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# Self-Organizing Linked List - PowerPoint PPT Presentation

Self-Organizing Linked List. 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

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Presentation Transcript

• 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

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

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Next, search for the following: C, C, E, E

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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

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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)

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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)

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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)

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