I read that it's possible to make quicksort run at O(nlogn)
the algorithm says on each step choose the median as a pivot
but, suppose we have this array:
10 8 39 2 9 20
which value will be the median?
In math if I remember correct the median is (39+2)/2 = 41/2 = 20.5
I don't have a 20.5 in my array though
thanks in advance
You can choose either of them; if you consider the input as a limit, it does not matter as it scales up.
We're talking about the exact wording of the description of an algorithm here, and I don't have the text you're referring to. But I think in context by "median" they probably meant, not the mathematical median of the values in the list, but rather the middle point in the list, i.e. the median INDEX, which in this cade would be 3 or 4. As coffNjava says, you can take either one.
The median is actually found by sorting the array first, so in your example, the median is found by arranging the numbers as 2 8 9 10 20 39 and the median would be the mean of the two middle elements, (9+10)/2 = 9.5, which doesn't help you at all. Using the median is sort of an ideal situation, but would work if the array were at least already partially sorted, I think.
With an even numbered array, you can't find an exact pivot point, so I believe you can use either of the middle numbers. It'll throw off the efficiency a bit, but not substantially unless you always ended up sorting even arrays.
Finding the median of an unsorted set of numbers can be done in O(N) time, but it's not really necessary to find the true median for the purposes of quicksort's pivot. You just need to find a pivot that's reasonable.
As the Wikipedia entry for quicksort says:
In very early versions of quicksort, the leftmost element of the partition would often be chosen as the pivot element. Unfortunately, this causes worst-case behavior on already sorted arrays, which is a rather common use-case. The problem was easily solved by choosing either a random index for the pivot, choosing the middle index of the partition or (especially for longer partitions) choosing the median of the first, middle and last element of the partition for the pivot (as recommended by R. Sedgewick).
Finding the median of three values is much easier than finding it for the whole collection of values, and for collections that have an even number of elements, it doesn't really matter which of the two 'middle' elements you choose as the potential pivot.
Related
I think the speed of quick sort is less efficient when arranging an array with duplicate data, right? when datatype is char, the bigger the array(over 100000), the closer it gets to the n^2 order.
and assuming there is no duplicate data, to get the best case of a quick sort where the first element is placed as a pivot, first elementsI think we can recursively change the first and intermediate elements by dividing the already aligned array like a merge sort. right? is there general best case?
Lomuto partition scheme, which scans from one end to the other during partition, is slower with duplicates. If all the values are the same, then each partition step splits it into sizes 1 and n-1, a worst case scenario.
Hoare partition scheme, which scans from both both ends towards each other until the indexes (or iterators or pointers) cross, is usually faster with duplicates. Even though duplicates result in more swaps, each swap occurs just after reading and comparing two values to the pivot and are still in the cache for the swap (assuming object size is not huge). As the number of duplicates increases, the splitting improves towards the ideal case where each partition step splits the data into two equal halves. I ran a benchmark sorting 16 million 64 bit integers: with random data, it took about 1.37 seconds, improving with duplicates and with all values the same, it took about about 0.288 seconds.
Another alternative is a 3 way partition, which splits a partition into elements < pivot, elements == pivot, elements > pivot. If all the elements are the same, it's done in O(n) time. For n elements with only k possible values, then time complexity is O(n ⌈log3(k)⌉), and since k is constant, the time complexity is still O(n).
Wiki links:
https://en.wikipedia.org/wiki/Quicksort#Repeated_elements
https://en.wikipedia.org/wiki/Dutch_national_flag_problem
Well, I want to use Quick sort on given 3 values, doesn't matter what values, how can I get to the worst case which is 9 operations?
Can anyone draw a tree and show how it show nlogn and n^2 operations? I've tried to find on the internet, but I still didnt manage to draw one properly to show that.
The worst case complexity of quick sort depends on the chosen pivot. If the pivot chosen is the leftmost or the rightmost element. Then the worst case complexity will occur in the following cases:
1) Array is already sorted in same order.
2) Array is already sorted in reverse order.
3) All elements are same (special case of case 1 and 2).
Since these cases occur very frequently, the pivot is chosen randomly. By choosing pivot randomly the chances of worst case are reduced.
The analysis of quicksort algorithm is explained in this blogpost by khan academy.
I need to derive clusters of integers from an input array of integers in such a way that the variation within the clusters is minimized. (The integers or data values in the array are corresponding to the gas usage of 16 cars running between cities. At the end I will derive 4 clusters from the 16 cars into based on the clusters of the data values.)
Constraints: always the number of elements is 16, no. of clusters is 4 and the size of
the cluster is 4.
One simple way I am planning to do is that I will sort the input array and then divide them into 4 groups as shown below. I think that I can also use k-means clustering.
However, the place where I stuck was as follows: The data in the array change over time. Basically I need to monitor the array for every 1 second and regroup/recluster them so that the variation within the cluster is minimized. Moreover, I need to satisfy the above constraint. For this, one idea I am getting is to select two groups based on their means and variations and move data values between the groups to minimize variation within the group. However, I am not getting any idea of how to select the data values to move between the groups and also how to select those groups. I cannot apply sorting on the array in every second because I cannot afford NlogN for every second. It would be great if you guide me to produce a simple solution.
sorted `input array: (12 14 16 16 18 19 20 21 24 26 27 29 29 30 31 32)`
cluster-1: (12 14 16 16)
cluster-2: (18 19 20 21)
cluster-3: (24 26 27 29)
cluster-4: (29 30 31 32)
Let me first point out that sorting a small number of objects is very fast. In particular when they have been sorted before, an "evil" bubble sort or insertion sort usually is linear. Consider in how many places the order may have changed! All of the classic complexity discussion doesn't really apply when the data fits into the CPUs first level caches.
Did you know that most QuickSort implementations fall back to insertion sort for small arrays? Because it does a fairly good job for small arrays and has little overhead.
All the complexity discussions are only for really large data sets. They are in fact proven only for inifinitely sized data. Before you reach infinity, a simple algorithm of higher complexity order may still perform better. And for n < 10, quadratic insertion sort often outperforms O(n log n) sorting.
k-means however won't help you much.
Your data is one-dimensional. Do not bother to even look at multidimensional methods, they will perform worse than proper one-dimensional methods (which can exploit that the data can be ordered)
If you want guaranteed runtime, k-means with possibly many iterations is quite uncontrolled.
You can't easily add constraints such as the 4-cars rule into k-means
I believe the solution to your task (because of the data being 1 dimensional and the constraints you added) is:
Sort the integers
Divide the sorted list into k even-sized groups
While working with Quick Sort algorithm I wondered whether any formula or some kind of stuff might be available for finding the no of passes that a particular set of values may take to completely sorted in ascending order.
Is there any formula to calculate the no of passes that a Quick Sort algorithm will take?
Any given set of values will have a different number of operations, based on pivot value selection method, and the actual values being sorted.
So...no, unless the approximations of 'between O(N log(N)) and O(N^2)' is good enough.
That one has to qualify the average versus worst case should be enough to show that the only way to determine the number of operations is to actually run the quicksort.
I've read that the pivote can be the median of 3 numbers, bottom, middle and top. But, could that generate overflow? What happens if the median returns a value larger than the array size?
I assume that the this choice is by assuming that they array values can't be longer than the array size.
I think I'm confused at what a pivote really is.
The pivot is just the value that you compare other values against - lower values go the left, higher to the right. The pivot can be chosen by taking any of the existing values in the array. If the array is completely unsorted, it won't matter which value you choose. If it is somewhat sorted, you should choose a value from the middle of the array.
UPDATE: Some reading informs me that a better pivot choice may be to choose the median value of 3 values in the array (such as middle, bottom and top or 3 random positions). Some people advocate taking the median of 5 values. The worst-case performance of quicksort occurs when pivot is close to the smallest or largest value in the array, and this tactic is intended to defend against that occurring. This is just an optimisation for certain kinds of data - it is not a necessity.