Just out of curiosity. Let's say I have a list which contain N elements(which will repeat) and a function which will return the frequency of these elements. I think the time complexity of this program should be O(N) right? As the function just need to loop through N and check whether the elements is already exist, if yes, +=, else = 1. Okay, so my friend and I have an argument as, how about if we need to multiple the elements with its frequency as well? And maybe divided by its total number? My friend think the complexity should be O(N^2) but it doesn't sounds right for me. What do you think, and why?
Thank you.
It depends on how you will record the frequencies. If you will be using an array they for each += you need to find the previous frequency value first, the complexity is quadratic. However if you maintain a hash table structure, on which access is instant on average, the complexity would be linear.
Related
My teacher made this argument and asked us what could be wrong with it.
for an array A of n distinct numbers. Since there are n! permutations of A,
we cannot check for each permutation whether it is sorted, in a total time which
is polynomial in n. Therefore, sorting A cannot be in P.
my friend thought that it just should be : therefore sorting a cannot be in NP.
Is that it or are we thinking to easily about it?
The problem with this argument is that it fails to adequately specify the exact problem.
Sorting can be linear-time (O(n)) in the number of elements to sort, if you're sorting a large list of integers drawn from a small pool (counting sort, radix sort).
Sorting can be linearithmic-time (O(nlogn)) in the number of elements to sort, if you're sorting a list of arbitrary things which are all totally ordered according to some ordering relation (e.g., less than or equal to on the integers).
Sorting based on a partial order (e.g. topological sorting) must be analyzed in yet another way.
We can imagine a problem like sorting whereby the sortedness of a list cannot be determined by comparing adjacent entries only. In the extreme case, sortedness (according to what we are considering to be sorting) might only be verifiable by checking the entire list. If our kind of sorting is designed so as to guarantee there is exactly one sorted permutation of any given list, the time complexity is factorial-time (O(n!)) and the problem is not in P.
That's the real problem with this argument. If your professor is assuming that "sorting" refers to sorting integers not in any particular small range, the problem with the argument then is that we do not need to consider all permutations in order to construct the sorted one. If I have a bag with 100 marbles and I ask you to remove three marbles, the time complexity is constant-time; it doesn't matter that there are n(n-1)(n-2)/6 = 161700, or O(n^3), ways in which you can accomplish this task.
The argument is a non-sequitur, the conclusion does not logically follow from the previous steps. Why doesn't it follow? Giving a satisfying answer to that question requires knowing why the person who wrote the argument thinks it is correct, and then addressing their misconception. In this case, the person who wrote the argument is your teacher, who doesn't think the argument is correct, so there is no misconception to address and hence no completely satisfying answer to the question.
That said, my explanation would be that the argument is wrong because it proposes a specific algorithm for sorting - namely, iterating through all n! permutations and choosing the one which is sorted - and then assumes that the complexity of the problem is the same as the complexity of that algorithm. This is a mistake because the complexity of a problem is defined as the lowest complexity out of all algorithms which solve it. The argument only considered one algorithm, and didn't show anything about other algorithms which solve the problem, so it cannot reach a conclusion about the complexity of the problem itself.
Probably a simple qustion, but I'm a beginner, and find time complexity difficult.
What is the time complexeity when traversing n linked lists with, say, one hundred(100) elements?
Thanks!
If the size of each list is limited by a constant number (e.g. every list contains no more than 100 elements), then the time complexity is linear: O(n).
More generally, if the size of each list is limited by some function of n, say, f(n), then the time complexity is O(n*f(n)).
Consider 2 sequences X[1..m] and Y[1..n]. The memoization algorithm would compute the LCS in time O(m*n). Is there any better algorithm to find out LCS wrt time? I guess memoization done diagonally can give us O(min(m,n)) time complexity.
Gene Myers in 1986 came up with a very nice algorithm for this, described here: An O(ND) Difference Algorithm and Its Variations.
This algorithm takes time proportional to the edit distance between sequences, so it is much faster when the difference is small. It works by looping over all possible edit distances, starting from 0, until it finds a distance for which an edit script (in some ways the dual of an LCS) can be constructed. This means that you can "bail out early" if the difference grows above some threshold, which is sometimes convenient.
I believe this algorithm is still used in many diff implementations.
If you know a priori an upper bound on the maximum size k you care about, you can force the LCS algorithm to exit early by adding an extra check in the inner loop. This means then when k << min(m,n) you can get small running times in spite of the fact you are doing LCS.
yes we could create a better algorithm than Order O(m*n)---
i.e O(min(m,n)). to find a length.....
just compare the diagonal elements.and whenever the increment is done suppose it occured in c[2,2] then increment all the value from c[2,2++] and c[2++,2] by 1..
and proceed till c[m,m]..(suppose m
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.