I’m having a hard time using O(n) principles to generalize the time complexity of an algorithm whose more specific time complexity is O(sum(a)) where a is an array of integers.
My intuition is that this time complexity should generalize to O(n) as you can think of this as a “linear” equation of ki values that occur n times where k is the integer value in the array, making it O(n)( k=1 for a straight up O(n) case).
But it doesn’t seem to be exactly the same as O(n) - the value of k could be much larger than n, and if all these k values are larger you have something that could be O(n^2) or O(n^3) depending on how large that value is.
Is this something to take into account for O(n) complexity where n is the length of the array? Should I actually be defining n as the sum of all elements in the array instead of the length of the array?
In general, what would be the best way to think about this?
Fundamentally, we want to describe the runtime of an algorithm based on the input. The "runtime" is a vague term, that is often swept under the rug. For example, the "runtime" of a sorting algorithm or a hashtable operation is measured in number of comparisons, but using "runtime" to mean the number of basic operations (which are also usually only vaguely defined) is also possible.
There are two choices (or simplifications) often made when calcuating runtime. The first, is to ignore the actual input, and to use the size of the input (measured somehow) instead. This size is usually denoted n. The second, is to use big-O notation to describe the worst case (or best case, or average, or amortized...).
Neither of these choices is always necessary, and sometimes, they won't make sense. To repeat, since this is the crux of the answer: describing runtimes in big-O of n is not the only way to describe runtimes and sometimes it makes no sense to do so.
For example, in the case of an algorithm that runs in O(sum(a)) time:
func f(a) {
t = 0
for x in a {
for i = 1..x {
t += 1
}
}
}
It's not useful to describe the runtime of this using the length of the input array a. It's not useful because the length of a doesn't say anything about the worst-case runtime.
Saying that t is incremented sum(a) times is a useful statement about the runtime of the program. It doesn't use big-O complexity notation.
And if you do want to express that in big-O notation, you can say that the runtime of this code is O(sum(a)). This blurs exactly what you're measuring in the runtime, because you can be including the cost of performing the statements other than incrementing t.
And going back to the example, you could (and if you were studying complexity classes, you probably would) say n is the size (in bits) of the input array. Then you could say something about the runtime (measured in basic operations): it's O(2^n), since the worst case input is an array with one element which takes the value 2^n-1 (*note).
*note: this ignores some technical details about how to encode an array using bits.
Related
Let's say that the algorithm involves iterating through a string character by character.
If I know for sure that the length of the string is less than, say, 15 characters, will the time complexity be O(1) or will it remain as O(n)?
There are two aspects to this question - the core of the question is, can problem constraints change the asymptotic complexity of an algorithm? The answer to that is yes. But then you give an example of a constraint (strings limited to 15 characters) where the answer is: the question doesn't make sense. A lot of the other answers here are misleading because they address only the second aspect but try to reach a conclusion about the first one.
Formally, the asymptotic complexity of an algorithm is measured by considering a set of inputs where the input sizes (i.e. what we call n) are unbounded. The reason n must be unbounded is because the definition of asymptotic complexity is a statement like "there is some n0 such that for all n ≥ n0, ...", so if the set doesn't contain any inputs of size n ≥ n0 then this statement is vacuous.
Since algorithms can have different running times depending on which inputs of each size we consider, we often distinguish between "average", "worst case" and "best case" time complexity. Take for example insertion sort:
In the average case, insertion sort has to compare the current element with half of the elements in the sorted portion of the array, so the algorithm does about n2/4 comparisons.
In the worst case, when the array is in descending order, insertion sort has to compare the current element with every element in the sorted portion (because it's less than all of them), so the algorithm does about n2/2 comparisons.
In the best case, when the array is in ascending order, insertion sort only has to compare the current element with the largest element in the sorted portion, so the algorithm does about n comparisons.
However, now suppose we add the constraint that the input array is always in ascending order except for its smallest element:
Now the average case does about 3n/2 comparisons,
The worst case does about 2n comparisons,
And the best case does about n comparisons.
Note that it's the same algorithm, insertion sort, but because we're considering a different set of inputs where the algorithm has different performance characteristics, we end up with a different time complexity for the average case because we're taking an average over a different set, and similarly we get a different time complexity for the worst case because we're choosing the worst inputs from a different set. Hence, yes, adding a problem constraint can change the time complexity even if the algorithm itself is not changed.
However, now let's consider your example of an algorithm which iterates over each character in a string, with the added constraint that the string's length is at most 15 characters. Here, it does not make sense to talk about the asymptotic complexity, because the input sizes n in your set are not unbounded. This particular set of inputs is not valid for doing such an analysis with.
In the mathematical sense, yes. Big-O notation describes the behavior of an algorithm in the limit, and if you have a fixed upper bound on the input size, that implies it has a maximum constant complexity.
That said, context is important. All computers have a realistic limit to the amount of input they can accept (a technical upper bound). Just because nothing in the world can store a yottabyte of data doesn't mean saying every algorithm is O(1) is useful! It's about applying the mathematics in a way that makes sense for the situation.
Here are two contexts for your example, one where it makes sense to call it O(1), and one where it does not.
"I decided I won't put strings of length more than 15 into my program, therefore it is O(1)". This is not a super useful interpretation of the runtime. The actual time is still strongly tied to the size of the string; a string of size 1 will run much faster than one of size 15 even if there is technically a constant bound. In other words, within the constraints of your problem there is still a strong correlation to n.
"My algorithm will process a list of n strings, each with maximum size 15". Here we have a different story; the runtime is dominated by having to run through the list! There's a point where n is so large that the time to process a single string doesn't change the correlation. Now it makes sense to consider the time to process a single string O(1), and therefore the time to process the whole list O(n)
That said, Big-O notation doesn't have to only use one variable! There are problems where upper bounds are intrinsic to the algorithm, but you wouldn't put a bound on the input arbitrarily. Instead, you can describe each dimension of your input as a different variable:
n = list length
s = maximum string length
=> O(n*s)
It depends.
If your algorithm's requirements would grow if larger inputs were provided, then the algorithmic complexity can (and should) be evaluated independently of the inputs. So iterating over all the elements of a list, array, string, etc., is O(n) in relation to the length of the input.
If your algorithm is tied to the limited input size, then that fact becomes part of your algorithmic complexity. For example, maybe your algorithm only iterates over the first 15 characters of the input string, regardless of how long it is. Or maybe your business case simply indicates that a larger input would be an indication of a bug in the calling code, so you opt to immediately exit with an error whenever the input size is larger than a fixed number. In those cases, the algorithm will have constant requirements as the input length tends toward very large numbers.
From Wikipedia
Big O notation is a mathematical notation that describes the limiting behavior of a function when the argument tends towards a particular value or infinity.
...
In computer science, big O notation is used to classify algorithms according to how their run time or space requirements grow as the input size grows.
In practice, almost all inputs have limits: you cannot input a number larger than what's representable by the numeric type, or a string that's larger than the available memory space. So it would be silly to say that any limits change an algorithm's asymptotic complexity. You could, in theory, use 15 as your asymptote (or "particular value"), and therefore use Big-O notation to define how an algorithm grows as the input approaches that size. There are some algorithms with such terrible complexity (or some execution environments with limited-enough resources) that this would be meaningful.
But if your argument (string length) does not tend toward a large enough value for some aspect of your algorithm's complexity to define the growth of its resource requirements, it's arguably not appropriate to use asymptotic notation at all.
NO!
The time complexity of an algorithm is independent of program constraints. Here is (a simple) way of thinking about it:
Say your algorithm iterates over the string and appends all consonants to a list.
Now, for iteration time complexity is O(n). This means that the time taken will increase roughly in proportion to the increase in the length of the string. (Time itself though would vary depending on the time taken by the if statement and Branch Prediction)
The fact that you know that the string is between 1 and 15 characters long will not change how the program runs, it merely tells you what to expect.
For example, knowing that your values are going to be less than 65000 you could store them in a 16-bit integer and not worry about Integer overflow.
Do problem constraints change the time complexity of algorithms?
No.
If I know for sure that the length of the string is less than, say, 15 characters ..."
We already know the length of the string is less than SIZE_MAX. Knowing an upper fixed bound for string length does not make the the time complexity O(1).
Time complexity remains O(n).
Big-O measures the complexity of algorithms, not of code. It means Big-O does not know the physical limitations of computers. A Big-O measure today will be the same in 1 million years when computers, and programmers alike, have evolved beyond recognition.
So restrictions imposed by today's computers are irrelevant for Big-O. Even though any loop is finite in code, that need not be the case in algorithmic terms. The loop may be finite or infinite. It is up to the programmer/Big-O analyst to decide. Only s/he knows which algorithm the code intends to implement. If the number of loop iterations is finite, the loop has a Big-O complexity of O(1) because there is no asymptotic growth with N. If, on the other hand, the number of loop iterations is infinite, the Big-O complexity is O(N) because there is an asymptotic growth with N.
The above is straight from the definition of Big-O complexity. There are no ifs or buts. The way the OP describes the loop makes it O(1).
A fundamental requirement of big-O notation is that parameters do not have an upper limit. Suppose performing an operation on N elements takes a time precisely equal to 3E24*N*N*N / (1E24+N*N*N) microseconds. For small values of N, the execution time would be proportional to N^3, but as N gets larger the N^3 term in the denominator would start to play an increasing role in the computation.
If N is 1, the time would be 3 microseconds.
If N is 1E3, the time would be about 3E33/1E24, i.e. 3.0E9.
If N is 1E6, the time would be about 3E42/1E24, i.e. 3.0E18
If N is 1E7, the time would be 3E45/1.001E24, i.e. ~2.997E21
If N is 1E8, the time would be about 3E48/2E24, i.e. 1.5E24
If N is 1E9, the time would be 3E51/1.001E27, i.e. ~2.997E24
If N is 1E10, the time would be about 3E54/1.000001E30, i.e. 2.999997E24
As N gets bigger, the time would continue to grow, but no matter how big N gets the time would always be less than 3.000E24 seconds. Thus, the time required for this algorithm would be O(1) because one could specify a constant k such that the time necessary to perform the computation with size N would be less than k.
For any practical value of N, the time required would be proportional to N^3, but from an O(N) standpoint the worst-case time requirement is constant. The fact that the time changes rapidly in response to small values of N is irrelevant to the "big picture" behaviour, which is what big-O notation measures.
It will be O(1) i.e. constant.
This is because for calculating time complexity or worst-case time complexity (to be precise), we think of the input as a huge chunk of data and the length of this data is assumed to be n.
Let us say, we do some maximum work C on each part of this input data, which we will consider as a constant.
In order to get the worst-case time complexity, we need to loop through each part of the input data i.e. we need to loop n times.
So, the time complexity will be:
n x C.
Since you fixed n to be less than 15 characters, n can also be assumed as a constant number.
Hence in this case:
n = constant and,
(maximum constant work done) = C = constant
So time complexity is n x C = constant x constant = constant i.e. O(1)
Edit
The reason why I have said n = constant and C = constant for this case, is because the time difference for doing calculations for smaller n will become so insignificant (compared to n being a very large number) for modern computers that we can assume it to be constant.
Otherwise, every function ever build will take some time, and we can't say things like:
lookup time is constant for hashmaps
I've seen coding problems that are similar to this:
int doSomething(String s)
Where it says in the problem description that s will contain at most one of every character, so s cannot be more than length 26. I think in this case, iterating over s would be constant time.
But I've also seen problems where inputs are constrained to a random large number, like 10^5, just to avoid stack overflows and other weird edge cases. If we are going to consider inputs that are constrained by constants to be constant complexity, shouldn't these inputs also be considered constant complexity?
But it doesn't make sense to me to consider s to be of O(n) complexity either, because there are many problems were people allocate char[26] arrays to hold every letter of the alphabet. How does if make sense to consider an input that we know will be less than or equal to 26 to be of greater complexity than an array of size 26?
The point of analyzing the complexity of algorithms is to estimate how long it will take to run it. If the problem you're trying to solve limits the maximum value of n to a constant, you can consider n to be a constant and you wouldn't be wrong. But would that be useful if you wanted to predict whether an algorithm that does 2^n operations will run in a few seconds for n = 26? On the other hand, if you had an algorithm that does n*m operations and m is at most 3, how useful would it be to include m in the complexity analysis?
Calculating complexity has focus on what is the most critical variable related to the running time. If the running time is dominant by the length of s, it is our main focus of analyzing complexity and that should be in bigO notation. And in that case, of course it's not a constant.
If the input is constrained to a large number like 10^5.
And if the algorithm is getting slower proportional to that input.
for example,
int sort(string s); //length of s is less than 10^5
In this case, depending on what sorting algorithm you use,
the running time will be proportional to the length of s
like O(n^2) or O(nlogn) if n is the length of s
In this case you cannot say it's constant because running time is very different as the length of s is changing.
But if the algorithm inside has nothing to do with the length of s, like it has constant calculation time, then you can say 10^5 constraint is just a constant.
When talking about complexity in general, things like O(3n) tend to be simplified to O(n) and so on. This is merely theoretical, so how does complexity work in reality? Can O(3n) also be simplified to O(n)?
For example, if a task implies that solution must be in O(n) complexity and in our code we have 2 times linear search of an array, which is O(n) + O(n). So, in reality, would that solution be considered as linear complexity or not fast enough?
Note that this question is asking about real implementations, not theoretical. I'm already aware that O(n) + O(n) is simplified to O(n)?
Bear in mind that O(f(n)) does not give you the amount of real-world time that something takes: only the rate of growth as n grows. O(n) only indicates that if n doubles, the runtime doubles as well, which lumps functions together that take one second per iteration or one millennium per iteration.
For this reason, O(n) + O(n) and O(2n) are both equivalent to O(n), which is the set of functions of linear complexity, and which should be sufficient for your purposes.
Though an algorithm that takes arbitrary-sized inputs will often want the most optimal function as represented by O(f(n)), an algorithm that grows faster (e.g. O(n²)) may still be faster in practice, especially when the data set size n is limited or fixed in practice. However, learning to reason about O(f(n)) representations can help you compose algorithms to have a predictable—optimal for your use-case—upper bound.
Yes, as long as k is a constant, you can write O(kn) = O(n).
The intuition behind is that the constant k doesn't increase with the size of the input space and at some point will be incomparably small to n, so it doesn't have much influence on the overall complexity.
Yes - as long as the number k of array searches is not affected by the input size, even for inputs that are too big to be possible in practice, O(kn) = O(n). The main idea of the O notation is to emphasize how the computation time increases with the size of the input, and so constant factors that stay the same no matter how big the input is aren't of interest.
An example of an incorrect way to apply this is to say that you can perform selection sort in linear time because you can only fit about one billion numbers in memory, and so selection sort is merely one billion array searches. However, with an ideal computer with infinite memory, your algorithm would not be able to handle more than one billion numbers, and so it is not a correct sorting algorithm (algorithms must be able to handle arbitrarily large inputs unless you specify a limit as a part of the problem statement); it is merely a correct algorithm for sorting up to one billion numbers.
(As a matter of fact, once you put a limit on the input size, most algorithms will become constant-time because for all inputs within your limit, the algorithm will solve it using at most the amount of time that is required for the biggest / most difficult input.)
I am currently trying to learn time complexity of algorithms, big-o notation and so on. However, some point confuses me a lot. I know that most of the time, the input size of an array or whatever we are dealing with determines the running time of the algorithm. Let's say I have an unsorted array with size N and I want to find the maximum element of this array without using any special algorithm. I just want to iterate over the array and find the maximum element. Since the size of my array is N, this process runs at O(N) or linear time. Let M is an integer that is the square root of N. So N can be written as the square of M that is M*M or M^2. So, I think there is nothing wrong if I want to replace N with M^2. I know that M^2 is also the size of my array so my big-o notation could be written as O(M^2). So, my new running time looks like running in quadratic time. Why does this happen?
You are correct, if it happens to be that you have some variable M such that M^2 ~= N is always true, you could say the algorithm runs in O(M^2).
But, note that now - the algorithm runs in quadratic related to M, and not quadratic time related to the input, it is still linear related to the size of the input.
The important thing here is defining linear/quadratic, etc.
More precisely, you have to detail linear/quadratic, etc. with respect to something (N or M for your example). The most natural choice is to study the complexity wrt. the size of the input (N for your example).
Another trap for big integers is that the size of n is log(n). So for instance if you loop over all smaller integers, your algorithm is not polynomial.
This question already has answers here:
Closed 12 years ago.
Possible Duplicate:
Plain english explanation of Big O
I'd imagine this is probably something taught in classes, but as I a self-taught programmer, I've only seen it rarely.
I've gathered it is something to do with the time, and O(1) is the best, while stuff like O(n^n) is very bad, but could someone point me to a basic explanation of what it actually represents, and where these numbers come from?
Big O refers to the worst case run-time order. It is used to show how well an algorithm scales based on the size of the data set (n->number of items).
Since we are only concerned with the order, constant multipliers are ignored, and any terms which increase less quickly than the dominant term are also removed. Some examples:
A single operation or set of operations is O(1), since it takes some constant time (does not vary based on data set size).
A loop is O(n). Each element in the data set is looped over.
A nested loop is O(n^2). A nested nested loop is O(n^3), and onward.
Things like binary tree searching are log(n), which is more difficult to show, but at every level in the tree, the possible number of solutions is halved, so the number of levels is log(n) (provided the tree is balanced).
Something like finding the sum of a set of numbers that is closest to a given value is O(n!), since the sum of each subset needs to be calculated. This is very bad.
It's a way of expressing time complexity.
O(n) means for n elements in a list, it takes n computations to sort the list. Which isn't bad at all. Each increase in n increases time complexity linearly.
O(n^n) is bad, because the amount of computation required to perform a sort (or whatever you are doing) will exponentially increase as you increase n.
O(1) is the best, as it means 1 computation to perform a function, think of hash tables, looking up a value in a hash table has O(1) time complexity.
Big O notation as applied to an algorithm refers to how the run time of the algorithm depends on the amount of input data. For example, a sorting algorithm will take longer to sort a large data set than a small data set. If for the sorting algorithm example you graph the run time (vertical-axis) vs the number of values to sort (horizontal-axis), for numbers of values from zero to a large number, the nature of the line or curve that results will depend on the sorting algorithm used. Big O notation is a shorthand method for describing the line or curve.
In big O notation, the expression in the brackets is the function that is graphed. If a variable (say n) is included in the expression, this variable refers to the size of the input data set. You say O(1) is the best. This is true because the graph f(n) = 1 does not vary with n. An O(1) algorithm takes the same amount of time to complete regardless of the size of the input data set. By contrast, the run time of an algorithm of O(n^n) increases with the square of the size of the input data set.
That is the basic idea, for a detailed explanation, consult the wikipedia page titled 'Big O Notation'.