I have been given a sorting algorithm assignment through school and our task is to review several sorting algorithms. One of the sections for the report is to do with "when are simple sorts faster".
The sorting algorithms i have are:
Bubble sort
Selection sort
Insertion
which are all O(n^2) average
Then i have the following O(n log n) algorithms:
Merge sort
Quick sort
and Radix sort O(kn)
I have ran several tests on unsorted and sorted data ranging from n entries of 10 up to 100,000 and always the complex sorts O(n log n) execute in a faster time.
I also tried using sorted data sets containing n elements where n = 10 up to n = 100,000
But still the O(n log n) algorithms were faster.
SO my question is, when are the simple sorts faster than the complex ones.
Thanks,
Chris.
The branching within inner loops of the sorts that have run time proportional to the square of the input size tends to be very simple.
Insertion sort usually works out best of all in this regard.
On the other hand, sorts with run time proportional to N log N for input of size N have more complicated branching.
The branching complexity shows up as a constant factor in the run time. So we have simple and complex run times of
S N^2 and C N log N
On older computers, S tended to be quite a bit smaller than C. Say just for fun you have S = 1 and C = 4, then you have an equation that can be solved:
1 N^2 = 4 N log N
The solution of interest is at N = 16 (logs are base 2). This is the size of input where the constant factor of better speed that the "simple" algorithm has is overcome by the better asymptotic speed. Inputs smaller than 16 will be sorted faster by the sinple algorithm! The power of the complex algorithm "kicks in" for N > 16.
HOWEVER, in modern computers - as you have discovered - faster branching at the hardware level and better compilers have made S and C much closer. Only for tiny embedded processors is the old folklore about N^2 sorts running faster clearly observable in my experience.
This really depends on the particular implementation of the algorithms. Typically, insertion sort will beat most of the O(n log n) sorts for n between 0 and low tens of elements, but the actual number depends on how insertion sort is implemented, how sorted the data is, how expensive comparisons are, etc. This is something where you pretty much have to run the particular implementations of the sorting algorithms against one another to determine which one is faster when.
Hope this helps!
First of all, radix sort is not O(nlogn). It is O(kn) where k is the maximum number of digits in any number.
O(n^2) sorts can be faster than O(nlogn) sorts when
Size of input is small. If your input contains ~100 (or even ~1000) numbers, modern processors can show O(n^2) sorts to be faster (or same as O(nlogn)).
For some cases an O(n^2) sort can stop in the middle of its sorting when it recognizes the array to be sorted which is difficult (but still possible) to implement in case of O(nlogn) sorts.
For example, in Insertion sort or Flagged Bubble sort, it can be possible to stop the algorithm if the array becomes sorted. This results in best case O(n) performance.
If the input is small or if the input is almost sorted then the "simple" sort algorithms work quite well.
To get the answer for this one you will first understand how Big O works.
10000 n^2 is O(n^2)
10 n^3 is O(n^3)
so lets say n is 2
then 10 n^3 will work way faster than 10000 n^2.
but assuming you don't know the length of n or n can be really big then 10000 n^2 will be better
my question is, when are the simple sorts faster than the complex ones.
It is hard to tell in general; too much depends on the implementation details, on the data and on the hardware, etc.
However, I can give you one practical example. Quicksort has an O(n*log(n)) complexity, insertion sort O(n2) on average, see Sorting in Wikipedia. Yet, it is still worth switching to insertion sort on tiny subarrays in quicksort, see Cutoff to insertion sort in Algorithms by Sedgewick & Wayne (p. 296 in the 4th Ed.). They claim that "any value between 5 and 15 is likely to work well in most situations." The resulting algorithm is better than either quicksort or insertion sort alone.
Related
What I seen:
First I have read these two other SO post
Why is Insertion sort better than Quick sort for small list of elements?
Is there ever a good reason to use Insertion Sort?
But the answers on there are not specific enough for me.
From answers from these two post they mainly pointed out Merge Sort and Quick Sort can be slow because of the extra overhead from the recursive function calls. But I am wondering how the specific threshold 7 get set?
My Question:
I want to know why the cut off is around 7 elements where quadratic sorting algorithm like Insertion Sort is faster than O(nlogn) sorting algorithm like Quick Sort or Merge Sort.
Use insertion sort on small subarrays. Mergesort has too much overhead for tiny subarrays.
Cutoff to insertion sort for ~ 7 elements.
I got this from Princeton lecture slide which I think is reputable enough source. see on the 11th slide under Mergesort: Practical Improvements section.
I will really appreciate it if your answer includes examples for mathematical proof.
Big-O only notes the factor that dominates as n gets large. It ignores constant factors and lesser terms, which pretty much always exist and are more significant when n is small. As a consequence, Big-O is near useless for comparing algorithms that will only ever need to work on tiny inputs.
For example, you can have an O(n log n) function with a time graph like t = 5n log n + 2n + 3, and an O(n^2) function whose time graph was like t = 0.5n^2 + n + 2.
Compare those two graphs, and you'll find that in spite of Big-O, the O(n^2) function would be slightly faster until n reaches about 13.
I have been learning big o efficiency at school as the "go to" method for describing algorithm runtimes as better or worse than others but what I want to know is will the algorithm with the better efficiency always outperform the worst of the lot like bubble sort in every single situation, are there any situations where a bubble sort or a O(n2) algorithm will be better for a task than another algorithm with a lower O() runtime?
Generally, O() notation gives the asymptotic growth of a particular algorithm. That is, the larger category that an algorithm is placed into in terms of asymptotic growth indicates how long the algorithm will take to run as n grows (for some n number of items).
For example, we say that if a given algorithm is O(n), then it "grows linearly", meaning that as n increases, the algorithm will take about as long as any other O(n) algorithm to run.
That doesn't mean that it's exactly as long as any other algorithm that grows as O(n), because we disregard some things. For example, if the runtime of an algorithm takes exactly 12n+65ms, and another algorithm takes 8n+44ms, we can clearly see that for n=1000, algorithm 1 will take 12065ms to run and algorithm 2 will take 8044ms to run. Clearly algorithm 2 requires less time to run, but they are both O(n).
There are also situations that, for small values of n, an algorithm that is O(n2) might outperform another algorithm that's O(n), due to constants in the runtime that aren't being considered in the analysis.
Basically, Big-O notation gives you an estimate of the complexity of the algorithm, and can be used to compare different algorithms. In terms of application, though, you may need to dig deeper to find out which algorithm is best suited for a given project/program.
Big O is gives you the worst cast scenario. That means that it assumes the input in in the worst possible It also ignores the coefficient. If you are using selection sort on an array that is reverse sorted then it will run in n^2 time. If you use selection sort on a sorted array then it will run in n time. Therefore selection sort would run faster than many other sort algorithms on an already sorted list and slower than most (reasonable) algorithms on a reverse sorted list.
Edit: sorry, I meant insertion sort, not selection sort. Selection sort is always n^2
Isn't Insertion sort O(n^2) > Quicksort O(n log n)...so for a small n, won't the relation be the same?
Big-O Notation describes the limiting behavior when n is large, also known as asymptotic behavior. This is an approximation. (See http://en.wikipedia.org/wiki/Big_O_notation)
Insertion sort is faster for small n because Quick Sort has extra overhead from the recursive function calls. Insertion sort is also more stable than Quick sort and requires less memory.
This question describes some further benefits of insertion sort. ( Is there ever a good reason to use Insertion Sort? )
Define "small".
When benchmarking sorting algorithms, I found out that switching from quicksort to insertion sort - despite what everybody was saying - actually hurts performance (recursive quicksort in C) for arrays larger than 4 elements. And those arrays can be sorted with a size-dependent optimal sorting algorithm.
That being said, always keep in mind that O(n...) only is the number of comparisons (in this specific case), not the speed of the algorithm. The speed depends on the implementation, e. g., if your quicksort function as or not recursive and how quickly function calls are dealt with.
Last but not least, big oh notation is only an upper bound.
If algorithm A requires 10000 n log n comparions and algorithm B requires 10 n ^ 2, the first is O(n log n) and the second is O(n ^ 2). Nevertheless, the second will (probably) be faster.
O()-notation is typically used to characterize performance for large problems, while deliberately ignoring constant factors and additive offsets to performance.
This is important because constant factors and overhead can vary greatly between processors and between implementations: the performance you get for a single-threaded Basic program on a 6502 machine will be very different from the same algorithm implemented as a C program running on an Intel i7-class processor. Note that implementation optimization is also a factor: attention to detail can often get you a major performance boost, even if all other factors are the same!
However, the constant factor and overhead are still important. If your application ensures that N never gets very large, the asymptotic behavior of O(N^2) vs. O(N log N) doesn't come into play.
Insertion sort is simple and, for small lists, it is generally faster than a comparably implemented quicksort or mergesort. That is why a practical sort implementation will generally fall back on something like insertion sort for the "base case", instead of recursing all the way down to single elements.
Its a matter of the constants that are attached to the running time that we ignore in the big-oh notation(because we are concerned with order of growth). For insertion sort, the running time is O(n^2) i.e. T(n)<=c(n^2) whereas for Quicksort it is T(n)<=k(nlgn). As c is quite small, for small n, the running time of insertion sort is less then that of Quicksort.....
Hope it helps...
Good real-world example when insertion sort can be used in conjunction with quicksort is the implementation of qsort function from glibc.
The first thing to point is qsort implements quicksort algorithm with a stack because it consumes less memory, stack implemented through macros directives.
Summary of current implementation from the source code (you'll find a lot of useful information through comments if you take a look at it):
Non-recursive
Chose the pivot element using a median-of-three decision tree
Only quicksorts TOTAL_ELEMS / MAX_THRESH partitions, leaving
insertion sort to order the MAX_THRESH items within each partition.
This is a big win, since insertion sort is faster for small, mostly
sorted array segments.
The larger of the two sub-partitions is always pushed onto the
stack first
What is MAX_THRESH value stands for? Well, just a small constant magic value which
was chosen to work best on a Sun 4/260.
How about binary insertion sort? You can absolutely search the position to swap by using binary search.
Is there a sort of an array that works in O(n*log(n)) worst case time complexity?
I saw in Wikipedia that there are sorts like that, but they are unstable, what does that mean? Is there a way to do in low space complexity?
Is there a best sorting algorithm?
An algorithm that requires only O(1) extra memory (so modifying the input array is permitted) is generally described as "in-place", and that's the lowest space complexity there is.
A sort is described as "stable" or not, according to what happens when there are two elements in the input which compare as equal, but are somehow distinguishable. For example, suppose you have a bunch of records with an integer field and a string field, and you sort them on the integer field. The question is, if two records have the same integer value but different string values, then will the one that came first in the input, also come first in the output, or is it possible that they will be reversed? A stable sort is one that guarantees to preserve the order of elements that compare the same, but aren't identical.
It is difficult to make a comparison sort that is in-place, and stable, and achieves O(n log n) worst-case time complexity. I've a vague idea that it's unknown whether or not it's possible, but I don't keep up to date on it.
Last time someone asked about the subject, I found a couple of relevant papers, although that question wasn't identical to this question:
How to sort in-place using the merge sort algorithm?
As far as a "best" sort is concerned - some sorting strategies take advantage of the fact that on the whole, taken across a large number of applications, computers spend a lot of time sorting data that isn't randomly shuffled, it has some structure to it. Timsort is an algorithm to take advantage of commonly-encountered structure. It performs very well in a lot of practical applications. You can't describe it as a "best" sort, since it's a heuristic that appears to do well in practice, rather than being a strict improvement on previous algorithms. But it's the "best" known overall in the opinion of people who ship it as their default sort (Python, Java 7, Android). You probably wouldn't describe it as "low space complexity", though, it's no better than a standard merge sort.
You can check out between mergesort, quicksort or heapsort all nicely described here.
There is also radix sort whose complexity is O(kN) but it takes full advantage of extra memory consumption.
You can also see that for smaller collections quicksort is faster but then mergesort takes the lead but all of this is case specific so take your time to study all 4 algorithms
For the question best algorithm, the simple answer is, it depends.It depends on the size of the data set you want to sort,it depends on your requirement.Say, Bubble sort has worst-case and average complexity both О(n2), where n is the number of items being sorted. There exist many sorting algorithms with substantially better worst-case or average complexity of O(n log n). Even other О(n2) sorting algorithms, such as insertion sort, tend to have better performance than bubble sort. Therefore, bubble sort is not a practical sorting algorithm when n is large.
Among simple average-case Θ(n2) algorithms, selection sort almost always outperforms bubble sort, but is generally outperformed by insertion sort.
selection sort is greatly outperformed on larger arrays by Θ(n log n) divide-and-conquer algorithms such as mergesort. However, insertion sort or selection sort are both typically faster for small arrays.
Likewise, you can yourself select the best sorting algorithm according to your requirements.
It is proven that O(n log n) is the lower bound for sorting generic items. It is also proven that O(n) is the lower bound for sorting integers (you need at least to read the input :) ).
The specific instance of the problem will determine what is the best algorithm for your needs, ie. sorting 1M strings is different from sorting 2M 7-bits integers in 2MB of RAM.
Also consider that besides the asymptotic runtime complexity, the implementation is making a lot of difference, as well as the amount of available memory and caching policy.
I could implement quicksort in 1 line in python, roughly keeping O(n log n) complexity (with some caveat about the pivot), but Big-Oh notation says nothing about the constant terms, which are relevant too (ie. this is ~30x slower than python built-in sort, which is likely written in C btw):
qsort = lambda a: [] if not a else qsort(filter(lambda x: x<a[len(a)/2], a)) + filter(lambda x: x == a[len(a)/2], a) + qsort(filter(lambda x: x>a[len(a)/2], a))
For a discussion about stable/unstable sorting, look here http://www.developerfusion.com/article/3824/a-guide-to-sorting/6/.
You may want to get yourself a good algorithm book (ie. Cormen, or Skiena).
Heapsort, maybe randomized quicksort
stable sort
as others already mentioned: no there isn't. For example you might want to parallelize your sorting algorithm. This leads to totally different sorting algorithms..
Regarding your question meaning stable, let's consider the following: We have a class of children associated with ages:
Phil, 10
Hans, 10
Eva, 9
Anna, 9
Emil, 8
Jonas, 10
Now, we want to sort the children in order of ascending age (and nothing else). Then, we see that Phil, Hans and Jonas all have age 10, so it is not clear in which order we have to order them since we sort just by age.
Now comes stability: If we sort stable we sort Phil, Hans and Jonas in the order they were before, i.e. we put Phil first, then Hans, and at last, Jonas (simply because they were in this order in the original sequence and we only consider age as comparison criterion). Similarily, we have to put Eva before Anna (both the same age, but in the original sequence Eva was before Anna).
So, the result is:
Emil, 8
Eva, 9
Anna, 9
Phil, 10 \
Hans, 10 | all aged 10, and left in original order.
Jonas, 10 /
To put it in a nutshell: Stability means that if two elements are equal (w.r.t. the chosen sorting criterion), the one coming first in the original sequence still comes first in the resulting sequence.
Note that you can easily transform any sorting algorithm into a stable sorting algorithm: If your original sequence holds n elements: e1, e2, e3, ..., en, you simply attach a counter to each one: (e1, 0), (e2, 1), (e3, 2), ..., (en, n-1). This means you store for each element its original position.
If now two elements are equal, you simply compare their counters and put the one with the lower counter value first. This increases runtime (and memory) by O(n), which is asymptotic no worsening since the best (comparison) sort algorithm needs already O(n lg n).
Is it theoretically possible to sort an array of n integers in an amortized complexity of O(n)?
What about trying to create a worst case of O(n) complexity?
Most of the algorithms today are built on O(nlogn) average + O(n^2) worst case.
Some, while using more memory are O(nlogn) worst.
Can you with no limitation on memory usage create such an algorithm?
What if your memory is limited? how will this hurt your algorithm?
Any page on the intertubes that deals with comparison-based sorts will tell you that you cannot sort faster than O(n lg n) with comparison sorts. That is, if your sorting algorithm decides the order by comparing 2 elements against each other, you cannot do better than that. Examples include quicksort, bubblesort, mergesort.
Some algorithms, like count sort or bucket sort or radix sort do not use comparisons. Instead, they rely on the properties of the data itself, like the range of values in the data or the size of the data value.
Those algorithms might have faster complexities. Here is an example scenario:
You are sorting 10^6 integers, and each integer is between 0 and 10. Then you can just count the number of zeros, ones, twos, etc. and spit them back out in sorted order. That is how countsort works, in O(n + m) where m is the number of values your datum can take (in this case, m=11).
Another:
You are sorting 10^6 binary strings that are all at most 5 characters in length. You can use the radix sort for that: first split them into 2 buckets depending on their first character, then radix-sort them for the second character, third, fourth and fifth. As long as each step is a stable sort, you should end up with a perfectly sorted list in O(nm), where m is the number of digits or bits in your datum (in this case, m=5).
But in the general case, you cannot sort faster than O(n lg n) reliably (using a comparison sort).
I'm not quite happy with the accepted answer so far. So I'm retrying an answer:
Is it theoretically possible to sort an array of n integers in an amortized complexity of O(n)?
The answer to this question depends on the machine that would execute the sorting algorithm. If you have a random access machine, which can operate on exactly 1 bit, you can do radix sort for integers with at most k bits, which was already suggested. So you end up with complexity O(kn).
But if you are operating on a fixed size word machine with a word size of at least k bits (which all consumer computers are), the best you can achieve is O(n log n). This is because either log n < k or you could do a count sort first and then sort with a O (n log n) algorithm, which would yield the first case also.
What about trying to create a worst case of O(n) complexity?
That is not possible. A link was already given. The idea of the proof is that in order to be able to sort, you have to decide for every element to be sorted if it is larger or smaller to any other element to be sorted. By using transitivity this can be represented as a decision tree, which has n nodes and log n depth at best. So if you want to have performance better than Ω(n log n) this means removing edges from that decision tree. But if the decision tree is not complete, than how can you make sure that you have made a correct decision about some elements a and b?
Can you with no limitation on memory usage create such an algorithm?
So as from above that is not possible. And the remaining questions are therefore of no relevance.
If the integers are in a limited range then an O(n) "sort" of them would involve having a bit vector of "n" bits ... looping over the integers in question and setting the n%8 bit of offset n//8 in that byte array to true. That is an "O(n)" operation. Another loop over that bit array to list/enumerate/return/print all the set bits is, likewise, an O(n) operation. (Naturally O(2n) is reduced to O(n)).
This is a special case where n is small enough to fit within memory or in a file (with seek()) operations). It is not a general solution; but it is described in Bentley's "Programming Pearls" --- and was allegedly a practical solution to a real-world problem (involving something like a "freelist" of telephone numbers ... something like: find the first available phone number that could be issued to a new subscriber).
(Note: log(10*10) is ~24 bits to represent every possible integer up to 10 digits in length ... so there's plenty of room in 2*31 bits of a typical Unix/Linux maximum sized memory mapping).
I believe you are looking for radix sort.