Calculate the time complexity of this code fragment if the function "process" has a complexity of O(logn)
void funct (int a[], int n)
{
int i=0
while (i < n){
process (a, n);
if (a[i]%2 == 0)
i = i*2+1;
else
i = i+1;
}
I tried to calculate the best and worst case for time complexity;
Worst case is when the "else" statement get called so it should just be:
Worst case : T(n) = O(nlogn)
I have some problems with the best case. I tried this way but i don't know if this is correct
Since in the "if" statement "i" get incremented by "2i+1" it should be
i=2^k-1
2^k < n+1
so k < log_2(n+1)
Is correct to say that the while loop get executed (log_2(n+1)-2)/2 times because this is the last possible value for which i < n ?
if so the time complexity is O(lognlogn) in best case?
The best case is if the sampled values in a are all even. In that case, the complexity is O(log(n)*log(n)), since the loop trip count is O(log(n)).
The worst case is if the sampled values in a are all odd. In that case, the complexity is O(n*log(n)), since the loop trip count is O(n).
Related
When looking at this code for example :
for (int i = 1; i < n; i*=2)
for (int j = 0; j < i; j +=2)
{
// some contstant time operations
}
Is it as simple as saying that because the outer loop is log and and inner loop is n , that combined the result is big(O) of nlogn ?
Here is the analysis of the example in the question. For simplicity I will neglect the increment of 2 in the inner loop and will consider it as 1, because in terms of complexity it does not matter - the inner loop is linear in i and the constant factor of 2 does not matter.
So we can notice, that the outer loop is producing is of values which are powers of 2 capped by n, that is:
1, 2, 4, 8, ... , 2^(log2 n)
these numbers are also the numbers that the "constant time operation" in the inner loop is running for each i.
So all we have to do is to sum up the above series. It is easy to see that these are geometric series:
2^0 + 2^1 + 2^2 + ... + 2^(log2 n)
and it has a well known solution:
(from Wiki )
We have a=1, r=2, and... well n_from_the_image =log n. We have a same name for different variables here, so it is a bit of a problem.
Now let's substitute and get that the sum equals
(1-2^((log2 n) + 1) / (1 - 2) = (1 - 2*n) / (1-2) = 2n-1
Which is a linear O(n) complexity.
Generally, we take the O time complexity to be the number of times the innermost loop is executed (and here we assume the innermost loop consists of statements of O(1) time complexity).
Consider your example. The first loop executes O(log N) times, and the second innermost loop executes O(N) times. If something O(N) is being executed O(log N) times, then yes, the final time complexity is just them multiplied: O(N log N).
Generally, this holds true with most nested loops: you can assume their big-O time complexity to be the time complexity of each loop, multiplied.
However, there are exceptions to this rule when you can consider the break statement. If the loop has the possibility of breaking out early, the time complexity will be different.
Take a look at this example I just came up with:
for(int i = 1; i <= n; ++i) {
int x = i;
while(true) {
x = x/2;
if(x == 0) break;
}
}
Well, the innermost loop is O(infinity), so can we say that the total time complexity is O(N) * O(infinity) = O(infinity)? No. In this case we know the innermost loop will always break in O(log N), giving a total O(N log N) time complexity.
I think the time complexity of this code will be O(n^2) but I am not sure so if someone can explain what will be the time complexity of this code it would be really helpful
int func2()
{
int i, j, k = 0;
scanf("%d", &n);
for (i = 1; i < n; i++)
{
i -= 1;
i *= 2;
k = k + i;
for (j = 1; j < n; j++);
}
}
It looks like an infinite loop to me, so the time complexity is O(infinity).
On the first iteration of the outer loop, i -= 1 will set i to 0. Multiplying by 2 leaves it still 0.
The loop iteration i++ will then increment i to 1, and the next iteration will repeat the above computations.
I am a beginner on time complexity but these are my views:-
The outer for loop is in the condition of an infinite loop as on the first iteration of the outer loop, execution starts with i=1.
On executing i -= 1 it will set i=0.
Executing i*=2, the value of i remains the same as 0.
On going in the increment phase, i is incremented and i=1.
So the same process occurs.
Thus the value of i remains the same causing it to run indefinitely.
Now, coming forward inside the outer for loop is a nested for (in the variable j) loop that is followed by a semicolon. This causes it to have a time complexity of O(1).
So the resultant overall time complexity can be expected to be O(infinity).
First, 'n' not declared here and input value is being assigned to it.
Second, technically, this code is an infinite loop (done ridiculously hard way) and for non-terminating, forever-running algorithms the time-complexity is 'Undefined' as by the principles of Algorithm Analysis, time-complexity is only computed for algorithms that perform a task with certainity to terminate.
In case if this would have been a terminating loop, time complexity of this function is O(n^2) would have been quadratic in nature due to nesting of for(;;) inside of another for(;;) with enclosed statements of O(1) - linear time complexity. The higher order complexity ( O(n^2) ) supersedes.
I don't know how to solve it's complexity and such. How to know if it is faster than other sorting algorithms?
I find difficulty finding it because I'm a little bit bad at math.
#include <stdio.h>
int func(int arr[])
{
int temp;
int numofarrays=9999;
for(int runtime=1; runtime<=numofarrays/2; runtime++)
{
for(int i=0; i<=numofarrays; i++)
{
if(arr[i] > arr[i+1])
{
temp=arr[i];
arr[i]=arr[i+1];
arr[i+1]=temp;
}
if(arr[numofarrays-i-1] > arr[numofarrays-i])
{
temp=arr[numofarrays-i-1];
arr[numofarrays-i-1]=arr[numofarrays-i];
arr[numofarrays-i]=temp;
}
}
}
for(int i=0; i<=9999; i++)
{
printf("%i\n",arr[i]);
}
}
int main()
{
int arr[10000];
for(int i=0; i<=9999; i++)
{
arr[i]=rand() % 10;
}
func(arr);
}
The big o notation is where the limit of steps of the code you write goes to infinity. Since the limit goes to infinity, we neglect the coefficients and look at the largest term.
for(int runtime=1; runtime<=numofarrays/2; runtime++)
{
for(int i=0; i<=numofarrays; i++)
{
Here, the largest term of the first loop is n, and the largest term of the second is n. But since loop 2 makes n turns for each turn of the first loop, we're multiplying n by n.
Result is O(n^2).
Here:
for(int runtime=1; runtime<=numofarrays/2; runtime++)
{
for(int i=0; i<=numofarrays; i++)
{
you have two nested for loops that both depends of the array size. So the complexity is O(N^2).
Further you ask:
How to know if it is faster than other sorting algorithms?
Well big-O does not directly tell you about execution time. For some values of N an O(N) implementation may be slower than an O(N^2) implementation.
Big-O tells you how the execution time increases as N increase. So you know that an O(N^2) will be slower than an O(N) when N gets above a apecific value but you can't know what that value is! It could be 1, 5, 10, 100, 1000, ...
Example:
The empirical approach is to time (t) your algorithm for a given input of n. Then do that experiment for larger and large n, say, 2^n for n = 0 to 20. Plot (2^n, t) on a graph. If you get a straight line, it's a linear algorithm. If you get something that grows faster, try graph ln(t) and see if you get a line, that would be an exponential algorithm, etc.
The analytical approach looks scary and sometimes math heavy, but it's doesn't really have to be. You decide what your expensive operation is, then you count how of many of those you do. Anything that runs in loops including recursive functions are the main candidates the run-time of that goes up as the loop runs more times:
Let's say we want count the number of swaps.
The inner loop swap numofarrays times which is O(n).
The outer loop runs numofarrays/2 times which is also O(n) as we drop the factor 1/2 as it's doesn't matter when n gets large.
This mean we do O(n) * O(n) = O(n^2) number of swaps.
Your print loop, is not doing any swaps so we consider them "free", but if determine that they are as expensive as swaps then we want to count those too. Your algorithm does numofarrays print operations which is O(n).
O(n^2) + O(n) is just O(n^2) as the larger O(n^2) dominates O(n) for sufficiently large n.
I would like to know exactly how to compute the big O of the second while when the number of repetitions keeps going down over time.
int duplicate_check(int a[], int n)
{
int i = n;
while (i > 0)
{
i--;
int j = i - 1;
while (j >= 0)
{
if (a[i] == a[j])
{
return 1;
}
j--;
}
}
return 0;
}
Still O(n^2) regardless of the smaller repetition.
The value you are computing is Sum of (n-k) for k = 0 to n.
This equates to (n^2 + n) / 2 which since O() ignores constants and minor terms is O(n^2).
Note you can solve this problem more efficiently by sorting the array O(nlogn) and then searching for two consecutive numbers that are the same O(n) so total O(nlogn)
Big O is an estimate/theoretical speed, it's not the exact calculation.
Like twain249 said, regardless, the time complexity is O(n^2)
BigO shows the worst case time complexity of an algorithm that means the maximum time an algorithm can take ever.It shows upper bound which indicates that whatever the input is time complexity will always be under that bound.
In your case the worst case will when i will iterate until 0 then complexity will be like:
for i=n j will run n-1 times for i=n-1 j will run n-2 times and so on.
adding all (n-1)+(n-2)+(n-3)+............(n-n)=(n-1)*(n)/2=n^2/2-n/2
after ignoring lower term that is n and constant that is 1/2 it becomes n^2.
So O(n^2) that's how it is computed.
I am pretty sure about my answer but today had a discussion with my friend who said I was wrong.
I think the complexity of this function is O(n^2) in average and worst case and O(n) in best case. Right?
Now what happens when k is not length of array? k is the number of elements you want to sort (rather than the whole array).
Is it O(nk) in best and worst case and O(n) in best case?
Here is my code:
#include <stdio.h>
void bubblesort(int *arr, int k)
{
// k is the number of item of array you want to sort
// e.g. arr[] = { 4,15,7,1,19} with k as 3 will give
// {4,7,15,1,19} , only first k elements are sorted
int i=k, j=0;
char test=1;
while (i && test)
{
test = 0;
--i;
for (j=0 ; j<i; ++j)
{
if ((arr[j]) > (arr[j+1]))
{
// swap
int temp = arr[j];
arr[j]=arr[j+1];
arr[j+1]=temp;
test=1;
}
} // end for loop
} // end while loop
}
int main()
{
int i =0;
int arr[] = { 89,11,15,13,12,10,55};
int n = sizeof(arr)/sizeof(arr[0]);
bubblesort(arr,n-3);
for (i=0;i<n;i++)
{
printf("%d ",arr[i]);
}
return 0;
}
P.S. This is not homework, just looks like one. The function we were discussing is very similar to Bubble sort. In any case, I have added homework tag.
Please help me confirm if I was right or not. Thank you for your help.
Complexity is normally given as a function over n (or N), like O(n), O(n*n), ...
Regarding your code the complexity is as you stated. It is O(n) in best case and O(n*n) in worst case.
What might have lead to misunderstanding in your case is that you have a variable n (length of array) and a variable k (length of part in array to sort). Of course the complexity of your sort does not depend on the length of the array but on the length of the part that you want to sort. So with respect to your variables the complexity is O(k) or O(k*k). But since normally complexity notation is over n you would say O(n) or O(n*n) where n is the length of the part to sort.
Is it O(nk) in best and worst case and O(n) in best case?
No, it's O(k^2) worst case and O(k) best case. Sorting the first k elements of an array of size n is exactly the same as sorting an array of k elements.
That's O(n^2), the outer while goes from k down to 1 (possibly stopping earlier for specific data, but we're talking worst case here), and the inner for goes from 0 to i (which in turn goes up to k), so multiplied they're k^2 in the worst case.
If you care about the best case, that's O(n) because the outer while loop only executes once then gets aborted.