CSE 4125: Distributed Database Systems Chapter - 3...Jubair | AUST 13 Other Applications of...

CSE 4125: Distributed Database Systems

Chapter – 6

Optimization of Access Strategies.

(part – C)

1 Jubair | AUST

Outline

• Semi-join programs.

• Full reducer.

Jubair | AUST 2

Semi-join Programs

R JNC=A S ↔ (R SJC=A PJA S) JNC=A S

Jubair | AUST 3

Semi-join Programs (contd.)

R JNC=A S ↔ (R SJC=A PJA S) JNC=A S

Jubair | AUST 4

S S’ = PJAS

S’ R

R’ = R SJ C=A S’

R’

Q = R’ JN C=A S

site 1 site 2

S’

S’

R’ R’


Cost of Semi-join program

• Cost of sending S’ to site – 2: TC1 = ?

Jubair | AUST 5



• Cost of sending S’ to site – 2: TC1 = C0+C1*size(A)*val(A[S])

Jubair | AUST 6




• Cost of computing R’ at site – 2: TC2 = ?

Jubair | AUST 7




• Cost of computing R’ at site – 2: TC2 = 0

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• Cost of sending R’ to site – 1: TC3 = ?

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• Cost of sending R’ to site – 1: TC3 = C0+C1*size(R)*card(R’)

Jubair | AUST 10






• Cost of computing Q at site – 1: TC4 = ?

Jubair | AUST 11






• Cost of computing Q at site – 1: TC4 = 0

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• Total cost TCSJ = TC1 + TC2 + TC3 + TC4

• If TCSJ < TCJN then semi-join program is profitable.

– Here TCJN is the cost of performing join without semi-join program.

Jubair | AUST 13

Other Applications of Semi-join Programs

• Semi-join programs can be used as fragment reducers (operations that can reduce cardinality of a relation).

– Similarly to unary operations.

• Full reducer:

– Chain of semi-joins.

Jubair | AUST 14

Full Reducer

Jubair | AUST 15

R

A B

1 a

2 b

3 c

S

B C

a x

b y

c z

T

C A

x 2

y 3

z 4

R

S

T A=A

B=B C=C

Full Reducer (contd.)

Jubair | AUST 16

R

A B

1 a

2 b

3 c

S

B C

a x

b y

c z

T

C A

x 2

y 3

z 4

R

S

T A=A

B=B C=C

R’ = R SJ A=A T

A B

2 b

3 c


Jubair | AUST 17

R

A B

1 a

2 b

3 c

S

B C

a x

b y

c z

T

C A

x 2

y 3

z 4

R’

S

T A=A

B=B C=C

R’ = R SJ A=A T

A B

2 b

3 c

S’ = S SJ B=B R’

B C

b y

c z


Jubair | AUST 18

R

A B

1 a

2 b

3 c

S

B C

a x

b y

c z

T

C A

x 2

y 3

z 4

R’

S’

T A=A

B=B C=C

R’ = R SJ A=A T

A B

2 b

3 c


B C

b y

c z

T’ = T SJ C=C S’

C A

y 3

z 4


Jubair | AUST 19

R

A B

1 a

2 b

3 c

S

B C

a x

b y

c z

T

C A

x 2

y 3

z 4

R’

S’

T’ A=A

B=B C=C

R’ = R SJ A=A T

A B

2 b

3 c


B C

b y

c z


C A

y 3

z 4

R” = R’ SJ A=A T’

A B

3 c


Jubair | AUST 20

R

A B

1 a

2 b

3 c

S

B C

a x

b y

c z

T

C A

x 2

y 3

z 4

R”

S’

T’ A=A

B=B C=C

R’ = R SJ A=A T

A B

2 b

3 c


B C

b y

c z


C A

y 3

z 4


A B

3 c

S” = S’ SJ B=B R”

B C

c z


Jubair | AUST 21

R

A B

1 a

2 b

3 c

S

B C

a x

b y

c z

T

C A

x 2

y 3

z 4

R”

S”

T’ A=A

B=B C=C

R’ = R SJ A=A T

A B

2 b

3 c


B C

b y

c z


C A

y 3

z 4


A B

3 c


B C

c z

T” = T’ SJ C=C S”

C A

z 4


Jubair | AUST 22

R

A B

1 a

2 b

3 c

S

B C

a x

b y

c z

T

C A

x 2

y 3

z 4

R”

S”

T” A=A

B=B C=C

R’ = R SJ A=A T

A B

2 b

3 c


B C

b y

c z


C A

y 3

z 4


A B

3 c


B C

c z


C A

z 4

R”’ = R” SJ A=A T”

Φ


Jubair | AUST 23

R

A B

1 a

2 b

3 c

S

B C

a x

b y

c z

T

C A

x 2

y 3

z 4

R”’

S”

T” A=A

B=B C=C

R’ = R SJ A=A T

A B

2 b

3 c


B C

b y

c z


C A

y 3

z 4


A B

3 c


B C

c z


C A

z 4

R”’ = R” SJ A=A T”

Φ

Additional Reading

• Length of full reducer.

• Tree queries.

• “Best” reducer.

Jubair | AUST 24

Practice Problems/ Questions

1. With an example, prove that the semi-join is not symmetric. [hint: page. 142]

2. Textbook exercise: 6.1, 6.4

Jubair | AUST 25

CSE 4125: Distributed Database Systems Chapter - 3...Jubair | AUST 13 Other Applications of...

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