The outer loop is constant, so the loop is the same as doing:
for (j = 0; j < N/2; j++)
{
idx = (0 * N/2) + j;
foo(arr[idx]);
idx = (1 * N/2) + j;
foo(arr[idx]);
}
Which actually makes the intent of the code a lot clearer also, but as you can see, the amount of operations scales linearly with N, so it's O(N) complexity rather than exponential growth. I can't explain it any better as it's 8.30am and I haven't slept, but I imagine you get my gist.