// Main Part 3 about Regular Expression Matching//=============================================object M3 {// Regular Expressionsabstract class Rexpcase object ZERO extends Rexpcase object ONE extends Rexpcase class CHAR(c: Char) extends Rexpcase class ALTs(rs: List[Rexp]) extends Rexp // alternatives case class SEQ(r1: Rexp, r2: Rexp) extends Rexp // sequencecase class STAR(r: Rexp) extends Rexp // star//the usual binary choice can be defined in terms of ALTsdef ALT(r1: Rexp, r2: Rexp) = ALTs(List(r1, r2))// some convenience for typing in regular expressionsimport scala.language.implicitConversions import scala.language.reflectiveCalls def charlist2rexp(s: List[Char]): Rexp = s match { case Nil => ONE case c::Nil => CHAR(c) case c::s => SEQ(CHAR(c), charlist2rexp(s))}implicit def string2rexp(s: String): Rexp = charlist2rexp(s.toList)implicit def RexpOps (r: Rexp) = new { def | (s: Rexp) = ALT(r, s) def % = STAR(r) def ~ (s: Rexp) = SEQ(r, s)}implicit def stringOps (s: String) = new { def | (r: Rexp) = ALT(s, r) def | (r: String) = ALT(s, r) def % = STAR(s) def ~ (r: Rexp) = SEQ(s, r) def ~ (r: String) = SEQ(s, r)}// (1) Complete the function nullable according to// the definition given in the coursework; this // function checks whether a regular expression// can match the empty string and Returns a boolean// accordingly.def nullable (r: Rexp) : Boolean = r match { case ZERO => false case ONE => true case CHAR(_) => false case ALTs(rs) => rs.exists(nullable) case SEQ(r1, r2) => nullable(r1) && nullable(r2) case STAR(_) => true}// (2) Complete the function der according to// the definition given in the coursework; this// function calculates the derivative of a // regular expression w.r.t. a character.def der (c: Char, r: Rexp) : Rexp = r match { case ZERO => ZERO case ONE => ZERO case CHAR(d) => if (c == d) ONE else ZERO case ALTs(rs) => ALTs(rs.map(der(c, _))) case SEQ(r1, r2) => if (nullable(r1)) ALT(SEQ(der(c, r1), r2), der(c, r2)) else SEQ(der(c, r1), r2) case STAR(r1) => SEQ(der(c, r1), STAR(r1))}// (3) Implement the flatten function flts. It// deletes 0s from a list of regular expressions// and also 'spills out', or flattens, nested // ALTernativeS.def flts(rs: List[Rexp]) : List[Rexp] = rs match { case Nil => Nil case ZERO::tl => flts(tl) case ALTs(rs1)::rs2 => rs1 ::: flts(rs2) case r::rs => r :: flts(rs) }// (4) Complete the simp function according to// the specification given in the coursework; this// function simplifies a regular expression from// the inside out, like you would simplify arithmetic // expressions; however it does not simplify inside // STAR-regular expressions.def simp(r: Rexp) : Rexp = r match { case ALTs(rs) => (flts(rs.map(simp)).distinct) match { case Nil => ZERO case r::Nil => r case rs => ALTs(rs) } case SEQ(r1, r2) => (simp(r1), simp(r2)) match { case (ZERO, _) => ZERO case (_, ZERO) => ZERO case (ONE, r2s) => r2s case (r1s, ONE) => r1s case (r1s, r2s) => SEQ(r1s, r2s) } case r => r}simp(ALT(ONE | CHAR('a'), CHAR('a') | ONE))// (5) Complete the two functions below; the first // calculates the derivative w.r.t. a string; the second// is the regular expression matcher taking a regular// expression and a string and checks whether the// string matches the regular expression.def ders (s: List[Char], r: Rexp) : Rexp = s match { case Nil => r case c::s => ders(s, simp(der(c, r)))}// main matcher functiondef matcher(r: Rexp, s: String) = nullable(ders(s.toList, r))// (6) Complete the size function for regular// expressions according to the specification // given in the coursework.def size(r: Rexp): Int = r match { case ZERO => 1 case ONE => 1 case CHAR(_) => 1 case ALTs(rs) => 1 + rs.map(size).sum case SEQ(r1, r2) => 1 + size(r1) + size (r2) case STAR(r1) => 1 + size(r1)}// some testing data//matcher(("a" ~ "b") ~ "c", "abc") // => true//matcher(("a" ~ "b") ~ "c", "ab") // => false// the supposedly 'evil' regular expression (a*)* b// val EVIL = SEQ(STAR(STAR(CHAR('a'))), CHAR('b'))//println(matcher(EVIL, "a" * 1000 ++ "b")) // => true//println(matcher(EVIL, "a" * 1000)) // => false// size without simplifications//println(size(der('a', der('a', EVIL)))) // => 28//println(size(der('a', der('a', der('a', EVIL))))) // => 58// size with simplification//println(simp(der('a', der('a', EVIL)))) //println(simp(der('a', der('a', der('a', EVIL)))))//println(size(simp(der('a', der('a', EVIL))))) // => 8//println(size(simp(der('a', der('a', der('a', EVIL)))))) // => 8// Python needs around 30 seconds for matching 28 a's with EVIL. // Java 9 and later increase this to an "astonishing" 40000 a's in// around 30 seconds.//// Lets see how long it takes to match strings with // 5 Million a's...it should be in the range of a // couple of seconds.def time_needed[T](i: Int, code: => T) = { val start = System.nanoTime() for (j <- 1 to i) code val end = System.nanoTime() "%.5f".format((end - start)/(i * 1.0e9))}//for (i <- 0 to 5000000 by 500000) {// println(s"$i ${time_needed(2, matcher(EVIL, "a" * i))} secs.") //}// another "power" test case //simp(Iterator.iterate(ONE:Rexp)(r => SEQ(r, ONE | ONE)).drop(100).next) == ONE// the Iterator produces the rexp//// SEQ(SEQ(SEQ(..., ONE | ONE) , ONE | ONE), ONE | ONE)//// where SEQ is nested 50 times.}