Parsers -

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Parsing
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Compiler
Program File
v = 5;
if (v>5)
x = 12 + v;
while (x !=3) {
x = x - 3;
v = 10;
}
......
Add v,v,5
cmp v,5
jmplt ELSE
THEN:
add x, 12,v
ELSE:
WHILE:
cmp x,3
...
Machine Code

5
Lexical
analyzer
parser
Compiler
Program
file
machine
code
Input String Output

6
Lexical analyzer:
• Recognizes the lexemes of the
input program file:
Keywords (if, then, else, while,…),
Integers,
Identifiers (variables), etc
•It is built with DFAs (based on the
theory of regular languages)

7
•Knows the grammar of the
programming language to be compiled
Parser:
•Constructs derivation (and derivation tree)
for input program file (input string)
•Converts derivation to machine code

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Example Parser
PROGRAM STMT_LIST
STMT_LIST STMT; STMT_LIST | STMT;
STMT EXPR | IF_STMT | WHILE_STMT
| { STMT_LIST }
EXPR EXPR + EXPR | EXPR - EXPR | ID
IF_STMT if (EXPR) then STMT
| if (EXPR) then STMT else STMT
WHILE_STMT while (EXPR) do STMT







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The parser finds the derivation
of a particular input file
10 + 2 * 5
Example
Parser
E -> E + E
| E * E
| INT
E => E + E
=> E + E * E
=> 10 + E*E
=> 10 + 2 * E
=> 10 + 2 * 5
Input string
derivation

10
E => E + E
=> E + E * E
=> 10 + E*E
=> 10 + 2 * E
=> 10 + 2 * 5
derivation derivation tree
10
E
2 5
E E
E E
+
*
mult a, 2, 5
add b, 10, a
machine codeDerivation trees
are used to build
Machine code
a
b

11
A simple (exhaustive) parser

12
grammar
Exhaustive Parser
input
string
derivation
We will build an exhaustive search parser
that examines all possible derivations

13
Example:
Exhaustive Parser
derivation




S
bSaS
aSbS
SSSInput string
?aabb
aabbFind derivation of string

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Exhaustive Search
||| bSaaSbSSS 
Phase 1:
All possible derivations of length 1
Find derivation
of aabb




S
bSaS
aSbS
SSS

15
Find derivation
of aabb
Cannot possibly produce aabb
Phase 1:




S
bSaS
aSbS
SSS

16
aSbS
SSS


Phase 1
In Phase 2,
explore the next step
of each derivation
from Phase 1

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Phase 2
aSbS
SSS

 SSSS
bSaSSSS
aSbSSSS
SSSSSS




Phase 1
abaSbS
abSabaSbS
aaSbbaSbS
aSSbaSbS




Find derivation
of aabb

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Phase 2
SSSS
aSbSSSS
SSSSSS



aaSbbaSbS
aSSbaSbS


Find derivation
of aabb
In Phase 3 explore
all possible derivations

19
Phase 2
SSSS
aSbSSSS
SSSSSS



aaSbbaSbS
aSSbaSbS


A possible derivation
of Phase 3
aabbaaSbbaSbS 
Find derivation
of aabb

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Final result of exhaustive search
Exhaustive Parser
derivation




S
bSaS
aSbS
SSSInput
string
aabb
aabbaaSbbaSbS 

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( -productions)
Suppose that the grammar does not have
productions of the form
A
BA  (unit productions)

Time Complexity

Since the are no -productions
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wxxxS k  21
|||| wxi 
For any derivation of a
string of terminals )(GLw 
it holds that for all i


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Since the are no unit productions
1. At most derivation steps are needed
to produce a string with at most
variables
||w
jx ||w
2. At most derivation steps are needed
to convert the variables of to the
string of terminals
||w
jx
w

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Therefore, at most derivation
steps are required to produce
||2 w
w
The exhaustive search requires at most
||2 w phases

25
Possible derivation choices
to be examined in phase 1: k
kSuppose the grammar has productions
at most

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Choices for phase 2: at most
2
kkk 
Choices of
phase 1
Number of
Productions
Choices for phase i: at most
ii
kkk  )1(
Choices of
phase i-1
Number of
Productions
In General

27
Total exploration choices for string :w
)( ||2||22 ww
kOkkk  
Extremely bad!!!
phase 1 phase 2 phase 2|w|
Exponential to the string length

29
There exist faster parsing algorithms
for specialized grammars
S-grammar: avA 
Symbol String of variables
),( X
appears once in a production
Each pair of variable, terminal
(a restricted version of Greinbach Normal form)
wX 

30
S-grammar example:
cS
bSSS
aSS



abccabcSabSSaSS 
Each string has a unique derivation

31
In the exhaustive search parsing
there is only one choice in each phase
For S-grammars:
Total steps for parsing string :w || w
Steps for a phase: 1

32
For general context-free grammars:
Next, we give a parsing algorithm
that parses a string in timew )|(| 3
wO
(this time is very close to the worst case
optimal since parsing can be used to solve
the matrix multiplication problem)

33
The CYK Parsing Algorithm
Input: • Arbitrary Grammar
in Chomsky Normal Form
G
• String
Output: Determine if )(GLw
w
Number of Steps: )|(| 3
wO
Can be easily converted to a Parser

34
Basic Idea
Denote by the set of variables
that generate a string
)(wF
w
Consider a grammar
In Chomsky Normal Form
G
)(wFX  wX
*
if

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Suppose that we have computed )(wF
Check if :)(wFS 
YES
NO
)(GLw 
)(GLw 
)(
*
wS 

36
and there is production
uvw 
)(uFX 
)(wF can be computed recursively:
)(vFY 
XYH 
Then )(wFH 
prefix suffix
If
)(
*
vY )(
*
uX 
and
)(
**
wuvuYXYH 
Write

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Examine all prefix-suffix
decompositions of w
1||1 
 w
vuw
2||2 
 w
vuw
11||
vuw w 

1
2
|w|-1
Length Set of Variables
that generate w
1H
2H
1|| wH
1||21)(  wHHHwF Result:

38
At the basis of the recursion
we have strings of length 1
}symbolgeneratethatVariables{)(  F
symbol X
Very easy to find

39
The whole algorithm can be implemented
with dynamic programming:
Remark:
First compute for smaller
substrings and then use this
to compute the result for larger
substrings of
)(wF 
w 
w

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• Grammar :G
bABB
aBBA
ABS
|
|



• Determine if )(GLaabbbw 
Example:

41
a a b b b
aa ab bb bb
aab abb bbb
aabb abbb
aabbb
aabbbDecompose the string
to all possible substrings
Length
1
2
3
4
5

a
{A}
a
{A}
b
{B}
b
{B}
b
{B}
aa ab bb bb
aab abb bbb
aabb abbb
aabbb
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bABBaBBAABS |,|, 
)(F

a
{A}
a
{A}
b
{B}
b
{B}
b
{B}
aa
{}
ab
{S,B}
bb
{A}
bb
{A}
aab abb bbb
aabb abbb
aabbb
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)(F
)(F

44
)(aaF
}{)( AaF 
AAX There is no production of form
aaprefix suffix
}{)( AaF 
Thus, {})( aaF
)(abF
}{)( AaF 
ABX There are two productions of form
abprefix suffix
}{)( BbF 
Thus, },{)( BSabF 
ABBABS  ,

a
{A}
a
{A}
b
{B}
b
{B}
b
{B}
aa
{}
ab
{S,B}
bb
{A}
bb
{A}
aab
{S,B}
abb
{A}
bbb
{S,B}
aabb abbb
aabbb
45

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)(aabF
}{)( AaF 
ASX There is no production of form
aabprefix suffix
},{)( BSabF 
ABX 
ABBABS  ,
There are 2 productions of form
Decomposition 1
},{1 BSH 

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)(aabF
{})( aaF
BX There is no production of form
aabprefix suffix
}{)( BbF 
Decomposition 2
}{2 H
},{}{},{)( 21 BSBSHHaabF 

a
{A}
a
{A}
b
{B}
b
{B}
b
{B}
aa
{}
ab
{S,B}
bb
{A}
bb
{A}
aab
{S,B}
abb
{A}
bbb
{S,B}
aabb
{A}
abbb
{S,B}
aabbb
{S,B} 48
)(aabbbF
)(GLaabbb
)(wFS 
Since

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Approximate time complexity:
)|(||)||(| 32
wOwwO 
Number of
substrings
Number of
Prefix-suffix
decompositions
for a string

Questions
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Email: Badawy@caiwave.net
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