I am trying to understand the subtle difference in the complexity of
each of the examples below.
Example A
int sum = 0;
for (int i = 1; i < N; i *= 2)
for (int j = 0; j < N; j++)
sum++;
My Analysis:
The first for loop goes for lg n times.
The inner loop is independent of outer loop and executes N times every time outer loop executes.
So the complexity must be:
n+n+n... lg n times
Therefore the complexity is n lg n.
Is this correct?
Example B
int sum = 0;
for (int i = 1; i < N; i *= 2)
for(int j = 0; j < i; j++)
sum++;
My Analysis:
The first for loop goes for lg n times.
The inner loop execution depends on outer loop.
So how do I calculate the complexity when no of times inner loop executes depends on outer loop?
Example C
int sum = 0;
for (int n = N; n > 0; n /= 2)
for (int i = 0; i < n; i++)
sum++;
I think example C and example B must have same complexity because no of times the inner loop executes depends on outer loop.
Is this correct?
In examples B and C, the inner loop executes 1 + 2 + ... + n/2 + n times. There happen to be lg n terms in this sequence, and that does mean that int i = 0 executes lg n times, however the sum for the statement(s) in the inner loop is 2n. So we get O(n + lg n) = O(n)
(a) Your analysis is correct
(b) The outer loop goes log(N) times. The inner loop goes in the sequence 1, 2, 4, 8, ... for log(N) times which is a geometric series and is equal to (approx) O(2^log(N)) or twice the amount of the highest multiple.
E.g. : 1 + 2 + 4 = (approx)2*4, 1 + 2 + 4 + 8 = (approx)2*8.
Hence the total complexity is O(2^log(N)) = O(N)
(c) This is same as (b) in reverse order
Fine Time complexity
I=1;
K=1;
While(k<n)
{
Stmt;
K=k+i;
I++;
}
Related
I wanted to confirm if I got the correct big-O for a few snippets of code involving for-loops.
for ( int a = 0; a < n; a ++)
for ( int b = a; b < 16 ; b ++)
sum += 1;
I think this one is O(16N) => O(N), but the fact that b starts at a rather than 0 in the second for-loop is throwing me off.
int b = 0;
for ( int a = 0; a < n ; a ++)
sum += a * n;
for ( ; b < n ; b ++)
sum++;
I want to say O(N^2) since there are nested for-loops where both loops go to n. However, b in the second loop uses the initialization from the outer scope, and I'm not sure if that affects the runtime.
for (int a = 0; a < (n * n); a ++)
sum++;
if (a % 2 == 1)
for (; a < (n * n * n); a ++)
sum++;
I got that the first for-loop is O(N^2) and the one under the if-statement is O(N^3), but I don't know how to account for the if-statement.
The first one is O(n*min(n, 16)) because the 16-for-loop counts as O(1) assuming n > 16. if n < 16 then it's O(n^2).
The second one is O(n) because after the first iteration b is already n.
The third one is O(n^3) because after a maximum of 2 iterations you reach the if statement and then a is incremented until n^3 which means that in the next outer for-loop iteration a<n*n is indeed true so it will exit out of the for loop entirely.
Hopefully this answers your questions, good luck!
I made a function that loops through the array and prints any two values of the array that can add up to a value K. The outer for loop is O(n), but the inner loop is a bit confusing to me if the runtime is a O(Log n) or O(n). can you help please? Thank you!!
int canMakeSum(int *array, int n, int key){
int i, j;
for(i = 0; i < n; i++){
for(j = (i+1); j < n; j++){
if(array[i]+array[j] == key){
printf("%d + %d = %d\n", array[i], array[j], key);
}
}
}
}
As others have already shown, the inner loop is still O(n); it's a mean of n/2 iterations, the values 1 through n distributed evenly over the iterations of the outer loop.
Yes, you can solve the problem in O(n log n).
First, sort the array; this is n log n. Now, you have a linear (O(n)) process to find all combinations.
lo = 0
hi = n-1
while lo < hi {
sum = array[lo] + array[hi]
if sum == k {
print "Success", array[lo], array[hi]
lo += 1
hi -= 1
}
else if sum < k // need to increase total
lo += 1
else // need to decrease total
hi -= 1
As the inner loops is dependent to the value of the outer loop, you can't find the complexity of the total porgram without analyzing the both with together. The complexity of the inner loop is n - i - 1.
If you want to compute the complexity of the program, you can sum over n - i -1 from i = 0 to i = n - 1. Hence, the total complexity is T(n) = (n - 1) + (n-2) + ... + 1 + 0 = (n-1)n/2 = \Theta(n^2) (as the statment in the inner loop has a constant complexity (\Theta(1))).
Although the inner loop decreases in the number of items it scans for each iteration in the outer loop, it would still be O(n). The overall time complexity is O(n^2).
Imagine you've an array of 25000 elements. At the starting point at i = 0 and j = 1, the number of elements that j will iterate through (worst case no matches to key) is 24999 elements. Which is a small difference from the total number of elements, so it is 'like' going through n elements.
how can I find the time complexity of inner loops that depend on i from an outer loop such as:
int sum = 0;
for(int i = 0; i < n * n; i++) {
for(int j = n - 1; j >= n - 1 - i; j--) {
sum = i + j;
System.out.println(sum);
}
}
I'm having such a hard time figuring out the time complexity for that. I know that if there is i we use the summation rule but in this case how is that going to be?
The inner loop performs 1 iteration at the start, increasing up to n*n iterations by the end.
i iterations of inner loop
- ------------------------
0 1
1 2
2 3
...
n*n-2 n*n-1
n*n-1 n*n
The total number of iterations of the inner loop is the sum of these:
or
The time complexity is therefore Θ(n4).
The inner loop changes with the outer loop, so it is dependent. You cannot directly find the cost of only the inner loop.
for(int i = 0; i < n * n; i++) { \\ \sum_{i=0}^{n^2}
for(int j = n - 1; j >= n - 1 - i; j--) { \\ \sum_{n-1-i}^{n-1}
sum = i + j; \\ 1
System.out.println(sum);
}
}
So, we have to solve
Note: there was a typo on the Latex image, and it is updated. Thanks to Anatolii
I understand how to use summation to extract the time complexity (and Big O) from linear for-loops using summation, but how would you use it for multiplication incremental loops to get O(logn). For example, the code below is O(nlogn), but I don't know why.
for (i = 0; i < n; i++)
for (j = 1; j < n; j*7)
/*some O(1) operations*/
Also, why is a while loop O(logn) and a do-while loop O(n^2).
At each iteration of the inner loop you perform j = j * 7 (I assume this is what you meant)
That is, at each iteration j = 7j
After n iterations, j = j*7*7*7*7*...*7*7 = j*(7 ^ n)
Let n be the number we want to reach and m the number of iterations, so:
n = j*7*7*7*...7 = j*(7 ^ m)
Let's take a log from both sides:
log(n) = log(j * (7 ^ m)) ~= m*log(7) = O(m)
So, as we can see - the inner loop runs O(log(n)) times.
#include <stdio.h>
int main() {
int N = 8; /* for example */
int sum = 0;
for (int i = 1; i <= N; i++)
for (int j = 1; j <= i*i; j++)
sum++;
printf("Sum = %d\n", sum);
return 0;
}
for each n value (i variable), j values will be n^2. So the complexity will be n . n^2 = n^3. Is that correct?
If problem becomes:
#include <stdio.h>
int main() {
int N = 8; /* for example */
int sum = 0;
for (int i = 1; i <= N; i++)
for (int j = 1; j <= i*i; j++)
for (int k = 1; k <= j*j; k++)
sum++;
printf("Sum = %d\n", sum);
return 0;
}
Then you use existing n^3 . n^2 = n^5 ? Is that correct?
We have i and j < i*i and k < j*j which is x^1 * x^2 * (x^2)^2 = x^3 * x^4 = x^7 by my count.
In particular, since 1 < i < N we have O(N) for the i loop. Since 1 < j <= i^2 <= N^2 we have O(n^2) for the second loop. Extending the logic, we have 1 < k <= j^2 <= (i^2)^2 <= N^4 for the third loop.
Inner to Outer loops, we execute up to N^4 times for each j loop, and up to N^2 times for each i loop, and up to N times over the i loop, making the total be of order N^4 * N^2 * N = N^7 = O(N^7).
I think the complexity is actually O(n^7).
The first loop executes N steps.
The second loop executes N^2 steps.
In the third loop, j*j can reach N^4, so it has O(N^4) complexity.
Overall, N * N^2 * N^4 = O(N^7)
For i = 1 inner loop runs 1^1 times, for i = 2inner loop runs 2^2 times .... and for i = N inner loop runs N^N times. Its complexity is (1^1 + 2^2 + 3^3 + ...... + N^N) of order O(N^3).
In second case, for i = N first inner loop iterates N^N times and hence the second inner loop(inner most) will iterate up to N * (N^N) * (N^N) times. Hence the complexity is of order N * N^2 * N^4, i.e, O(N^7).
Yes. In the first example, the i loop runs N times, and the inner j loop tuns i*i times, which is O(N^2). So the whole thing is O(N^3).
In the second example there is an additional O(N^4) loop (loop to j*j), so it is O(N^5) overall.
For a more formal proof, work out how many times sum++ is executed in terms of N, and look at the highest polynomial order of N. In the first example it will be a(N^3)+b(N^2)+c(N)+d (for some values of a, b, c and d), so the answer is 3.
NB: Edited re example 2 to say it's O(N^4): misread i*i for j*j.
Consider the number of times all loops will be called.
int main() {
int N = 8; /* for example */
int sum = 0;
for (int i = 1; i <= N; i++) /* Called N times */
for (int j = 1; j <= i*i; j++) /* Called N*N times for i=0..N times */
for (int k = 1; k <= j*j; k++) /* Called N^2*N^2 times for j=0..N^2 times and i=0..N times */
sum++;
printf("Sum = %d\n", sum);
return 0;
}
Thus sum++ statement is called O(N^4)*O(N^2)*O(N) times = O(N^7) and this the overall complexity of the program.
The incorrect way to solve this (although common, and often gives the correct answer) is to approximate the average number of iterations of an inner loop with its worst-case. Here, the inner loop loops at worst O(N^4), the middle loop loops at worst O(N^2) times and the outer loop loops O(N) times, giving the (by chance correct) solution of O(N^7) by multiplying these together.
The right way is to work from the inside out, being careful to be explicit about what's being approximated.
The total number of iterations, T, of the increment instruction is the same as your code. Just writing it out:
T = sum(i=1..N)sum(j=1..i^2)sum(k=1..j^2)1.
The innermost sum is just j^2, giving:
T = sum(i=1..N)sum(j=1..i^2)j^2
The sum indexed by j is a sum of squares of consecutive integers. We can calculate that exactly: sum(j=1..n)j^2 is n*(n+1)*(2n+1)/6. Setting n=i^2, we get
T = sum(i=1..N)i^2*(i^2+1)*(2i^2+1)/6
We could continue to compute the exact answer, by using the formula for sums of 6th, 4th and 2nd powers of consecutive integers, but it's a pain, and for complexity we only care about the highest power of i. So we can approximate.
T = sum(i=1..N)(i^6/3 + o(i^5))
We can now use that sum(i=1..N)i^p = Theta(N^{p+1}) to get the final result:
T = Theta(N^7)