CSCI 3130: Formal Languages and Automata Theory Tutorial 5 Hung Chun Ho Office: SHB 1026 Department of Computer Science & Engineering 1

CSCI 3130: Formal Languages and

Automata Theory

Tutorial 5Hung Chun Ho

Office: SHB 1026

Department of Computer Science & Engineering1

Agenda

• Cocke-Younger-Kasami (CYK) algorithm– Parsing CFG in normal form

• Pushdown Automata (PDA)– Design

2

CYK Algorithm

Bottom-up Parsing for normal form

3

Cocke-Younger-Kasami Algorithm

• Used to parse context-free grammar in Chomsky normal form (or simply normal form)

Every production is of type

1) X YZ

2) X a

3) S ε

Normal Form Example

S AB

A CC | a | c

B BC | b

C CB | BA | c

4

CYK Algorithm - Idea

• = Algorithm 2 in Lecture Note (10L8.pdf)• Idea: Bottom Up Parsing• Algorithm:

Given a string s of length NFor k = 1 to N

For every substring of length k Determine what variable(s) can derive it

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CYK Algorithm - Example

• CFG

• Parse abbc

S AB

A CC | a | c

B BC | b

C CB | BA | c

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CYK Algorithm – Idea (1)

• Idea: We parse the strings in this order:• Length-1 substring

abbcabbcabbcabbc

7



abbcabbcabbc

8



abbcabbc

• Length-4 substringabbc

• Done!

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• Idea: Parsing of longer substrings depends on parsing of shorter substrings

• Example: abb may be decomposed as– ab + b– a + bb

• If we know how to parse ab and b (or, a and bb) then we know how to parse abb

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CYK Algorithm – Substring

• Denote sub(i, j) := substring with start index = i and end index = j

• Example: For abbc, sub(2,4) = bbc• This notation is not to complicate things, but

just for the sake of convenience in the following discussion…

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CYK Algorithm – Table

• Each cell corresponds to a substring• Store variables deriving the substring

Substring of length = 3Starting with index = 2

i.e., sub(2,3) = bbc

ba b c

Length of S

ubstring

Start Index of Substring 12

CYK Algorithm – Simulation

• Base Case : length = 1– The possible choices of variable(s) can be known

by scanning through each production

a b b c

S AB

A CC | a | c

B BC | b

C CB | BA | c A B B A, C

　　　　　　

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　　　　　　A B B A, C


• Loop : length = 2– For each substring of length 2

• Decompose into shorter substrings• Check cells below it

S AB

A CC | a | c

B BC | b

C CB | BA | c

a b b c

ab Let’s parse this substring

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• For sub(1,2) = ab, it can be decomposed:– ab = a + b

= sub(1,1) + sub(2,2)– Possible choices: AB– Scan rules

a b b c

S AB

A CC | a | c

B BC | b

C CB | BA | c

: S

S

　　　　　　A B B A, C

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• For sub(2,3) = bb, it can be decomposed:– bb = b + b

= sub(2,2) + sub(3,3)– Possible choices: BB– Scan rules

a b b c

S AB

A CC | a | c

B BC | b

C CB | BA | c

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: ∅

∅

No suitable rules are found The CFG cannot parse this substring

　　　

S　　　A B B A, C


• For sub(3,4) = bc, it can be decomposed:– bc = b + c

= sub(3,3) + sub(4,4)– Possible choices: BA, BC– Scan rules

a b b c

S AB

A CC | a | c

B BC | b

C CB | BA | c

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: B, C

B, C

　　　

S　 ∅　　A B B A, C


• For sub(1,3) = abb:– abb = ab + b

= sub(1,2) + sub(3,3)– Possible choices: SB– Scan rules

a b b c

S AB

A CC | a | c

B BC | b

C CB | BA | c

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: ∅

No suitable variables found yetBut, there is another way to decompose the string

　　　

S　 ∅　B,

C　A B B A, C


• For sub(1,3) = abb:– abb = a + bb

= sub(1,1) + sub(2,3)– Possible choices: ∅– Scan rules

a b b c

S AB

A CC | a | c

B BC | b

C CB | BA | c

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Cant parse smaller substring Cant parse the string No need to scan rules

　　　

S　 ∅　B,

C　A B B A, C


• For sub(1,3) = abb:– abb = sub(1,1) + sub(2,3) gives no valid parsing– abb = sub(1,2) + sub(3,3) gives no valid parsing

• Cannot parse

a b b c

S AB

A CC | a | c

B BC | b

C CB | BA | c

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∅

　　　

S　 ∅　 B, C

A B B A, C


• For sub(2,4) = bbc:– bbc = sub(2,2) + sub(3,4)

• Possible choices: BB, BC

– bbc = sub(2,3) + sub(4,4)• Possible choices: ∅

a b b c

S AB

A CC | a | c

B BC | b

C CB | BA | c

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Variable: B

B

　∅　　

S　 ∅　B,

C　A B B A, C


• Finally, for sub(1,4) = abbc:– Possible choices:

•

– Variables:•

a b b c

S AB

A CC | a | c

B BC | b

C CB | BA | c

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AB

S

, SB, SC

This cell represents the original string, and it consists S abbc is in the language

　∅　 B　

S　 ∅　B,

C　A B B A, C

CYK Algorithm – Parse Tree

• abbc is in the language!• How to obtain the parse tree?

– Tracing back the derivations:• sub(1,4) is derived using SAB from sub(1,1) and

sub(2,4)• sub(1,1) is derived using Aa• sub(2,4) is derived using BBC from sub(2,2) and

sub(3,4)• …

• So, record also the used derivations!

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CYK Algorithm – Parse Tree

• Obtained from the table

a b b c

S　∅　 B　

S　 ∅　B,

C　A B B A, C

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CYK Algorithm – Conclusion

• A bottom up parsing algorithm– Dynamic Programming– Solution of a subproblem (parsing of a substring)

depends on that of smaller subproblems• Before employing CYK Algorithm, convert the

grammar into normal form– Remove ε-productions– Remove unit-productions

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CYK Algorithm – DetailedD = “On input w = w1w2…wn:

If w = ε, and S ε is rule, AcceptFor i = 1 to n: For each variable A: Test whether A b is a rule, where b = wi.

If so, place A in table(i, i).For l = 2 to n: For i = 1 to n – l + 1: Let j = i + l – 1, For k = i to j – 1: For each rule A BC:If table(i,k) contains B and table(k+1, j) contains C

Put A in table(i, j)If S is in table (1,n), accept. Otherwise, reject.”

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Pushdown Automata

NFA with infinite memory/states

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Pushdown Automata

• PDA ~= NFA, with a stack of memory• Transition:

– NFA – Depends on input– PDA – Depends on input and top of stack

• Push a symbol to stack• Pop a symbol to stack• Read a terminal on string

• Transitions are non-deterministic

(possibly ε)(possibly ε)

(possibly ε)

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Pushdown Automata and NFA

• Accept:– NFA – Go to an Accept state– PDA – Go to an Accept state

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PDA – Example 1

• Given the following language:

• Design a PDA for itL = {0i1j: i ≤ j ≤ 2i, i=0,1,…}, S = {0, 1}

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PDA – Example 1 - Idea

• Idea: The input has two sections– First half

• All ‘0’s

– Second half• All ‘1’s• #‘1 depends on #‘0’

– #‘0’ ≤ #‘1’ ≤ #‘0’ × 2

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PDA – Example 1 – Solution

• Solution:

q0

e,e/$

0,e/Xe,e/e q1

q2

e,$/e

1,X/e

1,X/X 1,X/eq3

L = {0i1j: i ≤ j ≤ 2i, i=0,1,…}, S = {0, 1}

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PDA – Example 1 – Explain

• Solution:

• Let’s try some string… w = 00111– See white board for simulation…

q0

e,e/$

0,e/Xe,e/e q1

q2

e,$/e

1,X/e

1,X/X 1,X/eq3

L = {0i1j: i ≤ j ≤ 2i, i=0,1,…}, S = {0, 1}

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• Solution:

• Indicates the start of parsing

q0

e,e/$

0,e/Xe,e/e q1

q2

e,$/e

1,X/e

1,X/X 1,X/eq3

L = {0i1j: i ≤ j ≤ 2i, i=0,1,…}, S = {0, 1}

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• Solution:

• This part saves information about #‘0’• # ‘X’ in stack = #‘0’

q0

e,e/$

0,e/Xe,e/e q1

q2

e,$/e

1,X/e

1,X/X 1,X/eq3

L = {0i1j: i ≤ j ≤ 2i, i=0,1,…}, S = {0, 1}

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• Solution:

• This part accounts for #‘1’– #‘0’ ≤ #‘1’ ≤ #‘0’ × 2

q0

e,e/$

0,e/Xe,e/e q1

q2

e,$/e

1,X/e

1,X/X 1,X/eq3

L = {0i1j: i ≤ j ≤ 2i, i=0,1,…}, S = {0, 1}

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• Solution:

• Consume one ‘X’ and eats one ‘1’

q0

e,e/$

0,e/Xe,e/e q1

q2

e,$/e

1,X/e

1,X/X 1,X/eq3

L = {0i1j: i ≤ j ≤ 2i, i=0,1,…}, S = {0, 1}

37


• Solution:

• Consume one ‘X’ and eats two ‘1’

q0

e,e/$

0,e/Xe,e/e q1

q2

e,$/e

1,X/e

1,X/X 1,X/eq3

L = {0i1j: i ≤ j ≤ 2i, i=0,1,…}, S = {0, 1}

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• Solution:

• Consume one ‘X’, and then– eats one ‘1’, or– eat two ‘1’

q0

e,e/$

0,e/Xe,e/e q1

q2

e,$/e

1,X/e

1,X/X 1,X/eq3

L = {0i1j: i ≤ j ≤ 2i, i=0,1,…}, S = {0, 1}

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• Solution:

• Indicates the end of parsing

q0

e,e/$

0,e/Xe,e/e q1

q2

e,$/e

1,X/e

1,X/X 1,X/eq3

L = {0i1j: i ≤ j ≤ 2i, i=0,1,…}, S = {0, 1}

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PDA – Example 2

• Given the following language:

• Design a PDA for it

L = { aibjckdl: i, j, k, l=0,1,…; i+k=j+l }, where the alphabet Σ= {a, b, c, d}

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PDA – Example 2 – Idea

• Idea:– Sequentially read (multiple) ‘a’, ‘b’, ‘c’ and ‘d’– Maintain:

• #‘a’ + #‘c’• #‘b’ + #‘d’

– If these numbers equal• Accept

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PDA – Example 2 – Solution

• Solution:


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e,e/$ q5q1

a,e/X

e,e/e

b,$/$Y

q2e,e/e

c,X/XX

q3e,e/e q4

e, $ /e

b,X/e

b,Y/YY

c,$/$X

c,Y/e

d,X/e

d,$/$Y

d,Y/YY


• Solution:


a b c d endstart

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e,e/$ q5q1

a,e/X

e,e/e

b,$/$Y

q2e,e/e

c,X/XX

q3e,e/e q4

e, $ /e

b,X/e

b,Y/YY

c,$/$X

c,Y/e

d,X/e

d,$/$Y

d,Y/YY


• Solution:

• Each X in stack = An extra a or c


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e,e/$ q5q1

a,e/X

e,e/e

b,$/$Y

q2e,e/e

c,X/XX

q3e,e/e q4

e, $ /e

b,X/e

b,Y/YY

c,$/$X

c,Y/e

d,X/e

d,$/$Y

d,Y/YY


• Solution:

• Each Y in stack = An extra b or d


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e,e/$ q5q1

a,e/X

e,e/e

b,$/$Y

q2e,e/e

c,X/XX

q3e,e/e q4

e, $ /e

b,X/e

b,Y/YY

c,$/$X

c,Y/e

d,X/e

d,$/$Y

d,Y/YY


• Solution:

• X and Y ‘cancel’ each other• The stack contains only X’s or only Y’s


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e,e/$ q5q1

a,e/X

e,e/e

b,$/$Y

q2e,e/e

c,X/XX

q3e,e/e q4

e, $ /e

b,X/e

b,Y/YY

c,$/$X

c,Y/e

d,X/e

d,$/$Y

d,Y/YY


• Solution:

• No X’s and no Y’s means– #a + #c = #b + #d Accept

e,e/$ q5q1

a,e/X

e,e/e

b,$/$Y

q2e,e/e

c,X/XX

q3e,e/e q4

e, $ /e

b,X/e

b,Y/YY

c,$/$X

c,Y/e

d,X/e

d,$/$Y

d,Y/YY


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Documents

CSCI 3130: Formal Languages and Automata Theory Tutorial 5 Hung Chun Ho Office: SHB 1026 Department of Computer Science & Engineering 1