what is the worst-case of quicksort? In my opinion, the worst case is when the min/max element is chosen in every step, e.g for the array 3,2,1,4 choosing 1,2,3,4 (each one was min of the reminder of the array when chosen) or 4,3,2,1 (each one was max) or 4,1,3,2 (4 was max, 1 was min, 3 was max, and 2 was both). In small arrays almost every coise is worst case (for 4 elements, for instance, there are more than those I've listed above), but for large arrays the differene between best/average case and worst case is "huge". By the way, I know this post is duplicate, but I didn't understand from the other posts whether the latter example I've given is correct too or only the former two examples were correct for some reason. Am I right? Is it correct? thanx:)
This is what you can get :
Worst case performance O(n2)
Best case performance O(n log n) (simple partition)
or O(n) (three-way partition and equal keys)
Average case performance : O(n log n)
Related
If an algorithm must make n-1 comparisons to find a certain element, then can we assume that best possible runtime of the algorithm is O(n)?
I know that the lower bound for sorting algorithms is nlogn but since we only return the found one element, I figured it would be possible to do better in terms of run time?
Thanks!
To find a certain element in an unsorted list you need O(n).
But if you sort the array (takes O(n log n) in general) you can find a certain element in O(log n).
So if you want to find often elements in the same list it is most likely worth to sort the list to then be able to find elements much more efficient.
If your array is unsorted and you find some element in it then in worst case Linear search algorithm make n-1 comparisons and time complexity will be O(n).
But if you want to reduce your time complexity then first sort your array and use Binary search algorithm it is take O(logn) in worst case.
So Binary search algorithm is more efficient then linear search.
For unsorted elements, worst case is when you have to go over all the elements, i.e., O(N). If you need many look-ups then you have several pre-processing alternatives that speed up all future accesses.
Option 1: put the elements in a standard hash table. Creating the hash table costs O(N), on average, and later pay O(1) on average for each lookup. This assumes that a reasonable hash-function can be created for this type of elements.
Most languages/libraries implement bucket-based hash-tables, which in pathological cases can put all elements in one bucket, costing O(N) per lookup.
Option 2: there are other hash-table implementations that don't suffer from pathological O(N) cases. The Robin Hood hashing (Wikipedia) (more at Programming.Guide) guarantees O(log N) lookup in the worst case, with average of O(1).
Option 3: another option is to sort elements in O(N log N) once, and then use binary-search to lookup in O(log N). Usually this is slower than Robin Hood hashing (Option 2).
Option 4: If the values are simple integers with limited range, with max-min around N, then it is possible to put the values in an array (list), such that array[value-min] will contain a count of how many times the value appears in the input. It costs O(N) to construct, and O(1) to lookup. Better, the constants for both preprocessing and lookup are significantly lower than in any other method.
Note: I didn't mention the O(N) counting-sort as an alternative to the general case of O(N log N) sorting (option 3), since if max(value)-min(value) is small enough for counting-sort, then option 4 is relevant and is simpler and faster.
If applicable, choose option 4, otherwise if you wish to invest time and code then choose option 2. If 4 isn't applicable, and 2 is not worth the effort in your case, then choose option 2 if you don't mind the pathological worst-case (never choose option 2 when an adversary may want to harm you in a DOS attack).
Your question has nothing to do with sorting, let alone linear search.
If you claim that n-1 comparisons are mandated, then your problem has certainly complexity Ω(n). But with that information alone, you can't guarantee O(n) because it is not said that these n-1 comparisons are sufficient, nor that the algorithm does not perform extra operations, for instance to decide which comparisons to perform. It could turn out that your algorithm is O(n³) with no chance to do better, but we can't tell.
Best case complexity: Ω(n).
Worst case complexity: unknown.
I have a question which I am trying to solve for personal understanding of comparing algorithms as follows. I am given n to be the number of objects and m be the number of keys. I want to find the ratios objects/keys (m/n) that CountingSort is faster in the worst case compared to QuickSort when quicksort chooses the last element as the pivot.
So I have that the worst case running time for CountingSort is O(n+k) and the case when quicksort chooses the last element as the pivot is Theta(n^2). I'm confused of how to approach this question and would welcome some guidance so I could reach a solution.
My idea is that it will be for when the number of keys is double the number of objects? So we have that the runtime for worst case of quicksort is O(n^2) so we want to have counting sort less than this. Therefore if we have rations of m/n < m^2/n this will be the case?
I've been trying to figure this out all day. Some other threads address this, but I really don't understand the answers. There are also many answers that contradict one another.
I understand that an algorithm will never take longer than the upper bound and never be faster than the lower bound. However, I didn't know an upper bound existed for best case time and a lower bound existed for worst case time. This question really threw me in a loop. I can't wrap my head around this... a given run time can have a different upper and lower bound?
For example, if someone asked: "Show that the worst-case running time of some algorithm on a heap of size n is Big Omega(lg(n))". How do you possibly get a lower bound, any bound for that matter, when given a run time?
So, in summation, an algorithm's worst case upper bound can be different than its worst case lower bound? How can this be? Once given the case, don't bounds become irrelevant? Trying to independent study algorithms and I really need to wrap my head around this first.
The meat of my accepted answer to that question is a function whose running time oscillates between n^2 and n^3 depending on whether n is odd. The point that I was trying to make is that sometimes bounds of the form O(n^k) and Omega(n^k) aren't sufficiently descriptive, even though the worst case running time is a perfectly well defined function (which, like all functions, is its own best lower and upper bound). This happens with more natural functions like n log n, which is Omega(n^k) but not O(n^k) for k ≤ 1, and O(n^k) but not Omega(n^k) for k > 1 (and hence not Theta(n^k) regardless of how we choose a constant k).
Suppose you write a program like this to find the smallest prime factor of an integer:
function lpf(n):
for i = 2 to n
if n%i == 0 then return i
If you run the function on the number 10^11 + 3, it will take 10^11 + 2 steps. If you run it on the number 10^11 + 4 it will take just one step. So the function's best-case time is O(1) steps and its worst-case time is O(n) steps.
Big O notation, describes efficiency in runtime iterations, generally based on size of an input data set.
The notation is written in its simplest form, ignoring multiples or additives, but keeping exponential form. If you have an operation of O(1) it is executed in constant time, no matter the input data.
However if you have something such as O(N) or O(log(N)), they will execute at different rates depending on input data.
The high and low bounds describe the largest and least iterations, respectively, that an algorithm can take.
Example: O(N), high bound is largest input data and low bound is smallest.
Extra sources:
Big O Cheat Sheet and MIT Lecture Notes
UPDATE:
Looking at the Stack Overflow question mentioned above, that algorithm is broken into three parts, where it has 3 possible types of runtime, depending on data. Now really, this is three different algorithms designed to handle for different data values. An algorithm is generally classified with just one notation of efficiency and that is of the notation taking the least time for ALL possible values of N.
In the case of O(N^2), larger data will take exponentially longer, and having a smaller number will proceed quickly. The algorithm determines how quickly a data set will be run, yet bounds are given depending on the range of data the algorithm is designed to handle.
I will try to explain it in the quicksort algorithm.
In quicksort you have an array and choose an element as pivot. The next step is to partition the input array into two arrays. The first one will contain elements < pivot and the second one elements > pivot.
Now assume you will apply quicksort on an already sorted list and the pivot element will always be the last element of the array. The result of partition will be an array of size n-1 and an array oft size 1 (the pivot element). This will result in a runtime of O(n*n). Now assume that the pivot element will always split the array in two equal sized array. In every step the array size will be cut in halves. This will result in O(n log n). I hope this example will make this a bit clearer for you.
Another well known sort algorithm is mergesort. Mergesort has always runtime of O(n log n). In mergesort you will cut the array down until only one element is left und will climb up the call stack to merge the one sized arrays and after that merge the array of size two and so on.
Let's say you implement a set using an array. To insert a element you simply put in the next available bucket. If there is no available bucket you increase the capacity of the array by a value m.
For the insert algorithm "there is no enough space" is the worse case.
insert (S, e)
if size(S) >= capacity(S)
reserve(S, size(S) + m)
put(S,e)
Assume we never delete elements. By keeping track of the last available position, put, size and capacity are Θ(1) in space and memory.
What about reserve? If it is implemented like [realloc in C][1], in the best case you just allocate new memory at the end of the existing memory (best case for reserve), or you have to move all existing elements as well (worse case for reserve).
The worst case lower bound for insert is the best case of
reserve(), which is linear in m if we dont nitpick. insert in
worst case is Ω(m) in space and time.
The worst case upper bound for insert is the worse case of
reserve(), which is linear in m+n. insert in worst case is
O(m+n) in space and time.
I am currently reading amortized analysis. I am not able to fully understand how it is different from normal analysis we perform to calculate average or worst case behaviour of algorithms. Can someone explain it with an example of sorting or something ?
Amortized analysis gives the average performance (over time) of each operation in
the worst case.
In a sequence of operations the worst case does not occur often in each operation - some operations may be cheap, some may be expensive Therefore, a traditional worst-case per operation analysis can give overly pessimistic bound. For example, in a dynamic array only some inserts take a linear time, though others - a constant time.
When different inserts take different times, how can we accurately calculate the total time? The amortized approach is going to assign an "artificial cost" to each operation in the sequence, called the amortized cost of an operation. It requires that the total real cost of the sequence should be bounded by the total of the amortized costs of all the operations.
Note, there is sometimes flexibility in the assignment of amortized costs.
Three methods are used in amortized analysis
Aggregate Method (or brute force)
Accounting Method (or the banker's method)
Potential Method (or the physicist's method)
For instance assume we’re sorting an array in which all the keys are distinct (since this is the slowest case, and takes the same amount of time as when they are not, if we don’t do anything special with keys that equal the pivot).
Quicksort chooses a random pivot. The pivot is equally likely to be the smallest key,
the second smallest, the third smallest, ..., or the largest. For each key, the
probability is 1/n. Let T(n) be a random variable equal to the running time of quicksort on
n distinct keys. Suppose quicksort picks the ith smallest key as the pivot. Then we run quicksort recursively on a list of length i − 1 and on a list of
length n − i. It takes O(n) time to partition and concatenate the lists–let’s
say at most n dollars–so the running time is
Here i is a random variable that can be any number from 1 (pivot is the
smallest key) to n (pivot is largest key), each chosen with probability 1/n,
so
This equation is called a recurrence. The base cases for the recurrence are T(0) = 1 and T(1) = 1. This means that sorting a list of length zero or one takes at most one dollar (unit of time).
So when you solve:
The expression 1 + 8j log_2 j might be an overestimate, but it doesn’t
matter. The important point is that this proves that E[T(n)] is in O(n log n).
In other words, the expected running time of quicksort is in O(n log n).
Also there’s a subtle but important difference between amortized running time
and expected running time. Quicksort with random pivots takes O(n log n) expected running time, but its worst-case running time is in Θ(n^2). This means that there is a small
possibility that quicksort will cost (n^2) dollars, but the probability that this
will happen approaches zero as n grows large.
Quicksort O(n log n) expected time
Quickselect Θ(n) expected time
For a numeric example:
The Comparison Based Sorting Lower Bound is:
Finally you can find more information about quicksort average case analysis here
average - a probabilistic analysis, the average is in relation to all of the possible inputs, it is an estimate of the likely run time of the algorithm.
amortized - non probabilistic analysis, calculated in relation to a batch of calls to the algorithm.
example - dynamic sized stack:
say we define a stack of some size, and whenever we use up the space, we allocate twice the old size, and copy the elements into the new location.
overall our costs are:
O(1) per insertion \ deletion
O(n) per insertion ( allocation and copying ) when the stack is full
so now we ask, how much time would n insertions take?
one might say O(n^2), however we don't pay O(n) for every insertion.
so we are being pessimistic, the correct answer is O(n) time for n insertions, lets see why:
lets say we start with array size = 1.
ignoring copying we would pay O(n) per n insertions.
now, we do a full copy only when the stack has these number of elements:
1,2,4,8,...,n/2,n
for each of these sizes we do a copy and alloc, so to sum the cost we get:
const*(1+2+4+8+...+n/4+n/2+n) = const*(n+n/2+n/4+...+8+4+2+1) <= const*n(1+1/2+1/4+1/8+...)
where (1+1/2+1/4+1/8+...) = 2
so we pay O(n) for all of the copying + O(n) for the actual n insertions
O(n) worst case for n operation -> O(1) amortized per one operation.
Is anybody able to give a 'plain english' intuitive, yet formal, explanation of what makes QuickSort n log n? From my understanding it has to make a pass over n items, and it does this log n times...Im not sure how to put it into words why it does this log n times.
Complexity
A Quicksort starts by partitioning the input into two chunks: it chooses a "pivot" value, and partitions the input into those less than the pivot value and those larger than the pivot value (and, of course, any equal to the pivot value have go into one or the other, of course, but for a basic description, it doesn't matter a lot which those end up in).
Since the input (by definition) isn't sorted, to partition it like that, it has to look at every item in the input, so that's an O(N) operation. After it's partitioned the input the first time, it recursively sorts each of those "chunks". Each of those recursive calls looks at every one of its inputs, so between the two calls it ends up visiting every input value (again). So, at the first "level" of partitioning, we have one call that looks at every input item. At the second level, we have two partitioning steps, but between the two, they (again) look at every input item. Each successive level has more individual partitioning steps, but in total the calls at each level look at all the input items.
It continues partitioning the input into smaller and smaller pieces until it reaches some lower limit on the size of a partition. The smallest that could possibly be would be a single item in each partition.
Ideal Case
In the ideal case we hope each partitioning step breaks the input in half. The "halves" probably won't be precisely equal, but if we choose the pivot well, they should be pretty close. To keep the math simple, let's assume perfect partitioning, so we get exact halves every time.
In this case, the number of times we can break it in half will be the base-2 logarithm of the number of inputs. For example, given 128 inputs, we get partition sizes of 64, 32, 16, 8, 4, 2, and 1. That's 7 levels of partitioning (and yes log2(128) = 7).
So, we have log(N) partitioning "levels", and each level has to visit all N inputs. So, log(N) levels times N operations per level gives us O(N log N) overall complexity.
Worst Case
Now let's revisit that assumption that each partitioning level will "break" the input precisely in half. Depending on how good a choice of partitioning element we make, we might not get precisely equal halves. So what's the worst that could happen? The worst case is a pivot that's actually the smallest or largest element in the input. In this case, we do an O(N) partitioning level, but instead of getting two halves of equal size, we've ended up with one partition of one element, and one partition of N-1 elements. If that happens for every level of partitioning, we obviously end up doing O(N) partitioning levels before even partition is down to one element.
This gives the technically correct big-O complexity for Quicksort (big-O officially refers to the upper bound on complexity). Since we have O(N) levels of partitioning, and each level requires O(N) steps, we end up with O(N * N) (i.e., O(N2)) complexity.
Practical implementations
As a practical matter, a real implementation will typically stop partitioning before it actually reaches partitions of a single element. In a typical case, when a partition contains, say, 10 elements or fewer, you'll stop partitioning and and use something like an insertion sort (since it's typically faster for a small number of elements).
Modified Algorithms
More recently other modifications to Quicksort have been invented (e.g., Introsort, PDQ Sort) which prevent that O(N2) worst case. Introsort does so by keeping track of the current partitioning "level", and when/if it goes too deep, it'll switch to a heap sort, which is slower than Quicksort for typical inputs, but guarantees O(N log N) complexity for any inputs.
PDQ sort adds another twist to that: since Heap sort is slower, it tries to avoid switching to heap sort if possible To to that, if it looks like it's getting poor pivot values, it'll randomly shuffle some of the inputs before choosing a pivot. Then, if (and only if) that fails to produce sufficiently better pivot values, it'll switch to using a Heap sort instead.
Each partitioning operation takes O(n) operations (one pass on the array).
In average, each partitioning divides the array to two parts (which sums up to log n operations). In total we have O(n * log n) operations.
I.e. in average log n partitioning operations and each partitioning takes O(n) operations.
There's a key intuition behind logarithms:
The number of times you can divide a number n by a constant before reaching 1 is O(log n).
In other words, if you see a runtime that has an O(log n) term, there's a good chance that you'll find something that repeatedly shrinks by a constant factor.
In quicksort, what's shrinking by a constant factor is the size of the largest recursive call at each level. Quicksort works by picking a pivot, splitting the array into two subarrays of elements smaller than the pivot and elements bigger than the pivot, then recursively sorting each subarray.
If you pick the pivot randomly, then there's a 50% chance that the chosen pivot will be in the middle 50% of the elements, which means that there's a 50% chance that the larger of the two subarrays will be at most 75% the size of the original. (Do you see why?)
Therefore, a good intuition for why quicksort runs in time O(n log n) is the following: each layer in the recursion tree does O(n) work, and since each recursive call has a good chance of reducing the size of the array by at least 25%, we'd expect there to be O(log n) layers before you run out of elements to throw away out of the array.
This assumes, of course, that you're choosing pivots randomly. Many implementations of quicksort use heuristics to try to get a nice pivot without too much work, and those implementations can, unfortunately, lead to poor overall runtimes in the worst case. #Jerry Coffin's excellent answer to this question talks about some variations on quicksort that guarantee O(n log n) worst-case behavior by switching which sorting algorithms are used, and that's a great place to look for more information about this.
Well, it's not always n(log n). It is the performance time when the pivot chosen is approximately in the middle. In worst case if you choose the smallest or the largest element as the pivot then the time will be O(n^2).
To visualize 'n log n', you can assume the pivot to be element closest to the average of all the elements in the array to be sorted.
This would partition the array into 2 parts of roughly same length.
On both of these you apply the quicksort procedure.
As in each step you go on halving the length of the array, you will do this for log n(base 2) times till you reach length = 1 i.e a sorted array of 1 element.
Break the sorting algorithm in two parts. First is the partitioning and second recursive call. Complexity of partioning is O(N) and complexity of recursive call for ideal case is O(logN). For example, if you have 4 inputs then there will be 2(log4) recursive call. Multiplying both you get O(NlogN). It is a very basic explanation.
In-fact you need to find the position of all the N elements(pivot),but the maximum number of comparisons is logN for each element (the first is N,second pivot N/2,3rd N/4..assuming pivot is the median element)
In the case of the ideal scenario, the first level call, places 1 element in its proper position. there are 2 calls at the second level taking O(n) time combined but it puts 2 elements in their proper position. in the same way. there will be 4 calls at the 3rd level which would take O(n) combined time but will place 4 elements into their proper position. so the depth of the recursive tree will be log(n) and at each depth, O(n) time is needed for all recursive calls. So time complexity is O(nlogn).