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Big O, how do you calculate/approximate it?
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Closed 5 years ago.
I'm studying algorithm's complexity and I'm still not able to determine the complexity of some algorithms ... Ok I'm able to figure out basic O(N) and O(N^2) loops but I'm having some difficult in routines like this one:
// What is time complexity of fun()?
int fun(int n)
{
int count = 0;
for (int i = n; i > 0; i /= 2)
for (int j = 0; j < i; j++)
count += 1;
return count;
}
Ok I know that some guys can calculate this with the eyes closed but I would love to to see a "step" by "step" how to if possible.
My first attempt to solve this would be to "simulate" an input and put the values in some sort of table, like below:
for n = 100
Step i
1 100
2 50
3 25
4 12
5 6
6 3
7 1
Ok at this point I'm assuming that this loop is O(logn), but unfortunately as I said no one solve this problem "step" by "step" so in the end I have no clue at all of what was done ....
In case of the inner loop I can build some sort of table like below:
for n = 100
Step i j
1 100 0..99
2 50 0..49
3 25 0..24
4 12 0..11
5 6 0..5
6 3 0..2
7 1 0..0
I can see that both loops are decreasing and I suppose a formula can be derived based on data above ...
Could someone clarify this problem? (The Answer is O(n))
Another simple way to probably look at it is:
Your outer loop initializes i (can be considered step/iterator) at n and divides i by 2 after every iteration. Hence, it executes the i/2 statement log2(n) times. So, a way to think about it is, your outer loop run log2(n) times. Whenever you divide a number by a base continuously till it reaches 0, you effectively do this division log number of times. Hence, outer loop is O(log-base-2 n)
Your inner loop iterates j (now the iterator or the step) from 0 to i every iteration of outer loop. i takes the maximum value of n, hence the longest run that your inner loop will have will be from 0 to n. Thus, it is O(n).
Now, your program runs like this:
Run 1: i = n, j = 0->n
Run 2: i = n/2, j = 0->n/2
Run 3: i = n/4, j = 0->n/4
.
.
.
Run x: i = n/(2^(x-1)), j = 0->[n/(2^(x-1))]
Now, runnning time always "multiplies" for nested loops, so
O(log-base-2 n)*O(n) gives O(n) for your entire code
Lets break this analysis up into a few steps.
First, start with the inner for loop. It is straightforward to see that this takes exactly i steps.
Next, think about which different values i will assume over the course of the algorithm. To start, consider the case where n is some power of 2. In this case, i starts at n, then n/2, then n/4, etc., until it reaches 1, and finally 0 and terminates. Because the inner loop takes i steps each time, then the total number of steps of fun(n) in this case is exactly n + n/2 + n/4 + ... + 1 = 2n - 1.
Lastly, convince yourself this generalizes to non-powers of 2. Given an input n, find smallest power of 2 greater than n and call it m. Clearly, n < m < 2n, so fun(n) takes less than 2m - 1 steps which is less than 4n - 1. Thus fun(n) is O(n).
Related
1) i=s=1;
while(s<=n)
{
i++;
s=s+i;
}
2) for(int i=1;i<=n;i++)
for(int j=1;j<=n;j+=i)
cout<<"*";
3) j=1;
for(int i=1;i<=n;i++)
for(j=j*i;j<=n;j=j+i)
cout<<"*";
can someone explain me the time complexity of these three codes?
I know the answers but I can't understand how it came
1) To figure this out, we need to figure out how large s is on the x'th iteration of the loop. Then we'll know how many iterations occur until the condition s > n is reached.
On the x'th iteration, the variable i has value x + 1
And the variable s has value equal to the sum of i for all previous values. So, on that iteration, s has value equal to
sum_{y = 1 .. x} (y+1) = O(x^2)
This means that we have s = n on the x = O(\sqrt{n}) iteration. So that's the running time of the loop.
If you aren't sure about why the sum is O(x^2), I gave an answer to another question like this once here and the same technique applies. In this particular case you could also use an identity
sum_{y = 1 .. x} y = y choose 2 = (y+1)(y) / 2
This identity can be easily proved by induction on y.
2) Try to analyze how long the inner loop runs, as a function of i and n. Since we start at one, end at n, and count up by i, it runs n/i times. So the total time for the outer loop is
sum_{i = 1 .. n} n/i = n * sum_{i = 1 .. n} 1 / i = O(n log n)
The series sum_{i = 1 .. n} 1 / i is called the harmonic series. It is well-known that it converges to O(log n). I can't enclose here a simple proof. It can be proved using calculus though. This is a series you just have to know. If you want to see a simple proof, you can look on on wikipedia at the "comparison test". The proof there only shows the series is >= log n, but the same technique can be used to show it is <= O(log n) also.
3.) This looks like kind of a trick question. The inner loop is going to run once, but once it exits with j = n + 1, we can never reenter this loop, because no later line that runs will make j <= n again. We will run j = j * i many times, where i is a positive number. So j is going to end up at least as large as n!. For any significant value of n, this is going to cause an overflow. Ignoring that possibility, the code is going to perform O(n) operations in total.
x=1;
While(x<n)
{
x=x + n/100;
}
I'm trying to figure out if it's o(n) or o(1). Because no matter what we put in n's place I think the loop will go just 10 times.
lets say n=1.1
then it will go for 10 times and if n=1.2 loop will go on for 17 times
and if n=2 it will go on for 50 times and when n>=101 loop will be repeated 100 times even if n=10^10000 else you can figure out
Unfortunately you're wrong it it being O(n) or O(1) and this is immediately clear by the fact that it can't be O(1), because it takes different numbers of iterations for varying values of n(even looking at n = 1,2,3,4,5), and it can't be O(n) because it doesn't grow linearly.
Even through a bit of manual calculation you can see clearly that it won't always run 10 times. Examine the following short python program:
def t(n):
x = 1
c = 0
while x < n:
c += 1
x += n/100
return c
a = []
for i in range(10000):
a += [i/100 + 1]
with open("out.csv","w") as f:
for i in a:
f.write(str(i) + "," + str(t(i)) + "\n")
Using Excel or some other application you can easily trend the number of iterations taken to see the following curve:
It is immediately clear at this point that the number of iterations taken is logarithmic in the range {0:100} with any n < 1 taking 0 iterations and n > 100 taking 100 operations. So while Big-O notation wasn't my best subject, I would guess that the time complexity is thus O(log(n)).
This is the code I need to analyse:
i = 1
while i < n
do
j = 0;
while j <= i
do
j = j + 1
i = 2i
So, the first loop should run log(2,n) and the innermost loop should run log(2,n) * (i + 1), but I'm pretty sure that's wrong.
How do I use a theta notation to prove it?
An intuitive way to think about this is to see how much work your inner loop is doing for a fixed value of outer loop variable i. It's clearly as much as i itself. Thus, if the value of i is 256, then then you will do j = j + 1 that many times.
Thus, total work done is the sum of the values that i takes in the outer loop's execution. That variable is increasing much rapidly to catch up with n. Its values, as given by i = 2i (it should be i = 2*i), are going to be like: 2, 4, 8, 16, ..., because we start with 2 iterations of the inner loop when i = 1. This is a geometric series: a, ar, ar^2 ... with a = 1 and r = 2. The last term, as you figured out will be n and there will be log2 n terms in the series. And that is simple summation of a geometric series.
It doesn't make much sense to have a worst case or a best case for this algorithm because there are no different permutations of the input which is just a number n in this case. Best case or worst case are relevant when a particular input (e.g. a particular sequence of numbers) affects the running time of the algorithm.
The running time then is the sum of geometric series (a.(r^num_terms - 1)/(r-1)):
T(n) = 2 + 4 + ... 2^(log2 n)
= 2 . (2^log2 n - 1)
= 2 . (n - 1)
⩽ 3n = O(n)
Thus, you can't be doing work that is more than some constant multiple of n. Hence, the running time of this algorithm is O(n).
You can't be doing some work that is less than some (other) constant multiple of n, since you have to go through the increment in inner loop as shown above. Thus, the running time of this algorithm is also ≥ c.n i.e. it is Ω(n).
Together, this means that running time of this algorithm is Θ(n).
You can't use i in your final expression; only n.
You can easily see that the inner loop executes i times each time it is reached. And it sounds like you've figured out the different values that i can have. So add up those values, and you have the total amount of work.
Hello I am having trouble finding the running time for this algorithm with following assumptions that s = O(logN) and the running time of Random_Integer is constant time:
1 express N-1 as 2^t * u where u is odd:
2 for i <-- to s do
3 a <-- Random_Integer(2, N-2);
4 if EuclidGCD(a, N) not equal to 1 then
5 return false;
6 x sub 0 <-- a^u mod N;
7 for j <-- 1 to t do
8 x sub j <-- x^2 sub j-1 mod N;
9 if x sub j = 1 and x sub j-1 not equal to 1 and x sub j-1 not equal to N -1 then
10 return false;
11 if x sub t not equal to one then
12 return false;
13 return true;
Starting from the inner loop the exponential modulous operation takes n^3 time and the loop runs for n iterations giving a total of n^4. Then working my way to the outter loop, we have another exponential modulous operation which takes again n^3 time and then the EuclidGCD also take n^3 time. Finally the outter loop also runs for n iterations. I believe that these values are correct but, im confused as to how I get a total running time. I'm also confused on if these two nested loops running time should be multiplied together and if the method call for ExtendedEuclid within the outer loop should be multiplied with the running time for the outter loop. I hope this is clear thanks for any help.
The inner loop is n^4 (the slowest part inside the outer loop) and runs once for every iteration of the outer loop, which is EDIT: logn times, so n^4logn.
HOWEVER...
Depending on how often return false is reached early, it may only be n^5 in the worst case, e.g. if almost all of the time you return false on the first iteration then you've only spent n^3-n^4 of work, so you'd have an average of O(n^3) or O(n^4) (depending on which return false it was) instead.
So I was looking at this code from a textbook:
for (int i=0; i<N; i++)
for(int j=i+1; j<N; j++)
The author stated that the inner for-loop iterates for exactly N*(N-1)/2 times but gives no basis for how he arrived to such an equation. I understand N*(N-1) but why divide by 2? I ran the code myself and sure enough when N is 10, the inner loop iterates 45 times (10*9/2).
I messed around with the code myself and tried the following (assigned only i to j):
for (int i=0; i<N; i++)
for(int j=i; j<N; j++)
With N = 10, this results in 55. So I'm having trouble understanding the underlying math here. Sure I could just plug in all the values and bruteforce my way through the problem, but I feel there is something essential and very simple I'm missing. How would you come up with an equation for describing the for loop I just constructed? Is there a way to do it without relying on the outputs? Would really appreciate any help thanks!
Think about what happens each time the outer loop iterates. The first time, i == 0, so the inner loop starts at 1 and runs to N-1, which is N-1 iterations in total. The next time through the outer loop, i has incremented to 1, so the inner loop starts at 2 and runs up to N-1, for a total of N-2 iterations. And that pattern continues: the third time through the outer loop, you get N-3 iterations, the fourth time through, N-4, etc. When you get to the last iteration of the outer loop, i == N-1, so the inner loop starts with j = N and stops immediately. So that's zero iterations.
The total number of iterations is the sum of all these numbers:
(N-1) + (N-2) + (N-3) + ... + 1 + 0
To look at it another way, this is just the sum of the positive integers from 1 to N-1. The result of this sum is called the (N-1)th triangular number, and Wikipedia explains how you can find that the formula for the n'th triangular number is n(n+1)/2. But here you have the (N-1)th triangular number, so if you set n=N-1, you get
(N-1)(N-1+1)/2 = N(N-1)/2
You're looking at nested loops where the outer one runs N times and the inner one (N-1). You're in effect adding up the sum of 1 + 2 + 3 + ....
The N * (N+1) / 2 is a "classic" formula in mathematics. Young Carl Gauss, later a famous mathematician, was given in-class busywork: Adding up the numbers from 1 to 100. The teacher expected to keep the kids busy for an hour but Carl came up with the answer almost immediately: 5050. He explained: 1 + 100; 2 + 99; 3 + 98; 4 + 97; and so on up to 50 + 51. That's 50 sums of 101 each. You could also see that as (100 / 2) * (100 + 1); that's where the /2 comes from.
As for why it's (N-1) instead of the (N+1) I mentioned... that could have to do with starting from 1 rather than 0, that would drop one iteration from the inner loop, I think.
Look at how many times the inner (j) loop runs for each value of i. When N = 10, the outer (i) loop runs 10 times, and the j loop should run 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 times. Now you just add up those numbers to see how many times the inner loop runs. You can sum the numbers from 0 to N-1 with the formula N(N-1)/2. This is a very slight modification of a well-known formula for adding the numbers from 1 to N.
For a visual aid, you can see why 1 + 2 + 3 + ... + n = n * (n+1) / 2
If you count the iterations of the inner loop, you get:
1 2 3 4 5 6 7 8 9 10
To get the total for an arbitrary number of iterations, you can "wrap" the numbers around like this:
0 1 2 3 4
9 8 7 6 5
Now, if we add each of those columns, the all add to 9 (N-1), and there are 5 (N/2) columns. It's pretty obvious that for any even N, we'd still get N/2 columns that each added up to (N-1). As such, when the total number of iterations is even, the total number of iterations is always (N/2)(N-1), which (thanks to the commutative property) we can rewrite as N(N-1)/2.
If we did the same for an odd number of iterations, we'd have one "odd" column that couldn't be paired. In this case, we can ignore the '0' since we know it won't affect the overall sum in any case. For example, let's consider N=9 instead of N=10. For that, we get:
1 2 3 4
8 7 6 5
This gives us (N-1)/2 columns (9-1=8, 8/2=4) that each add up to N, so the sum will be N*(N-1)/2. Even though we've arrived at it slightly differently, this is an exact match for the formula above for when N is even. Again, it seems pretty obvious that this would remain true regardless of the number of columns we used (i.e., total number of iterations).
For any N (odd or even), the sum of the numbers from 0 through N-1 is N*(N-1)/2.