Skip lists - BGUds162/wiki.files/07-skip-list.pdf · 2017-02-27 · Skip list Each node stores 4...

Skip lists

Skip lists

Expectation — example

Consider a lottery with 1000 tickets and the following prizes.

1 ticket wins 100$.

9 ticket win 10$ each.

90 ticket win 5$ each.

The remaining tickets do not have prizes.

The cost of a ticket is 1$. Is it worthwhile to participate?

The sum of the prizes is

100 · 1 + 10 · 9 + 5 · 90 + 0 · 900 = 640 < 1000

The average win is

100 · 1 + 10 · 9 + 5 · 90 + 0 · 900

1000= 0.64 < 1

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The previous computation can be done using probabilitytheory.

The probability to win 100$ is 1/1000.




The expected win is

100 · 1

1000+ 10 · 9

1000+ 5 · 90

1000+ 0 · 900

1000= 0.64

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Probability space

A probability space is a pair (Ω,P) whereΩ is a set of outcomes. Ω is called sample space.P : Ω→ [0, 1] is a function that satisfies∑

x∈Ω P(x) = 1. P is called probability measure.

An event is a subset of Ω.For an event A ⊆ Ω, Pr[A] =

∑x∈A P(x).

Example

The lottery example can be modeled by the followingprobability space.

Ω = 1, 2, . . . , 1000.P(x) = 1/1000 for all x ∈ Ω.

A1 = 1 (the event of winning 100$)A2 = 2, 3, . . . , 9 (the event of winning 10$), etc.Pr[A1] = 1/1000, Pr[A2] = 9/1000.

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Expectation

A random variable is a function X : Ω→ R .

The expectation of a random variable X is

E [X ] =∑k

k · Pr[X = k].

Example

X (y) =

100 if y = 1

10 if 2 ≤ y ≤ 10

5 if 11 ≤ y ≤ 100

0 otherwise

E [X ] = 100 · 1

1000+ 10 · 9

1000+ 5 · 90

1000+ 0 · 900

1000= 0.64

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Suppose we toss a coin 3 times. What is the expected numberof times it lands on “head”?

Ω = HHH,HHT,HTH,HTT,THH,THT,TTH,TTTP(x) = 1/8 for every x ∈ Ω.

Let X (y) = number of ‘H’s in y .For example, X (HHH) = 3, X (HHT) = 2.

E [X ] = 0 · 18

+ 1 · 38

+ 2 · 38

+ 3 · 18

= 32.

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Indicator random variable

An indicator random variable for an event A is a randomvariable X such that

X (y) =

1 if y ∈ A

0 otherwise

If X is an indicator random variable for an event A,

E [X ] = 0 · Pr[X = 0] + 1 · Pr[X = 1] = Pr[A]

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Linearity of expectation

Theorem

For random variables X1, . . . ,Xk , E [∑k

i=1 Xi ] =∑k

i=1 E [Xi ].

Example

Suppose we toss a coin 3 times. What is the expected numberof times it lands on “head”?

Let X = number of times the coin lands on “head”.

Let Xi be an indicator random variable for the event thatthe coin lands on “head” in the i -th toss.

X = X1 + X2 + X3, so

E [X ] = E [X1] + E [X2] + E [X3] =1

2+

1

2+

1

2=

3

2

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Binomial distribution

A random variable X with binomial distribution withparameters n, p is the number of times a coin lands on “head”when the coin is tossed n times, and the probability to land on“head” is p.

Ω = all H/T strings of length n.

P(y) = pk(1− p)n−k where k in the number of ‘H’s in y .

X (y) = number of ‘H’s in y .

Theorem

E [X ] = np.

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Geometric distribution

A random variable X with geometric distribution is the numberof coin tosses when a coin is tossed until the first “head”.

Ω = H,TH,TTH, . . ..P(H) = 1

2, P(TH) = 1

4, P(TTH) = 1

8, . . .

X (y) = length of y .

Theorem

E [X ] = 2.

E [X ] = 1 · 1

2+ 2 · 1

4+ 3 · 1

8+ · · · =

1

2+

1

4+

1

8+ · · · 1

+1

4+

1

8+ · · · 1

2

+1

8+ · · · 1

4...

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Dynamic set ADT

A dynamic set ADT is a structure that stores a set ofelements. Each element has a (unique) key and satellite date.The structure supports the following operations.

Search(S , k) Return the element whose key is k .

Insert(S , x) Add x to S .

Delete(S , x) Remove x from S (the operation receives apointer to x).

Minimum(S) Return the element in S with smallest key.

Maximum(S) Return the element in S with largest key.

Successor(S , x) Return the element in S with smallest keythat is larger than x .key.

Predecessor(S , x) Return the element in S with largest keythat is smaller than x .key.

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Skip list

Skip list is an implementation of dynamic set with thefollowing complexities:

Insert: Θ(log n) expected.Delete: Θ(1) expected.Minimum,Maximum,Successor,Predecessor: Θ(1) (worstcase).

Skip lists

Skip list

A skip list for a set S consists of sorted linked lists S0, S1, . . . , Sh

such that:

Each list Si contains dummy elements −∞ and ∞.

S0 contains all the elements of S .

Si is a sublist of Si−1.

Sh contains only the elements −∞ and ∞.

12-∞

-∞

-∞

-∞

-∞

-∞

17 20 25 31 38 39 44 50 55 ∞

12 17 25 31 38 44 55 ∞

17 25 55 ∞

17 25 55 ∞

17 ∞

∞

20

20

20

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Skip list

Each node stores 4 pointers: next, prev, below, above.

The structure stores a pointer S .topleft to the firstelement in Sh.

For efficient Minimum/Maximum, the structure storespointers to the first/last element in S0.

12-∞

-∞

-∞

-∞

-∞

-∞

17 20 25 31 38 39 44 50 55 ∞

12 17 25 31 38 44 55 ∞

17 25 55 ∞

17 25 55 ∞

17 ∞

∞

20

20

20

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Searching

Assume that Find(S , k) is a procedure that returns theelement in S with largest key that is at most k .

Search(S , k)

(1) p ← Find(S , k)(2) if p.key = k(3) return p(4) else(5) return NULL

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Find

Find(S , k)

(1) p ← S .topleft(2) while p.below 6= NULL(3) p ← p.below(4) while p.next.key ≤ k(5) p ← p.next(6) return p // p.key is largest key that is ≤ k

12-∞

-∞

-∞

-∞

-∞

-∞

17 20 25 31 38 39 44 50 55 ∞

12 17 25 31 38 44 55 ∞

17 25 55 ∞

17 25 55 ∞

17 ∞

∞

20

20

20

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Find

Find(S , k)

(1) p ← S .topleft(2) while p.below 6= NULL(3) p ← p.below(4) while p.next.key ≤ k(5) p ← p.next(6) return p // p.key is largest key that is ≤ k

12-∞

-∞

-∞

-∞

-∞

-∞

17 20 25 31 38 39 44 50 55 ∞

12 17 25 31 38 44 55 ∞

17 25 55 ∞

17 25 55 ∞

17 ∞

∞

20

20

20

Find(S,50)

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InsertionInsert(S , x)

(1) p ← Find(S , x .key)(2) Insert x after p(3) while random() < 1/2(4) while p.above = NULL(5) p ← p.prev(6) p ← p.above(7) Insert a copy y of x after p(8) if p = S .topleft crate a new top list

12-∞

-∞

-∞

-∞

-∞

-∞

17 20 25 31 38 39 44 50 55 ∞

12 17 25 31 38 44 55 ∞

17 25 55 ∞

17 25 55 ∞

17 ∞

∞

20

20

20

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12-∞

-∞

-∞

-∞

-∞

-∞

17 20 25 31 38 39 44 50 55 ∞

12 17 25 31 38 44 55 ∞

17 25 55 ∞

17 25 55 ∞

17 ∞

∞

20

20

20

40

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12-∞

-∞

-∞

-∞

-∞

-∞

17 20 25 31 38 39 44 50 55 ∞

12 17 25 31 38 44 55 ∞

17 25 55 ∞

17 25 55 ∞

17 ∞

∞

20

20

20

40

40

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12-∞

-∞

-∞

-∞

-∞

-∞

17 20 25 31 38 39 44 50 55 ∞

12 17 25 31 38 44 55 ∞

17 25 55 ∞

17 25 55 ∞

17 ∞

∞

20

20

20

40

40

40

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Deletion

Delete(S , x)

(1) while x 6= NULL(2) Remove x from its list(3) if the list becomes empty, delete the list(4) x ← x .above

12-∞

-∞

-∞

-∞

-∞

-∞

17 20 25 38 39 44 50 55 ∞

12 17 25 38 44 55 ∞

17 25 55 ∞

17 25 55 ∞

17 ∞

∞

20

20

20

31x31

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Deletion

Delete(S , x)


12-∞

-∞

-∞

-∞

-∞

-∞

17 20 25 38 39 44 50 55 ∞

12 17 25 38 44 55 ∞

17 25 55 ∞

17 25 55 ∞

17 ∞

∞

20

20

20

31

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Deletion

Delete(S , x)


12-∞

-∞

-∞

-∞

-∞

-∞

17 20 25 38 39 44 50 55 ∞

12 17 25 38 44 55 ∞

17 25 55 ∞

17 25 55 ∞

17 ∞

∞

20

20

20

Skip lists

The height of skip list

Let S be a skip list with n elements.

The probability that an element x is in the list Si is 1/2i .

The probability that Si has at least one element(excluding ±∞) is at most n · 1/2i .

The probability that S3 log n has at least one element is atmost n/23 log n = n/n3 = 1/n2.

The probability that S3 log n doesn’t have elements is atleast 1− 1/n2.

In other words, with high probability, the height of theskip list is O(log n).

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Space complexity

The expected space for the ±∞ elements is O(log n).

The probability that an element x is in the list Si is 1/2i .

The size of Si (excluding ±∞) has binomial distributionwith parameters n and 1/2i .

Therefore, E [|Si |] = n/2i .

The expected size of the skip list is

O(log n) +∞∑i=0

n

2i= O(log n) + n

∞∑i=0

1

2i= O(log n) + 2n

The expected space of a skip list is Θ(n).

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Time complexity – search/insertion

Find(S , k)

(1) p ← S .topleft(2) while p.below 6= NULL(3) p ← p.below(4) while p.next.key ≤ k(5) p ← p.next(6) return p

17 25

17

20

TTH40

40

Find(S,30)The expected number of iterations of the while loop inline 2 is O(log n).The number of iterations of while loop in line 4 on onelist Si has geometric distribution, so the expected numberof iterations is 2.The expected total number of iterations of the while loopin line 4 is O(log n).The expected time is O(log n).

Skip lists

Time complexity — deletion

The time of deletion is linear in the height of the tower ofthe deleted element.

The height of a tower has geometric distribution.

The expected height of a tower is 2.

The expected time of Delete is Θ(1).

Skip lists

Skip lists - BGUds162/wiki.files/07-skip-list.pdf · 2017-02-27 · Skip list Each node stores 4...

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