12 – analisis amortized

7/31/2019 12 Analisis Amortized

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12 Analisis Amortized

Mata Kuliah: Perancangan dan Analisis Algoritma

Dosen: Anny Yuniarti, S.Kom, M.Comp.Sc


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Introduction

5/28/2012Jurusan Teknik Informatika FTIF ITS2

Amortized analysis is a tool for analyzing algorithms that perform a

sequence of similar operations.

Instead of bounding the cost of the sequence of operations bybounding the actual cost of each operation separately, an amortizedanalysis can be used to provide a bound on the actual cost of theentire sequence.

One reason this idea can be effective is that it may be impossible ina sequence of operations for all of the individual operations to run intheir known worst-case time bounds. While some operations areexpensive, many others might be cheap.

Amortized analysis is not just an analysis tool, however; it is also away of thinking about the design of algorithms, since the designof an algorithm and the analysis of its running time are often closelyintertwined.


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Amortized analysis


Analyze a sequenceof operations on a datastructure.

Goal:Show that although some individualoperations may be expensive, on average the cost

per operation is small.

Averagein this context does not mean that wereaveraging over a distribution of inputs.

No probability is involved.

Were talking about average cost in the worst case.


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Organization


Well look at 3 methods: aggregate analysis

accounting method

potential method

The aggregate method, though simple, lacks theprecision of the other two methods. In particular, theaccounting and potential methods allow a specific

amortized cost to be allocated to each operation.


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Aggregate analysis


Show that for all n, a sequence of n operations takeworst-case time T(n) in total

In the worst case, the average cost, or amortized

cost , per operation is T(n)/n.

The amortized cost applies to each operation, evenwhen there are several types of operations in the

sequence.


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Stack Example

3 ops:

Push(S,x) Pop(S)Multi-

pop(S,k)

Worst-casecost:

O(1) O(1)O(min(|S|,k)

= O(n)

Amortized cost: O(1) per op

PUSH(S, x)

1 top[S] top[S] + 1

2 S[top[S]] x


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Stack Example

3 ops:


pop(S,k)

Worst-casecost:

O(1) O(1)O(min(|S|,k)

= O(n)


STACK-EMPTY(S)

1 iftop[S] = 0

2 then return TRUE

3 else return FALSE

POP(S)

1 if STACK-EMPTY(S)

2 then error "underflow"

3 elsetop[S] top[S] - 1

4 returnS[top[S] + 1]


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Stack Example

3 ops:


pop(S,k)

Worst-casecost:

O(1) O(1)O(min(|S|,k)

= O(n)


MULTIPOP(S, k)

1 while not STACK-EMPTY(S) and k 0

2 do POP(S)

3 k k - 1


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Stack Example : Aggregate Analysis


Sequence of n push, pop, Multipop operations Worst-case cost of Multipop is O(n)

Have n operations

Therefore, worst-case cost of sequence is O(n2)

Observations Each object can be popped only once per time that its

pushed

Have


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Incrementing a Binary Counter

k-bit Binary Counter: A[0..k1]

INCREMENT(A)1. i0

2. while i < length[A] andA[i] = 1

3. do A[i] 0 reset a bit

4. ii + 1

5. if i < length[A]

6.then

A[i]

1set a bit

10 2][

k

i

iiAx


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k-bit Binary Counter

Ctr A[4] A[3] A[2] A[1] A[0]

0 0 0 0 0 0

1 0 0 0 0 1

2 0 0 0 1 0

3 0 0 0 1 1

4 0 0 1 0 0

5 0 0 1 0 1

6 0 0 1 1 0

7 0 0 1 1 1

8 0 1 0 0 0

9 0 1 0 0 1

10 0 1 0 1 0

11 0 1 0 1 1

INCREMENT(A)1. i02. while i < length[A]

andA[i] = 1

3. do A[i] 0

4. ii + 1

5.if i < length[A]

6. then A[i]1


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k-bit Binary Counter

Ctr A[4] A[3] A[2] A[1] A[0] Cost

0 0 0 0 0 0 0

1 0 0 0 0 1 1

2 0 0 0 1 0 3

3 0 0 0 1 1 4

4 0 0 1 0 0 7

5 0 0 1 0 1 8

6 0 0 1 1 0 10

7 0 0 1 1 1 11

8 0 1 0 0 0 15

9 0 1 0 0 1 16

10 0 1 0 1 0 18

11 0 1 0 1 1 19

INCREMENT(A)1. i02. while i < length[A]

andA[i] = 1

3. do A[i] 0

4. ii + 1

5.if i < length[A]

6. then A[i]1

The running cost forflipping bits


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INCREMENT(A)1. i0

2. while i < length[A] andA[i] = 1

3. do A[i] 0

4. ii + 15. if i < length[A]

6. then A[i]1


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Worst-case analysis

Consider a sequence of ninsertions.The worst-case time to execute oneinsertion is Q(k). Therefore, the worst-

case time for ninsertions is n Q(k) =Q(nk).WRONG! In fact, the worst-case costfor ninsertions is only Q(n) Q(nk).

Lets see why. Note: Youll be correctIf youd said O(nk).But, its an overestimate.


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Tighter analysis

Ctr A[4] A[3] A[2] A[1] A[0] Cost

0 0 0 0 0 0 0

1 0 0 0 0 1 1

2 0 0 0 1 0 3

3 0 0 0 1 1 4

4 0 0 1 0 0 7

5 0 0 1 0 1 8

6 0 0 1 1 0 10

7 0 0 1 1 1 11

8 0 1 0 0 0 15

9 0 1 0 0 1 16

10 0 1 0 1 0 18

11 0 1 0 1 1 19

A[0] flipped every op n

A[1] flipped every 2 opsn/2



A[i] flipped every 2iopsn/2i

Total cost of noperations

i k never 0


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Tighter analysis (continued)

)(

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1

2

1

2

0

1

0

nO

nnn

n

ii

k

i

i

Cost of nincrements

Thus, the average cost of each incrementoperation is O(n)/n= O(1).


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Accounting method


Assign different charges to different operations. Some are charged more than actual cost.

Some are charged less.

Amortized cost= amount we charge.

When amortized cost > actual cost, store thedifference on specific objectsin the data structure ascredit.

Use credit later to pay for operations whose actualcost > amortized cost.

Differs from aggregate analysis: In the accounting method, different operations can have

different costs.

In aggregate analysis, all operations have same cost.


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Accounting method


Need credit to never go negative. Otherwise, have a sequence of operations for which the

amortized cost is not an upper bound on actual cost.

Amortized cost would tell us nothing.

Let ci=actual cost of ith operation ,=amortized cost of ith operation .

ic

n

ii

n

ii

cc

11


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A Simple Example: Accounting

3 ops:

Push(S,x) Pop(S) Multi-pop(S,k)

Assignedcost: 2 0 0

Actual cost: 1 1 min(|S|,k)

Push(S,x) pays for possible later pop of x.

1

1

1

1

1

1

1

1

1

0

1

1

0

0

0


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Stack Example: Accounting Methods

When pushing an object, pay $2 $1 pays for the push $1 is prepayment for it being popped by either

pop or Multipop

Since each object has $1, which is credit, thecredit can never go negative Therefore, total amortized cost = O(n), is an

upper bound on total actual cost


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Example:

Accounting analysis of INCREMENT

Charge an amortized cost of $2 every time a bitis set from 0 to 1

$1 pays for the actual bit setting. $1 is stored for later re-setting (from 1 to 0).

At any point, every 1 bit in the counter has $1 onit that pays for resetting it. (reset is free)

0 0 0 1$1 0 1$1 00 0 0 1$1 0 1$11$1

0 0 0 1$11$10 0

Cost = $2

Cost = $2


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Incrementing a Binary Counter

When Incrementing,

Amortized cost for line 3 = $0

Amortized cost for line 6 = $2 Amortized cost for INCREMENT(A) = $2

Amortized cost for nINCREMENT(A) = $2n=O(n)

INCREMENT(A)1. i02. while i < length[A] andA[i] = 13. do A[i] 0 reset a bit4. ii +15. if i< length[A]6. then A[i]1 set a bit


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The potential method


The potential method, or physicist's view, defines a function that

maps a data structure onto a real valued, nonnegative potential.The potential stored in the data structure may be released to pay forfuture operations.

The potential is associated with the data structure as a whole ratherthan with specific objects within the data structure.

Let us refer to the data structure at time i(or alternatively, just afteroperation i), as Di.

The potential function (Di) maps Dionto a real value. Again, denote

the actual cost of operation ias ci.

In the potential method, the amortized cost of operation iis equal tothe actual cost plus the increase in potential due to the operation:


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The potential method


Ensure that the total amortized cost of a sequence isalways an upper bound on the total actual cost


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Example: Stack operations


A potential function on a stack is (Di) = the numberof items in the stack after operationi.

For the empty stack D0 with which we start, we have(D0) = 0.

If the ith operation on a stack containing sobjects isa PUSH operation, then the potential difference is

(Di) - (Di-1) = (s+ 1)s = 1

The amortized cost of this PUSH operation is


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Example: Stack operations(contd)


The amortized cost of each of the three operations isO(1), and thus the total amortized cost of asequence of noperations is O(n).


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Example: binary counter


The potential of the counter after the ith INCREMENT

operation to be bi, the number of 1s in the counterafter theith operation.

Suppose that the ith INCREMENT operation resets tibits. Theactual cost of the operation is therefore at most ti+ 1, since in

addition to resetting tibits, it sets at most one bit to 1.

If bi= 0, then the ith operation resets all kbits, and so bi-1 = ti=k.

If bi> 0, then bi= bi-1 - ti+ 1.

In either case, bi bi-1 - ti+ 1, and the potential difference is


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Example: binary counter (contd)


The amortized cost is therefore

If the counter starts at zero, then (D0) = 0. Since

(Di) 0 for all i, the total amortized cost of asequence of nINCREMENT operations is an upperbound on the total actual cost, and so the worst-casecost of nINCREMENT operations is O(n).


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Comparing the accounting and potential

methods


The accounting method charges each operation acertain amount according to its type (e.g., insert ordelete), putting the focus on each operation'sprepayment to offset future expensive operations.

The potential method instead puts the focus on theeffect of a particular operation at a particular point intime, based on the effects of the operation on the

data structure and the corresponding change inpossibility for future operations to incur expense.


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Comparing the accounting and potential

methods


The accounting method stores credit in particular

areas of a data structure, such as nodes of a tree oritems in a stack, whereas the potential methodgenerally considers properties of the entire datastructure (such as how many nodes are in a tree, orhow many items are in the stack) to compute thepotential.

12 – analisis amortized

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