I'm not entirely sure what amortized complexity means. Take a balanced binary search tree data structure (e.g. a red-black tree). The cost of a normal search is naturally log(N) where N is the number of nodes. But what is the amortized complexity of a sequence of m searches in, let's say, ascending order. Is it just log(N)/m?
Well you can consider asymptotic analysis as a strict method to set a upper bound for the running time of algorithms, where as amortized analysis is a some what liberal method.
For example consider an algorithm A with two statements S1 and S2. The cost of executing S1 is 10 and S2 is 100. Both the statements are placed inside a loop as follows.
n=0;
while(n<100)
{
if(n % 10 != 0)
{
S1;
}
else
{
s2;
}
n++;
}
Here the number of times S1 executed is 10 times the count of S2. But asymptotic analysis will only consider the facts that S2 takes a time of 10 units and it is inside a loop executing 100 times. So the upper limit for execution time is of the order of 10 * 100 = 1000. Where as amortized analysis averages out the number of times the statements S1 and S2 are executed. So the upper time limit for execution is of the order of 200. Thus amortized analysis gives a better estimate of the upper limit for executing an algorithm.
I think it is mlog(N) because you have to do m search operations (each time from root node downto target node), while the complexity of one single operation is log(N).
EDIT: #user1377000 you are right, I have mistaken amortized complexity from asymptotic complexity. But I don't think it is log(N)/m... because it is not guaranteed that you can finished all m search operations in O(logN) time.
What is amortized analysis of algorithms?
I think this might help.
In case of a balanced search tree the amortized complexity is equal to asymptotic one. Each search operation takes O(logn) time, both asymptotic and average. Therefore for m searches the average complexity will be O(mlogn).
Pass in the items to be found all at once.
You can think of it in terms of divide-and-conquer.
Take the item x in the root node.
Binary-search for x into your array of m items.
Partition the array into things less than x and greater than x. (Ignore things equal to x, since you already found it.)
Recursively search for the former partition in your left child, and for the latter in your right child.
One worst case: your array of items is just the list of things in the leaf nodes. (n is roughly 2m.) You'd have to visit every node. Your search would cost lg(n) + 2*lg(n/2) + 4*lg(n/4) + .... That's linear. Think of it as doing smaller and smaller binary searches until you hit every element in the array once or twice.
I think there's also a way to do it by keeping track of where you are in the tree after a search. C++'s std::map and std::set return iterators which can move left and right within the tree, and they might have methods which can take advantage of an existing iterator into the tree.
Related
In Algorithms, 4th edition by Robert Sedgewick, the time complexity table for different algorithms is given as:
Based on this table, the searching time complexity of a BST is N, and of binary search in and of itself is logN.
What is the difference between the two? I have seen explanations about these separately and they made sense, however, I can't seem to understand why the searching time complexity of a BST isn't logN, as we are searching by continually breaking the tree in half and ignoring the other parts.
From binary-search-trees-bst-explained-with-examples
...on average, each comparison allows the operations to skip about half of the tree, so that each lookup, insertion or deletion takes time proportional to the logarithm of the number of items stored in the tree, O(log n) . However, some times the worst case can happen, when the tree isn't balanced and the time complexity is O(n) for all three of these functions.
So, you kind of expect log(N) but it's not absolutely guaranteed.
the searching time complexity of a BST is N, and of binary search in and of itself is logN. What is the difference between the two?
The difference is that a binary search on a sorted array always starts at the middle element (i.e. the median when n is odd). This cannot be guaranteed in a BST. The root might be the middle element, but it doesn't have to be.
For instance, this is a valid BST:
10
/
8
/
5
/
2
/
1
...but it is not a balanced one, and so the process of finding the value 1 given the root of that tree, will include visiting all its nodes. If however the same values were presented in a sorted list (1,2,5,8,10), a binary search would start at 5 and never visit 8 or 10.
Adding self-balancing trees to the table
We can extend the given table with self-balancing search trees, like AVL, and then we get this:
implementation
search
insert
delete
sequential search (unordered list)
𝑁
𝑁
𝑁
binary search (ordered array)
lg𝑁
𝑁
𝑁
BST
𝑁
𝑁
𝑁
AVL
lgN
lgN
lgN
during my classroom i asked this question to my teacher and he couldn't answer that's why i am asking here.
i asked that during a code , what if we have a loop to run from 1 to 10 , does the complexity would be O(1) {big O of 1} . heanswered yes. so here's the question what if i have written a loop to run from 1 to 1 million .is it sill O(1)? or is it O(n) or something else?
pseudo code -
for i in range(1,1 million):
print("hey")
what is the time complexity for that loop
now , if you think the answer is O(n) , how can you say it to be O(n) , because O(n) is when complexity is linear.
and what is the silver lining? when a code gets O(1) and O(n) .
like if i would have written a loop for 10 or 100 or 1000 or 10000 or 100000. when did it transformed from O(1) to O(n).
By definition, O(10000000) and O(1) are equal, Let me quickly explain what complexity means.
What we try to represent with the abstraction of time (and space) complexity isn't how fast a program will run, it what is the growth in runtime (or space) given the growth in input length.
For instance, given a loop with a fixed number of iterations (lets say 10), it doesnt matter if your input will be 1 long or 10000000000000, because your loop will ALWAYS run the same number of iteration therefore, no growth in runtime (even if that 10 iterations may take 1 week to run, it will always be 1 week).
but, if your algorithm's steps are dependent in your input length, that means the longer your input, the longer your algorithm's steps, the question is, how much more steps?
in summary, time (and space) complexity is an abstraction, its not here to tell us how long things will take, its simply here to tell us how the growth in time will be given growth in input, O(1) == O(10000000), because its not about how long it will take, its about the change in the runtime, O(1) algorithm can take 10 years, but it will always take 10 years, even for very large input length.
I think you are confusing the term. Time complexity for a given algorithm is given by the relationship between change in execution time with respect to change in input size.
If you are running a fixed loop from 1 to 10, but doing something in each iteration, then that counts as O(10), or O(1), meaning that it will take the same time each run.
But, as soon as the number of iterations starts depending on the number of elements or tasks, then a loop becomes O(n), meaning that the complexity becomes linear. The more the tasks, proportionally more the time.
I hope that clears some things up. :-)
I'm currently taking algorithms and data structure. After nearly two months of studying, I still find time complexity extremely confusing.
I was told (by my professor) that if I big-omega and big-O of some program aren't equal, big-theta doesn't exist.
I now literally question everything I've learned so far. I'll take BubbleSort as an example, with big-omega(n),big-theta(n^2) and big-O(n^2). Big-theta indeed does exist (and it makes sense when i analyze it).
Can anyone explain to me whether my professor is wrong or perhaps I misunderstood something?
There exists big O, Big ϴ, and Big Ω.
O is the upper bound
Ω is the lower bound
ϴ exists if and only only if O = Ω
In essence, Big-O is the most useful in that it tells us the WORST a function can behave.
Big-Ω indicates the BEST it can behave.
When WORST = BEST, you get ALWAYS. That's Big-ϴ. When a function always behaves the same.
Example (optimized bubble sort, then has boolean flag for when a swap occurs):
bubbleSort() ∈ Ω(n)
bubbleSort() ∈ O(n^2)
This means, the best bubble sort can do is linear time. But, in the worst case, it DEGRADES to quadratic time. Reason, on a pre-sorted list, bubble sort behaves quiet well, as it does one loop through the list (n iterations) and exits. Whereas, on a list in descending order, it would do roughly n(n-1)/2 iterations, which is proportional to n^2. Depending on the input, bubble sort behaves DRASTICALLY (different order or magnitude) differently.
Whereas:
mergeSort() ∈ Ω(n * log n)
mergeSort() ∈ O(n * log n)
This means, in the best case merge sort is in n * log n time, and in the worst case. It is ALWAYS n * log n. That is because, no matter what the input is, merge sort will always recursively divide the list in half-size sub-arrays, and put them back together once each sub-array is of size 1. However, you can only break something in half so many times (log2(n)) times. Then, the merge() routine that is O(n) is called once per mergeSort() which occurs log2(n) times. So, for ALL executions of mergeSort() you get n * log2(n).
Therefore we can make a STRONGER statement and say that:
mergeSort() ∈ ϴ(n * log n)
We can only make such definitive statements (use big theta) if the runtime of a function is bounded above and below by functions of the exact same order of magnitude.
How I remember it:
ϴ is an end all be all, whereas O, and Ω are simply limits.
lets take this problem in 2 ways:
First Way:
we say that O(bubble sort) is O(n^2) and Ω(bubble sort) is Ω(n).
The above statement holds true because, when the elements are laid down in array in sorted or almost sorted manner then bubble sort will have an equation of running time in form of -> an + b, where a and b are constants (a is positive).
but when elements are laid down in array in not so sorted manner or very unsorted manner, then the running time will be in form -> an^2 + bn + c. where a,b,c are constants (a is positive).
thus, bubble sort ∈ O(n^2) and bubble sort ∈ Ω(n).
PS: differentiating what will constitute to almost sorted manner and what will constitute to not so sorted manner is talk for another day, but you can google it you will find it easily.
Now, Second way:
Rather than taking Big-O and Big-Ω of bubble sort we divide bubble sort into its worst case and best case.
now we say what is Big-O of worst case of bubble sort we get. O(n^2), i.e,
worst case of bubble sort ∈ O(n^2).
but now since we have specified that we are dealing with only worst case then
Big-Ω of worst case of bubble sort will also be Ω(n^2).
Thus in this case ϴ(n^2) exists. because O = Ω.
Using similar approach, we can see that the best case of bubble sort will have ϴ(n).
Note: worst case will be in form: an^2 + bn + c.
and best case will be in form: an + b.
-------------------------------------------------------
In second way we bifurcated the worst case and best case then only we were able to reach to ϴ notations for both the cases.
Bubble sort as a whole had different polynomial equations as its running time with different degree that is why it did not have a ϴ notation.
Ive always been a bit confused on this, possibly due to my lack of understanding in compilers. But lets use python as an example. If we had some large list of numbers called numlist and wanted to get rid of any duplicates, we could use a set operator on the list, example set(numlist). In return we would have a set of our numbers. This operation to the best of my knowledge will be done in O(n) time. Though if I were to create my own algorithm to handle this operation, the absolute best I could ever hope for is O(n^2).
What I don't get is, what allows a internal operation like set() to be so much faster then an external to the language algorithm. The checking still needs to be done, don't they?
You can do this in Θ(n) average time using a hash table. Lookup and insertion in a hash table are Θ(1) on average . Thus, you just run through the n items and for each one checking if it is already in the hash table and if not inserting the item.
What I don't get is, what allows a internal operation like set() to be so much faster then an external to the language algorithm. The checking still needs to be done, don't they?
The asymptotic complexity of an algorithm does not change if implemented by the language implementers versus being implemented by a user of the language. As long as both are implemented in a Turing complete language with random access memory models they have the same capabilities and algorithms implemented in each will have the same asymptotic complexity. If an algorithm is theoretically O(f(n)) it does not matter if it is implemented in assembly language, C#, or Python on it will still be O(f(n)).
You can do this in O(n) in any language, basically as:
# Get min and max values O(n).
min = oldList[0]
max = oldList[0]
for i = 1 to oldList.size() - 1:
if oldList[i] < min:
min = oldList[i]
if oldList[i] > max:
max = oldList[i]
# Initialise boolean list O(n)
isInList = new boolean[max - min + 1]
for i = min to max:
isInList[i] = false
# Change booleans for values in old list O(n)
for i = 0 to oldList.size() - 1:
isInList[oldList[i] - min] = true
# Create new list from booleans O(n) (or O(1) based on integer range).
newList = []
for i = min to max:
if isInList[i - min]:
newList.append (i)
I'm assuming here that append is an O(1) operation, which it should be unless the implementer was brain-dead. So with k steps each O(n), you still have an O(n) operation.
Whether the steps are explicitly done in your code or whether they're done under the covers of a language is irrelevant. Otherwise you could claim that the C qsort was one operation and you now have the holy grail of an O(1) sort routine :-)
As many people have discovered, you can often trade off space complexity for time complexity. For example, the above only works because we're allowed to introduce the isInList and newList variables. If this were not allowed, the next best solution may be sorting the list (probably no better the O(n log n)) followed by an O(n) (I think) operation to remove the duplicates.
An extreme example, you can use that same extra-space method to sort an arbitrary number of 32-bit integers (say with each only having 255 or less duplicates) in O(n) time, provided you can allocate about four billion bytes for storing the counts.
Simply initialise all the counts to zero and run through each position in your list, incrementing the count based on the number at that position. That's O(n).
Then start at the beginning of the list and run through the count array, placing that many of the correct value in the list. That's O(1), with the 1 being about four billion of course but still constant time :-)
That's also O(1) space complexity but a very big "1". Typically trade-offs aren't quite that severe.
The complexity bound of an algorithm is completely unrelated to whether it is implemented 'internally' or 'externally'
Taking a list and turning it into a set through set() is O(n).
This is because set is implemented as a hash set. That means that to check if something is in the set or to add something to the set only takes O(1), constant time. Thus, to make a set from an iterable (like a list for example), you just start with an empty set and add the elements of the iterable one by one. Since there are n elements and each insertion takes O(1), the total time of converting an iterable to a set is O(n).
To understand how the hash implementation works, see the wikipedia artcle on hash tables
Off hand I can't think of how to do this in O(n), but here is the cool thing:
The difference between n^2 and n is sooo massive that the difference between you implementing it and python implementing is tiny compared to the algorithm used to implement it. n^2 is always worse than O(n), even if the n^2 one is in C and the O(n) one is in python. You should never think that kind of difference comes from the fact that you're not writing in a low level language.
That said, if you want to implement your own, you can do a sort then remove dups. the sort is n*ln(n) and the remove dups in O(n)...
There are two issues here.
Time complexity (which is expressed in big O notation) is a formal measure of how long an algorithm takes to run for a given set size. It's more about how well an algorithm scales than about the absolute speed.
The actual speed (say, in milliseconds) of an algorithm is the time complexity multiplied by a constant (in an ideal world).
Two people could implement the same removal of duplicates algorithm with O(log(n)*n) complexity, but if one writes it in Python and the other writes it in optimised C, then the C program will be faster.
This question already has answers here:
What is a plain English explanation of "Big O" notation?
(43 answers)
Closed 9 years ago.
What is Big O notation? Do you use it?
I missed this university class I guess :D
Does anyone use it and give some real life examples of where they used it?
See also:
Big-O for Eight Year Olds?
Big O, how do you calculate/approximate it?
Did you apply computational complexity theory in real life?
One important thing most people forget when talking about Big-O, thus I feel the need to mention that:
You cannot use Big-O to compare the speed of two algorithms. Big-O only says how much slower an algorithm will get (approximately) if you double the number of items processed, or how much faster it will get if you cut the number in half.
However, if you have two entirely different algorithms and one (A) is O(n^2) and the other one (B) is O(log n), it is not said that A is slower than B. Actually, with 100 items, A might be ten times faster than B. It only says that with 200 items, A will grow slower by the factor n^2 and B will grow slower by the factor log n. So, if you benchmark both and you know how much time A takes to process 100 items, and how much time B needs for the same 100 items, and A is faster than B, you can calculate at what amount of items B will overtake A in speed (as the speed of B decreases much slower than the one of A, it will overtake A sooner or later—this is for sure).
Big O notation denotes the limiting factor of an algorithm. Its a simplified expression of how run time of an algorithm scales with relation to the input.
For example (in Java):
/** Takes an array of strings and concatenates them
* This is a silly way of doing things but it gets the
* point across hopefully
* #param strings the array of strings to concatenate
* #returns a string that is a result of the concatenation of all the strings
* in the array
*/
public static String badConcat(String[] Strings){
String totalString = "";
for(String s : strings) {
for(int i = 0; i < s.length(); i++){
totalString += s.charAt(i);
}
}
return totalString;
}
Now think about what this is actually doing. It is going through every character of input and adding them together. This seems straightforward. The problem is that String is immutable. So every time you add a letter onto the string you have to create a new String. To do this you have to copy the values from the old string into the new string and add the new character.
This means you will be copying the first letter n times where n is the number of characters in the input. You will be copying the character n-1 times, so in total there will be (n-1)(n/2) copies.
This is (n^2-n)/2 and for Big O notation we use only the highest magnitude factor (usually) and drop any constants that are multiplied by it and we end up with O(n^2).
Using something like a StringBuilder will be along the lines of O(nLog(n)). If you calculate the number of characters at the beginning and set the capacity of the StringBuilder you can get it to be O(n).
So if we had 1000 characters of input, the first example would perform roughly a million operations, StringBuilder would perform 10,000, and the StringBuilder with setCapacity would perform 1000 operations to do the same thing. This is rough estimate, but O(n) notation is about orders of magnitudes, not exact runtime.
It's not something I use per say on a regular basis. It is, however, constantly in the back of my mind when trying to figure out the best algorithm for doing something.
A very similar question has already been asked at Big-O for Eight Year Olds?. Hopefully the answers there will answer your question although the question asker there did have a bit of mathematical knowledge about it all which you may not have so clarify if you need a fuller explanation.
Every programmer should be aware of what Big O notation is, how it applies for actions with common data structures and algorithms (and thus pick the correct DS and algorithm for the problem they are solving), and how to calculate it for their own algorithms.
1) It's an order of measurement of the efficiency of an algorithm when working on a data structure.
2) Actions like 'add' / 'sort' / 'remove' can take different amounts of time with different data structures (and algorithms), for example 'add' and 'find' are O(1) for a hashmap, but O(log n) for a binary tree. Sort is O(nlog n) for QuickSort, but O(n^2) for BubbleSort, when dealing with a plain array.
3) Calculations can be done by looking at the loop depth of your algorithm generally. No loops, O(1), loops iterating over all the set (even if they break out at some point) O(n). If the loop halves the search space on each iteration? O(log n). Take the highest O() for a sequence of loops, and multiply the O() when you nest loops.
Yeah, it's more complex than that. If you're really interested get a textbook.
'Big-O' notation is used to compare the growth rates of two functions of a variable (say n) as n gets very large. If function f grows much more quickly than function g we say that g = O(f) to imply that for large enough n, f will always be larger than g up to a scaling factor.
It turns out that this is a very useful idea in computer science and particularly in the analysis of algorithms, because we are often precisely concerned with the growth rates of functions which represent, for example, the time taken by two different algorithms. Very coarsely, we can determine that an algorithm with run-time t1(n) is more efficient than an algorithm with run-time t2(n) if t1 = O(t2) for large enough n which is typically the 'size' of the problem - like the length of the array or number of nodes in the graph or whatever.
This stipulation, that n gets large enough, allows us to pull a lot of useful tricks. Perhaps the most often used one is that you can simplify functions down to their fastest growing terms. For example n^2 + n = O(n^2) because as n gets large enough, the n^2 term gets so much larger than n that the n term is practically insignificant. So we can drop it from consideration.
However, it does mean that big-O notation is less useful for small n, because the slower growing terms that we've forgotten about are still significant enough to affect the run-time.
What we now have is a tool for comparing the costs of two different algorithms, and a shorthand for saying that one is quicker or slower than the other. Big-O notation can be abused which is a shame as it is imprecise enough already! There are equivalent terms for saying that a function grows less quickly than another, and that two functions grow at the same rate.
Oh, and do I use it? Yes, all the time - when I'm figuring out how efficient my code is it gives a great 'back-of-the-envelope- approximation to the cost.
The "Intuitition" behind Big-O
Imagine a "competition" between two functions over x, as x approaches infinity: f(x) and g(x).
Now, if from some point on (some x) one function always has a higher value then the other, then let's call this function "faster" than the other.
So, for example, if for every x > 100 you see that f(x) > g(x), then f(x) is "faster" than g(x).
In this case we would say g(x) = O(f(x)). f(x) poses a sort of "speed limit" of sorts for g(x), since eventually it passes it and leaves it behind for good.
This isn't exactly the definition of big-O notation, which also states that f(x) only has to be larger than C*g(x) for some constant C (which is just another way of saying that you can't help g(x) win the competition by multiplying it by a constant factor - f(x) will always win in the end). The formal definition also uses absolute values. But I hope I managed to make it intuitive.
It may also be worth considering that the complexity of many algorithms is based on more than one variable, particularly in multi-dimensional problems. For example, I recently had to write an algorithm for the following. Given a set of n points, and m polygons, extract all the points that lie in any of the polygons. The complexity is based around two known variables, n and m, and the unknown of how many points are in each polygon. The big O notation here is quite a bit more involved than O(f(n)) or even O(f(n) + g(m)).
Big O is good when you are dealing with large numbers of homogenous items, but don't expect this to always be the case.
It is also worth noting that the actual number of iterations over the data is often dependent on the data. Quicksort is usually quick, but give it presorted data and it slows down. My points and polygons alogorithm ended up quite fast, close to O(n + (m log(m)), based on prior knowledge of how the data was likely to be organised and the relative sizes of n and m. It would fall down badly on randomly organised data of different relative sizes.
A final thing to consider is that there is often a direct trade off between the speed of an algorithm and the amount of space it uses. Pigeon hole sorting is a pretty good example of this. Going back to my points and polygons, lets say that all my polygons were simple and quick to draw, and I could draw them filled on screen, say in blue, in a fixed amount of time each. So if I draw my m polygons on a black screen it would take O(m) time. To check if any of my n points was in a polygon, I simply check whether the pixel at that point is green or black. So the check is O(n), and the total analysis is O(m + n). Downside of course is that I need near infinite storage if I'm dealing with real world coordinates to millimeter accuracy.... ...ho hum.
It may also be worth considering amortized time, rather than just worst case. This means, for example, that if you run the algorithm n times, it will be O(1) on average, but it might be worse sometimes.
A good example is a dynamic table, which is basically an array that expands as you add elements to it. A naïve implementation would increase the array's size by 1 for each element added, meaning that all the elements need to be copied every time a new one is added. This would result in a O(n2) algorithm if you were concatenating a series of arrays using this method. An alternative is to double the capacity of the array every time you need more storage. Even though appending is an O(n) operation sometimes, you will only need to copy O(n) elements for every n elements added, so the operation is O(1) on average. This is how things like StringBuilder or std::vector are implemented.
What is Big O notation?
Big O notation is a method of expressing the relationship between many steps an algorithm will require related to the size of the input data. This is referred to as the algorithmic complexity. For example sorting a list of size N using Bubble Sort takes O(N^2) steps.
Do I use Big O notation?
I do use Big O notation on occasion to convey algorithmic complexity to fellow programmers. I use the underlying theory (e.g. Big O analysis techniques) all of the time when I think about what algorithms to use.
Concrete Examples?
I have used the theory of complexity analysis to create algorithms for efficient stack data structures which require no memory reallocation, and which support average time of O(N) for indexing. I have used Big O notation to explain the algorithm to other people. I have also used complexity analysis to understand when linear time sorting O(N) is possible.
From Wikipedia.....
Big O notation is useful when analyzing algorithms for efficiency. For example, the time (or the number of steps) it takes to complete a problem of size n might be found to be T(n) = 4n² − 2n + 2.
As n grows large, the n² term will come to dominate, so that all other terms can be neglected — for instance when n = 500, the term 4n² is 1000 times as large as the 2n term. Ignoring the latter would have negligible effect on the expression's value for most purposes.
Obviously I have never used it..
You should be able to evaluate an algorithm's complexity. This combined with a knowledge of how many elements it will take can help you to determine if it is ill suited for its task.
It says how many iterations an algorithm has in the worst case.
to search for an item in an list, you can traverse the list until you got the item. In the worst case, the item is in the last place.
Lets say there are n items in the list. In the worst case you take n iterations. In the Big O notiation it is O(n).
It says factualy how efficient an algorithm is.