San Diego State University

In a sequence of operations on a data structure often the worst case can not occur in each operation

Assume we have a list of n = 2k items: a1, a2, ..., an

Will perform n accesses on the list

Will access item a1 n/2 times

Worst case access requires n compares

Thus worst case cost for n access is O(n*n)?????

At most 1 + n/2 accesses require n comparisons!

Once item a1 is accessed it will not make it to the end of the list

Let A[0..k-1] be an array of bits representing a number X

A[0] - low order bit, so

length[A] = k

Start with X = 0

Increment (A)

- J = 0
- while J < length[A] and A[J] == 1 do
- A[J] = 0
- J = J + 1
- end while
- if J < length[A] then
- A[J] = 1
- end if

Count bits flipped

Worst Case

- Increment flips k bits in worst case
- Sequence of n Increment operations takes O(nk)

T(n) = all work done in worst case in sequence of n operations

Amortized cost per operation is T(n)/n

X A[4] A[3] A[2] A[1] A[0] Total 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

A[0] flips each time Increment is called n

A[1] flips every other time

A[1] flips every fourth time

A[J] flips every 2**J time

Total number of flips is:

So amortized cost of each operation is 2 = O(1)

Assign an

Amortized cost may be more or less than the actual cost

If amortized cost is more than the actual cost of the operation assign the difference to part of the data structure as a credit

No negative credit allowed

Total amortized cost is >= total worst case cost

Amortized cost of setting bit to 1 2 units

- 1 unit to pay for setting bit to 1
- 1 unit stored with bit

Amortized cost of setting bit to 0 0 units

- Only a 1-bit is set to 0,
- All 1-bits have credit of one unit
- This pays for setting bit to 0

J = 0 while J < length[A] and A[J] == 1 do A[J] = 0 Cost 0 J = J + 1 end while if J < length[A] then A[J] = 1 Cost 2 end ifend Increment

X A[4] A[3] A[2] A[1] A[0] Amortized cost 0 0 0 0 0 0 0 1 0 0 0 0 1 (1) 2 2 0 0 0 1 (1) 0 4 3 0 0 0 1 (1) 1 (1) 6 4 0 0 1 (1) 0 0 8 5 0 0 1 (1) 0 1 (1) 10 6 0 0 1 (1) 1 (1) 0 12 7 0 0 1 (1) 1 (1) 1 (1) 14 8 0 1 (1) 0 0 0 16 9 0 1 (1) 0 0 1 (1) 18

Assume we have a list of n = 2k items: a1, a2, ..., an

Will perform n accesses on the list

Will access item a1 n/2 times

Amortized costs:

- First access of a
- initial location of a <= n
- All other access of a
- 1
- Accessing a non-a item
- actual cost + 1 <= n +1
- assign credit to a

a, b, c, d, e Amortized cost b, a(1), c, d, e accessed b 2 c, b, a(2), d, e accessed c 4 a, c, b, d, e accessed a 1 e a(1), c, b, d accessed e 6 a, c, b, d, e accessed a 1 a, c, b, d, e accessed a 1

(Amortized cost of all accesses of a) <= n + n/2 - 1 = 3n/2 - 1

(Amortized cost of accessing all non-a items) <= n*(n+1)/2

(Total Cost) <= (4n + n*n)/2 -1, (Average Cost/access) <= 4 + n/2

Assign an

Amortized cost may be more or less than the actual cost

If amortized cost is more than the actual cost of the operation assign the difference the entire data structure as potential energy

ck= actual cost of operation k

= amortized cost of operation k

Dk = the state of the data structure after applying k'th operation to Dk

= potential associated with Dk

So if >=0 then is an upper bound on total cost of the algorithm

Potential = number of 1's in the counter

if the k'th operation sets tk bits to 0 the actual cost is tk + 1

=1 - tk

so

= 2

How to find ?

Table -> pointer to a table

Number -> number of items in the table

Size -> size of the table

AddToTable(x)

- if Number == Size then
- allocate NewTable with size 2*Size
- insert all items from Table to newTable
- free Table
- Table = NewTable
- Size = 2 * Size

- end if
- insert x into Table
- Number = Number + 1

Amortized Cost per AddToTable 3 inserts

Table after moving from size 4 to size 8

Perform 4 AddToTable operations

X X X X X X X X Y(2) X X X X Y(2) Y(2) X X X X Y(2) Y(2) Y(2) X X X X Y(2) Y(2) Y(2) Y(2)

1/2 the table has 2 credits

One credit per item to pay for the move to next size table

X X X X Y Y Y Y

Policy

- When table becomes full move to table twice the size
- When table contracts to 1/4 full, move to table 1/2 size

Let load = (Size of table) / (number of items in the table)

Amortized Cost

- Inserting when load >= 1/2
- 3 units
- Inserting when load < 1/2
- 0 units
- Deleting when load > 1/2
- 0 units
- Deleting when load <= 1/2
- 2 units

X X X X X X X X Y(2) X X X X X X X (1) X X (1) (1) X X

Amortized Costs per operation over n operations

linked List Dynamic Array Insertion 1 call to new lg(n)/n call to new set two links 3 data moves

Deletion after find two links 3 data moves 1 delete lg(n)/n call to new