What would be the worst time complexity big O notation for the following pseudocode? (assuming the function call is an O(1)) I'm very new to big O notation so I'm unsure of an answer but I was thinking O(log(n)) because the while loop parameters multiplied by 2 each time or would that just be O(loglog(n))? Or am I wrong on both counts? Any input/help is appreciated, I'm trying to grasp the concept of big O notation for worst time complexity which I just started learning. Thanks!
i ← 1
while(i<n)
doSomething(...)
i ← i * 2
done
If i is doubling every time, then the number of times the loop will execute is the number of times you can double i before reaching n. Or to write it mathematically, if x is the number of times the loop will execute we have 2^x <= n. Solving for x gives x <= log_2(n). Therefore the number of times the loop will execute is O(log(n))
i is growing exponentially, thus loop will be done in logarithmic time, O(log(n))
O(log(n)) is correct when you want to state the time complexity of that algorithm in terms of the number n. However in computer science complexity is often stated in the size of the input, i.e. the number of bits. Then your algorithm would be linear, i.e. in O(k) where k is the input size.
Typically, other operations like addition are also said to be linear not logarithmic. A logarithmic complexity usually means that an algorithm does not have to consider the complete input. (E.g. binary search).
If this is part of an exercise or you want to discuss complexity of algorithms in a computer science context this difference is important.
Also, if one would want to be really pedantic: comparison on large integers is not a constant time operation, and if you are considering the usual integer types, the algorithm is basically constant time as it only needs up to 32 or 64 iterations.
Related
i have a question about O-notation. (big O)
In my code, i am using a for loop to iterate through an array of users.
The for loop has if-statements that makes it break out of the loop, if the rigth user is found.
My question is how i measure the O-notation?
Is the O-notation is O(N) as i loop through all the users in the array?
Or is the O-notation O(1), as the loop breaks and never runs again?
O notation defines an "order of" relationship between an amount of work (however measured) and the number of items processed (usually 'n'). So "O(n)" means "in direct proportion to the number of items n". "O(1)" means simply "constant". If a loop processes every item once then the amount of work is intuitively in direct proportion to n, but let's say that your exit condition gets hit on average half way through, we might be tempted to say that this is O(n/2), but instead we still say that it is O(n) because the relationship to n is still direct/linear. Similarly if you were to assess the relationship to be O(7n^3 + 2n), you'd say the relationship was simply O(n^3) because n^3 is the term that dominates as n grows large.
The answer to your specific question is therefore O(n) because the number of iterations is in direct proportion to n. All that this says is that if N user records take M milliseconds to process, 2N should take about 2M milliseconds.
It is probably worth noting that O notation is strictly concerned with worst case and not the average cost of algorithms (although I have started to find that it is quite common for people to use it in the latter sense). It is always a good idea to specify to avoid ambiguity.
Big O notation answers the following two questions:
If there are N data elements, how many steps will the algorithm take?
How will the performance of the algorithm change if the number of data elements increases?
Best-case scenario in your case is that the user you are searching for is found at the first index. Time complexity in this case would be O(1) because number of steps taken by the algorithm are constant and do not change if the number of elements in the array are changed.
The worst-case scenario is that your loop will have to iterate over all the users. That makes the time complexity to be O(N) because number of steps taken by the algorithm will be directly proportional to the number of elements in the array.
Big O notation generally refers to the worst-case scenario, so you can say that the time complexity in your case is O(N).
Best case complexity of for loop is O(1) and worst case complexity is O(N). In linear search best case is O(N) and worst case is O(N). It also depends on the approach followed by you to solve problem. Like for(int i = n; i>1; i=i/2) in this case complexity is O(log(N). Complexity of if else condition is O(1).
I am having trouble fully understanding this question:
Two O(n2) algorithms will always take the same amount of time for a given vale of n. True or false? Explain.
I think the answer is false, because from my understanding, I think that the asymptotic time complexity only measures the two algorithms running at O(n2) time, however one algorithm might take longer as perhaps it might have additional O(n) components to the algorithm. Like O(n2) vs (O(n2) + O(n)).
I am not sure if my logic is correct. Any help would be appreciated.
Yes, you are right. Big Oh notation depicts the upper bound of time complexity. There might some extra constant term c or smaller term of n like O(n) added to it which won't be considered for time complexity.
Moreover,
for i = 0 to n
for j = 0 to n
// some constant time operation
end
end
And
for i = 0 to n
for j = i to n
// some constant time operation
end
end
Both of these are O(n^2) asymptotically but won't take same time.
The concept of big Oh analysis is not to calculate the precise amount of time a program takes to execute, it's not about counting how many times a loop iterates. Rather it indicates the algorithm's growth rate with n.
The answer is correct but the explanation is lacking.
For one, the big O notation allows arbitrary constant factors. so both n2 and 100*n2 are in O(n2) but clearly the second is always larger.
Another reason is that the notation only gives an upper bound so even a runtime of n is in O(n2) so one of the algorithms may in fact be linear.
Question
Hi I am trying to understand what order of complexity in terms of Big O notation is. I have read many articles and am yet to find anything explaining exactly 'order of complexity', even on the useful descriptions of Big O on here.
What I already understand about big O
The part which I already understand. about Big O notation is that we are measuring the time and space complexity of an algorithm in terms of the growth of input size n. I also understand that certain sorting methods have best, worst and average scenarios for Big O such as O(n) ,O(n^2) etc and the n is input size (number of elements to be sorted).
Any simple definitions or examples would be greatly appreciated thanks.
Big-O analysis is a form of runtime analysis that measures the efficiency of an algorithm in terms of the time it takes for the algorithm to run as a function of the input size. It’s not a formal bench- mark, just a simple way to classify algorithms by relative efficiency when dealing with very large input sizes.
Update:
The fastest-possible running time for any runtime analysis is O(1), commonly referred to as constant running time.An algorithm with constant running time always takes the same amount of time
to execute, regardless of the input size.This is the ideal run time for an algorithm, but it’s rarely achievable.
The performance of most algorithms depends on n, the size of the input.The algorithms can be classified as follows from best-to-worse performance:
O(log n) — An algorithm is said to be logarithmic if its running time increases logarithmically in proportion to the input size.
O(n) — A linear algorithm’s running time increases in direct proportion to the input size.
O(n log n) — A superlinear algorithm is midway between a linear algorithm and a polynomial algorithm.
O(n^c) — A polynomial algorithm grows quickly based on the size of the input.
O(c^n) — An exponential algorithm grows even faster than a polynomial algorithm.
O(n!) — A factorial algorithm grows the fastest and becomes quickly unusable for even small values of n.
The run times of different orders of algorithms separate rapidly as n gets larger.Consider the run time for each of these algorithm classes with
n = 10:
log 10 = 1
10 = 10
10 log 10 = 10
10^2 = 100
2^10= 1,024
10! = 3,628,800
Now double it to n = 20:
log 20 = 1.30
20 = 20
20 log 20= 26.02
20^2 = 400
2^20 = 1,048,576
20! = 2.43×1018
Finding an algorithm that works in superlinear time or better can make a huge difference in how well an application performs.
Say, f(n) in O(g(n)) if and only if there exists a C and n0 such that f(n) < C*g(n) for all n greater than n0.
Now that's a rather mathematical approach. So I'll give some examples. The simplest case is O(1). This means "constant". So no matter how large the input (n) of a program, it will take the same time to finish. An example of a constant program is one that takes a list of integers, and returns the first one. No matter how long the list is, you can just take the first and return it right away.
The next is linear, O(n). This means that if the input size of your program doubles, so will your execution time. An example of a linear program is the sum of a list of integers. You'll have to look at each integer once. So if the input is an list of size n, you'll have to look at n integers.
An intuitive definition could define the order of your program as the relation between the input size and the execution time.
Others have explained big O notation well here. I would like to point out that sometimes too much emphasis is given to big O notation.
Consider matrix multplication the naïve algorithm has O(n^3). Using the Strassen algoirthm it can be done as O(n^2.807). Now there are even algorithms that get O(n^2.3727).
One might be tempted to choose the algorithm with the lowest big O but it turns for all pratical purposes that the naïvely O(n^3) method wins out. This is because the constant for the dominating term is much larger for the other methods.
Therefore just looking at the dominating term in the complexity can be misleading. Sometimes one has to consider all terms.
Big O is about finding an upper limit for the growth of some function. See the formal definition on Wikipedia http://en.wikipedia.org/wiki/Big_O_notation
So if you've got an algorithm that sorts an array of size n and it requires only a constant amount of extra space and it takes (for example) 2 n² + n steps to complete, then you would say it's space complexity is O(n) or O(1) (depending on wether you count the size of the input array or not) and it's time complexity is O(n²).
Knowing only those O numbers, you could roughly determine how much more space and time is needed to go from n to n + 100 or 2 n or whatever you are interested in. That is how well an algorithm "scales".
Update
Big O and complexity are really just two terms for the same thing. You can say "linear complexity" instead of O(n), quadratic complexity instead of O(n²), etc...
I see that you are commenting on several answers wanting to know the specific term of order as it relates to Big-O.
Suppose f(n) = O(n^2), we say that the order is n^2.
Be careful here, there are some subtleties. You stated "we are measuring the time and space complexity of an algorithm in terms of the growth of input size n," and that's how people often treat it, but it's not actually correct. Rather, with O(g(n)) we are determining that g(n), scaled suitably, is an upper bound for the time and space complexity of an algorithm for all input of size n bigger than some particular n'. Similarly, with Omega(h(n)) we are determining that h(n), scaled suitably, is a lower bound for the time and space complexity of an algorithm for all input of size n bigger than some particular n'. Finally, if both the lower and upper bound are the same complexity g(n), the complexity is Theta(g(n)). In other words, Theta represents the degree of complexity of the algorithm while big-O and big-Omega bound it above and below.
Constant Growth: O(1)
Linear Growth: O(n)
Quadratic Growth: O(n^2)
Cubic Growth: O(n^3)
Logarithmic Growth: (log(n)) or O(n*log(n))
Big O use Mathematical Definition of complexity .
Order Of use in industrial Definition of complexity .
I have a question, what does it mean to find the big-o order of the memory required by an algorithm?
Like what's the difference between that and the big o operations?
E.g
a question asks
Given the following pseudo-code, with an initialized two dimensional array A, with both dimensions of size n:
for i <- 1 to n do
for j <- 1 to n-i do
A[i][j]= i + j
Wouldn't the big o notation for memory just be n^2 and the computations also be n^2?
Big-Oh is about how something grows according to something else (technically the limit on how something grows). The most common introductory usage is for the something to be how fast an algorithm runs according to the size of inputs.
There is nothing that says you can't have the something be how much memory is used according to the size of the input.
In your example, since there is a bucket in the array for everything in i and j, the space requirements grow as O(i*j), which is O(n^2)
But if your algorithm was instead keeping track of the largest sum, and not the sums of every number in each array, the runtime complexity would still be O(n^2) while the space complexity would be constant, as the algorithm only ever needs to keep track of current i, current j, current max, and the max being tested.
Big-O order of memory means how does the number of bytes needed to execute the algorithm vary as the number of elements processed increases. In your example, I think the Big-O order is n squared, because the data is stored in a square array of size nxn.
The big-O order of operations means how does the number of calculations needed to execute the algorithm vary as the number of elements processed increases.
Yes you are correct the space and time complexity for the above pseudo code is n^2.
But for the below code the space or memory complexity is 1 and but time complexity is n^2.
I usually go by the assignments etc done within the code which gives you the memory complexity.
for i <- 1 to n do
for j <- 1 to n-i do
A[0][0]= i + j
I honestly never heard of "big O for memory" but I can easily guess it is only loosely relater to the computation time - probably only setting a lower bound.
As an example, it is easy to design an algorithm which uses n^2 memory and n^3 computation, but i think it is impossible to do the other way round - you cannot process n^2 data with n complexity computationally.
Your algorithm has complexity 1/2 * n^ 2, thus O(n^2)
If A is given to your algorithm, then the space complexity is O(1). Iterating over an existing 2D array and writing values to existing memory locations uses no additional memory.
However, if the algorithm allocates A, then the space complexity is O(n2).
The time complexity is O(n2) either way.
What would be the BigO time of this algorithm
Input: Array A sorting n>=1 integers
Output: The sum of the elements at even cells in A
s=A[0]
for i=2 to n-1 by increments of 2
{
s=s+A[i]
}
return s
I think the function for this equation is F(n)=n*ceiling(n/2) but how do you convert that to bigO
The time complexity for that algorithm would be O(n), since the amount of work it does grows linearly with the size of the input. The other way to look at it is that loops over the input once - ignore the fact that it only looks at half of the values, that doesn't matter for Big-O complexity).
The number of operations is not proportional to n*ceiling(n/2), but rather n/2 which is O(n). Because of the meaning of big-O, (which includes the idea of an arbitrary coefficient), O(n) and O(n/2) are absolutely equivalent - so it is always written as O(n).
This is an O(n) algorithm since you look at ~n/2 elements.
Your algorithm will do N/2 iterations given that there are N elements in the array. Each iteration requires constant time to complete. This gives us O(N) complexity, and here's why.
Generally, the running time of an algorithm is a function f(x) from the size of data. Saying that f(x) is O(g(x)) means that there exists some constant c such that for all sufficiently large values of x f(x) <= cg(x). Easy to see how this applies in our case: if we assume that each iteration takes a unit of time, obviously f(N) <= 1/2N.
A formal manner to obtain the exact number of operations and your algorithm's order of growth: