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Chapter 10

Amortized Analysis

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A sequence of operations: OP1, OP2, OPm

OPi : several pops (from the stack) andone push (into the stack)

ti : time spent by OPithe average time per operation:

An example push and pop

!

!m

i

it

m 1ave

1t

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Example: a sequence of push and pop

p: pop , u: push

i 1 2 3 4 5 6 7 8OPi 1u 1u 2p 1u 1u 1u 2p 1p

1u 1u 1u

ti 1 1 3 1 1 1 3 2

tave = (1+1+3+1+1+1+3+2)/8

= 13/8

= 1.625

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Another example: a sequence of pushand pop

p: pop , u: push

tave = (1+2+1+1+1+1+6+1)/8

= 14/8

= 1.75

i 1 2 3 4 5 6 7 8

OPi 1u 1p 1u 1u 1u 1u 5p 1u

1u 1u

ti 1 2 1 1 1 1 6 1

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Amortized time and potential function

ai = ti + * *i i 1ai : amortized timeofO i

*i :potentialf nctionofthe stackafterO i

* *i i 1: changeofthe potential

a tii

m

ii

m

ii

m

i! ! !

! 1 1 1

1( )* *

! !

t ii

m

m1

0* *

If * *m 0 u 0, then a ii

m

!

1

represents an upper

boundof t ii

m

!

1

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Suppose that before we execute Opi , there are kelements in the stack and Opi consists of n pops and1 push.

Amortized analysis of thepush-and-pop sequence

0haveWe

stactheine e entsof#:

0u**

*

m

i

2

)1()1(

1Then,

1

1

1

!

!

**!

!*

!

!*

-kk-nn

ta

nk

nt

k

iiii

i

i

i

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By observation, at most m pops and mpushes are executed in m operations. Thus,

2.tThen,

2/)(haveWe

ave

1

e

!!

m

i

ima

2.tave e

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Skew heaps

1

50

13 20

10

16 25

5

12

30

19

40 14

Two skew heaps

Step 1: Merge the right paths.

5 right heavy nodes: yellow

meld: merge + swapping

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1

5

13

20

10

16

25

50

12

30

19

40

14

Step 2: Swap the children along the right path.

No right heavy node

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Amortized analysis of skew heaps

meld: merge + swapping

operations on a skew heap:

find-min(h): find the min of a skew heap h.

insert(x, h): insert x into a skew heap h.

delete-min(h): delete the min from a skew heap h.

meld(h1, h2): meld two skew heaps h1 and h2.

The first three operations can be implementedby melding.

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Potential function of skew heaps

wt(x): # of descendants of node x, includingx.

heavy node x: wt(x) " wt(p(x))/2, where

p(x) is the parent node of x.

light node : not a heavy node

potential function *i: # of right heavy nodes

of the skew heap.

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light nodese log2n

# heavy= 3e log2n

possible heavy nodes# of nodes: n

Any path in an n-node tree contains atmost

log2n light nodes.

The number ofright heavy nodes attached to

the left path is atmost log2n .

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Amortized time

heap: h1

# of nodes: n1

heap: h2

# of nodes: n2

# lighte log2n1# heavy= k1

# lighte log2n2# heavy= k2

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ai=ti+*i*i-1

ti : timespent by OPitie 2+ log2n1+k1+ log2n2+k2

(2 countsthe roots of h1and h2)

e 2+2 log2n+k

1+k2where n=n1+n2

*i*i-1= k3-(k1+k2) e log2nk1k2ai=ti+*i*i-1

e2+2 log2n+k1+k2+ log2nk1k2=2+3 log2n

ai= O(log2n)

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AVL-treesheight balance of node v:hb(v)= (height of right subtree) (height of left subtree)

The hb(v) of every node never differ by more than 1.

M

I O

E K QN

G LC J P R

B D F

-1

-1

-1

0

0

0 0 0

0

0

0

0

00

0

+1

Fig. An AVL-Tree with Height Balance Labeled

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Add a new node A.

Before insertion, hb(B)=hb(C)=hb(E)=0

hb(I){0 the first nonzero from leaves.

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Amortized analysis of AVL-trees

Consider a sequence ofm insertions on anempty AVL-tree.

T0: an empty AVL-tree.

Ti: the tree after the ith insertion.

Li: the length of the critical path involved in the ithinsertion.

X1: total # of balance factor changing from 0 to +1or -1 during these m insertions (the total cost for

rebalancing)

X1= Lii

m

!

1

, we wantto find X1.

Val(T): # ofunbalanced node in T

(height balance{0)

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Case 1Case 1 : Absorption

Val(Ti)=Val(Ti-1)+(Li1)

The tree height is not increased, we need notrebalance it.

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Case 2.1Case 2.1 single rotation

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Case 2Case 2 : Rebalancing the tree

D

B F

E GA C

D

B

F

E

G

A C

D

B

F

E G

A

C

left

rotation

right

rotation

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Case 2.1Case 2.1 single rotation

After a right rotation on the subtree rooted atD:

Val(Ti)=Val(Ti-1)+(Li-2)

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Case 2.2Case 2.2 double rotation

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Case 2.2Case 2.2 double rotation

Val(Ti)=Val(Ti-1)+(Li-2)

After a left rotation on the subtree rooted atB and a right rotation on the subtree rootedat F:

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Case 3Case 3 : Height increase

Val(Ti)=Val(Ti-1)+Li

Li is the height of the root. 0

-1

-1

-1

root

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Amortized analysis of X1X2: # ofabsorptions in case1

X3: # ofsingle rotations in case 2

X4: # of double rotations in case 2

X5: # of height increases in case 3Val(Tm) = Val(T0)+ Li

i

m

!

1

X22(X3+X4)

=0+X1X22(X3+X4)

Val(Tm) e0.618m (proved by Knuth)X1= Val(Tm)+2(X2+X3+X4)X2e0.618m+2m= 2.618m

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A self-organizing sequentialsearch heuristics

3 methods for enhancing the performance ofsequential search

(1) Transpose Heuristics:Query Sequence

B B

D D B

A D A B

D D A B

D D A B

C D A C B

A A D C B

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(2) Move-to-the- ront euristics:uery equence

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(3) ount euristics:( ecreasing order bythecount)uery equence

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Analysis of the move-to-the-

front heuristics interword comparison: unsuccessful

comparison

intraword comparison: successful comparison pairwise independent property:

For any sequence S and all pairs P and Q, # ofinterword comparisons of P and Q is exactly # of

interword comparisons made for the subsequenceof S consisting of only Ps and Qs.

(See the example on the next page.)

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e.g.Query Sequence (A, B) comparison

C CA A C

C C A

B B C A

C C B AA A C B

# ofcomparisons made between A and B: 2

Pairwise independent

property in move-to-the-front

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Considerthesubsequenceconsisting of A and B:

Query Sequence (A, B) comparisonA A

B B A

A A B

# ofcomparisons made between A and B: 2

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Query equence

(A, ) 0 1 1

(A, ) 0 1 1 0 1

(B, ) 0 0 1 1

0 1 1 2 1 2

hereare 3 distinctinter ord co parisons:

(A, B), (A, ) and (B, )

We can consider them separately and thenadd them up.

the total number of interword comparisons:

0+1+1+2+1+2 = 7

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Theorem for the move-to-the-

front heuristicsCM(S): # of comparisons of the move-to-the-

front heuristics

CO(S): # of comparisons of the optimal staticordering

CM

(S)e 2CO(S)

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Proof

Proof:

InterM(S): # of interword comparisons of the move tothe front heuristics

InterO(S): # of interword com

parisons of the optim

alstatic ordering

Let S consist of a As and b Bs, a b.

The optimal static ordering: BA

InterO(S) = aInterM(S) e 2a

InterM(S) e 2InterO(S)

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Consider any sequence consisting ofmore than twoitems. Because of the pairwise independent property,we have InterM(S) e 2InterO(S)

IntraM(S): # of intraword comparisons of the move-to-the-front heuristics

IntraO(S): # of intraword comparisons of the optimalstatic ordering

IntraM(S) = Intra

O(S)

InterM(S) + IntraM(S) e 2InterO(S) + IntraO(S)

CM(S) e 2CO(S)

Proof(cont.)

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The count heuristics

The count heuristics has a similar result:

CC(S) e 2CO(S), where CC(S) is the cost

of the count heuristics

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The transposition heuristics

The transposition heuristics does not possess thepairwise independent property.

We can not have a similar upper bound for the cost

of the transposition heuristics.

e.g.onsider pairs ofdistinct items independently.Query equence A B A

(A, B) 0 1 1

(A, ) 0 1 1 0 1

(B, ) 0 0 1 1

0 1 1 2 1 2

# ofinter ord comparisons: 7 (notcorrect)

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thecorrect interword comparisons:Query Sequence C A C B C A

Data Ordering C AC CA CBA CBA CAB

Number ofInterword

Comparisons0 1 1 2 0 2

6