Title: Propositional Logic Syntax Acquisition Using Induction and Self-Organization
1Propositional 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.
24Experiments Evolution Communicative Success
guessing game 5 agents, 15000 games about
propositional formulas La, b, c, r, l, u
constructed using ?, ?, ?, ?, ?
25Guessing 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
26Guessing 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
27Guessing 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