Propositional Logic Syntax Acquisition Using Induction and Self-Organization - PowerPoint PPT Presentation

1 / 27
About This Presentation
Title:

Propositional Logic Syntax Acquisition Using Induction and Self-Organization

Description:

competing meanings: meaning2, ..., meaningN. The score of a sentence (or ... decreases scores of rules competing meanings. hearer's meaning speaker's meaning ... – PowerPoint PPT presentation

Number of Views:23
Avg rating:3.0/5.0
Slides: 28
Provided by: carbonVide1
Category:

less

Transcript and Presenter's Notes

Title: Propositional Logic Syntax Acquisition Using Induction and Self-Organization


1
Propositional Logic Syntax Acquisition Using
Induction and Self-Organization
  • Josefina Sierra Santibáñez
  • Departamento de Lenguajes y Sistemas Informáticos
  • Universidad Politécnica de Cataluña
  • Spain

2
  • Language Game Guessing (Steels 1999)
  • Speaker chooses a formula from a propositional
    language, generates a sentence for expressing the
    formula and communicates the sentence to hearer.
  • Hearer tries to interpret the sentence generated
    by the speaker. If it can parse it using its
    lexicon and grammar, it extracts a meaning.
  • The speaker communicates the formula it had in
    mind to the hearer. They adjust their grammars to
    become sucessful in future language games.
  • Success speakers formula ? hearers
    meaning

3
  • Goals of the Experiments
  • Observe the evolution of
  • The communicative success average of sucess-ful
    language games in the last ten language ga-mes
    played by the agents.
  • The internal grammars constructed by the
    individual agents.
  • The external language used by the population.

4
  • Definite clause grammarsemantic, score, use
  • s(right, 0.25, 20) ? right
  • s(light, 0.70, 50) ? light
  • s(P,Q, S, 12) ? c1(P,S1,C1), s(Q,S2,C2), S is
    S1?S2?0.01
  • c1(not, 0.80, 55) ? not
  • s(P,Q,R,S,3) ? c2(P,S1,C1),s(Q,S2,C2),s(R,S3,C3)
    , S is S1?S2?S3?0.01
  • c2(and, 0.50, 35) ? and
  • Formula Meaning
    Sentence
  • right ? light and, right, light
    andrightlight
  • light not, light
    notlight

5
  • Invention
  • Generates a sentence E for a meaning M
  • If M is atomic, it invents a new word E.
  • If M is a list, it tries to construct an
    expression for each of the elements in M using
    the agents grammar.
  • If it cannot construct an expression for an
    element using its grammar, it invents a new
    expression.
  • It concatenates the expressions associated with
    the elements of M randomly in order to construct
    a sen-tence E for the whole meaning M.
  • Adds a new rule to the grammar s(M, 0.01, 0) ?
    E

6
  • Adoption
  • Communication fails because
  • The hearer cannot parse the speakers sentence
  • The speaker communicates the formula it had in
    mind to the hearer.
  • The hearer adopts an association between that
    formula and the sentence used by the speaker.
  • s(M, 0.01, 0) ? E
  • The hearer can parse the sentence, but its
    interpre-tation is not consistent with the
    speakers meaning.
  • Hearer decrase the scores of used associations.
  • It may adopt an association between the formula
    and the sentence used by the speaker.

7
  • Induction
  • The agents use some induction mechanisms to
    extract
  • generalisations from the grammar rules learnt
    so far.
  • The induction rules used in the experiments are
    based
  • on the following rules proposed in (Kirby
    2002)
  • Simplification
  • Chunk
  • They are applied whenever the agents invent or
    adopt
  • a new association.

8
  • Simplification scores (Vogt 05) and use counters
  • r1 ? n(m1, S2, C2) ? e1
  • r2 ? left(m1, S1, C1) ? e1
  • Rule r2 is replaced with rule r3, of the form
  • left(X, S3, 0) ? n(X,S,C), S3 is S?0.01
  • or
  • letf(X.., S3, 0) ? n(X,S,C), S3 is S?T?0.01
  • depending on whether S1 was a constant or a
    variable.
  • In the second case, S1 is T?sr1 is the
    arithmetic ex-pression on the right hand side of
    r1.

9
  • Simplification example 1
  • r1 ? s(right, 0.25, 16) ? right
  • r2 ? s(and,light,right, 0.10, 6) ?
    andlightright
  • Rule r2 is replaced with rule r3
  • r3 ? S(and,light,R, S, 0) ? andlight, s(R, S3,
    C3), S is S3?0.01
  • r4 ? s(light, 0.70, 25) ? light
  • Rule r3 is replaced with rule r5
  • r5 ? s(and,Q,R, S, 0) ? and, s(Q, S2, C2), s(R,
    S3, C3), S is S2?S3?0.01

10
  • Simplification example 2
  • r1 ? s(right, 0.25, 16) ? right
  • r6 ? s(or,light,right, 0.10, 6) ?
    orlightright
  • Rule r6 is replaced with rule r7
  • r7 ? S(or,light,R, S, 0) ? orlight, s(R, S3,
    C3), S is S3?0.01
  • r4 ? s(light, 0.70, 25) ? light
  • Rule r7 is replaced with rule r8
  • r8 ? s(or,Q,R, S, 0) ? or, s(Q, S2, C2), s(R,
    S3, C3), S is S2?S3?0.01

11
  • Simplification example 3
  • r1 ? s(right, 0.25, 16) ? right
  • r9 ? s(or,light,right, 0.10, 6) ?
    lightorright
  • Rule r9 is replaced with rule r10
  • r10 ? S(or,light,R, S, 0) ? lightor, s(R, S3,
    C3), S is S3?0.01
  • r4 ? s(light, 0.70, 25) ? light
  • Rule r10 is replaced with rule r11
  • r11 ? s(or,Q,R, S, 0) ? s(Q, S2, C2), or, s(R,
    S3, C3), S is S2?S3?0.01

12
  • Chunk I scores (Vogt 05) and use counters
  • r1 ? left(f(m1), S1, C1) ? right(e1)?,
  • r2 ? left(f(m2), S2, C2) ? right(e2)?,
  • A new category symbol n is created and rules
    added
  • n(m1, 0.01, 0) ? e1 n(m2, 0.01,
    0) ? e2
  • Rules r1 and r2 are replaced with rule r3, of the
    form
  • left(f(X), S3, 0) ? right?(n(X, S, C)), S3 is
    S?0.01
  • or
  • left(f(X), S3, 0) ? right?(n(X,S,C)), S3 is
    S?T?0.01

13
  • Chunk I example 1
  • r1 ? s(and,Q,R, S, 18) ? and, s(Q, S2, C2),
    s(R, S3, C3), S is S2?S3?0.10
  • r2 ? s(or,Q,R, S, 23) ? or, s(Q, S2, C2), s(R,
    S3, C3), S is S2?S3?0.30
  • The following new rules are added to grammar
  • c2(and, 0.01, 0) ? and c2(or, 0.01,
    0) ? or
  • Rules r1 and r2 are replaced with rule r3
  • r3 ? s(P,Q,R, S, 0) ? c2(P, S1, C1), s(Q, S2,
    C2), s(R, S3, C3), S is S1?S2?S3?0.01

14
  • Chunk I example 2
  • r1 ? s(and,Q,R, S, 18) ? and, s(Q, S2, C2),
    s(R, S3, C3), S is S2?S3?0.10
  • r2 ? s(or,Q,R, S, 23) ? s(Q, S2, C2), or, s(R,
    S3, C3), S is S2?S3?0.30
  • Chunk cannot be applied to r1 and r2, because
    they
  • place the expressions associated with the
    connectives
  • and and or in different positions in the
    sentence.

15
  • Chunk I example 3
  • r1 ? s(and,Q,R, S, 18) ? s(R, S3, C3), and,
    s(Q, S2, C2), S is S2?S3?0.10
  • r2 ? s(or,Q,R, S, 23) ? s(Q, S2, C2), or, s(R,
    S3, C3), S is S2?S3?0.30
  • Chunk cannot be applied to r1 and r2, because
    they
  • place the expressions associated with the
    arguments
  • of the binay connective, Q and R, in different
    posi-
  • tions in the sentence.
  • Rules must agree on the positions of the expres-
  • sions associated with the connectives and their
  • arguments in the sentence.

16
  • Chunk II scores (Vogt 05) and use counters
  • r1 ? left(f(X), S1, C1) ? right?(n(X, S, C)),
  • r2 ? left(f(m1), S2, C2) ? right?(e1),
  • Rule r2 is replaced with the following rule
  • n(m1, 0.01, 0) ? e1
  • right?(X) is the result of removing the
    arithmetic ex-
  • pression S3 is S? of the right hand side of
    r1.

17
  • Chunk II example 1
  • r1 ? s(P,Q,R, S, 25) ? c2(P, S1, C1), s(Q, S2,
    C2), s(R, S3, C3), S is S1?S2?S3?0.20
  • r2 ? s(iff,Q,R, S, 33) ? iff, s(Q, S2, C2),
    s(R, S3, C3), S is S2?S3?0.50
  • Rule r2 is replaced with the following rule
  • c2(iff, 0.01, 0) ? iff

18
  • Chunk II example 2
  • r1 ? s(P,Q,R, S, 25) ? c2(P, S1, C1), s(Q, S2,
    C2), s(R, S3, C3), S is S1?S2?S3?0.20
  • r2 ? s(iff,Q,R, S, 33) ? s(Q, S2, C2), s(R, S3,
    C3), iff, S is S2?S3?0.50
  • Chunk cannot be applied, because rule r1 places
    the
  • expresion associated with the connective in first
    po-
  • sition in the sentence and rule r2 places the
    expres-
  • sion associated with the connective third
    position.

19
  • Need for Coordination The agents must reach
  • agreements on how to
  • name propositional constants and connectives
  • a1 if ? if
    a2 if ? si
  • order the expressions associated with the
    different components of non-atomic meanings
    consistently
  • a1 not ? un, 2pos not, right ?
    rightnot
  • a2 not ? un, 1pos not, right ?
    notright
  • a1 if ? bin, 2pos, inv if,right,light ?
    lightifright
  • a2 if ? bin, 2pos, noinv if,right,light ?
    rightiflight

20
  • Self-organization Coordinate agents grammars
  • The agents construct a shared external language
    and
  • prefer using the rules in that language over the
    rest
  • in the rules in their individual grammars.
  • The scores of the rules indicate the agents
    preferences
  • meaning ? sentence1 highest score
  • competing sentences sentence2, ,
    sentenceN
  • sentence ? meaning1 highest score
  • competing meanings meaning2, ,
    meaningN
  • The score of a sentence (or meaning) is computed
    at
  • generation (parsing) multiplying the scores of
    the rules
  • involved (Vogt 2005).

21
  • Score of a sentence (meaning) example
  • s(right, 0.25, 20) ? right c1(if,
    0.50, 10) ? if
  • s(light, 0.70, 50) ? light c2(if,
    0.10, 15) ? si
  • s(P,Q,R,S,5) ? c1(P,S1,C1),s(Q,S2,C2),s(R,S3,C3)
    , S is S1?S2?S3?0.10
  • s(P,Q,R,S,3) ? c2(P,S1,C1), s(R,S3,C3),s(Q,S2,C2
    ), S is S1?S2?S3?0.01
  • Meaning if, right, light ? Generation
  • Sentence ifrightlight score
    0.50?0.25?0.70?0.10
  • Comp senten silightright score
    0.10?0.25?0.70?0.01

22
  • Coordination takes place at the third stage of a
    lan-guage game when the speaker communicates the
    meaning it had in mind to the hearer.
  • hearers meaning ? speakers meaning
  • Speaker increases scores of rules ? sentence
  • decreases scores of rules ? competing
    sentences
  • Hearer increases scores of rules ? meaning
  • decreases scores of rules ? competing
    meanings
  • hearers meaning ? speakers meaning
  • Speaker and hearer decrease scores of rules they
    used for generating and interpreting the sentence.

23
  • Reinforcement and Inhibition
  • The rules used successfully are reinforced.
  • The rules used for generating competing sentences
    or competing meanings are inhibited.
  • The rules used for updating scores of grammar
    rules (Steels 1999) replace the original score S
    with
  • S1 ? maximum(1, S µ) if the score is
    increased
  • S2 ? minimum(1, S ? µ) if the score is
    decreased
  • Purging the rules that have been used more than
    30 times and have scores ? 0.01 are removed from
    the agents grammars.

24
Experiments Evolution Communicative Success
guessing game 5 agents, 15000 games about
propositional formulas La, b, c, r, l, u
constructed using ?, ?, ?, ?, ?
25
Guessing negation ?
a1 s(not,Y, R) ? 2, f, s(Y,Q), R is Q?1
a2 s(X,Y, R) ? 2, c1(X,P), s(Y,Q), R is P?Q?1 c1(not, 1) ? f
a3 s(not,Y, R) ? 2, f, s(Y,Q), R is Q?1
a4 s(not,Y, R) ? 2, f, s(Y,Q), R is Q?1
a5 s(X,Y, R) ? 2, c1(X,P), s(Y,Q), R is P?Q?1 c1(not, 1) ? f
26
Guessing conjuction ? (commut)
a1 s(and,Y,Z, T) ? 3, dyp, s(Z,Q), s(Y,R), T is Q?R?1
a2 s(X,Y,Z,T) ? 3, c2(X,P),s(Z,Q),s(Y,R), T is P?Q?R c2(and, 1) ? dyp
a3 s(X,Y,Z,T) ? 3, c1(X,P),s(Y,Q),s(Z,R), T is P?Q?R c1(and, 1) ? dyp
a4 s(X,Y,Z,T) ? 3, c4(X,P),s(Y,Q),s(Z,R), T is P?Q?R c4(and, 1) ? dyp
a5 s(X,Y,Z,T) ? 3, c4(X,P),s(Z,Q),s(Y,R), T is P?Q?R c4(and, 1) ? dyp
27
Guessing implication ? (non-com)
a1 s(X,Y,Z,T) ? 1, c1(X,P),s(Y,Q),s(Z,R), T is P?Q?R c1(if, 1) ? bqi
a2 s(X,Y,Z,T) ? 1, c3(X,P),s(Y,Q),s(Z,R), T is P?Q?R c3(if, 1) ? bqi
a3 s(X,Y,Z,T) ? 1, c2(X,P),s(Y,Q),s(Z,R), T is P?Q?R c2(if, 1) ? bqi
a4 s(if,Y,Z, T) ? 1, bqi, s(Y,Q), s(Z,R), T is Q?R?1
a5 s(if,Y,Z, T) ? 1, bqi, s(Y,Q), s(Z,R), T is Q?R?1
Write a Comment
User Comments (0)
About PowerShow.com