Self-Organizing Linked List
1
Motivation
• Searching for an element in a linked list takes O(N)
worst case time
– If we are lucky, and the element is at the
beginning, then search is fast
• Our goal is to try to keep elements that are likely to
be searched in the near future close to the head
– Methods vary in how they determine the nodes
that need to be close to the head
– The worst case search time in still O(N), but the
average time can be much better
– All linked list features, except search remain
unchanged
2
Common Methods for Self-Organization
• Move to front
– After an element is found in a search, move it to
the front
• Transpose
– After an element is found in a search, move it one
step closer to the front
• Count
– Order the nodes by the number of times an
element is searched
• Ordering
– Order by some other criterion, such as
lexicographic order
3
Example
Element
searched
or inserted
Plain
Move to Transpose
front
Count
Lexical
ordering
A
A
A
A
A
A
C
AC
AC
AC
AC
AC
B
ACB
ACB
ACB
ACB
ABC
C
ACB
CAB
CAB
C-1AB
ABC
D
ACBD
CABD
CABD
C-1ABD
ABCD
A
ACBD
ACBD
ACBD
C-1A-1BD
ABCD
D
ACBD
DACB
ACDB
C-1A-1D-1B ABCD
A
ACBD
?
?
?
?
C
ACBD
?
?
?
?
A
ACBD
?
?
?
?
Next, search for the following: C, C, E, E
4
Average Case Analysis
• Usually, the time taken is neither the worst case time
nor the best case time
– Average case analysis assumes a probability
distribution for the set of possible inputs and gives
the expected time
• tav =  input i probability(i)time(i)
– Average search time in a linked list
• tav = probability(key is not present)  n +
 i=1n probability(key is in node i)  i
• tav = n  [1 -  i=1n probability(key is in node i)] +
 i=1n probability(key is in node i)  i
5
Average Case Analysis Example - 1
• Probability(i) = 1/n, 1 < i < n
tav = n  [1 -  i=1n probability(key is in node i)] +
 i=1n probability(key is in node i)  i
= n  [1 -  i=1n 1/n] +  i=1n 1/n  i
= n  [0] + (1/n)  i=1n i = (1/n) n(n+1)/2 = (n+1)/2
= O(n)
6
Average Case Analysis Example - 2
• Probability(i) = 1/(2n), 1 < i < n
tav = n  [1 -  i=1n probability(key is in node i)] +
 i=1n probability(key is in node i)  i
= n  [1 -  i=1n 1/(2n)] +  i=1n 1/(2n)  i
= n  [1/2] + (1/(2n))  i=1n i = n/2 + (1/2n) n(n+1)/2
= n/2 + (n+1)/4 = O(n)
7
Average Case Analysis Example - 3
• Prob(1) = 0.5, prob(2) = 0.3, prob(3) = 0.2
tav = n  [1 -  i=1n probability(key is in node i)] +
 i=1n probability(key is in node i)  i
= n[1 - ] + 0.51 + 0.32 + 0.23 +  i=4n 0i
= n[0] + 0.5 + 0.6 + 0.6 + 0 = 1.7 = O(1)
8