Chapter 10
Amortized Analysis
10 -1
An example– push and pop

A sequence of operations: OP1, OP2, … OPm
OPi : several pops (from the stack) and
one push (into the stack)
ti : time spent by OPi
the average time per operation:
t ave 
1
m
t

m
i
i 1
10 -2

Example: a sequence of push and pop
p: pop , u: push
i
O Pi
1
1u
2
1u
3
2p
1u
4
1u
5
1u
6
1u
7
2p
1u
8
1p
1u
ti
1
1
3
1
1
1
3
2
tave = (1+1+3+1+1+1+3+2)/8
= 13/8
= 1.625
10 -3

Another example: a sequence of push
and pop
p: pop , u: push
i
O Pi
1
1u
2
1p
1u
3
1u
4
1u
5
1u
6
1u
7
5p
1u
8
1u
ti
1
2
1
1
1
1
6
1
tave = (1+2+1+1+1+1+6+1)/8
= 14/8
= 1.75
10 -4
Amortized time and potential function
ai = ti +  i   i 1
ai : amortized time of OPi
 i : potential function of the stack after OPi
 i   i 1 : change of the potential
m
m
m
i 1
i 1
m
i 1
 a i   t i   (  i   i 1 )
  ti   m   0
i 1
m
If  m   0  0, then  a i represents an upper
i 1
m
bound of  t i
i 1
10 -5
Amortized analysis of the
push-and-pop sequence

 i : # of elements
We have

in the stack
m  0  0
Suppose that before we execute Opi , there are k
elements in the stack and Opi consists of n pops and
1 push.
 i 1  k
ti  n  1
Then,
i  k  n 1
a i  t i   i   i 1
 ( n  1)  ( k-n  1) -k
2
10 -6
m

We have (  a i ) / m  2
i 1
Then,

t ave  2.
By observation, at most m pops and m
pushes are executed in m operations. Thus,
t ave  2.
10 -7
Skew heaps
meld: merge + swapping

1
1
5
50
10
19
13
12
50
5
20
40
30
14
19
16
10
25
Two skew heaps
40
13
12
16
30
14
20
Step 1: Merge the right paths.
5 right heavy nodes: yellow
25
10 -8
Step 2: Swap the children along the right path.
1
5
50
10
12
14
20
19
13
30
40
16
No right heavy node
25
10 -9
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 implemented
by melding.
10 -10
Potential function of skew heaps




wt(x): # of descendants of node x, including
x.
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.
10 -11

Any path in an n-node tree contains at most
log2n light nodes.
light nodes  log2n
# heavy=k3 log2n
possible heavy nodes
# of nodes: n

The number of right heavy nodes attached to
the left path is at most log2n .
10 -12
Amortized time
# light  log2n1
# heavy = k1
heap: h1
# of nodes: n1
# light  log2n2
# heavy = k2
heap: h2
# of nodes: n2
10 -13
a i = t i +  i   i-1
t i : tim e spent by O P i
t i  2+ log 2 n 1 + k 1 + log 2 n 2 + k 2
(“2” counts the roots of h 1 and h 2 )
 2+ 2log 2 n+ k 1 + k 2
w here n= n 1 + n 2
 i   i-1 = k 3 -(k 1 + k 2 )  log 2 n k 1  k 2
a i = t i +  i   i-1
 2+ 2log 2 n+ k 1 + k 2 + log 2 n k 1  k 2
= 2+ 3log 2 n
 a i = O (log 2 n)
10 -14
AVL-trees
height 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.
-1
M
-1
+1
I
O
0
0
E
K
0
-1
C
G
0
0
B
D
N
0
J
0
0
0
L
Q
0
0
P
R
0
F
Fig. An AVL-Tree with Height Balance Labeled
10 -15

Add a new node A.
-2
M
Critical
node
+1
-2
O
I
Critical
path
G
C
B
D
J
L
0
0
P
R
0
0
-1
0
0
-1
-1
Q
N
K
E
0
0
0
-1
F
0
A
Before insertion, hb(B)=hb(C)=hb(E)=0
hb(I)0 the first nonzero from leaves.
10 -16
Amortized analysis of AVL-trees

Consider a sequence of m insertions on an
empty AVL-tree.




T0: an empty AVL-tree.
Ti: the tree after the ith insertion.
Li: the length of the critical path involved in the ith
insertion.
X1: total # of balance factor changing from 0 to +1
or -1 during these m insertions (the total cost for
rebalancing)
m
X 1 =  L i , w e w ant to find X 1 .
i 1
V al(T ): # of unbalanced node in T
(height balance  0)
10 -17
Case 1 : Absorption

The tree height is not increased, we need not
rebalance it.
1
0
Li
0
0
-1
0
newly
added
-1
-1
Ti-1
Ti
Val(Ti)=Val(Ti-1)+(Li1)
10 -18
Case 2.1 Single rotation
-1
D
0
B
Li
E(n)
A(n)
0
C(n)
E(n): height of tree E is n
0
newly added
10 -19
Case 2 : Rebalancing the tree
D
B
A
F
D
B
A
C
C
E
G
right
rotation
left
rotation
G
E
F
B
A
D
C
F
E
G
10 -20
Case 2.1 Single rotation

After a right rotation on the subtree rooted at
D:
0
B
0
D
A(n)
-1
-1
C(n)
E(n)
Val(Ti)=Val(Ti-1)+(Li-2)
10 -21
Case 2.2 Double rotation
-1
F
0
B
0
G(n)
D
A(n)
C(n-1)
E(n-1)
newly
added
10 -22
Case 2.2 Double rotation

After a left rotation on the subtree rooted at
B and a right rotation on the subtree rooted
at F:
0
D
0
1
B
A(n)
F
C(n-1)
E(n-1)
G(n)
Val(Ti)=Val(Ti-1)+(Li-2)
10 -23
Case 3 : Height increase
0
Li
A(n)
-1
root
0
-1
0
-1
root
newly
added

Li is the height of the root.
Val(Ti)=Val(Ti-1)+Li
10 -24
Amortized analysis of X1
X 2:
X 3:
X 4:
X 5:
#
#
#
#
of
of
of
of
ab so rp tio n s in case 1
sin g le ro tatio n s in case 2
d o u b le ro tatio n s in case 2
h eig h t in creases in case 3
m
V al(T m ) = V al(T 0 )+  L i  X 2  2 (X 3 + X 4 )
i 1
= 0 + X 1  X 2  2 (X 3 + X 4 )
V al(T m )  0 .6 1 8 m (p ro v ed b y K n u th )
 X 1 = V al(T m )+ 2 (X 2 + X 3 + X 4 ) X 2
 0 .6 1 8 m + 2 m
= 2 .6 1 8 m
10 -25
A self-organizing sequential
search heuristics

3 methods for enhancing the performance of
sequential search
(1 ) T ransp o se H euristics:
Q u ery
S eq u en ce
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
10 -26
(2 ) M o ve-to -the-F ro nt H euristics:
Q u ery
S eq u en ce
B
B
D
D B
A
A D B
D
D A B
D
D A B
C
C D A B
A
A C D B
10 -27
(3 ) C o unt H euristics: (d ecreasing o rd er b y the co unt)
Q u ery
S eq u en ce
B
B
D
B D
A
B D A
D
D B A
D
D B A
A
D A B
C
D A B C
A
D A B C
10 -28
Analysis of the move-to-thefront heuristics
 interword comparison: unsuccessful
comparison
 intraword comparison: successful comparison
 pairwise independent property:
 For any sequence S and all pairs P and Q, # of
interword comparisons of P and Q is exactly # of
interword comparisons made for the subsequence
of S consisting of only P’s and Q’s.
(See the example on the next page.)
10 -29
Pairwise independent
property in move-to-the-front
e.g .
Q u ery
S eq u en ce
C
C
A
A C
C
C A
B
B C A
C
C B A
A
A C B
(A , B ) co m p ariso n


# o f co m p ariso n s m ad e b etw een A an d B : 2
10 -30
C o n sid er th e su b seq u en ce co n sistin g o f A an d B :
Q uery
S equence
(A , B ) com parison
A
A
B
B A

A
A B

# o f co m p ariso n s m ad e b etw een A an d B : 2
10 -31
Query
Sequence
(A, B)
(A, C)
(B, C)
C
0
0
0
A
0
1
1
C
B
1
C
1
0
1
0
1
1
2
1
A
1
1
2
There are 3 distinct interword comparisons:
(A, B), (A, C) and (B, C)
 We can consider them separately and then
add them up.
the total number of interword comparisons:
0+1+1+2+1+2 = 7
10 -32
Theorem for the move-to-thefront heuristics
CM(S): # of comparisons of the move-to-thefront heuristics
CO(S): # of comparisons of the optimal static
ordering
CM (S) 2CO(S)
10 -33
Proof


Proof:
InterM(S): # of interword comparisons of the move to
the front heuristics
InterO(S): # of interword comparisons of the optimal
static ordering
Let S consist of a A’s and b B’s, a  b.
The optimal static ordering: BA
InterO(S) = a
 InterM(S)  2InterO(S)
InterM(S)  2a
10 -34
Proof





(cont.)
Consider any sequence consisting of more than two
items. Because of the pairwise independent property,
we have InterM(S)  2InterO(S)
IntraM(S): # of intraword comparisons of the moveto-the-front heuristics
IntraO(S): # of intraword comparisons of the optimal
static ordering
IntraM(S) = IntraO(S)
InterM(S) + IntraM(S)  2InterO(S) + IntraO(S)
 CM(S)  2CO(S)
10 -35
The count heuristics

The count heuristics has a similar result:
CC(S)  2CO(S), where CC(S) is the cost
of the count heuristics
10 -36
The transposition heuristics


The transposition heuristics does not possess the
pairwise independent property.
We can not have a similar upper bound for the cost
of the transposition heuristics.
e.g .
C o n sid er p airs o f d istin ct item s in d ep en d en tly.
Q u ery
S eq u en ce
C
(A , B )
A
C
0
(A , C )
0
(B , C )
0
0
1
1
B
C
1
1
A
1
0
0
1
1
1
2
1
# o f in terw o rd co m p ariso n s: 7 (n o t co rrect)
1
2
10 -37
th e co rrect in terw o rd co m p ariso n s:
Q uery S equence
C
A
C
B
D ata O rdering
C
AC
CA
CBA CBA CAB
1
1
2
N u m b er o f In terw o rd 0
C o m p ariso n s
C
0
A
2

6
10 -38