The question is to set up a recurrence relation to find the value given by the algorithm. The answer should be in teta() terms.
foo = 0;
for int i=1 to n do
for j=ceiling(sqrt(i)) to n do
for k=1 to ceiling(log(i+j)) do
foo++
Not entirely sure but here goes.
Second loop executes 1 - sqrt(1) + 2 - sqrt(2) + ... + n - sqrt(n) = n(n+1)/2 - n^1.5 times => O(n^2) times. See here for a discussion that sqrt(1) + ... + sqrt(n) = O(n^1.5).
We've established that the third loop will get fired O(n^2) times. So the algorithm is asymptotically equivalent to something like this:
for i = 1 to n do
for j = 1 to n do
for k = 1 to log(i+j) do
++foo
This leads to the sum log(1+1) + log(1+2) + ... + log(1+n) + ... + log(n+n). log(1+1) + log(1+2) + ... + log(1+n) = log(2*3*...*(n+1)) = O(n log n). This gets multiplied by n, resulting in O(n^2 log n).
So your algorithm is also O(n^2 log n), and also Theta(n^2 log n) if I'm not mistaken.
Related
I have a algorithm, the pseudo code below:
def foo(n):
if n == 0
return;
// Loop below take O(N)
for(i=0; i<n:i++){
....
}
foo(n-1):
The idea is that each recursion takes n time, and there are n recursions.
The total time should be like 1 + 2 3 + 4 +5 + ... +n
Can it be proved as O(n*n)?
Yes, it is O(n^2).
The sum of n natural numbers is: n * (n+1) / 2, link. Which is different to n^2 by a constant factor, so O(n * (n+1) / 2) == O(n^2)
First, you have n iterations in the for loop, then the function will repeat with n-1, n-2, ..., 0.
It's easy to see that n + (n-1) + (n-2) + ... + 1 = (n+1) * n/2 = (n^2 + n)/2 = O(n^2).
To evaluate Big O, that is, the complexity of the worst case, remember you have to ignore the all the coeficients, constants and lower power terms:
(n^2 + n)/2 = (1/2) * (n^2 + n)
O( (1/2) * (n^2 + n) ) = O(n^2 + n) = O(n^2)
Computing complexity and Big o of an algorithm
T(n) = 5n log n + 3log n + 2 // to the base 2 big o = o(n log n)
for(int x = 0,i = 1;i <= N;i*=2)
{
for(int j = 1 ;j <= i ;j++)
{
x++;
}
}
The Big o expected was linear where as mine is logarithmic
Your Big-Oh analysis is not correct. While it is true that the outer loop is executed log n times, the inner loop is linear in i at each iteration.
If you count the total number of iterations of the inner loop, you will see that the whole thing is linear:
The inner loop will do 1 + 2 + 4 + 8 + 16 + ... + (the last power of 2 <= N) iterations. This sum will be between N and 2*N, which makes the whole loop linear.
Let me explain why your analysis is wrong.
It is clear that inner loop will execute 1 + 2 + 4 + ... + 2^k times where k is the biggest integer which satisfies equation . This implies that upper bound for k is
Without loss of generality we can take upper bound for k and assume that k is integer, complexity equals to 1 + 2 + 4 + ... + = which is geometric series so it is equal to
Therefore in O notation it is O(n)
First, you should notice that your analysis is not logarithmic! As N \log N is not logarithmic.
Also, the time complexity is T(n) = sum_{j = 0}^{log(n)} 2^j (as the value of i duplicated each time). Hence, T(n) = 2^(log(N) + 1) - 1 = 2N - 1 = \Theta(N).
This is pseudocode. I tried to calculate the time complexity of this function as this answer said. It should be like:
n + n/3 + n/9 + ...
Maybe the time complexity is something like O(nlog(n)) I guess? Or the log(n) should be log(n) base 3? Someone said the time complexity is O(n), which is totally unacceptable for me.
j = n
while j >= 1 {
for i = 1 to j {
x += 1
}
j /= 3
}
The algorithm will run in:
n + n/3 + n/9 + ... = series ~= O(3/2 * n) = O(n)
since 3/2 is a constant. Here the k-th loop will run in n/3k steps.
Please notice the crucial difference from the linked question, where the outer loop runs n times and that is fixed.
What is the time complexity for the following function?
for(int i = 0; i < a.size; i++) {
for(int j = i; j < a.size; i++) {
//
}
}
I think it is less than big O n^2 because we arent iterating over all of the elements in the second for loop. I believe the time complexity comes out to be something like this:
n[ (n) + (n-1) + (n-2) + ... + (n-n) ]
But when I solve this formula it comes out to be
n^2 - n + n^2 - 2n + n^2 - 3n + ... + n^2 - n^2
Which doesn't seem correct at all. Can somebody tell me exactly how to solve this problem, and where I am wrong.
That is O(n^2). If you consider the iteration where i = a.size() - 1, and you work your way backwards (i = a.size() - 2, i = a.size - 3, etc), you are looking at the following sum of number of iterations, where n = a.size.
1 + 2 + 3 + 4 + ... + n
The sum of this series is n(n+1)/2, which is O(n^2). Note that big-O notation ignores constants and takes the highest polynomial power when it is applied to a polynomial function.
It will run for:
1 + 2 + 3 + .. + n
Which is 1/2 n(n+1) which give us O(n^2)
The Big-O notation will only keep the dominant term, neglecting constants too
The Big-O is only used to compare algorithms on the same variation of a problem using the same complexity analysis standard, if and only if the dominant terms are different.
If the dominant terms are the same, you need to compare Big-Theta or Time complexity, which will show minor differences.
Example
A
for i = 1 .. n
for j = i .. n
..
B
for i = 1 .. n
for j = 1 .. n
..
We have
Time(A) = 1/2 n(n+1) ~ O(n^2)
Time(B) = n^2 ~ O(n^2)
O(A) = O(B)
T(A) < T(B)
Analysis
To visualize how we got 1 + 2 + 3 + .. n:
for i = 1 .. n:
print "(1 + "
sum = 0
for j = i .. n:
sum++
print sum") + "
will print the following:
(1+n) + (1+(n-1)) + .. + (1+3) + (1+2) + (1+1) + (1+0)
n+1 + n + n-1 + .. + 3 + 2 + 1
1 + 2 + 3 + .. + n + n+1
1/2 n(n+1) + (n+1)
1/2 n^2 + 1/2 n + n + 1
1/2 n^2 + 3/2 n + 1
Yes, the number of iterations is strictly less than n^2, but it's still Θ(n^2). It will eventually be greater than n^k for any k<2, and it will eventually be less than n^k for any k>2.
(As a side note, computer scientists often say big-O when they really mean big-theta (Θ). It's technically correct to say that almost every algorithm you've seen has O(n!) running time; all reasonably algorithms have running times that grow no more quickly than n!. But it's not really useful to say that the complexity is O(n!) if it's also O(n log n), so by some kind of Gricean maxim we assume that when someone says an algorithm's complexiy is O(f(x)) that f(x) is as small as possible.)
I'm wondering what is the big O notation for each of the below statement:
Sum = 0;
for i=1 to N^2 do:
for j=1 to N do:
'Sum += 1;
Sum = 0 is O(1) for sure, because it will only be executed once.
But I'm confused by the second statement, should it be O(N) because it's the first loop? or it should be O(N^2) because N^2 is a quadratic function about variable N?
The first loop is O(N2) because it executes N2 steps. Each of those steps executes the inner loop, which involves N steps, so there are N2 * N or N3 steps, and the algorithm is O(N3).
You'll be looping through N three rounds..so i say: O(n^3)
Algorithm:
Sum = 0; ~ 1
for i=1 to N^2 do: ~ 1+2N^2
for j=1 to N do: ~ (1+2N) * N^2
'Sum += 1; ~ 1 * N * N^2
Time Complexity:
Time = 1 + 1+2N^2 + (1+2N)*N^2 + 1 * N * N^2
Time = 2 + 2N^2 + N^2 + 2N^3 + N^3
Time = 2 + 3N^2 + 3N^3
O(Time) = O(2 + 3N^2 + 3N^3) ~ O(N^3)