In Merge Sort, the running time of Divide step for every element is considered theta(1) i.e constant time in CLRS.
My confusion is :
Let divide is an elementary operation and which takes constant time C1.
Now, Consider the pesudocode
Now, When MERGE-SORT function called for the first time then it will take C1 to divide the input (Note:- We are not considering Time taken by Merge function)
And this MERGE-SORT will divide the input lg(n) time ( or we can say it will call itself lg((n)) times )
So the total time taken in dividing inputs should be C1 * lg(n)
that is theta(lg(n)).
But In CLRS :
If we consider that then Divide Step will require theta(1) time ( as whole)
NOTE: lg is log to the base 2
P.S. - Sorry in advance because my English is not upto that mark. Edits are welcome :)
One step of division indeed takes constant time O(1).
But contrary to your thinking, there are Θ(N) divisions in total (1+2+4+8+...N/2).
Anyway this is not accounted for as the total merging workload is Θ(N Log N).
The basic understanding of Merge Sort is as such - Consider we have two halves of an array which are already sorted. We can merge these two halves in O(n) time.
Mathematically speaking, the function Merge Sort calls itself twice for two of its halves (The DIVIDE step), and then merges these two (The CONQUER step).
Now, if we solve this recurrence relation (by using appropriate mathematical analysis), we get the relation as follows:
This clearly indicates that the DIVIDE step that takes O(1) time is called n times, while the total merging workload of merging all the smaller halves of various sizes is O(n logn), which is the upper limit. Hence the complexity of O(n logn) (Thanks to Yves Daoust for this clarification).
However, the DIVIDE step takes O(1) time only, which is called O(n) times. Thus, the workload for division is O(n).
Related
Consider the problem:
Although merge sort runs in Θ(nlgn) worst-case time and insertion sort runs in Θ(n^2) worstcase
time, the constant factors in insertion sort makes it faster for small n. Thus, it makes sense to use insertion sort within merge sort when subproblems become sufficiently small.
Consider a modification for merge sort in which n/k sublists of length k are sorted using insertion sort and then merged using the standard merging mechanism, where k is a value to be determined.
Question: Show that the sublists can be merged in Θ(n lg(n/k)) worst-case time:
My solution:
To merge n/k sublists into n/2k it takes Θ(n) times
To merge n/2k sublists into n/4k it takes Θ(n) times
...
To merge 2 sublists into 1 it takes Θ(n) times
Then I was struggling with further steps and I had a look at the solution:
We have lg(n/k) such merges, so merging n/k sublists into one list takes Θ(n lg(n/k)) worst-case time.
I have two questions:
1)How do they end up with lg(n/k) merges? Please, clarify the calculations?
2)Why is the final result Θ(n lg(n/k))?
You seem to be pretty close to the actual answer. I believe the phrasing of the answer you looked up is what makes it harder for you to understand, because I do not think that the total number of individual merges required is lg(n/k). What I believe the answer refers to is the number of merging steps required until we end up with the sorted list.
Instead of the answer, however, let's continue building onto your reasoning. Merging two lists of length k has O(k) time complexity. To merge n/k such lists into n/(2k) lists, we will do n/(2k) merges with complexity O(k) each, resulting in an overall O(n) complexity, as you mentioned.
You may extend this logic to the next step, where n/(2k) lists are merged into n/(4k), and state that the second step has O(n) complexity, as well. In fact, each merging step, will take O(n) time.
The next thing to do here is estimating how many of these merging steps we have. We started with n/k lists, and after the first step we obtained n/(2k) lists. After that, at each step, the number of lists is halved, until there is only 1 list left, which will be our result. (i.e. sorted list) Well, how many times do you think we have to divide n/k by 2, until we end up with 1? That is exactly what log(n/k) means, isn't it? So, there will be log(n/k) of such merging steps, each of which takes O(n).
Consequently, the entire procedure will have a time complexity of O(nlog(n/k)).
Question:
Here's a modification of quick sort: whenever we have ten items or fewer in a sublist, we sort the sublist using selection sort rather than further recursing with quicksort. Does this change the big-oh time complexity of quicksort? Explain.
In my opinion the big-oh time complexity would change. We know that selection sort is O(n^2) and therefore sorting the sublist of ten items or fewer would take O(n^2). Until we get to a sublist that has ten or fewer items we would use quicksort and keep partitioning the list. So in the end we would have O( nlogn + n^2) which is O(n^2).
Am I correct? If not, could someone explain why?
The reason that the time complexity is actually unaffected is that 10 is a constant term. No matter how large the total array is, it always takes a constant amount of time to sort subarrays of size 10 or less. If you are sorting a list with one million elements, that constant 10 is going to play a very small role in the actual time it takes to sort the list (most of the time will be spent partitioning the original array into subarrays recursively).
If sorting a list of 10 elements takes constant time, partitioning the array at each recursive step is linear, and you end up with log n subarrays of 10 items or fewer, you end up with O(n log n + log n), which is the same as O(n log n).
Saying that selection sort is O(n^2) means that the running time of the algorithm increases quadratically with the size of the input. Running selection sort on an array with a constant number of elements will always take constant time on its own, but if you were to compare the running time of selection sort on arrays of varying sizes, you would see a quadratic increase in the running time as input size varies linearly.
The big O complexity does not change. Please read up on the Master Method (aka Master Theorem) https://en.wikipedia.org/wiki/Master_theorem
If you think through the algorithm as the size of the sorting list grows exceptionally large the time to sort the final ten in any given recursion substree will make insignificant contributions to overall running time.
I'm trying to find out which is the Theta complexity of this algorithm.
(a is a list of integers)
def sttr(a):
for i in xrange(0,len(a)):
while s!=[] and a[i]>=a[s[-1]]:
s.pop()
s.append(i)
return s
On the one hand, I can say that append is being executed n (length of a array) times, so pop too and the last thing I should consider is the while condition which could be executed probably 2n times at most.
From this I can say that this algorithm is at most 4*n so it is THETA(n).
But isn't it amortised analysis?
On the other hand I can say this:
There are 2 nested cycles. The for cycle is being executed exactly n times. The while cycle could be executed at most n times since I have to remove item in each iteration. So the complexity is THETA(n*n).
I want to compute THETA but don't know which of these two options is correct. Could you give me advice?
The answer is THETA(n) and your arguments are correct.
This is not amortized analysis.
To get to amortized analysis you have to look at the inner loop. You can't easily say how fast the while will execute if you ignore the rest of the algorithm. Naive approach would be O(N) and that's correct since that's the maximum number of iterations. However, since we know that the total number of executions is O(N) (your argument) and that this will be executed N time we can say that the complexity of the inner loop is O(1) amortized.
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.
I know there are quite a bunch of questions about big O notation, I have already checked:
Plain english explanation of Big O
Big O, how do you calculate/approximate it?
Big O Notation Homework--Code Fragment Algorithm Analysis?
to name a few.
I know by "intuition" how to calculate it for n, n^2, n! and so, however I am completely lost on how to calculate it for algorithms that are log n , n log n, n log log n and so.
What I mean is, I know that Quick Sort is n log n (on average).. but, why? Same thing for merge/comb, etc.
Could anybody explain me in a not too math-y way how do you calculate this?
The main reason is that Im about to have a big interview and I'm pretty sure they'll ask for this kind of stuff. I have researched for a few days now, and everybody seem to have either an explanation of why bubble sort is n^2 or the unreadable explanation (for me) on Wikipedia
The logarithm is the inverse operation of exponentiation. An example of exponentiation is when you double the number of items at each step. Thus, a logarithmic algorithm often halves the number of items at each step. For example, binary search falls into this category.
Many algorithms require a logarithmic number of big steps, but each big step requires O(n) units of work. Mergesort falls into this category.
Usually you can identify these kinds of problems by visualizing them as a balanced binary tree. For example, here's merge sort:
6 2 0 4 1 3 7 5
2 6 0 4 1 3 5 7
0 2 4 6 1 3 5 7
0 1 2 3 4 5 6 7
At the top is the input, as leaves of the tree. The algorithm creates a new node by sorting the two nodes above it. We know the height of a balanced binary tree is O(log n) so there are O(log n) big steps. However, creating each new row takes O(n) work. O(log n) big steps of O(n) work each means that mergesort is O(n log n) overall.
Generally, O(log n) algorithms look like the function below. They get to discard half of the data at each step.
def function(data, n):
if n <= constant:
return do_simple_case(data, n)
if some_condition():
function(data[:n/2], n / 2) # Recurse on first half of data
else:
function(data[n/2:], n - n / 2) # Recurse on second half of data
While O(n log n) algorithms look like the function below. They also split the data in half, but they need to consider both halves.
def function(data, n):
if n <= constant:
return do_simple_case(data, n)
part1 = function(data[n/2:], n / 2) # Recurse on first half of data
part2 = function(data[:n/2], n - n / 2) # Recurse on second half of data
return combine(part1, part2)
Where do_simple_case() takes O(1) time and combine() takes no more than O(n) time.
The algorithms don't need to split the data exactly in half. They could split it into one-third and two-thirds, and that would be fine. For average-case performance, splitting it in half on average is sufficient (like QuickSort). As long as the recursion is done on pieces of (n/something) and (n - n/something), it's okay. If it's breaking it into (k) and (n-k) then the height of the tree will be O(n) and not O(log n).
You can usually claim log n for algorithms where it halves the space/time each time it runs. A good example of this is any binary algorithm (e.g., binary search). You pick either left or right, which then axes the space you're searching in half. The pattern of repeatedly doing half is log n.
For some algorithms, getting a tight bound for the running time through intuition is close to impossible (I don't think I'll ever be able to intuit a O(n log log n) running time, for instance, and I doubt anyone will ever expect you to). If you can get your hands on the CLRS Introduction to Algorithms text, you'll find a pretty thorough treatment of asymptotic notation which is appropriately rigorous without being completely opaque.
If the algorithm is recursive, one simple way to derive a bound is to write out a recurrence and then set out to solve it, either iteratively or using the Master Theorem or some other way. For instance, if you're not looking to be super rigorous about it, the easiest way to get QuickSort's running time is through the Master Theorem -- QuickSort entails partitioning the array into two relatively equal subarrays (it should be fairly intuitive to see that this is O(n)), and then calling QuickSort recursively on those two subarrays. Then if we let T(n) denote the running time, we have T(n) = 2T(n/2) + O(n), which by the Master Method is O(n log n).
Check out the "phone book" example given here: What is a plain English explanation of "Big O" notation?
Remember that Big-O is all about scale: how much more operation will this algorithm require as the data set grows?
O(log n) generally means you can cut the dataset in half with each iteration (e.g. binary search)
O(n log n) means you're performing an O(log n) operation for each item in your dataset
I'm pretty sure 'O(n log log n)' doesn't make any sense. Or if it does, it simplifies down to O(n log n).
I'll attempt to do an intuitive analysis of why Mergesort is n log n and if you can give me an example of an n log log n algorithm, I can work through it as well.
Mergesort is a sorting example that works through splitting a list of elements repeatedly until only elements exists and then merging these lists together. The primary operation in each of these merges is comparison and each merge requires at most n comparisons where n is the length of the two lists combined. From this you can derive the recurrence and easily solve it, but we'll avoid that method.
Instead consider how Mergesort is going to behave, we're going to take a list and split it, then take those halves and split it again, until we have n partitions of length 1. I hope that it's easy to see that this recursion will only go log (n) deep until we have split the list up into our n partitions.
Now that we have that each of these n partitions will need to be merged, then once those are merged the next level will need to be merged, until we have a list of length n again. Refer to wikipedia's graphic for a simple example of this process http://en.wikipedia.org/wiki/File:Merge_sort_algorithm_diagram.svg.
Now consider the amount of time that this process will take, we're going to have log (n) levels and at each level we will have to merge all of the lists. As it turns out each level will take n time to merge, because we'll be merging a total of n elements each time. Then you can fairly easily see that it will take n log (n) time to sort an array with mergesort if you take the comparison operation to be the most important operation.
If anything is unclear or I skipped somewhere please let me know and I can try to be more verbose.
Edit Second Explanation:
Let me think if I can explain this better.
The problem is broken into a bunch of smaller lists and then the smaller lists are sorted and merged until you return to the original list which is now sorted.
When you break up the problems you have several different levels of size first you'll have two lists of size: n/2, n/2 then at the next level you'll have four lists of size: n/4, n/4, n/4, n/4 at the next level you'll have n/8, n/8 ,n/8 ,n/8, n/8, n/8 ,n/8 ,n/8 this continues until n/2^k is equal to 1 (each subdivision is the length divided by a power of 2, not all lengths will be divisible by four so it won't be quite this pretty). This is repeated division by two and can continue at most log_2(n) times, because 2^(log_2(n) )=n, so any more division by 2 would yield a list of size zero.
Now the important thing to note is that at every level we have n elements so for each level the merge will take n time, because merge is a linear operation. If there are log(n) levels of the recursion then we will perform this linear operation log(n) times, therefore our running time will be n log(n).
Sorry if that isn't helpful either.
When applying a divide-and-conquer algorithm where you partition the problem into sub-problems until it is so simple that it is trivial, if the partitioning goes well, the size of each sub-problem is n/2 or thereabout. This is often the origin of the log(n) that crops up in big-O complexity: O(log(n)) is the number of recursive calls needed when partitioning goes well.