2-12 Hashing MA Juncslabcms.nju.edu.cn/.../f/f7/2019-2-12_Hashing.pdf · Hashing: the idea Q: When...

Problem Solving2-12 Hashing

MA Jun

Institute of Computer Software

May 14, 2020

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Contents

1 Hash-table: Basic Idea

2 Hash Functions

3 Collision Resolution

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Applications of Hashing

There are many applications of hashing (not limited to hash table),including modern day cryptography hash functions. Some of theseapplications are listed below:

Message DigestPassword VerificationData Structures (Programming Languages)Compiler OperationRabin-Karp AlgorithmLinking File Name and Path Together

https://www.geeksforgeeks.org/applications-of-hashing/

MA Jun (Institute of Computer Software) Problem Solving May 14, 2020 1 / 34

https://www.geeksforgeeks.org/applications-of-hashing/

Contents


2 Hash Functions


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Hash-table: Basic Idea

Many applications require a Dynamic Set that supports only thedictionary operations Insert, Search, and Delete.

Searching for an element in a hash table can take as long as searching foran element in a linked list Θ(n) time in the worst case.

Under reasonable assumptions, the average time to search for an elementin a hash table is Θ(1).


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Hashing: the idea

Q : When should we use hash table?What situations is the hash table suitable for?

Key Space

x

Very large, but only a smallpart is used in an applica-tion at a certain time.

· · ·

· · ·

E [0]E [1]

E [m − 1]

Feasible size

Hash FunctionE [k]

Index distributionCollision handling


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Hashing: the idea

Q : What is a Collision? When does it take place?

Hash FunctionKey Space

· · ·

· · ·

E [0]E [1]

E [m − 1]

xyE [k]


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Hashing: the idea



· · ·

· · ·

E [0]E [1]

E [m − 1]

xy

E [k]


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Hashing: the idea



· · ·

· · ·

E [0]E [1]

E [m − 1]

xyE [k]


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Collision

Q : Given n keys and m slots, what is the expected number of keyshashed into the same slot?


xy

· · ·

· · ·

E [0]E [1]

E [m − 1]

E [k]

It depends ...


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Model & Assumption

ModelInserting n keys: sequence of n s independent trails.Each insertion corresponds to a value from {0, 1, · · · , m − 1}.

AssumptionFor each insertion, the result is Uniformly Distributed, i.e. each valuefrom {0, 1, · · · , m − 1} is equally picked.

In hashing n items into a hash table of size m, the expected number of itemsthat hash to any one location is α = n/m (α: loading factor)


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Empty location

After inserting n items into m locations

Q : What is the probability of a given location is empty?

(1 − 1m )n

Q : What is the expected number of empty locations?

Xi ={

1 , location i is empty0 , otherwise

X =∑

1≤i≤mXi

E (X ) = E (∑

1≤i≤mXi) =

∑1≤i≤m

E (Xi) =∑

1≤i≤m(1 − 1/m)n =

m(1 − 1/m)n


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After inserting n items into a hashtable with n locations

Q : What is the expected number of items in a given location?

Yi ={

1 , the ith item in the given location0 , otherwise

Y =∑

1≤i≤nYi

E (Y ) = E (∑

1≤i≤nYi) =

∑1≤i≤n

E (Yi) =∑

1≤i≤n1/n = 1

Expected number of empty locations: n(1 − 1/n)n = n/e ≈ 0.368n

PARADOX?collisions!


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Yi ={


Y =∑

1≤i≤nYi

E (Y ) = E (∑

1≤i≤nYi) =

∑1≤i≤n

E (Yi) =∑

1≤i≤n1/n = 1


PARADOX?

collisions!


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Yi ={


Y =∑

1≤i≤nYi

E (Y ) = E (∑

1≤i≤nYi) =

∑1≤i≤n

E (Yi) =∑

1≤i≤n1/n = 1


PARADOX?collisions!


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Expected number of collisions


Q : What is the expected number of collisions?

E (collisions) = n − E (occupied locations)

= n − (m − E (empty locations))

= n − (m − m (1 − 1/m)n)

= n − m + m (1 − 1/m)n

Example:When n = 100, m = 100, then E (collision) ≈ 37


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Q : What is the expected number of items needed to fullfill all mlocations?

Xi : the number of items to be added to increase thenumber of occupied locations from i − 1 to i .Given i − 1 occupied locations,

the success probabilityfor each trial is (m − i + 1)/m

Thus, E (Xi) = m/(m − i + 1)

E (X ) =m∑

i=1E (Xi) =

m∑i=1

m/(m − i + 1)

= mm∑

i=11/(m − i + 1)

= mm∑

j=11/j = Θ(m lg m)

· · ·

# = i − 1


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Xi : the number of items to be added to increase thenumber of occupied locations from i − 1 to i .

Given i − 1 occupied locations,

the success probabilityfor each trial is (m − i + 1)/m

Thus, E (Xi) = m/(m − i + 1)

E (X ) =m∑

i=1E (Xi) =

m∑i=1

m/(m − i + 1)

= mm∑

i=11/(m − i + 1)

= mm∑

j=11/j = Θ(m lg m)

· · ·

# = i − 1


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Xi : the number of items to be added to increase thenumber of occupied locations from i − 1 to i .Given i − 1 occupied locations,

the success probabilityfor each trial is (m − i + 1)/mThus, E (Xi) = m/(m − i + 1)

E (X ) =m∑

i=1E (Xi) =

m∑i=1

m/(m − i + 1)

= mm∑

i=11/(m − i + 1)

= mm∑

j=11/j = Θ(m lg m)

· · ·

# = i − 1


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Xi : the number of items to be added to increase thenumber of occupied locations from i − 1 to i .Given i − 1 occupied locations, the success probabilityfor each trial is (m − i + 1)/m

Thus, E (Xi) = m/(m − i + 1)

E (X ) =m∑

i=1E (Xi) =

m∑i=1

m/(m − i + 1)

= mm∑

i=11/(m − i + 1)

= mm∑

j=11/j = Θ(m lg m)

· · ·

# = i − 1


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Xi : the number of items to be added to increase thenumber of occupied locations from i − 1 to i .Given i − 1 occupied locations, the success probabilityfor each trial is (m − i + 1)/mThus, E (Xi) = m/(m − i + 1)

E (X ) =m∑

i=1E (Xi) =

m∑i=1

m/(m − i + 1)

= mm∑

i=11/(m − i + 1)

= mm∑

j=11/j = Θ(m lg m)

· · ·

# = i − 1


Contents


2 Hash Functions


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Hash Functions

Hashing: the idea

Hash Function

Index distributionCollision handling

Key Space

Very large, but only a smallpart is used in an applica-tion at a certain time.

x· · ·

· · ·

E [0]E [1]

E [m − 1]

E [k]

In feasible size

Hash Function is IMPORTANT!


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Hash Functions

Two typical hash functions

Division Method: map a key k into one of m slots by taking theremainder of k divided by m.

h(k) = k mod m

Multiplication Method:Step-1: multiply the k by a constant A in the range 0 < A < 1 andextract the fractional part of kA.Step-2: multiply the fractional part by m.


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Hash Functions


Division Method: map a key k into one of m slots by taking the remainderof k divided by m:

h(k) = k mod m

Q : Why should we avoid m = 2p?Since if m = 2p, then h(k) is just the p lowest-order bits of k.Unless we know that all low-order p-bit patterns are equally likely, weare better off designing the hash function to depend on all the bits ofthe key.

A prime not too close to an exact power of 2 is often a good choice for m.

Exercise 11.3-3


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Hash Functions



h(k) = k mod m

Q : Why should we avoid m = 2p?

Since if m = 2p, then h(k) is just the p lowest-order bits of k.Unless we know that all low-order p-bit patterns are equally likely, weare better off designing the hash function to depend on all the bits ofthe key.


Exercise 11.3-3


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Hash Functions



h(k) = k mod m

Q : Why should we avoid m = 2p?Since if m = 2p, then h(k) is just the p lowest-order bits of k.

Unless we know that all low-order p-bit patterns are equally likely, weare better off designing the hash function to depend on all the bits ofthe key.


Exercise 11.3-3


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Hash Functions



h(k) = k mod m

Q : Why should we avoid m = 2p?Since if m = 2p, then h(k) is just the p lowest-order bits of k.Unless we know that all low-order p-bit patterns are equally likely, weare better off designing the hash function to depend on all the bits ofthe key.


Exercise 11.3-3


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Hash Functions


Multiplication Method:Step-1: multiply the k by a constant A in the range 0 < A < 1 andextract the fractional part of kA.Step-2: multiply the fractional part by m

h(k) = ⌊m(kA mod 1)⌋

Q : Why do we usually choose m = 2p?


Contents


2 Hash Functions


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Collision Resolution

Collision


xy

· · ·

· · ·

E [0]E [1]

E [m − 1]

E [k]


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Chaining (Closed Addressing) Open Addressing


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Collision Resolution Chaining

Collision Resolution by Chaining


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Collision Resolution by Chaining: Unsuccessful SearchQ : What is the average cost of an unsuccessful search?

For j = 0, 1, 2, ..., m − 1, the average length of thelist at E [j] is n/m = α.

Any key that is not in the table is equally likely to hash to any of them addresses.The average cost to determine that the key is not in the list E [h(k)]is the cost to search to the end of the list, which is α.Total cost is Θ(1 + α)


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Any key that is not in the table is equally likely to hash to any of them addresses.

The average cost to determine that the key is not in the list E [h(k)]is the cost to search to the end of the list, which is α.Total cost is Θ(1 + α)


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Any key that is not in the table is equally likely to hash to any of them addresses.The average cost to determine that the key is not in the list E [h(k)]is the cost to search to the end of the list, which is α.

Total cost is Θ(1 + α)


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Any key that is not in the table is equally likely to hash to any of them addresses.The average cost to determine that the key is not in the list E [h(k)]is the cost to search to the end of the list, which is α.Total cost is Θ(1 + α)


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Collision Resolution by Chaining: Successful SearchQ : What is the average cost of an successful search?

xi : the ith element inserted into the table.The probability of that xi is searched is 1/n.For a specific xi , the number of elements examined in a successfulsearch is t + 1I t: the number of elements inserted into the same list as xi , after xi

has been inserted.I The probability of that xj is inserted into the same list of xi is 1/m.I The expected number of elements examined for a successful search

of xi is (1 +n∑

j=i+1

1m )


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xi : the ith element inserted into the table.

The probability of that xi is searched is 1/n.For a specific xi , the number of elements examined in a successfulsearch is t + 1I t: the number of elements inserted into the same list as xi , after xi


of xi is (1 +n∑

j=i+1

1m )


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xi : the ith element inserted into the table.The probability of that xi is searched is 1/n.

For a specific xi , the number of elements examined in a successfulsearch is t + 1I t: the number of elements inserted into the same list as xi , after xi


of xi is (1 +n∑

j=i+1

1m )


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xi : the ith element inserted into the table.The probability of that xi is searched is 1/n.For a specific xi , the number of elements examined in a successfulsearch is t + 1

I t: the number of elements inserted into the same list as xi , after xihas been inserted.

I The probability of that xj is inserted into the same list of xi is 1/m.I The expected number of elements examined for a successful search

of xi is (1 +n∑

j=i+1

1m )


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has been inserted.

I The probability of that xj is inserted into the same list of xi is 1/m.I The expected number of elements examined for a successful search

of xi is (1 +n∑

j=i+1

1m )


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has been inserted.I The probability of that xj is inserted into the same list of xi is 1/m.

I The expected number of elements examined for a successful searchof xi is (1 +

n∑j=i+1

1m )


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of xi is (1 +n∑

j=i+1

1m )


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Collision Resolution by Chaining: Successful SearchQ : What is the average cost of a successful search?

Average number of elements examined for an successful search is

1n

n∑i=1

(1 +n∑

j=i+11m )

= 1 + 1nm (

n∑j=i+1

1)

= 1 + 1nm (

n∑i=1

n − i)

= 1 + 1nm (n2 −

n∑i=1

i)

= 1 + 1nm (n2 − n(n+1)

2 )

= 1 + n2m − n

2nm

= 1 + α − α2n

Average cost for a successful search is 2 + α − α2n = Θ(1 + α)


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Collision Resolution by Chaining: Successful SearchQ : What is the average cost of a successful search?

Average number of elements examined for an successful search is

1n

n∑i=1

(1 +n∑

j=i+11m ) = 1 + 1

nm (n∑

j=i+11)

= 1 + 1nm (

n∑i=1

n − i)

= 1 + 1nm (n2 −

n∑i=1

i)

= 1 + 1nm (n2 − n(n+1)

2 )

= 1 + n2m − n

2nm

= 1 + α − α2n

Average cost for a successful search is 2 + α − α2n = Θ(1 + α)


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Collision Resolution by Chaining: Example

JAVA 7 HashMap

1 static int hash( int h) {2 h^= (h>>>20)^(h>>>12);3 return h^(h>>>7)^(h>>>4);4 }

In Java 7, after calculating hash from hashfunction, if more then one element has samehash than they are searched by linear search, soit’s complexity is O(n).In Java 8, that search is performed by binarysearch so the complexity will become log n.


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Collision Resolution Open Addressing

Open Addressing

All elements are stored in the hash table, no linked list isused. So, α ≤ 1.

Collision is settled by “rehashing”:I A function used to get a new hashing address for each

collided addressI The hash table slots are probed successively, until a valid

location is found.The probing sequence can be seen as a permutation of(0, 1, 2, · · · , m − 1)→⟨h(k, 0), h(k, 1), · · · , h(k, m − 1)⟩


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Open Addressing

All elements are stored in the hash table, no linked list isused. So, α ≤ 1.Collision is settled by “rehashing”:

I A function used to get a new hashing address for eachcollided address

I The hash table slots are probed successively, until a validlocation is found.

The probing sequence can be seen as a permutation of(0, 1, 2, · · · , m − 1)→⟨h(k, 0), h(k, 1), · · · , h(k, m − 1)⟩


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Open Addressing

All elements are stored in the hash table, no linked list isused. So, α ≤ 1.Collision is settled by “rehashing”:I A function used to get a new hashing address for each

collided address

I The hash table slots are probed successively, until a validlocation is found.



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Open Addressing



location is found.



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Open Addressing



location is found.The probing sequence can be seen as a permutation of(0, 1, 2, · · · , m − 1)→⟨h(k, 0), h(k, 1), · · · , h(k, m − 1)⟩


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Open Addressing: Commonly Used Probings

Linear Probing: (primary clustering may occur)Given an auxiliary hash function h′, the hash function is:

h(k, i) = (h′(k) + i) mod m, (i = 0, 1, ..., m − 1)

Quadratic Probing: (secondary clustering may occur)Given auxiliary function h′ and nonzero auxiliary constant c1 and c2, thehash function is:

h(k, i) = (h′(k) + c1i + c2i2) mod m, (i = 0, 1, ..., m − 1)

Double Hashing:Given auxiliary functions h1 and h2, the hash function is:

h(k, i) = (h1(k) + ih2(k)) mod m, (i = 0, 1, ..., m − 1)


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Linear Probing: (primary clustering may occur)Given an auxiliary hash function h′, the hash function is:

h(k, i) = (h′(k) + i) mod m, (i = 0, 1, ..., m − 1)

Quadratic Probing: (secondary clustering may occur)Given auxiliary function h′ and nonzero auxiliary constant c1 and c2, thehash function is:

h(k, i) = (h′(k) + c1i + c2i2) mod m, (i = 0, 1, ..., m − 1)

Double Hashing:Given auxiliary functions h1 and h2, the hash function is:

h(k, i) = (h1(k) + ih2(k)) mod m, (i = 0, 1, ..., m − 1)MA Jun (Institute of Computer Software) Problem Solving May 14, 2020 23 / 34

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Linear Probing: an Example

0 17761

1776

2

1776

3

1776

4

1776

5

1776

6

1776

7

1776

10551492

1918

Hasing function: h(x) = 5x mod 8

Rehasing function: rh(j) = (j + 1) mod 8

1812hashing

1812rehashing

1812

1945

hashing

1945rehashing

1945


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0 17761

1776

2

1776

3

1776

4

1776

5

1776

6

1776

7

1776

10551492

1918



1812

hashing

1812rehashing

1812

1945

hashing

1945rehashing

1945


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0 17761

1776

2

1776

3

1776

4

1776

5

1776

6

1776

7

1776

10551492

1918



1812hashing

1812rehashing

1812

1945

hashing

1945rehashing

1945


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0 17761

1776

2

1776

3

1776

4

1776

5

1776

6

1776

7

1776

10551492

1918



1812hashing

1812rehashing

1812

1945

hashing

1945

rehashing1945


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0 17761

1776

2

1776

3

1776

4

1776

5

1776

6

1776

7

1776

10551492

1918



1812hashing

1812rehashing

1812

1945

hashing

1945rehashing

1945


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Q : How to evaluate the goodness of a probing?

AssumptionEach key is equally likely to have any of the m! permutationsof (1, 2, · · · , m − 1) as its probe sequence.

Both linear and quadratic probing have only m distinct probesequences, as determined by the first probe.

h(k, i) = (h′(k) + i) mod m, (i = 0, 1, ..., m − 1)h(k, i) = (h′(k) + c1i + c2i2) mod m, (i = 0, 1, ..., m − 1)

Double hashing improves over linear or quadratic probing in thatΘ(m2) probe sequences are used.

h(k, i) = (h1(k) + ih2(k)) mod m, (i = 0, 1, ..., m − 1)


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Open Addressing: Deleting Element

Q : Why Open Addressing is not suitable for situations where itemsmight be deleted?

The probing chain might be broken if an item is deleted.How to deal with it?


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The probing chain might be broken if an item is deleted.

How to deal with it?


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The probing chain might be broken if an item is deleted.How to deal with it?


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Open Addressing: Unsuccessful SearchQ : Assuming uniform hashing, what is the average number of probesin an unsuccessful search?

X : the number of probes made in an unsuccessful search

E (X ) =∞∑

i=0i · Pr(X = i)

=∞∑

i=0i · (Pr(X ≥ i) − Pr(X ≥ i + 1))

=∞∑

i=1Pr(X ≥ i)


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Open Addressing: Unsuccessful Search

Q : How to compute Pr(X ≥ i)?

Ai : the ith probe occurs at an occupied slotX ≥ i = A1 ∩ A2 ∩ · · · ∩ Ai−1

Pr(X ≥ i) = Pr(A1 ∩ A2 ∩ · · · ∩ Ai−1)= Pr(A1) · Pr(A2|A1) · Pr(A3|A1 ∩ A2) · · · Pr(Ai−1|A1 ∩ A2 ∩ · · · ∩ Ai−2)

= nm · n−1

m−1 · · · n−i+2m−i+2

≤ ( nm )i−1

= αi−1

Then, E (x) =∞∑

i=1Pr(X ≥ i) ≤

∞∑i=1

αi−1 =≤∞∑

i=0αi = 1

1−α


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Q : How to compute Pr(X ≥ i)?Ai : the ith probe occurs at an occupied slot

X ≥ i = A1 ∩ A2 ∩ · · · ∩ Ai−1


= nm · n−1

m−1 · · · n−i+2m−i+2

≤ ( nm )i−1

= αi−1

Then, E (x) =∞∑

i=1Pr(X ≥ i) ≤

∞∑i=1

αi−1 =≤∞∑

i=0αi = 1

1−α


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Q : How to compute Pr(X ≥ i)?Ai : the ith probe occurs at an occupied slotX ≥ i = A1 ∩ A2 ∩ · · · ∩ Ai−1


= nm · n−1

m−1 · · · n−i+2m−i+2

≤ ( nm )i−1

= αi−1

Then, E (x) =∞∑

i=1Pr(X ≥ i) ≤

∞∑i=1

αi−1 =≤∞∑

i=0αi = 1

1−α


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Open Addressing: Inserting an Element

Q : What is the average cost for inserting an element into a tablewith load factor α?

Proof.An element is inserted only if there is room in the table, and thusα < 1.Inserting a key requires an unsuccessful search followed by placing thekey into the first empty slot found.Thus, the expected number of probes is at most 1/(1 − α)


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Open Addressing: Insertion/Unsuccessful Search

11−α


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Open Addressing: Successful Search

Q : What is the average number of probes in a successful search ofan element in a table with load factor α?


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Open Addressing: Successful SearchQ : what is the average number of probes in a successful search foran element in a table with load factor α?

Proof.

A search for a key k reproduces the same probe sequence as whenthe element with key k was inserted.If k is the (i+1)st key inserted into the table, the expected number ofprobes in a search for k is at most 1/(1 − i/m) = m/(m − i)The expected number of probes is:

1n

n−1∑i=0

mm−i = m

nn−1∑i=0

1m−i = 1

α

m∑k=m−n+1

1k

≤ 1α

∫ mm−n

1x dx = 1

α ln mm−n

= 1α ln 1

1−α


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Proof.A search for a key k reproduces the same probe sequence as whenthe element with key k was inserted.

If k is the (i+1)st key inserted into the table, the expected number ofprobes in a search for k is at most 1/(1 − i/m) = m/(m − i)The expected number of probes is:

1n

n−1∑i=0

mm−i = m

nn−1∑i=0

1m−i = 1

α

m∑k=m−n+1

1k

≤ 1α

∫ mm−n

1x dx = 1

α ln mm−n

= 1α ln 1

1−α


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Proof.A search for a key k reproduces the same probe sequence as whenthe element with key k was inserted.If k is the (i+1)st key inserted into the table, the expected number ofprobes in a search for k is at most 1/(1 − i/m) = m/(m − i)

The expected number of probes is:1n

n−1∑i=0

mm−i = m

nn−1∑i=0

1m−i = 1

α

m∑k=m−n+1

1k

≤ 1α

∫ mm−n

1x dx = 1

α ln mm−n

= 1α ln 1

1−α


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Proof.A search for a key k reproduces the same probe sequence as whenthe element with key k was inserted.If k is the (i+1)st key inserted into the table, the expected number ofprobes in a search for k is at most 1/(1 − i/m) = m/(m − i)The expected number of probes is:

1n

n−1∑i=0

mm−i = m

nn−1∑i=0

1m−i = 1

α

m∑k=m−n+1

1k

≤ 1α

∫ mm−n

1x dx = 1

α ln mm−n

= 1α ln 1

1−α


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Open Addressing: Successful Search

1α ln 1

1−α

For your reference:α = 0.5: 1.387α = 0.9: 2.559


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Thank You!Questions?

Office [email protected]


2-12 Hashing MA Juncslabcms.nju.edu.cn/.../f/f7/2019-2-12_Hashing.pdf · Hashing: the idea Q: When...

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