| Introduction to Algorithms | October 25, 2005 |
| Massachusetts Institute of Technology | 6.046J/18.410J |
| Professors Erik D. Demaine and Charles E. Leiserson | Handout 17 |
Lecture Notes on Skip Lists
Lecture 12 — October 26, 2005
Erik Demaine
•Balanced tree structures we know at this point: red-black trees, B-trees, treaps.
•Could you implement them right now? Probably, with time. . . but without looking up any details?
•Skip lists are a simple randomized structure you’ll never forget.
Starting from scratch
•Initial goal: just searches — ignore updates (Insert/Delete) for now
•Simplest data structure: linked list
•Sorted linked list: (n) time
•2 sorted linked lists:
–Each element can appear in 1 or both lists
–How to speed up search?
–Idea: Express and local subway lines
–Example: 14 , 23, 34 , 42 , 50, 59, 66, 72 , 79, 86, 96 , 103, 110, 116, 125
(What is this sequence?)
–Boxed values are “express” stops; others are normal stops
–Can quickly jump from express stop to next express stop, or from any stop to next normal stop
–Represented as two linked lists, one for express stops and one for all stops:
| 14 | 34 | 42 | 72 | 96 |
14
23 
34 
42 
50 
59 
66 
72 
79 
86 
96 
103
110
116
125
23 34 42 50 59 66 72 79 86 96 103110116125
–Every element is in bottom linked list (L2); some elements also in top linked list (L1)
–Link equal elements between the two levels
–To search, first search in L1 until about to go too far, then go down and search in L2
| 2 | Handout 17: Lecture Notes on Skip Lists | |||||||
| – Cost: | + |L2| = L | n | ||||||
| L | + | |||||||
| – Minimized when | | | 1| | |L1| | | 1| | |L1| | |||
| n | ||||||||
| |L1| = |L1| | ||||||||
| = | |L1|2 = n | |||||||
| = | |L1| = | 8 | ||||||
| n | 8 | |||||||
| = search cost = 2 | n | |||||||
| – Resulting 2-level structure: | ||||||||
| sqrt n | ||||||||
| sqrt n | sqrt n | sqrt n | sqrt n | |||||
| • 3 linked lists: 3 · 83 n | ||||||||
| • k linked lists: k · 8k n | ||||||||
| • lg n linked lists: lg n · lg8n n = lg n · n1/ lg n = (lg n) | ||||||||
| =2 | ||||||||
– Becomes like a binary tree:
| 14 | 79 | |||||||||||||
| 14 | 50 | 79 | 110 | |||||||||||
| 14 | 34 | 50 | 66 | 79 | 96 | 110 | 125 | |||||||
| 14 | 23 | 34 | 42 | 50 | 59 | 66 | 72 | 79 | 86 | 96 | 103 | 110 | 116 | 125 |
–(In fact, a level-linked B+-tree; see Problem Set 5.)
–Example: Search for 72
Level 1: 14 too small, 79 too big; go down 14
Level 2: 14 too small, 50 too small, 79 too big; go down 50Level 3: 50 too small, 66 too small, 79 too big; go down 66Level 4: 66 too small, 72 spot on
| Handout 17: Lecture Notes on Skip Lists | 3 |
Insert
•New element should certainly be added to bottommost level (Invariant: Bottommost list contains all elements)
•Which other lists should it be added to?
(Is this the entire balance issue all over again?)
•Idea: Flip a coin
–With what probability should it go to the next level?
–To mimic a balanced binary tree, we’d like half of the elements to advance to the next- to-bottommost level
–So, when you insert an element, flip a fair coin
–If heads: add element to next level up, and flip another coin (repeat)
•Thus, on average:
–1/2 the elements go up 1 level
–1/4 the elements go up 2 levels
–1/8 the elements go up 3 levels
–Etc.
•Thus, “approximately even”
Example
•Get out a real coin and try an example
•You should put a special value -* at the beginning of each list, and always promote this special value to the highest level of promotion
•This forces the leftmost element to be present in every list, which is necessary for searching
.. . many coins are flipped . . .
(Isn’t this easy?)
•The result is a skip list.
•It probably isn’t as balanced as the ideal configurations drawn above.
•It’s clearly good on average.
•Claim it’s really really good, almost always.
| 4 | Handout 17: Lecture Notes on Skip Lists |
Analysis: Claim of With High Probability
•Theorem: With high probability, every search costs O(lg n) in a skip list with n elements
•What do we need to do to prove this? [Calculate the probability, and show that it’s high!]
•We need to define the notion of “with high probability”; this is a powerful technical notion, used throughout randomized algorithms
•Informal definition: An event occurs with high probability if, for any → 1, there is an appropriate choice of constants for which E occurs with probability at least 1 - O(1/n )
•In reality, the constant hidden within O(lg n) in the theorem statement actually depends on c.
•Precise definition: A (parameterized) event E occurs with high probability if, for any→1, E occurs with probability at least 1 - c /n , where c is a “constant” depending only on .
•The term O(1/n ) or more precisely c /n is called the error probability
•The idea is that the error probability can be made very very very small by setting to something big, e.g., 100
Analysis: Warmup
•Lemma: With high probability, skip list with n elements has O(lg n) levels
•(In fact, the number of levels is (log n), but we only need an upper bound.)
•Proof:
– Pr{element x is in more than c lg n levels} = 1/2c lg n = 1/nc
– Recall Boole’s inequality / union bound:
Pr{E1 = E2 = · · · = Ek } Pr{E1} + Pr{E2} + · · · + Pr{Ek}
–Applying this inequality:
Pr{any element is in more than c lg n levels} n · 1/nc = 1/nc-1
–Thus, error probability is polynomially small and exponent ( = c - 1) can be made arbitrarily large by appropriate choice of constant in level bound of O(lg n)
| Handout 17: Lecture Notes on Skip Lists | 5 |
Analysis: Proof of Theorem
•Cool idea: Analyze search backwards—from leaf to root
–Search starts at leaf (element in bottommost level)
–At each node visited:
If node wasn’t promoted higher (got TAILS here), then we go [came from] leftIf node was promoted higher (got HEADS here), then we go [came from] up
–Search stops at root of tree
•Know height is O(lg n) with high probability; say it’s c lg n
•Thus, the number of “up” moves is at most c lg n with high probability
•Thus, search cost is at most the following quantity:
How many times do we need to flip a coin to get c lg n heads?
• Intuitively, (lg n)
Analysis: Coin Flipping
•Claim: Number of flips till c lg n heads is (lg n) with high probability
•Again, constant in (lg n) bound will depend on
•Proof of claim:
–Say we make 10c lg n flips
–When are there at least c lg n heads?
1 c lg n 1 9c lg n
–Pr{exactly c lg n heads} = c lg n · 2 · 2
| HHHTTTordersvs. HTHTHT | heads | tails | |||||||||||
| – Pr{at most c lg n heads} 10cclglgnn · | 21 | 9c lg n | |||||||||||
| overestimateon orders | tails | ||||||||||||
| – Recall bounds on xy : | xy | x xy e xy | x | ||||||||||
| 6 | Handout 17: Lecture Notes on Skip Lists |
– Applying this formula to the previous equation:
| Pr{at most c lg n heads} | 10cclglgnn | 21 9c lg n | |||
| e ·c10lgcnlg n c lg n · 21 | 9c lg n | ||||
| = | (10e)c lg n · | 21 9c lg n | |||
| = | 2lg(10e)·c lg n · 21 9c lg n | ||||
| = | 2(lg(10e)-9)c lg n | ||||
| = | 2- lg n | ||||
| = | 1/n | ||||
–The point here is that, as 10 *, = 9 - lg(10e) *, independent of (for all) c
•End of proof of claim and theorem
Acknowledgments
This lecture is based on discussions with Michael Bender at SUNY Stony Brook.
