1

Comparing the Comparing the performance performance

of Ganter’s algorithm and of Ganter’s algorithm and ELL’s one for Galois Lattices ELL’s one for Galois Lattices

Building Building Fatma BAKLOUTIFatma BAKLOUTI

Gérard LEVYGérard LEVY

Richard EMILIONRichard EMILION

Workshop On Symbolic Data Analysis 6/05/2004

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Plan Galois Lattices

Two algorithms :

Ganter

ELL

Experimental performance analysis

Conclusion and perspectives

3

Galois Lattices Using Galois Lattice (mathematical structure) for solving Data

Mining problems.

References :

Birkhoff’s Lattice Theory: 1940, 1973

Barbut & Monjardet : 1970

Wille : 1982

Chein, Norris, Ganter, Bordat, …

Diday, Duquenne, …

Emilion, Lévy, Diday, Lambert

Basic Concepts :

Context, Galois connection, Concept.

4

Context = (O, A, I) : O : finite set of examples A : finite set of attributes I : binary relationbinary relation between O and A, (I O x A)

Example :

A a b c

1 1 1 1

2 1 1

3 1 1

O

Galois Lattices - Definition

5

Galois connection

Oi O and Ai A, we define f et g like this :

f : P(O) P(A) f(Oi) = {a A / (o,a) I, o Oi} intent

g: P(A) P(O) g(Ai) = {o O / (o,a) I, a Ai} extent

f et g are decreasing applications

h =g · f and k = f · g, are :

Increasing O1 O2 h (O1) h (O2)

Extensive O1 h (O1)

Idempotent h (O1) = h · h (O1)

h and k are closure operators.

(f,g) = Galois connection between P(O) and P(A)

Galois Lattices - Definition

6

Concept

Oi O et Ai A,

(Oi, Ai) is a concept iff Oi is the extent of Ai and Ai is the intent of Oi

Oi = g (Ai) and Ai = f(Oi)

L ={(Oi, Ai) P(O)P(A) / Oi= g(Ai) et Ai = f(Oi)} : concepts set.

L: ordered set by the relationship ≤

(O1, A1) ≤ (O2, A2) iff O1 O2 (or A2 A1).

Galois Lattice

T=(L, ≤) an ordered set of concepts.

Galois Lattices - Definition

7

Concept: Example

O1 = {6,7} f(O1)= {a,c} intent

A1 = {a,c} g(A1)= {1,2,3,4,6,7} extent

Remark: h(O1)= g · f(O1)= g(A1) ≠ O1

({6,7} , {a,c}) L

({1,2,3,4,6,7}, {a,c}) L

Because:

h({1,2,3,4,6,7}) = g · f({1,2,3,4,6,7})

= g ({a,c})

= {1,2,3,4,6,7}

Galois Lattices - Definition

a b c d e f g h1 1 1 1 1 1 1 12 1 1 1 1 1 1 13 1 1 1 1 1 1 14 1 1 1 1 15 1 1 1 1 16 1 1 1 17 1 1 1

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1234567, a

123456, ab123467, ac

12345, abd 12346, abc 12356, abe

1236, abce1235, abde1234, abcd

236, abceh123, abcde135, abdeg

1247, acf

124, abcdf

12, abcdef 13, abcdeg 23, abcdeh

3, abcdegh1, abcdefg 2, abcdefh

Ø, abcdefgh

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Ganter AlgorithmLexicographic order

1111

0

x1

x2

x3

x4

0 1

0

0

0

11

1 1 1 1

1 1 1 1 11

0

0 0

01 00 0 0

0000 0001 0010 0011 0111 1000

a (a1, …, ai-1, ai, bi+1, …, bn)

a+ (a1, …, ai-1, 1+ai, 0, …, 0 )

a* (a1, …, ai-1, bi, bi+1,…, bn)

k (a+) = f (g (a+)) = y

If a+ y a* y closest closed of a

a = a*

aa ( 0 1 1 1 )

a+

a+ ( 1 0 0 0 )

a*

a* ( 1 1 1 1 )

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Ganter Algorithma (a1, …, ai-1, ai, bi+1, …, bn)

a+ (a1, …, ai-1, 1+ai ,0, … ,0 )

a* (a1, …, ai-1, bi, bi+1,…, bn)

a+( 1, 2, 0, 0, 0 )

a*( 1, 3, 5, 2, 1 )

a ( 1, 1, 5, 2, 1 ) 0 x5 10 x4 20 x3 5 0 x2 30 x1 3

k (a+) = f (g (a+)) = ?

Example :

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Context : < I, F, d >

T = <F, , , ≤>

Tj= <Fj, j, j, ≤j> for all j de J, J = [1,n]

d: I F

di = (di1,…, dij,…, din) : description of the individual i relatively to the attributes j of J.

di1 dij din

dk1 dkj dkn

1

i

kInd

ivid

uals

I

1 2 j n

x I

f (x) = d(i) i x Intent

z F

g (z) = { i I | z ≤ d(i) } Extent

Generalized Galois Lattices

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ELL Algorithm For two disjoint subsets X0 and K of I, ELL lists all the closed sets

of I obtained by extending X0 with some elements of K. Let X0 and K Ø and i0 K.

1) h (X0 U {i0}) = { i I : f (X0) f (i0) ≤ f (i) }

= X0 U A (X0 h (X0) h (X0 U {i0}) )

where

A = { i I \ X0: f (X0) f (i0) ≤ f (i) }2) If a closed set contains X0 and i0, then it also contains A.

Hence, if A K then (X0 U A) is the smallest closed set containing X0 and i0

and contained within X0 U K.

X = X0 U A, K = K / A3) If a closed set contains X0 and does not contain i0, then it also does not

contain any element of the set.

R = { i K: f (X0) f (i) ≤ f (i0) }, K = K / R

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ELL AlgorithmGL = Ø.

Procedure: Closed (X0, K)

Var i0: element of I, z, z0: elements of F; X, A, R: subsets of I; begin

z0 = f (X0);if K ≠ Ø then

begin

choose an element i 0 of K;

z = z0 f (i0); A = {i I \ X0: z ≤ f (i)};if A K thenbegin

X = X0 U A; insert node (X, z) in ELL;Closed (X, K\A);

end;

R = {i K: z0 f (i) ≤ f (i0)}; Closed (X0, K \ R);end;

end;

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Example X0 = Ø and K = I = {1, 2, 3, 4, 5}

We choose the element i0 = 1 of K; f (X0) = z = 1F = (3, 2, 3) z = z0 f (i0) = (3, 2, 3) (0, 1, 1) = (0, 1, 1) h (X) = { i I : f (X0) f (i0) ≤ f (i) } = {1, 4} A = { i I \ X0: z ≤ f (i)}= {1, 4} A K X = {1, 4} and z = (0, 1, 1), (X, z) Closed item pair X = X0 A = Ø {1, 4}

X0 = {1, 4} and K = {2, 3, 5} We choose the element i0 = 2 of K; f (X0) = z = (0, 1, 1) z = z0 f (i0) = (0, 1, 1) (2, 0, 1) = (0, 0, 1) A = {i I - X0: z ≤ f (i)}= {2, 3} K X = X0 A = {1, 4, 2, 3} X = {1, 4, 2, 3} and z = (0, 0, 1), (X, z) Closed item

pair ….

1 2 3

1 0 1 1

2 2 0 1

3 3 0 3

4 1 2 3

5 2 1 0

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Total number of closed pairs (X, z) of lattice T =GL (C) =14. pair (1)= x = {1,4,} z = {0,1,1,} pair (2)= x = {1,4,3,2,} z = {0,0,1,} pair (3)= x = {1,4,3,2,5,} z = {0,0,0,} pair (4)= x = {1,4,5,} z = {0,1,0,} pair (5)= x = {2,3,} z = {2,0,1,} pair (6)= x = {2,3,5,} z = {2,0,0,} pair (7)= x = {2,3,5,4,} z = {1,0,0,} pair (8)= x = {2,3,4,} z = {1,0,1,} pair (9)= x = {5,} z = {2,1,0,} pair (10)= x = {5,4,} z = {1,1,0,} pair (11)= x = {3,} z = {3,0,3,} pair (12)= x = {3,4,} z = {1,0,3,} pair (13)= x = {4,} z = {1,2,3,} pair (14)= x = {} z = {3,2,3,}

The case X0 = Ø isn’t treated by the algorithm so the test must be added. In this example:

X = Ø f (X) = z = 1F = (3, 2, 3) g (z) = {i I/ z ≤ d (i)} = Ø X = Ø and z = 1F = (3, 2, 3), (X, z): Closed item pair

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Experimental Performance Analysis

M N Time ELL Time Ganter CN

20 40 2640,6 36573,50 367261

20 50 4222,00 88149,90 541192

20 60 5575,00 170220,30 670195

20 70 8384,60 324807,60 881592

20 80 9581,40 499229,50 845229

20 90 11007,90 714281,20 940133

20 100 12525,10 1018060,80 983606

30 10 143,90 223,30 17222

30 20 7953,20 22231,10 1243216

30 30 76332,90 415517,1 8899601

30 40 276812,7 2369779,5 19891269

• Time in ms M: number of individuals

• CN : Closed Number N : number of attributes

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Experimental Performance Analysis

0.00

500000.00

1000000.00

1500000.00

2000000.00

2500000.00

40 50 60 70 80 90 100 10 20 30 40

20 20 20 20 20 20 20 30 30 30 30

Data context (M, N)

Tim

e (m

s) ELL

Ganter

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Other Data types …x y

1 [1,1] [3,4]

2 [1,1] [0,2]

3 [1,3] [0,4]

4 [0,1] [0,2]

5 [0,0] [3,3]

b1 [0,1] [0,4]

Intent:

f (x) = d(i) i x

x = {1,3}

x1 x3 = [1,1] [1,3] = [1,1]

y1 y3 = [3,4] [0,4] = [3,4]

Extent :

g (z) = {i I | z d(i)}

z = [1,1] [3,4]

g (z) = {1,3}

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x z1 z2

1 [0,1] [0,4]

2 {1,3} [1,1] [3,4]

3 {1,2,3,4} [1,1] 4 {1,2,3,4,5} 5 {1,3,5} [3,3]

6 {2,3,4} [1,1] [0,2]

7 {3} [1,1] [0,4]

8 {4} [0,1] [0,2]

9 {4,5} [0,0] 10 {5} [0,0] [3,3]

Closed pairs :

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; ([0,1] [0,4])

{3}; ([1,1] [0,4]) {4} ; ([0,1] [0,2]) {5} ; ([0,0] [3,3])

{1,3} ; ([1,1] [3,4]) {4,5} ; ([0,0] )

{1,3,5} ; ( [3,3]) {2,3,4} ; ([1,1] [0,2])

{1,2,3,4} ; ([1,1] )

{1,2,3,4,5} ; ( )

1

7 108

2 9

5 6

3

4

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Problems Large data volume:Large data volume:

Partition data on different server nodes

Process in parallel locally

Group results on one (client) node

Post-process

Our tool:

SDDS (Scalable Distributed Data Structures )

Conclusion and perspectives

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Solutions Column-sharing Row-sharing

«Parallel Algorithms for General Galois lattices building»

Fatma BAKLOUTI and Gérard LEVY

4th Int. Workshop on Distributed Structures (WDAS 2003)

Conclusion and perspectives

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Conclusion and perspectives

Generalized Galois Lattices.

Problem of large data base can be perhaps

resolved in our way.

Sharing context into two subsets.

Possibility of building different architectures

for station’s networks.

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Thank you for Your Attention