If you changed the random quicksort algorithm to repeatedly randomly pick a pivot and run partition until it finds a "good" pivot what would be the worst case cost of the algorithm? If we kept track of the pivots used so far so that we never use the same one twice for the same array.
In the worst case, the random pivot always is the maximum or minimum possible value, in which case the runtime is O(n^2).
For a more detailed analysis, look here.
The expected value of the runtime is still O(n * log n) though.
Is there any significance of selecting a random pivot over last element for quick select?
I can still find the required element by always selecting the last element as pivot. Will it effect the runtime.
The choice of pivot has a huge influence on the runtime of quickselect on a given array.
In deterministic quickselect, if you always choose the last element as your pivot, imagine what happens if you try to select out of an always-sorted list. Your pivot will always be the worst possible pivot and will only eliminate one number from the array, leading to an Θ(n2) runtime.
In randomized quickselect, it's still technically possible that the runtime will be Θ(n2) if the algorithm always makes the worst possible choice on each recursive call, but this is extraordinarily unlikely. The runtime will be O(n) with high probability, and there is no "killer" input.
In other words, quickselect with a deterministically-chosen pivot will always have at least one "killer" input that forces it to run in time Θ(n2), while quickselect with a random pivot has no killer inputs and has excellent average time guarantees. The analyses are totally different as a result.
Hope this helps!
I've been comparing the run times of various pivot selection algorithms. Surprisingly the simplest one where the first element is always chosen is the fastest. This may be because I'm filling the array with random data.
If the array has been randomized (shuffled) does it matter? For example picking the medium of 3 as the pivot is always(?) better than picking the first element as the pivot. But this isn't what I've noticed. Is it because if the array is already randomized there would be no reason to assume sortedness, and using the medium is assuming there is some degree of sortedness?
The worst case runtime of quicksort is O(n²). Quicksort is only in average case a fast sorting algorithm.
To reach a average runtime of O(n log n) you have to choose a random pivot element.
But instead of choosing a random pivot element, you can shuffle the list and choose the first element.
To see that this holds you can look at this that way: lets say all elements are in a specific order. Shuffling means you use a random permutation on the list of elements, so a random element will be at the first position and also on all other positions. You can also see it by shuffling the list by randomly choose one of all elements for the first element, then choosing randomly one element of the other (not yet coosen elements) for the second element, and so on.
If your list is already a random generated list, you can directly choose the first element as pivot, without shuffling again.
So, choosing the first element is the fastest one because of the random generated input, but choosing the thrid or the last will also as fast as choosing the first.
All other ways to choose a pivot element have to compute something (a median or a random number or something like this), but they have no advantage over a random choice.
A substantially late response, but I believe it will add some additional info.
Surprisingly the simplest one where the first element is always chosen
is the fastest.
This is actually not surprisingly at all, since you mentioned that you test the algorithm with the random data. In the reality, a percentage of almost-sorted and sorted data is much greater than it would statistically be expected. Take for example the chronological data, when you collect it into the log file some elements can be out of order, but most of them are already sorted. Unfortunately, the Quicksort implementation that takes first (or last) element as a pivot is vulnerable to such input and it degenerates into O(n^2) complexity because in the partition step you divide your array into two halves of size 1 and n-1 and therefore you get n partitions instead of log n, on average.
That's why people decided to add some sort of randomization that would make a probability of getting the problematic input as minimum as possible. There are three well-known approaches:
shuffle the input - to quote Robert Sedgewick, "the probability of getting O(n^2) performance with such approach is lower than the probability that you will be hit by a thunderstrike" :)
choose the pivot element randomly - Wikipedia says that in average, expected number of comparisons in this case is 1.386 n log n
choose the pivot element as a median of three - Wikipedia says that in average, expected number of comparisons in this case is 1.188 n log n
However, randomization costs. If you shuffle the input array, that is O(n) which is dominated by O(nlogn), but you need to take in the account the cost of invoking random(..) method n times. With your simple approach, that is avoided and it is thus faster.
See also:
Worst case for Quicksort - when can it occur?
I'm trying to understand the Select algorithm and I came across a good pivot VS a bad pivot . I can see that the algorithm is using Partition algorithm for separating the bigger elements on the right of
the pivot , and the smaller elements on the left of the pivot .
But what does it mean a bad pivot ?
How can a bad pivot throw the total run time into O(n^2) ?
Thanks
The selection algorithm will be fast if it can discard huge chunks of the array at each step. A good pivot is one that for some definition of "a lot" causes the algorithm to discard "a lot" of the array elements. A bad pivot is one for which the algorithm discards very title of the array.
In the worst case, the pivot might be the largest or smallest element of the array. If this happens, then the algorithm will partition the elements in a way where one group of values is empty, since there will be either no elements less than the pivot or no elements greater than the pivot. This partitioning step takes time O(n), and will have to be run O(n) times, since each iteration decreases the size of the array by one. This degrades the algorithm runtime to O(n2). Interestingly, this is also the way to get quick sort to degenerate to time O(n2).
Hope this helps!
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).