// Main Part 3 about Regular Expression Matching

//=============================================

object M3 {

// Regular Expressions

abstract class Rexp

case object ZERO extends Rexp

case object ONE extends Rexp

case class CHAR(c: Char) extends Rexp

case class ALTs(rs: List[Rexp]) extends Rexp      // alternatives 

case class SEQ(r1: Rexp, r2: Rexp) extends Rexp   // sequence

case class STAR(r: Rexp) extends Rexp             // star

//the usual binary choice can be defined in terms of ALTs

def ALT(r1: Rexp, r2: Rexp) = ALTs(List(r1, r2))

import 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 = ???

// (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 = ???

// (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] = ???

// (4) Complete the simp function according to

// the specification given in the coursework description; 

// 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. Use the _.distinct and 

// flts functions.

def simp(r: Rexp) : Rexp = ???

// (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 = ???

def matcher(r: Rexp, s: String): Boolean = ???

// (6) Complete the size function for regular

// expressions according to the specification 

// given in the coursework.

def size(r: Rexp): Int = ???

// 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'))

matcher(EVIL, "a" * 1000 ++ "b")   // => true

matcher(EVIL, "a" * 1000)          // => false

// size without simplifications

size(der('a', der('a', EVIL)))             // => 28

size(der('a', der('a', der('a', EVIL))))   // => 58

// size with simplification

size(simp(der('a', der('a', EVIL))))           // => 8

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

// 30 seconds.

//

// Lets see how long it really 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(50).next()) == ONE

// the Iterator produces the rexp

//

//      SEQ(SEQ(SEQ(..., ONE | ONE) , ONE | ONE), ONE | ONE)

//

//    where SEQ is nested 50 times.

*/

}