I am lost on these code fragments and finding a hard time to find any other similar examples.
//Code fragment 1
sum = 0;
for(i = 0;i < n; i++)
for(J=1;j<i*i;J++)
for(K=0;k<j;k++)
sum++;
I'm guessing it is O(n^4) for fragment 1.
//Code fragment 2
sum = 0;
for( 1; i < n; i++ )
for (j =1;,j < i * i; j++)
if( j % i == 0 )
for( k = 0; k < j; k++)
sum++;
I am very lost on this one. Not sure how does the if statement affects the loop.
Thank you for the help head of time!
The first one is in fact O(n^5). The sum++ line is executed 1^4 times, then 2^4 times, then 3^4, and so on. The sum of powers-of-k has a term in n^(k+1) (see e.g. Faulhaber's formula), so in this case n^5.
For the second one, the way to think about it is that the inner loop only executes when j is a multiple of i. So the second loop may as well be written for (j = 1; j < i * i; j+=i). But this is the same as for (j = 1; j < i; j++). So we now have a sequence of cubes, rather than powers-of-4. The highest term is therefore n^4.
I'm fairly sure the 1st fragment is actually O(n^5).
Because:
n times,
i^2 times, where i is actually half of n (average i for the case, since for each x there is a corresponding n-x that sum to 2n) Which is therefore n^2 / 4 times. (a times)
Then, a times again,
and when you do: n*a*a, or n*n*n/4*n*n/4 = n^5 / 16, or O(n^5)
I believe the second is O(4), because:
It's iterated n times.
Then it's iterated n*n times, (literally n*n/4, but not in O notation)
Then only 1/n are let through by the if (I can't remember how I got this)
Then n*n are repeated.
So, n*n*n*n*n/n = n^4.
With a sum so handy to compute, you could run these for n=10, n=50, and so on, and just look which of O(N^2), O(N^3), O(N^4), O(N^6) is a better match. (Note that the index for the inner-most loop also runs to n*n...)
First off I agree with your assumption for the first scenario. Here is my breakdown for the second.
The If statement will cause the last loop to run only half of the time since an odd value for i*i will only lead to the third inner loop for prime values that i*i can be decomposed into. Bottom line in big-O we ignore constants so I think your looking at O(n^3).
Related
function(n):
{
for (i = 1; i <= n; i++)
{
for (j = 1; j <= n / 2; j++)
output("")
}
}
Now I have calculated the time complexity for the first for loop which is O(n). Now the second for loop shows j <= n / 2 so any given n I put, for example, a range of [1,2,...,10] will give me O(log(n)) since it will continuously give me a series of n,n/2,n/4,n/8 .... K.
So if we wanted to compare the relationship it must look something like this 2^n = k.
My question is will it give me O(log(n))?
The correct summation according to the code is:
So, it's not O(log n). It's O(n^2).
No, it does not give you o(logn).
The first for loop is O(n). The second loop is O(n) as well, as the number of iterations grows as a function of n (the growth rate is linear).
It would be the same even by changing the second loop to something like
for (j=1; j<=n/2000; j++)
or in general if you replace the denominator with any constant k.
To conclude, the time compexity is quadratic, i.e., O(n^2)
What is the running time complexity of fun()?
int fun(int n)
{
int count = 0;
for (int i = n; i > 0; i =i-2)
for (int j = 2; j < i; j=j*j)
for (int k=j; k>0; k=k/2)
count += 1;
return count;
}
is it O(n * lglgn * lglglgn)?
----
Edit:
j = loglog(i) times, big iterative value of j can be almost n(for example n=17,max(j)=16)
k= log(j), since the max value of j is at most n. then the max iterative times can be log(n)
so we can say that the big O of this question is O(n* lglgn * lgn)
Since the value of j and k depends on the previous iterative value (i, j), maybe there is better tight answer to this question.
We need to count this carefully, because the number of iterations of each inner loop depends in a non-trivial way on the outer loop's variable. Simply averaging for each loop and multiplying the results together will not give the right answer.
The outer loop runs O(n) times, because i counts from n down to 0 in constant steps.
The middle loop's values of j are 2, then 2*2 = 4, then 4*4 = 16, and on the m'th iteration, j = 2^2^m. The last iteration will be when 2^2^m >= i, in which case m >= log log i. So this runs O(log log i) times.
The innermost loop runs O(log j) times, because on the m'th iteration, k = j / 2^m. The last iteration will be when k <= 1, in which case m >= log j. So this runs O(log j) times.
However, it is not correct to multiply these together to get O(n * log log n * log log log n) - because i is not n on every iteration, and j is not log log n on every iteration. This gives an upper bound, but not a tight one. To calculate the true time complexity, you will need to write it as a double-summation, and simplify it algebraically.
As a simpler example to think about, consider the following code:
for(i = 1; i < n; i *= 2) {
for(j = 0; j < i; j += 1) {
// do something
}
}
The outer loop runs O(log n) times, and the inner loop runs O(i) times, but the overall complexity is actually O(n). To see this, count how many times // do something is reached; the first time the outer loop iterates it'll be 1, then it'll be 2, then 4, then 8, and so on up to n. This is a geometric progression with a sum <= 2n, giving a total number of steps which is O(n).
Note that if we naively multiply the two loops' complexities we get O(n log n) instead, which is an upper bound, but not a tight one.
Using Big O notation:
With the first outer loop we get O(N/2). You have a loop of N items and you are reducing it by 2 everytime, though you get a total of N/2 loops.
With the outter loop we get O(Log(I)).
With the most inner loop we have O(Log(J)), because you are dividing by 2 your iterator on every loop.
If we multiply the three complexities because they are nested:
O(N/2)*O(Log(I)) + O(Log(j)) ~ O(N/2*Log(I)*Log(J)) ~ O(N/2*Log^2(N)) ~ O(N*Log^2(N)).
We get a linearithmic complexity: O(N*Log^2(N))
I need to calculate the time complexity of the following loop:
for (i = 1; i < n; i++)
{
statements;
}
Assuming n = 10,
Is i < n; control statement going to run for n time and i++; statement for n-1 times? And knowing that i = 1; statement is going to run for a unit of time.
Calculating the total time complexity of the three statements in for-loop yields 1+n+n-1 = 2n and the loop with its statements yields 2n+n-1 = 3n-1 = O(n).
Are my calculations correct to this point?
Yes, your calculations are correct, a for loop like such would have O(n) notation.
Similarly, you could make a calculation like such:
for(int i = 0; i <n*2; i++){
//calculations
}
in this case, the for loop would have a big O notation of O(n^2) (you get the idea)
This loop takes O(n^2) time; math function = n^n This way you can calculate how long your loop need for n 10 or 100 or 1000
This way you can build graphs for loops and such.
as DAle mentioned in the comments the big O notation is not affected by calculations within the loop, only the loop itself.
So these are the for loops that I have to find the time complexity, but I am not really clearly understood how to calculate.
for (int i = n; i > 1; i /= 3) {
for (int j = 0; j < n; j += 2) {
... ...
}
for (int k = 2; k < n; k = (k * k) {
...
}
For the first line, (int i = n; i > 1; i /= 3), keeps diving i by 3 and if i is less than 1 then the loop stops there, right?
But what is the time complexity of that? I think it is n, but I am not really sure.
The reason why I am thinking it is n is, if I assume that n is 30 then i will be like 30, 10, 3, 1 then the loop stops. It runs n times, doesn't it?
And for the last for loop, I think its time complexity is also n because what it does is
k starts as 2 and keeps multiplying itself to itself until k is greater than n.
So if n is 20, k will be like 2, 4, 16 then stop. It runs n times too.
I don't really think I am understanding this kind of questions because time complexity can be log(n) or n^2 or etc but all I see is n.
I don't really know when it comes to log or square. Or anything else.
Every for loop runs n times, I think. How can log or square be involved?
Can anyone help me understanding this? Please.
Since all three loops are independent of each other, we can analyse them separately and multiply the results at the end.
1. i loop
A classic logarithmic loop. There are countless examples on SO, this being a similar one. Using the result given on that page and replacing the division constant:
The exact number of times that this loop will execute is ceil(log3(n)).
2. j loop
As you correctly figured, this runs O(n / 2) times;
The exact number is floor(n / 2).
3. k loop
Another classic known result - the log-log loop. The code just happens to be an exact replicate of this SO post;
The exact number is ceil(log2(log2(n)))
Combining the above steps, the total time complexity is given by
Note that the j-loop overshadows the k-loop.
Numerical tests for confirmation
JavaScript code:
T = function(n) {
var m = 0;
for (var i = n; i > 1; i /= 3) {
for (var j = 0; j < n; j += 2)
m++;
for (var k = 2; k < n; k = k * k)
m++;
}
return m;
}
M = function(n) {
return ceil(log(n)/log(3)) * (floor(n/2) + ceil(log2(log2(n))));
}
M(n) is what the math predicts that T(n) will exactly be (the number of inner loop executions):
n T(n) M(n)
-----------------------
100000 550055 550055
105000 577555 577555
110000 605055 605055
115000 632555 632555
120000 660055 660055
125000 687555 687555
130000 715055 715055
135000 742555 742555
140000 770055 770055
145000 797555 797555
150000 825055 825055
M(n) matches T(n) perfectly as expected. A plot of T(n) against n log n (the predicted time complexity):
I'd say that is a convincing straight line.
tl;dr; I describe a couple of examples first, I analyze the complexity of the stated problem of OP at the bottom of this post
In short, the big O notation tells you something about how a program is going to perform if you scale the input.
Imagine a program (P0) that counts to 100. No matter how often you run the program, it's going to count to 100 as fast each time (give or take). Obviously right?
Now imagine a program (P1) that counts to a number that is variable, i.e. it takes a number as an input to which it counts. We call this variable n. Now each time P1 runs, the performance of P1 is dependent on the size of n. If we make n a 100, P1 will run very quickly. If we make n equal to a googleplex, it's going to take a little longer.
Basically, the performance of P1 is dependent on how big n is, and this is what we mean when we say that P1 has time-complexity O(n).
Now imagine a program (P2) where we count to the square root of n, rather than to itself. Clearly the performance of P2 is going to be worse than P1, because the number to which they count differs immensely (especially for larger n's (= scaling)). You'll know by intuition that P2's time-complexity is equal to O(n^2) if P1's complexity is equal to O(n).
Now consider a program (P3) that looks like this:
var length= input.length;
for(var i = 0; i < length; i++) {
for (var j = 0; j < length; j++) {
Console.WriteLine($"Product is {input[i] * input[j]}");
}
}
There's no n to be found here, but as you might realise, this program still depends on an input called input here. Simply because the program depends on some kind of input, we declare this input as n if we talk about time-complexity. If a program takes multiple inputs, we simply call those different names so that a time-complexity could be expressed as O(n * n2 + m * n3) where this hypothetical program would take 4 inputs.
For P3, we can discover it's time-complexity by first analyzing the number of different inputs, and then by analyzing in what way it's performance depends on the input.
P3 has 3 variables that it's using, called length, i and j. The first line of code does a simple assignment, which' performance is not dependent on any input, meaning the time-complexity of that line of code is equal to O(1) meaning constant time.
The second line of code is a for loop, implying we're going to do something that might depend on the length of something. And indeed we can tell that this first for loop (and everything in it) will be executed length times. If we increase the size of length, this line of code will do linearly more, thus this line of code's time complexity is O(length) (called linear time).
The next line of code will take O(length) time again, following the same logic as before, however since we are executing this every time execute the for loop around it, the time complexity will be multiplied by it: which results in O(length) * O(length) = O(length^2).
The insides of the second for loop do not depend on the size of the input (even though the input is necessary) because indexing on the input (for arrays!!) will not become slower if we increase the size of the input. This means that the insides will be constant time = O(1). Since this runs in side of the other for loop, we again have to multiply it to obtain the total time complexity of the nested lines of code: `outside for-loops * current block of code = O(length^2) * O(1) = O(length^2).
The total time-complexity of the program is just the sum of everything we've calculated: O(1) + O(length^2) = O(length^2) = O(n^2). The first line of code was O(1) and the for loops were analyzed to be O(length^2). You will notice 2 things:
We rename length to n: We do this because we express
time-complexity based on generic parameters and not on the ones that
happen to live within the program.
We removed O(1) from the equation. We do this because we're only
interested in the biggest terms (= fastest growing). Since O(n^2)
is way 'bigger' than O(1), the time-complexity is defined equal to
it (this only works like that for terms (e.g. split by +), not for
factors (e.g. split by *).
OP's problem
Now we can consider your program (P4) which is a little trickier because the variables within the program are defined a little cloudier than the ones in my examples.
for (int i = n; i > 1; i /= 3) {
for (int j = 0; j < n; j += 2) {
... ...
}
for (int k = 2; k < n; k = (k * k) {
...
}
}
If we analyze we can say this:
The first line of code is executed O(cbrt(3)) times where cbrt is the cubic root of it's input. Since i is divided by 3 every loop, the cubic root of n is the number of times the loop needs to be executed before i is smaller or equal to 1.
The second for loop is linear in time because j is executed
O(n / 2) times because it is increased by 2 rather than 1 which
would be 'normal'. Since we know that O(n/2) = O(n), we can say
that this for loop is executed O(cbrt(3)) * O(n) = O(n * cbrt(n)) times (first for * the nested for).
The third for is also nested in the first for, but since it is not nested in the second for, we're not going to multiply it by the second one (obviously because it is only executed each time the first for is executed). Here, k is bound by n, however since it is increased by a factor of itself each time, we cannot say it is linear, i.e. it's increase is defined by a variable rather than by a constant. Since we increase k by a factor of itself (we square it), it will reach n in 2log(n) steps. Deducing this is easy if you understand how log works, if you don't get this you need to understand that first. In any case, since we analyze that this for loop will be run O(2log(n)) time, the total complexity of the third for is O(cbrt(3)) * O(2log(n)) = O(cbrt(n) *2log(n))
The total time-complexity of the program is now calculated by the sum of the different sub-timecomplexities: O(n * cbrt(n)) + O(cbrt(n) *2log(n))
As we saw before, we only care about the fastest growing term if we talk about big O notation, so we say that the time-complexity of your program is equal to O(n * cbrt(n)).
I'm practising algorithm complexity and I came across this code online but I cannot figure out the order of growth for it. Any ideas?
int counter= 0;
for (int i = 0; i*i < N; i++)
for (int j = 0; j*j < 4*N; j++)
for (int k = 0; k < N*N; k++)
counter++;
Take it one step (or loop in this case) at a time:
The first loop increments i as long its square is lower than N, so this must be O(sqrt N), because int(sqrt(N)) or int(sqrt(N)) - 1 is the largest integer value whose square is lower than N;
The same holds for the second loop. We can ignore the 4 because it is a constant, and we do not care about those when dealing with big-oh notation. So the first two loops together are O(sqrt N)*O(sqrt N) = O(sqrt(N)^2) = O(N). You can multiply the complexities because the loops are nested, so the second loop will fully execute for each iteration of the first;
The third loop is obviously O(N^2), because k goes up to the square of N.
So the whole thing has to be O(N) * O(N^2) = O(N^3). You can usually solve problems like this by figuring out the complexity of the first loop, then the second, then the first two and so on.
Sqrt n x 2 Sqrt n x n ^ 2
Which gives:
O n^3
Explanation:
For the first loop, square root both sides of the equation
i^2 = n
For the second loop, square root both sides of the equation
j^2 = 4n^2
The third loop is straight forward.