110 lines
2.2 KiB
Python
110 lines
2.2 KiB
Python
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import logging
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import re
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from language.types import Language
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logger = logging.getLogger("elvish-rules")
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permitted_starting_clusters = [
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"ch",
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"cl",
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"cr",
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"cw",
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"cy",
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"dh",
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"dj",
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"dr",
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"dw",
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"dy",
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"fl",
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"fn",
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"fr",
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"fw",
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"fy",
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"gl",
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"ll",
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"ly",
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"ml",
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"mw",
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"my",
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"ny",
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"rh",
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"ry",
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"sh",
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"sl",
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"sm",
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"sn",
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"st",
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"sv",
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"sw",
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"sy",
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"th",
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"tr",
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"tw",
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"ty",
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"vy",
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"wh",
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"wr",
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"wy",
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"yh",
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]
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def cannot_start_with_two_consonants(language: Language, word: str) -> bool:
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found = re.compile(r"(^[bcdfghklmnpqrstvwxz]{2})").search(word)
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if not found:
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return True
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first, second = found.group(1)
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if first == second and first != "l":
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logger.debug(f"{word} starts with a repeated consonant.")
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return False
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if found.group(1) not in permitted_starting_clusters:
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logger.debug("f{word} cannot start with {first}{second}")
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return False
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return True
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def too_many_vowels(language: Language, word: str) -> bool:
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found = re.compile(r"[" + "".join(language.vowels.members) + "]{4}").findall(word)
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if found == []:
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return True
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logger.debug(f"{word} has too many contiguous vowels: {found}")
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return False
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def too_many_consonants(language: Language, word: str) -> bool:
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found = re.compile(r"[bcdfghklmnprstvw]{3}").findall(word)
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if found == []:
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return True
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logger.debug(f"{word} has too many contiguous consonants: {found}")
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return False
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def cannot_have_just_repeated_vowels(language: Language, word: str) -> bool:
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if len(word) == 1:
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return True
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uniq = {letter for letter in word}
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if len(uniq) > 1:
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return True
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logger.debug(f"{word} consists of only one repeated letter.")
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return False
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def must_have_a_vowel(language: Language, word: str) -> bool:
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for vowel in language.vowels.members:
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if vowel in word:
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return True
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logger.debug(f"{word} does not contain a vowel.")
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return False
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rules = {
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must_have_a_vowel,
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too_many_vowels,
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too_many_consonants,
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cannot_have_just_repeated_vowels,
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cannot_start_with_two_consonants,
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}
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