Talk:Constraint-based lexical selection module

Rule formats

Note: Felipe doesn't like "skip".

can't say I do either … sounds like a command rather than a constraint --unhammer 11:42, 18 November 2011 (UTC)

also, in the <or>, should we read them as independent of each other? that's a bit confusing since otherwise they're all required and have a certain order --unhammer 11:42, 18 November 2011 (UTC)

The regular expression for the OR below is:

(nazi<adj>[0-9A-Za-z <>]*|totalitari<adj>[0-9A-Za-z <>]*|feixista<adj>[0-9A-Za-z <>]*|franquista<adj>[0-9A-Za-z <>]*|militar<adj>[0-9A-Za-z <>]*|fiscal<adj>[0-9A-Za-z <>]*)

- Francis Tyers 11:53, 18 November 2011 (UTC)

1

  <rule>
    <remove lemma="règim" tags="n.*">
      <acception lemma="diet" tags="n.*"/>
    </remove>
    <or>
      <skip lemma="nazi" tags="adj.*"/>
      <skip lemma="totalitari" tags="adj.*"/>
      <skip lemma="feixista" tags="adj.*"/>
      <skip lemma="franquista" tags="adj.*"/>
      <skip lemma="militar" tags="adj.*"/>
      <skip lemma="fiscal" tags="adj.*"/>
    </or>
  </rule>

2

  <rule>
    <remove lemma="règim" tags="n.*">
      <acception lemma="diet" tags="n.*"/>
    </remove>
    <or>
      <pattern lemma="nazi" tags="adj.*"/>
      <pattern lemma="totalitari" tags="adj.*"/>
      <pattern lemma="feixista" tags="adj.*"/>
      <pattern lemma="franquista" tags="adj.*"/>
      <pattern lemma="militar" tags="adj.*"/>
      <pattern lemma="fiscal" tags="adj.*"/>
    </or>
  </rule>

3

  <rule>
    <remove-from lemma="règim" tags="n.*">
      <translation lemma="diet" tags="n.*"/>
    </remove-from>
    <pattern>
      <or>
        <pattern-item lemma="nazi" tags="adj.*"/>
        <pattern-item lemma="totalitari" tags="adj.*"/>
        <pattern-item lemma="feixista" tags="adj.*"/>
        <pattern-item lemma="franquista" tags="adj.*"/>
        <pattern-item lemma="militar" tags="adj.*"/>
        <pattern-item lemma="fiscal" tags="adj.*"/>
      </or>
    </pattern>
  </rule>

  <rule c="la dona dels seus somnis">
    <select-for lemma="dona" tags="n.*">
      <translation lemma="wife" tags="n.*"/>
    </select>
    <pattern>
      <pattern-item lemma="de" tags="pr.*"/>
      <pattern-item lemma="*" tags="det.pos.*"/>
      <pattern-item lemma="somni" tags="n.*"/>
    </pattern>
  </rule>

4

  <rule>
    <target lemma="règim" tags="n.*">
      <remove lemma="diet" tags="n.*"/>
    </target>
…
  </rule>

  <rule c="la dona dels seus somnis">
    <target lemma="dona" tags="n.*">
      <select lemma="wife" tags="n.*"/>
    </target>
…
  </rule>

Old application strategy

The following is an inefficient implementation of the rule application process:

# s	("prova" n)	("event" n)	(-3 "guanyador") (-2 "de") 
#
# tipus = "select";
# centre = "^prova<n>.*"
# tl_patro = ["^event<n>.*"]
# sl_patro = {-3: "^guanyador<", -2: "^de<"}

CLASS Rule: 
        tipus = enum('select', 'remove')
        centre = '';
        tl_patro = [];
        sl_patro = {};


rule_table = {}; # e.g. rule_table["estació"] = [rule1, rule2, rule3];
i = 0

DEFINE ApplyRule(rule, lu): 
    

    FOREACH target IN lu.tl: 
        SWITCH rule.tipus:
            'select': 
                 IF target NOT IN rule.tl_patro:
                     DELETE target
            'remove': 
                 IF target IN rule.tl_patro:
                     DELETE target



FOREACH pair(sl, tl) IN sentence:
   
    FOREACH centre IN rule_table: 

        IF centre IN sl: 

            FOREACH rule IN rule_table[centre]: 

                matched = False   

                FOREACH context_item IN rule_table[centre][rule]: 

                  IF context_item in sentence: 
                      matched = True
                  ELSE:
                      matched = False
                
                # If all of the context items have matched, and none of them have not matched
                # if a rule matches break and continue to the pair. 
                IF matched == True:
 
                      sentence[i] = ApplyRule(rule_table[centre][rule], sentence[i])
                      break 

    i = i + 1