3 ms·

Not sure if you are trolling, but here we go... In the case at hand, correctness of the validator expression V clearly means "V determines well-formedness of a

by ac42 7y ago

Not sure if you are trolling, but here we go... In the case at hand, correctness of the validator expression V clearly means "V determines well-formedness of any regular expressions" which is clearly not implied by "V is well-formed" (a much weaker statement because ".*" is well-formed but matches everything). Therefore, when applying V to itself, we only learn if a weak requirement for V's correctness holds. Similar, perhaps, to validating whether a given number is a Gödel number of a well-formed logical statement rather than assessing the verity of the logical statement it encodes. Also, I am not saying whether it is or is not possible to build such a regular expression. Rather, the question just doesn't tick the boxes of either, the Liar's Paradox, nor Gödel's Incompleteness result -- contrary to what was suggested. So you could still be right but for different reasons.

ukj 7y ago
You are axiomatically assuming that the proposition "V determines well-formedness of any regular expressions" to be true. I am asking you to prove that. Constructively.
- ac42 7y ago
  :-)
  - ukj 7y ago
    So we agree then that when the system says "I am free of errors", some (most?) of the time it's a lie ? ;) All models are wrong - some are useful.
- malisper 7y ago
  It's not possible to do this for regular expressions, but it is possible to do it for context free grammars. You can write a context-free grammar in Backus-Naur form that recognizes all context-free grammars in Backus-Naur form: <syntax> ::= <rule> | <rule> <syntax> <rule> ::= <opt-whitespace> "<" <rule-name> ">" <opt-whitespace> "::=" <opt-whitespace> <expression> <line-end> <opt-whitespace> ::= " " <opt-whitespace> | "" <expression> ::= <list> | <list> <opt-whitespace> "|" <opt-whitespace> <expression> <line-end> ::= <opt-whitespace> <EOL> | <line-end> <line-end> <list> ::= <term> | <term> <opt-whitespace> <list> <term> ::= <literal> | "<" <rule-name> ">" <literal> ::= '"' <text1> '"' | "'" <text2> "'" <text1> ::= "" | <character1> <text1> <text2> ::= '' | <character2> <text2> <character> ::= <letter> | <digit> | <symbol> <letter> ::= "A" | "B" | "C" | "D" | "E" | "F" | "G" | "H" | "I" | "J" | "K" | "L" | "M" | "N" | "O" | "P" | "Q" | "R" | "S" | "T" | "U" | "V" | "W" | "X" | "Y" | "Z" | "a" | "b" | "c" | "d" | "e" | "f" | "g" | "h" | "i" | "j" | "k" | "l" | "m" | "n" | "o" | "p" | "q" | "r" | "s" | "t" | "u" | "v" | "w" | "x" | "y" | "z" <digit> ::= "0" | "1" | "2" | "3" | "4" | "5" | "6" | "7" | "8" | "9" <symbol> ::= "|" | " " | "!" | "#" | "$" | "%" | "&" | "(" | ")" | "*" | "+" | "," | "-" | "." | "/" | ":" | ";" | ">" | "=" | "<" | "?" | "@" | "[" | "\" | "]" | "^" | "_" | "`" | "{" | "}" | "~" <character1> ::= <character> | "'" <character2> ::= <character> | '"' <rule-name> ::= <letter> | <rule-name> <rule-char> <rule-char> ::= <letter> | <digit> | "-" I got this from Wikipedia: https://en.wikipedia.org/wiki/Backus%E2%80%93Naur_form#Further_examples https://en.wikipedia.org/wiki/Backus%E2%80%93Naur_form#Furth...
  - ukj 7y ago
    Sure. Total functional programming (provably terminating code) comes at the cost of Turing-completeness. The choice exists. https://en.wikipedia.org/wiki/Total_functional_programming https://en.wikipedia.org/wiki/Total_functional_programming Observe though that BNF is just a notation. Parsers for BNF are not implemented in BNF. Bison. Yacc.

by ac42 7y ago

ukj 7y ago

You are axiomatically assuming that the proposition "V determines well-formedness of any regular expressions" to be true. I am asking you to prove that. Constructively.

ac42 7y ago
:-)
- ukj 7y ago
  So we agree then that when the system says "I am free of errors", some (most?) of the time it's a lie ? ;) All models are wrong - some are useful.
malisper 7y ago
It's not possible to do this for regular expressions, but it is possible to do it for context free grammars. You can write a context-free grammar in Backus-Naur form that recognizes all context-free grammars in Backus-Naur form: <syntax> ::= <rule> | <rule> <syntax> <rule> ::= <opt-whitespace> "<" <rule-name> ">" <opt-whitespace> "::=" <opt-whitespace> <expression> <line-end> <opt-whitespace> ::= " " <opt-whitespace> | "" <expression> ::= <list> | <list> <opt-whitespace> "|" <opt-whitespace> <expression> <line-end> ::= <opt-whitespace> <EOL> | <line-end> <line-end> <list> ::= <term> | <term> <opt-whitespace> <list> <term> ::= <literal> | "<" <rule-name> ">" <literal> ::= '"' <text1> '"' | "'" <text2> "'" <text1> ::= "" | <character1> <text1> <text2> ::= '' | <character2> <text2> <character> ::= <letter> | <digit> | <symbol> <letter> ::= "A" | "B" | "C" | "D" | "E" | "F" | "G" | "H" | "I" | "J" | "K" | "L" | "M" | "N" | "O" | "P" | "Q" | "R" | "S" | "T" | "U" | "V" | "W" | "X" | "Y" | "Z" | "a" | "b" | "c" | "d" | "e" | "f" | "g" | "h" | "i" | "j" | "k" | "l" | "m" | "n" | "o" | "p" | "q" | "r" | "s" | "t" | "u" | "v" | "w" | "x" | "y" | "z" <digit> ::= "0" | "1" | "2" | "3" | "4" | "5" | "6" | "7" | "8" | "9" <symbol> ::= "|" | " " | "!" | "#" | "$" | "%" | "&" | "(" | ")" | "*" | "+" | "," | "-" | "." | "/" | ":" | ";" | ">" | "=" | "<" | "?" | "@" | "[" | "\" | "]" | "^" | "_" | "`" | "{" | "}" | "~" <character1> ::= <character> | "'" <character2> ::= <character> | '"' <rule-name> ::= <letter> | <rule-name> <rule-char> <rule-char> ::= <letter> | <digit> | "-" I got this from Wikipedia: https://en.wikipedia.org/wiki/Backus%E2%80%93Naur_form#Further_examples https://en.wikipedia.org/wiki/Backus%E2%80%93Naur_form#Furth...
- ukj 7y ago
  Sure. Total functional programming (provably terminating code) comes at the cost of Turing-completeness. The choice exists. https://en.wikipedia.org/wiki/Total_functional_programming https://en.wikipedia.org/wiki/Total_functional_programming Observe though that BNF is just a notation. Parsers for BNF are not implemented in BNF. Bison. Yacc.