diff -r 59eeb22c9229 -r fced9a61c881 assignment2021scala/main3/re.scala
--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/assignment2021scala/main3/re.scala	Mon Nov 08 23:17:51 2021 +0000
@@ -0,0 +1,151 @@
+// 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
+
+
+// some convenience for typing regular expressions
+
+//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.
+
+*/
+
+}