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 Amortized analysis is technique to find
minimum time required to perform a order of

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Data-structure operations when a set of inputs
effects the next order of input we look at amortized analysis. We assume that a
order of operation can cost higher than the total cost. Amortized analysis is
not applicable everywhere it is applied whenever the order in one problem
effecting the other order in that problem. For calculating the worst-case
running time of any kind of data-structure operation in the order, p; if the order
contains m operations, then mp is an upper bound for the total running time.
This is technically correct but it may come with a surprisingly output and give
a wrong or unexpected result. Now from above paragraph we concluded the main
ideas of our topic is: Amortized analysis applied when a order of data
structure is effecting next order. Amortized analysis is itself an upper bound
which results as an average performance for each operation in worst case time.
Amortized analysis is exercised with the whole cost of order of operations. It does
not deals
with the cost of a single operation in that order. Let’s
suppose that
the amortized cost of insertion into a splay tree with m items is O
(log m), so when I insert ’54’ into this tree, the cost will be O
(log m). In fact, inserting ’54’ might require O
(m) operations. It is only appropriate to say, when I insert
m items into a tree, the average time for
each operation will be O (log m).


Amortization in finance
means to pay off a debt i.e. loans and
mortgage, by smaller payments made over a period of time.

A method aggregate analysis which is now known as
Amortized analysis is a technique which was introduced by Robert Tarjan.
According to him and as he wrote in 1985 in his paper that we can surprisingly
achieve upper bound and lower bound for many varieties of algorithm. This
technique is used by many researchers. Robert Tarjan reveals that by using this
technique we can obtain “self-adjusting” data structures that are efficient but
must have amortized complexity low.
Amortization plays a vital
role in the analysis of many other standard algorithms and data
structures, including

maximum flow (In optimization theory, maximum flow problems involve
finding a feasible flow through a single-source, single-sink flow
network that is maximum)

 Fibonacci heaps In a field of computer science,is
a data structure for priority
queue operations, consisting of a collection
of heap-ordered trees. It has a
better amortized running time than
many other priority queue data structures including the binary heap and binomial heap. Michael L. Fredman and Robert
E. Tarjan developed Fibonacci heaps in 1984
and published them in a scientific journal in 1987. They named Fibonacci heaps
after the Fibonacci numbers, which are
used in their running time analysis.

 Dynamic arrays In a field of computer science, is a array that is growable array, resizable
array, dynamic table, mutable array, or array list is a random
access, variable-size list data
structure that allows elements to be added or
removed. It is supplied with standard libraries in many modern mainstream
programming languages. Dynamic arrays overcome a limit of static arrays, which have a fixed capacity that needs to be specified
at allocation.

Comparison to other analysis techniques:

Worst-case analysis can give overly negative
(pessimistic) bounds for orders of
operations, because this type of analysis could not tackle
with the operations on same kind of data structure. Amortized analysis may result
into more than the
practical worst-case bound by observing such kind
of interactions into account. The bound that is given
by amortized analysis is, a worst-case
bound on the average time per
operation; a specific operation in a order of
data-structure may have more cost
than this bound, but the minimum cost on whole of
the operations in any valid order will always perform
within the bound. Amortized analysis is same as average case
analysis, however it is for average cost over a order of operations.

of Amortized analysis:

We are going to discuss three methods:

Aggregate method

Potential method

Accounting method


The aggregate method, where the all over running time for a order
of operations is analyzed.

Ø  In aggregate analysis,
we will proof that for every m, a order of m operations takes

             Worst-case time T (m) in

Ø  In the worst
case, the average cost, or amortized-cost, per operation is therefore T(m)/m.
Note that this amortized cost applies to each operation, even when there are
several types of operations in the order.

The other two methods are:

Ø  The accounting

Ø  The potential


                       In our first example we will
discuss to insert a new element we can do two operation push and pop let assume
the cost in O(1).


 pushes object x onto
stack S.

the top element/object of stack S and will return
back a popped object. Now make sure that you are not calling pop on an empty

Ø  What will happen
if you call pop operation on empty stack?

               Surely, it will give us an

Since every operation that we can
implement on stack takes O (1) time. The total cost of a order of number of
PUSH and POP operations is therefore n, and the actual running time for number
of operations is Q (n).

Now let we add one more
functionality/operation in stack operation of MULTIPOP (S,k),
where S is stack and k is a number of objects. MULTIPOP (S,k) perform
will do some simple task it will removes the k top objects of
stack S, it will result as empty stack if the
objects are fewer than k. For sure, we will always suppose that k is
positive number negative can never ever be taken if this happens the MULTIPOP operation
will not change the stack and leaves the stack same as it was previous. In the
following pseudocode, the operation STACK-EMPTY returns TRUE if there are no
objects currently on the stack, and FALSE otherwise.

The top 4 objects
are popped when we call MULTIPOP(S,4). The next
operation is MULTIPOP(S,8) which empties the stack since there
were fewer than 8 objects remaining.

According to you what in the
running time of Multipop(S,k)? The time analysis is linear for numbers of POP operations
executed, hence we can assume that the time analysis of multipop for every PUSH
and POP operation is 1.


The while loop will run
till stack is not empty and k is not equal zero that only the when we call
multipop with higher number even than it is working for each iteration of while
loop pop operation will be called as you can see in line 2. Thus, the total cost
of MULTIPOP is Q(sk),now if we recall the amortized analysis we can see
clearly the function give us linear time as whole but a part of order is
expensive than whole. Let us consider a order of
numbers of PUSH, POP, and MULTIPOP operations on an empty stack. The worst-case
cost of a MULTIPOP operation in the order s O(n), since the stack size is
minimum upto n. The worst-case time of any stack operation is therefore O(n),
and hence a order of n operations costs O(n2), since we may have
O(n) MULTIPOP operations costing O(n) each. Although this analysis is correct,
the O(n2) result, which we obtained by considering the worst-case
cost of each operation on each functionality, is not tightly bound. Using
aggregate analysis, we can obtain a better upper bound that considers the
entire order of n operations. In fact, although a single MULTIPOP operation can
be expensive, any order of n PUSH, POP, and MULTIPOP operations on an initially
empty stack can cost at most O(n).We can pop each object from the stack at most
once for each time we have pushed it onto the stack. Therefore, the number of
times that POP can be called on a nonempty stack, including calls within
MULTIPOP, is at most the number of PUSH operations, which is at most n. For any
value of n, any order of n PUSH, POP, and MULTIPOP operations takes a total of
O (n) time. The average cost of an operation is O(n)/n= O(1). In aggregate
analysis, we assign the amortized cost of each operation to be the average
cost. In this example, therefore, all three stack operations have an amortized
cost of O(1).We emphasize again that although we have just shown that the
average cost, and hence the running time, of a stack operation is O(1). We
actually showed a worst-case bound
of O(n) on a order of n operations. Div this total cost by n yielded the
average cost per operation, or the amortized cost.

Incrementing a
binary counter:

The worst case running
time occurs when all i bits are flipped, so increment (A) has running time O(i).
In a order of n increment operations, few increments will cause that many bits
to flip. Indeed, bit 0 flips with every increment bit 1 flips with every 2nd
increment bit 2 flips with every 4th increment

Total number of bit
flips in n increment operations is

n + n/2 + n/4 + … +
n/2i < n(1/(1-1/2))= 2n So total cost of the order is O(n). Amortized cost per operation is O (n)/n = O(1) The cost of each INCREMENT operation is linear in the number of bits flipped. As with the stack example, analysis produces a bound that is correct but not tight. The Accounting Method:               The accounting method is a method of amortized analysis based on accounting. Note, however, that this does not surety such analysis will be directly manifest; often, choosing the correct parameters for the accounting method requires as much knowledge of the problem and the complexity bounds one is endeavoring to prove as the other two methods., different operations are assigned with different charges, with some operations charged more or less than they actually cost. We call the amount we charge an operation its amortized cost.               When the amortized cost of an operation surpasses its actual cost, we assign the difference to specific objects in the data structure as credit. Credit can help pay for later operations when actual cost is greater than amortized cost.               Our credit must never be negative not just because banks use this but because if we let the credit goes negative, we can't guarantee that our amortized cost is upper bound on an actual cost for some orders. So we want a guarantee that for any order of operations, the amortized cost always gives an upper bound on the actual cost of that order.              Thus, we can view the amortized cost of an operation as being split between its actual cost and credit that is either deposited or used up. Different operations may have different amortized costs. This method differs from aggregate analysis, in which all operations have the same amortized cost.   Table Resizing In data structures like dynamic array, heap, stack, hash table (to name a few) we used the idea of resizing the underlying array, when the current array was either full or reached a chosen threshold. During resizing we allocate an array twice as large and move the data over to the new array. So, if an array is full, the cost of insertion is linear; if an array is not full, insertion takes a constant time. To derive an amortized cost we start with an array of size 1 and perform n insertions.               We must choose the amortized costs of operations carefully. If we want to show that the average cost per operation of the worst case is small by analyzing with amortized costs, we must guarantee that for any order of operations, the amortized cost always gives an upper bound on the actual cost of that order.. Moreover, as in aggregate analysis, this relationship must hold for all orders of operations. If we denote the actual cost of the ith operation by ci and the amortized cost of the ith operation by c yi, we require that the submission of amortized cost c (yi) is greater or equal to actual cost which is ci for all orders. In some cases, the actual cost is greater than amortized cost.                         Stack operations   To illustrate the accounting method of amortized analysis, let us return to the stack example. Recall that the actual costs of the operations were PUSH 1 , POP 1 , MULTIPOP min(k, s) where k is the argument supplied to MULTIPOP and s is the stack size when it is called. Let us assign the following amortized costs: PUSH 2 , POP 0 , MULTIPOP 0 . Note that the amortized cost of MULTIPOP is a constant (0), whereas the actual cost is variable. Here, all three amortized costs are constant. In general, the amortized costs of the operations under consideration may differ from each other, and they may even differ asymptotically. We shall now show that we can pay for any order of stack operations by charging the amortized costs. Suppose we use a dollar bill to represent each unit of cost. We start with an empty stack between the stack data structure and a stack of plates in a cafeteria. When we push a plate on the stack, we use 1 dollar to pay the actual cost of the push and are left with a credit of 1 dollar (out of the 2 dollars charged), which we leave on top of the plate. At any point in time, every plate on the stack has a dollar of credit on it. The dollar stored on the plate serves as prepayment for the cost of popping it from the stack. When we execute a POP operation, we charge the operation nothing and pay its actual cost using the credit stored in the stack. To pop a plate, we take the dollar of credit off the plate and use it to pay the actual cost of the operation. Thus, by charging the PUSH operation a little bit more, we can charge the POP Operation nothing costs of the operations under consideration may differ from each other, and they may even differ asymptotically. Moreover, we can also charge MULTIPOP operations nothing. To pop the first plate, we take the dollar of credit off the plate and use it to pay the actual cost of a POP operation. To pop a second plate, we again have a dollar of credit on the plate to pay for the POP operation, and so on. Thus, we have always charged enough up front to pay for MULTIPOP operations. In other words, since each plate on the stack has 1 dollar of credit on it, and the stack always has a nonnegative number of plates, we have ensured that the amount of credit is always nonnegative. Thus, for any order of n PUSH, POP, andMULTIPOP operations, the total amortized cost is an upper bound on the total actual cost. Since the total amortized cost is O.n/, so is the total actual cost. Incrementing a binary counter As another illustration of the accounting method, we analyze the INCREMENT operation on a binary counter that starts at zero. As we observed earlier, the running time of this operation is proportional to the number of bits flipped, which we shall use as our cost for this example. Let us once again use a dollar bill to represent each unit of cost (the flipping of a bit in this example). For the amortized analysis, let us charge an amortized cost of 2 dollars to set a bit to 1. When a bit is set, we use 1 dollar (out of the 2 dollars charged) to pay for the actual setting of the bit, and we place the other dollar on the bit as credit to be used later when we flip the bit back to 0. At any point in time, every 1 in the counter has a dollar of credit on it, and thus we can charge nothing to reset a bit to 0; we just pay for the reset with the dollar bill on the bit. Now we can determine the amortized cost of INCREMENT. The cost of resetting the bits within the while loop is paid for by the dollars on the bits that are reset. The INCREMENT procedure sets at most one bit, in line 6, and therefore the amortized cost of an INCREMENT operation is at most 2 dollars. The number of 1s in the counter never becomes negative, and thus the amount of credit stays nonnegative at all times. Thus, for n INCREMENT operations, the total amortized cost is O(n) which bounds the total actual cost. Potential Method: this method of amortized analysis shows the prepaid work as "potential energy," or just "potential," which can be released to pay for future operations. We associate the potential with data structure completely Instead of specific objects within the data structure. Working: View the bank account as the       potential energy (as in physics) of      the dynamic set. Start with an initial data structure D0. Operation i transforms Di–1 to Di.  The cost of operation i is ci. Define a potential function F : {Di} ® R, such that F(D0 ) = 0 and F(Di ) ³ 0 for all i. The amortized cost ?i with respect to F is defined to be ?i = ci + F(Di) – F(Di–1). Like the accounting method, but think of the credit as potential stored with the entire data structure.  Accounting method stores credit with specific objects while potential method stores potential in the data structure as a whole.  Can release potential to pay for future operations Most flexible of the amortized analysis methods ). ?i = ci + F(Di) – F(Di–1) If  DFi > 0, then ?i >
ci.  Operation i stores work in the data structure for later use.

 If  DFi < 0, then ?i < ci.  The data structure delivers up stored work to help pay for operation i.   The total amortized cost of n operations is Summing both sides telescopically.   since F(Dn) ³ 0 and F(D0 ) = 0.     Stack Example: Define:      f(Di) = #items in stack            Thus, f(D0)=0. Plug in for operations: Push:         ?i = ci + f(Di) - f(Di-1) = 1 +    j    -   (j-1) = 2 Pop:           ?i = ci + f(Di) - f(Di-1) = 1 +  (j-1) -   j = 0 Multi-pop: ?i = ci + f(Di) - f(Di-1) = k' + (j-k') -   j                 k'=min(|S|,k) = 0 Binary Counter Define the potential of the counter after the ith operation by F(Di) = bi, the number of 1's in the counter after the ith operation. Note:  F(D0 ) = 0,  F(Di) ³ 0 for all i. Example: 0   0   0   1   0    1    0  0   0   0   1$1 0   1$1 0  Accounting method Assume ith INCREMENT resets ti bits (in line 3). Actual cost ci = (ti + 1) Number of 1's after ith operation:  bi = bi–1 – ti + 1 The amortized cost of the i th INCREMENT is ?i = ci + F(Di) – F(Di–1) = (ti + 1) + (1 - ti) = 2 Therefore, n INCREMENTs cost Q(n) in the worst case. Examples:1 A sequence of n operations is performed on a data structure in which the kth operation costs k if k is an exact power of 2, and 1 otherwise. Use aggregate analysis to determine the amortized cost per operation. Solution: In a sequence of n operations there are blog2 nc+1 exact powers of 2, namely 1,2,4,...,2 blog2 n c . The total cost of these is a geometric sum ? blog2 n c I =0 2 I = 2 blog2 nc+1 ?1 ? 2 log2 n+1 = 2n. The rest of the operations are cheap, each having a cost of 1, and there are less than n such operations. The total cost of all operations is thus T(n) ? 2n+n = 3n = O(n), which means O(1) amortized cost per operation. Example:2 We want not only to increment a counter but also to reset it to zero (i.e., make all bits in it 0). Counting the time to examine or modify a bit as  (1), show how to implement a counter as an array of bits so that any sequence of n INCREMENT and RESET operations takes time O(n) on an initially zero counter We introduce a new field A:max to hold the index of the high-order 1 in A. Initially, A:max is set to 1, since the low-order bit of A is at index 0, and there are initially no 1's in A. The value of A:max is updated as appropriate when the counter is incremented or reset, and we use this value to limit how much of A must be looked at to reset it. By controlling the cost of RESET in this way, we can limit it to an amount that can be covered by credit from earlier INCREMENTs. As for the counter in the book, we assume that it costs $1 to flip a bit. In addition, we assume it costs $1 to update A:max. Setting and resetting of bits by INCREMENT will work exactly as for the original counter in the book: $1 will pay to set one bit to 1; $1 will be placed on the bit that is set to 1 as credit; the credit on each 1 bit will pay to reset the bit during incrementing. In addition, we'll use $1 to pay to update max, and if max increases, we'll place an additional $1 of credit on the new high-order 1. (If max doesn't increase, we can just waste that $1—it won't be needed.) Since RESET manipulates bits at positions only up to A:max, and since each bit up to there must have become the high-order 1 at some time before the high-order 1 got up to A:max, every bit seen by RESET has $1 of credit on it. So the zeroing of bits of A by RESET can be completely paid for by the credit stored on the bits. We just need $1 to pay for resetting max. Thus charging $4 for each INCREMENT and $1 for each RESET is sufficient, so the sequence of n INCREMENT and RESET operations takes O(n) time.  

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