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CS 3240 – Chapter 3 How would you delete all C++ files from a directory from the command line? How about all PowerPoint files that start with the letter a? PowerPoint file names that contain the string 3240? CS 3240 - Regular Languages and Grammars 2 *.cpp a*.ppt *3240*.ppt These are wildcard expressions Not bona fide regular expressions CS 3240 - Regular Languages and Grammars 3 Language Machine Grammar Regular Finite Automaton Regular Expression, Regular Grammar Context-Free Pushdown Automaton Context-Free Grammar Recursively Enumerable Turing Machine Unrestricted PhraseStructure Grammar CS 3240 - Introduction 4 Text patterns that represent regular languages We’ll show shortly that for every regular expression there is a finite automaton that accepts that language And vice-versa The operators are: () * + xy (Grouping) (Kleene Star) (Union) (Concatenation) CS 3240 - Regular Languages and Grammars 5 1) Specify base case(s) 2) Show how to generate other elements Rules that use what’s in the set already Example: Non-negative multiples of 5, F 1) 0, 5 is in F 2) For x, y in F, then x + y is in F Alternate definition: 1) 0 is in F 2) For x in F, so is x + 5 CS 3240 - Regular Languages and Grammars 6 Base cases: The empty set: ∅ or ( ) The empty string: λ Any letter in Σ Recursive rules: Given regular expressions r, r1, r2: (r) r* r1 + r2 r1r2 (Grouping) (Kleene Star) (Union) (Concatenation) CS 3240 - Regular Languages and Grammars 7 All strings beginning with a: a(a + b)* All strings containing aba: (a + b)*aba(a + b)* All strings of even length: ((a + b)(a + b))* = (aa + ba + ab + bb)* = ((a + b)2)* All strings of odd length: (a+b)((a + b)2)* Valid decimal integers in C: (1+2+3+4+5+6+7+8+9)(0+1+2+3+4+5+6+7+8+9)* CS 3240 - Regular Languages and Grammars 8 Put anything you want on an edge Use an “else” branch as well [0-9] (if-branch) ~[0-9] or [^(0-9)] or else (Decimal integers) CS 3240 - Regular Languages and Grammars 9 (b*ab*ab*ab* + b) * = b* (ab*ab*ab*) * = b* + (b*ab*ab*ab*) * (a(a+bb) *) * ((a + b)a) * CS 3240 - Regular Languages and Grammars 10 L(∅) =∅ L(λ) = λ L(c) = c, for c ∊ Σ L((r)) = L(r) L(r*) = L(r)* L(r1 + r2) = L(r1) ∪ L(r2) L(r1r2) = L(r1)L(r2) CS 3240 - Regular Languages and Grammars 11 r+s = s+r (r+s)+t = r+(s+t) r+r = r r+∅=r (rs)t = r(st) rλ = λr = r r ∅ = ∅r = ∅ r(s+t) = rs+rt (r+s)t = rt+st CS 3240 - Regular Languages and Grammars 12 1. For every regular expression there is an associated NFA that accepts the same language And therefore a DFA, by conversion 2. For every FA (either NFA or DFA) there is a regular expression that represents the same language CS 3240 - Regular Languages and Grammars 13 We will show how to convert each element of the definition of regular expressions to an NFA This is sufficient! And shows the convenience of recursive definitions (review slide 7 now) because if we can give a machine for every case in the definition of REs, we are done! CS 3240 - Regular Languages and Grammars 14 • Empty Language • Empty String CS 3240 - Regular Languages and Grammars • Single Character 15 CS 3240 - Regular Languages and Grammars 16 Just draw the lambdas from a new start state to the start states of each machine Remove the start notation from the original start states (No need to have a new final state) CS 3240 - Regular Languages and Grammars 17 CS 3240 - Regular Languages and Grammars 18 1) Just draw a lambda from each final state of the first machine to the start state of the second machine 2) remove the acceptability of those final states of the first machine CS 3240 - Regular Languages and Grammars 19 CS 3240 - Regular Languages and Grammars 20 We need to do two things: 1) Add the empty string, if needed 2) Loop from each final state back to the start state Procedure: 1) If the empty string is not accepted, create a new start state which accepts, and connect to the original start state with λ 2) Add a λ-edge from each final state to the original (or the new) start state CS 3240 - Regular Languages and Grammars 21 Draw NFAs for the REs on slides 8 and 9 CS 3240 - Regular Languages and Grammars 22 First remove all jails Then, if needed, convert the DFA to an equivalent NFA with A start state with no incoming edges A single final state with no outgoing edges Will need lambda transitions for this Then “eliminate” all but the start and final states Without changing the language accepted Using GTGs… CS 3240 - Regular Languages and Grammars 23 Allow regular expressions on the edges Accepts a* + a*(a+b)c* [Note: (c*)* = c*] CS 3240 - Regular Languages and Grammars 24 If the start state has an incoming edge (even if it’s a loop), create a new start state with a lambda transition to the old start state: CS 3240 - Regular Languages and Grammars 25 If there is more than one final state, or if the single final state has an outgoing edge (even if it’s a loop), create a new final state and link to it with a lambda transition from each final state: CS 3240 - Regular Languages and Grammars 26 “Remove” each intermediate state, one at a time: 1. Combine each incoming path with each outgoing path (only “through” paths; not loops) 2. Determine the regular expression equivalent to the combined path through the current state 3. Add an edge with that RE between the incoming state and the outgoing state 4. Repeat until all intermediate states vanish CS 3240 - Regular Languages and Grammars 27 CS 3240 - Regular Languages and Grammars 28 To eliminate 2: • 1-2-1: af*b • 1-2-3: af*c • 3-2-1: df*b • 3-2-3: df*c CS 3240 - Regular Languages and Grammars 29 To eliminate 1: • 0-1-3: (e+af*b)*(h+af*c) • 3-1-3: (i+df*b)(e+af*b)*(h+af*c) CS 3240 - Regular Languages and Grammars 30 Eliminate 3 (Final Result): (e+af*b) *(h+af*c)(g+df*c+(i+df*b)(e+af*b) *(h+af*c))* CS 3240 - Regular Languages and Grammars 31 CS 3240 - Regular Languages and Grammars 32 Find a regular expression for the language containing all strings that do not contain the substring aa CS 3240 - Regular Languages and Grammars 33 See bypass.doc Shows different possibilities by eliminating states in different orders But the REs obtained are equivalent ▪ Meaning they represent the same language CS 3240 - Regular Languages and Grammars 34 Language Machine Grammar Regular Finite Automaton Regular Expression, Regular Grammar Context-Free Pushdown Automaton Context-Free Grammar Recursively Enumerable Turing Machine Unrestricted PhraseStructure Grammar CS 3240 - Introduction 35 There is a natural correspondence between FAs and grammars Right-linear Grammars “Linear” means there is at most one variable on the right-hand side of the rule “Right-linear” means the variable occurs as the last entry in the rule: ▪ A → abC CS 3240 - Regular Languages and Grammars 36 The variables represent states The right-hand side contains the character(s) on the edge, optionally followed by the target state The accepting states have a lambda rule A → aB | bC | λ B → aA | bD C → aD | bA D → aC | bB CS 3240 - Regular Languages and Grammars 37 Go to an accepting state with no out-edges A→b CS 3240 - Regular Languages and Grammars 38 S → aaS | bbS | abA | baA | λ A → aaA | bbA | abS | baS a GTG CS 3240 - Regular Languages and Grammars 39 Construct a regular grammar for the language denoted by aab*a 1. First build a GTG 2. Then map to a right-linear grammar CS 3240 - Regular Languages and Grammars 40 S → Xa X → Xb | aa How did I come up with this? CS 3240 - Regular Languages and Grammars 41 If you have the single variable only at the left ends, you have a left-linear grammar This is also a regular grammar We will show how to convert between rightlinear and left-linear grammars We will use two facts to establish the process: If L is regular, so is LR (Section 2.3, exercise 12) L(GR) = L(G)R (obvious, but on next slide…) CS 3240 - Regular Languages and Grammars 42 GR means you reverse the right-hand sides of each rule in a grammar, G The language generated is L(G)R (the reverse of L(G)) S → abS | X X → bX | λ S → Sba | X X → Xb | λ (ab)*b* b*(ba)* CS 3240 - Regular Languages and Grammars 43 1. 2. Convert the right-linear grammar to a GTG “Reverse” the GTG (a la Section 2.3, #12) Ensure a single final state (use λ if needed) Interchange the role of the start and final states Reverse all arrows 3. 4. Convert the reversed GTG to a right-linear grammar Reverse the right-hand sides of each rule to obtain the left-linear grammar CS 3240 - Regular Languages and Grammars 44 A → aB B → abA | b (rev) C → bB B → aA A → baB | λ (rev) ba(baa)* CS 3240 - Regular Languages and Grammars C → Bb B → Aa A → Bab | λ (aab)*ab 45 Reverse the grammar, G, obtaining rightlinear grammar, GR, for L(G)R Convert to GTG Reverse the GTG Convert to Right-linear CS 3240 - Regular Languages and Grammars 46 CS 3240 - Regular Languages and Grammars 47