amortized analysis we calculate average time over sequence of data structure
over all operation which is performed in analysis. Amortized analysis always
show average cost minimum but in case if nested operation are performed then
amortized perform expensive. Amortized analysis is totally different from
average case analysis. This analysis ensure that average progress of each function
fall into worst case.
this paper we will discuss three types of techniques which are used in
amortized analysis. First is Aggregate
analysis second is Accounting method
and last is Potential method. We
will use two types of examples for explaining these techniques like MULTILOP and Binary counter.
Aggregate analysis define that for
all n input values a sequence of operation performed and it take total time in
worst case time T(n). On this worst case time we will perform amortized
analysis and then average case time will be T(n)/n. It is compulsory to perform
amortized analysis on each operation even there are different type of operation
in a sequence we have to perform aggregate analysis on each operation.
this example we analyze the stack that is augmented with a new operation. Here
explain two fundamental operation, each has O(1) time.
PUSH (S.x) Pushes object x into Stack
POP(S) pop the top object and return
it. On empty stack calling pop generate an error.
Since there is time given for each
operation which is 1. So each PUSH and POP operation take time 1 if there are n
PUSH and POP operation then it take (n) time.
From above example calling it return
(a) state from above picture. Then the top four objects are displayed by
calling this MULTIPOP(S, 4) show result in (b) from above picture. Then it will
empty the stack by calling MULTIPOP(S, 7) show in (c) from above picture.
We analyze that total time will be 1
for each PUSH POP function. Every time of calling While function POP and PUSH
produce a single time call and total cost of MULTIPOP is min(S, k). Let we
analyze that n POP and PUSH calls and MULTIPOP operations on an empty stack.
This will produce O (n) worst case
time on MULTIPOP operation in sequence. Hence sequence of n operation will cost
Now using aggregate analysis we can
get better upper bound then this. On every call of MULTIPOP there is a single
execution of POP and PUSH function and it will be worst case if there are n
inputs. In aggregate analysis total cost of an operation O(n)/n=O(1). Here cost
of all operation will be O(1) and here worst case bound will be O(n) by
amortize analysis on n operations.
Increment a binary counter:
Here another example of aggregate
analysis let a binary counter that implement k-bit binary counter that count
upward from zero. We use an array A 0….k-1 of bit where k-1.Initialy stored
variable x is zero and counter having A0 is starting and Ak-1 is ending
If above figure an 8bit counter run
from 0 to 16 by above increment operation. Those bits that are flip to achieve
next value are highlighted. The running cost of flipping shows in right. You
can see that total cost will always less than twice the total numbers of
INCREMENT operations. The total number of INCREMENT sequence are
The worst case time of sequence of
all increment operations is O (n). So average cost of amortized will be O (n)/n=O
The Counting Method:
accounting method we assign different charges to every different operation,
some operations charged along with less or more cost which they actually have.
We called this charges on operation amortized
cost. When amortized cost of any operation exceeded from original cost we
assign an object in data structure as credit.
By this method we can check the cost of every operation either exceeding or
nothing change happen. We also called this as deposit or used up. This method
is different from amortized analysis, which have same cost for all operations.
We have to take carefully amortized cost if we take average case time by using
aggregate method we have to ensure that that cost will provide upper bound on
total cost of operation. Denote cost of ith operation by ci and
amortized cost of ith operation by ci^.
For all sequence of n operations. Total credit cost that
stores in data structure is less that or between amortized or actual cost.
recall the previous stack operation’s cost
MULTIPOP min (k, s)
Here k is argument passed to MULTIPOP
and s is size of stack when this function is called. Let us assigning the
Here amortized cost of each
operations are zero but actual cost is some value here amortized cost can be
same and different from each other even asymptotically. Now we take example of
plates using stack and use one dollar as a cost for push and pay to credit, the
plate into the stack. On multipop we have to pay off credit from pop plates. In
other word for pop we have to pay one dollar to credit so that pop value always
will be non-negative number of plates and we ensure that stack value will be
non-negative. So n operation of PUSH POP and MULTIPOP will amortized cost will
be upper bound O (n).
Incrementing binary counter:
In accounting method we assume a
increment operation binary counter start from zero. We have observed earlier
that cost is proportional to total bit flipped. Let we use dollar bill example
in this example. For example in amortized analysis we use 2 dollar for set a
bit to 1. When a bit flipped it will cost 1 bit and kept other dollar for bit
flipped from zero to one. At any point there is one dollar on a counter to
credit on, So we have nothing to do to reset a bit we just have to pay just one
dollar bill on the bit. Thus for n operations there is only O(n) amortized cost
which bounds the actual cost.
accounting method we use credit for prepaid work but in potential we use just
“Potential Energy” or just “Potential” for future use pay work operations. We
use potential with hole in data structure rather than specifying object in data
Potential method working:
initial data structure
actual cost of ith operations
Di-1 after applying operation
? (Di) Potential method representation
w.r.t ith operation represent
? (D0)=0 and ?(Di )>0 for all i
– ?(Di-1)=? ?ci >0
then Ci Ci I store work is data structure for future use
? ?i<0 then Ci < Ci^ store work for pay future The amortized cost of each function is therefore its actual cost plus the change in potential due to operation. Stack Operation: We define a potential function ? that is total objects in stack. For empty stack Do we start ? (Do) =0 this number in stack can never negative and ith operation have non-negative potential energy thus Sine amortized cost of these operations is O(1) and for n operations cost will be O(n). Since we have already say that ? (Di)>= ? (Do) total cost is upper
bound to actual cost. Worst case of n operation is O(n).