I ran into an interview question recently. no additional info is given into question (maybe default implementation should be used...)
n arbitrary sequences of insert and remove operations on empty min heap
(location for delete element is known) has amortized cost of:
A) insert O(1), remove O(log n)
B) insert O(log n), remove O(1)
The option (B) is correct.
I'm surprized when see answer sheet. i know this is tricky, maybe empty heap, maybe knowing location of elements for delete,... i dont know why (A) is false? Why (B) is true?
When assigning amortized costs to operations on a data structure, you need to ensure that, for any sequence of operations performed, that the sum of the amortized costs is always at least as big as the sum of the actual costs of those operations.
So let's take Option 1, which assigns an amortized cost of O(1) to insertions and an amortized cost of O(log n) to deletions. The question we have to ask is the following: is it true that for any sequence of operations on an empty binary heap, the real cost of those operations is upper-bounded by the amortized cost of those operations? And in this case, the answer is no. Imagine that you do a sequence purely of n insertions into the heap. The actual cost of performing these operations can be Θ(n log n) if each element has to bubble all the way up to the top of the heap. However, the amortized cost of those operations, with this accounting scheme, would be O(n), since we did n operations and pretended that each one cost O(1) time. Therefore, this amortized accounting scheme doesn't work, since it will let us underestimate the work that we're doing.
On the other hand, let's look at Option 2, where we assign O(log n) as our amortized insertion cost and O(1) as our amortized remove cost. Now, can we find a sequence of n operations where the real cost of those operations exceeds the amortized costs? In this case, the answer is no. Here's one way to see this. We've set the amortized cost of an insertion to be O(log n), which matches its real cost, and so the only way that we could end up underestimating the total is with our amortized cost of a deletion (O(1)), which is lower than the true cost of a deletion. However, that's not a problem here. In order for us to be able to do a delete operation, we have to have previously inserted the element that we're deleting. The combined real cost of the insertion and the deletion is O(log n) + O(log n) = O(log n), and the combined amortized cost of the insertion and the deletion is O(log n) + O(1) = O(log n). So in that sense, pretending that deletions are faster doesn't change our overall cost.
A nice intuitive way to see why the second approach works but the first one doesn't is to think about what amortized analysis is all about. The intuition behind amortization is to charge earlier operations a bit more so that future operations appear to take less time. In the case of the second accounting scheme, that's exactly what we're doing: we're shifting the cost of the deletion of an element from the binary heap back onto the cost of inserting that element into the heap in the first place. In that way, since we're only shifting work backwards, the sum of the amortized costs can't be lower than the sum of the real costs. On the other hand, in the first case, we're shifting work forward in time by making deletions pay for insertions. But that's a problem, because if we do a bunch of insertions and then never do the corresponding deletions we'll have shifted the work to operations that don't exist.
Because the heap is initially empty, you can't have more deletes than inserts.
An amortized cost of O(1) per deletion and O(log N) per insertion is exactly the same as an amortized cost of O(log N) for both inserts and deletes, because you can just count the deletion cost when you do the corresponding insert.
It does not work the other way around. Since you can have more inserts than deletes, there might not be enough deletes to pay the cost of each insert.
Related
So a regular binary heap has an operation extract_min which is O(log(n)) worst time. Suppose the amortized cost of extract_min is O(1). Let n be the size of the heap
So a sequence where we have n extract_min operations performed and it initially contained n elements. Does this mean that the entire sequence would be processed in O(n) time since each operation is O(1)?
Lets get this out of the way first: Removing ALL the elements in a heap via extract_min operations takes O(N log N) time.
This is a fact, so when you ask "Does constant amortized time extract_min imply linear time for removing all the elements?", what you are really asking is "Can extract_min take constant amortized time even though it takes O(N log N) time to extract all the elements?"
The answer to this actually depends on what operations the heap supports.
If the heap supports only the add and extract_min operations, then every extract_min that doesn't fail (in constant time) must correspond to a previous add. We can then say that add takes amortized O(log N) time, and extract_min take amortized O(1) time, because we can assign all of its non-constant costs to a previous add.
If the heap supports an O(N) time make_heap operation (amortized or not), however, then its possible to perform N extract_min operations without doing anything else that adds up to O(N log N) time. The whole O(N log N) cost would then have to be assigned to the N extract_min operations, and we could not claim that extract_min takes amortized constant 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.
I am reading the basics of splay trees. The amortized cost of an operation is O(log n) over n operations. Some rough basic idea is that when you access a node, you splay it i.e. you take it to root so next time this is quickly accessed and also if the node was deep, it enhances balance-ness of tree.
I don't understand how the tree can perform amortized O(log n) for this sample input:
Say a tree of n nodes is already built. My next n operations are n reads. I access a deep node say at depth n. This takes O(n). True that after this access, the tree will become balanced. But say every time I access the most current deep node. This will never be less than O(log n). then how we can ever compensate for the first costly O(n) operation and bring the amortized cost of each read as O(log n)?
Thanks.
Assuming your analysis is correct and the operations are O(log(n)) per access and O(n) the first time...
If you always access the bottommost element (using some kind of worst-case oracle), a sequence of a accesses will take O(a*log(n) + n). And thus the amortized cost per operation is O((a*log(n) + n)/a)=O(log(n) + n/a) or just O(log(n)) as the number of accesses grows large.
This is the definition of asymptotic average-case performance/time/space, also called "amortized performance/time/space". You are accidentally thinking that a single O(n) step means all steps are at least O(n); one such step is only a constant amount of work in the long run; the O(...) is hiding what's really going on, which is taking the limit of [total amount of work]/[queries]=[average ("amortized") work per query].
This will never be less than O(log n).
It has to be in order to get O(log n) average performance. To get intuition, the following website may be good: http://users.informatik.uni-halle.de/~jopsi/dinf504/chap4.shtml specifically the image http://users.informatik.uni-halle.de/~jopsi/dinf504/splay_example.gif -- it seems that while performing the O(n) operations, you move the path you searched scrunching it towards the top of the tree. You probably only have a finite number of such O(n) operations to perform until the entire tree is balanced.
Here's another way to think about it:
Consider an unbalanced binary search tree. You can spend O(n) time balancing it. Assuming you don't add elements to it*, it takes O(log(n)) amortized time per query to fetch an element. The balancing setup cost is included in the amortized cost because it is effectively a constant which, as demonstrated in the equations in the answer, disappears (is dwarfed) by the infinite amount of work you are doing. (*if you do add elements to it, you need a self-balancing binary search tree, one of which is a splay tree)
As a simple example, in a specific implementation of the dynamic array, we double the size of the array each time it fills up.
Because of this, array reallocation may be required, and in the worst case an insertion may require O(n).
However, a sequence of n insertions can always be done in O(n) time, because the rest of the insertions are done in constant time, so n insertions can be completed in O(n) time.
The amortized time per operation is therefore O(n) / n = O(1).
--from Wiki
But in another book :Each doubling takes O(n) time, but happens so rarely that its amortized time is still O(1).
Hope somebody could tell me does the rare situation infer the Wiki explanation? Thanks
Amortized basically means average per number of operations.
So, if you have an array of n, you need to insert n+1 items until you'll need the reallocation.
So, you've done n inserts which each one of them took O(1), and another insert that took O(n), so in total you've got n+1 actions that cost you 2n operations .
2n / n+1 ~= 1
That's why the Amortized time is still O(1)
Yes, these two statements say the same thing, Wiki just explains it more thoroughly.
Is there one type of set-like data structure supporting merging in O(logn) time and k-th element search in O(logn) time? n is the size of this set.
You might try a Fibonacci heap which does merge in constant amortized time and decrease key in constant amortized time. Most of the time, such a heap is used for operations where you are repeatedly pulling the minimum value, so a check-for-membership function isn't implemented. However, it is simple enough to add one using the decrease key logic, and simply removing the decrease portion.
If k is a constant, then any meldable heap will do this, including leftist heaps, skew heaps, pairing heaps and Fibonacci heaps. Both merging and getting the first element in these structures typically take O(1) or O(lg n) amortized time, so O( k lg n) maximum.
Note, however, that getting to the k'th element may be destructive in the sense that the first k-1 items may have to be removed from the heap.
If you're willing to accept amortization, you could achieve the desired bounds of O(lg n) time for both meld and search by using a binary search tree to represent each set. Melding two trees of size m and n together requires time O(m log(n / m)) where m < n. If you use amortized analysis and charge the cost of the merge to the elements of the smaller set, at most O(lg n) is charged to each element over the course of all of the operations. Selecting the kth element of each set takes O(lg n) time as well.
I think you could also use a collection of sorted arrays to represent each set, but the amortization argument is a little trickier.
As stated in the other answers, you can use heaps, but getting O(lg n) for both meld and select requires some work.
Finger trees can do this and some more operations:
http://en.wikipedia.org/wiki/Finger_tree
There may be something even better if you are not restricted to purely functional data structures (i.e. aka "persistent", where by this is meant not "backed up on non-volatile disk storage", but "all previous 'versions' of the data structure are available even after 'adding' additional elements").