Compilers Final exam 2013 Model answer

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Cairo University
Faculty of Computers and Information
Final Exam
Department: CS
Course Name: Compilers Date: Tuesday 4-June-2013
Course Code: CS419 Duration: 2 hours
Instructor(s): Dr. Hussien Sharaf Total Marks: 60
Question 1 [12 marks]
For each of the following languages do the following:
i. Understand the languages without any assistance;
ii. Write sample words for the language where you show the smallest possible word.
iii. Construct a regular expression that represents the language.
Note: Draw and fill the following table in your answer sheet:
Sample Regular Expression
a. Σ ={a, b} Only words with exactly two consecutive a's [2 marks]
Solution:
Sample = {aa, aab, baa, baab, bbaabbb …..}
b*aab*
b. Σ ={a, b} Only words such that { an
bm
| n is even and m is odd} where n and m indicate number
of “a”s and “b”s respectively. [3 marks]
Solution:
Sample = {b,aab,aabbb,..}
(aa)* b(bb)* + (aa)* a (bb)* b
c. Σ ={a, b} Words where “a” appears ONLY in even positions or doesn’t appear at all. [3 marks]
Solution:
Sample = { Λ b ba bbb bbba baba …..}
b+(bb+ba)*b*
(b(a+b))* + b*
(ba)*(bb)*(ba)*
b* + (b(bb)* (a+b) )*
d. Σ ={a, b} Construct a regular expression for all strings that have exactly one double letter in
them. “One double letter” implies two equal touching letters; triples or more are excluded. [4 marks]
Solution:
Sample = { aa, baa, …..}
(b + Λ)(ab)*aa(ba)*(b + Λ) + (a + Λ)(ba)*bb(ab)*(a + Λ)
Question 2 [8 marks]
For each of the following languages do the following:
i. Understand the languages without any assistance;
ii. Write sample words for the language where you show the smallest possible word.
iv. Draw a deterministic finite automaton that represents the language and handles incorrect
inputs.

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a. Σ = {a, b} words where “a” appears ONLY in the second position or doesn’t appear at all. [3 marks]
Solution:
Sample = { Λ b ba bbb bab babbb …..}
b. Σ = {a, b} Words that begin and ending with the same letter. [5 marks]
Solution:
Sample = { Λ b ba bbb bab babbb …..}
Question 3 [10 marks]
An if-statement indicated by L is a language where multiple executable instructions “S” can be used only
by embedding them inside curly brackets pair where execution starts with the inner curly brackets pair;
i.e. {{{S1}S2}S3} means executing s1, s2 then s3. One statement can only be embedded inside one curly
bracket pair where the statement “S” should always stick to the closing curly bracket. Finally a curly
bracket pair must always contain a statement i.e. { {S} } is NOT valid.
Sample for if-statement in L:
if ((c4) OR (c5 AND c6))
{{{s1}s2}s3}
else
{{s5}s6}
a. Understand the language without any assistance and check if the following grammar represents
the conditional statements “C” indicated by L. If NO then explain why? If YES then show left
derivation for “((C4) OR (C5 AND C6))”; Do each step in a separate line and show rule
number. [3 marks]
R1: C (C)
R2: C C AND C
R3: C C OR C
R4: C c1|c2|c3|c4|c5|c6| …|c10
Note: Draw and fill the following table in your answer sheet:
a
b
b
-+1 2
a,b
+3
a
a
-+1
b
4
a
b
3
a
+2
b
b
a
a
5
b
+4

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Derivation Rule Number
Solution:
YES
Derivation Rule Number
C (C) R1
(C OR C) R3
((C) OR C) R1
((c4) OR C) R4
((c4) OR (C)) R1
((c4) OR (C AND C)) R2
((c4) OR (c5 AND C)) R4
((c4) OR (c5 AND c6)) R4
b. Write a CFG for the instructions “S” similar to the CFG of the “C”. Can you avoid using
Lambda? [4 marks]
Solution:
R1: S {ST}
R2: S
R3: T s1|s2|s3|s4|s5|s6| …|s10
must be used to recursion of S
c. Can language L be described using regular expressions? Why? [3 marks]
Solution:
No.
Because L have nested structures brackets and parenthesis that must be balanced. RE cannot
describe balanced structures such as an
bn
Question 4 [30 marks]
Ζ delta is a language that allows for nested if-statements L, executable expressions “E” and conditional
statements “C”.
CFG:
R1: Z E
R2: Z L
R3: L if C Z
R4: L if C Z else Z
R5: C (C)
R6: C N and N
R7: E id = N * N
R10: N→ 0
R11: N→ 1
Sample for a Z statement:
if (0 and 1) if (0 and 0) id = 1*1
a. Specify which pair of rule(s) contains the left factoring and rewrite all the CFG with numbers
(after removing left factoring) factoring? Hint: if you need to introduce a new variable then use
letter M. [3 marks]
Solution:
R3 and R4
CFG:
R1: Z →E
R2: Z →L

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R3: L→ if C Z M
R4:M→
R5:M→ else Z
R6: C→ (C)
R7: C→N and N
R8: E →id = N * N
R9: N→ 0
R10: N→ 1
b. Construct a parsing table for Z. Note that the set of terminals is {if, else, (,),id=,*, and, 0,1} and
set of non-terminals is {Z, E, L, M, C, N} [3 marks]
Solution:
If else ( ) and id = * 0 1 $
Z R2 R1
E R8
L R3
M R5 R4 R4 R4
C R6 R7 R7
N R9 R10
c. Show the top-down parsing steps of the sample above using the parsing table constructed in the
previous section. [16 marks]
Solution:
Stack Input Parser action
Z if (0 and 1) if (0 and 0) id = 1*1
$
R2
L .. R3
if C Z M if .. Match
C Z M (0 and 1) .. R6
(C) Z M (0 and 1) .. Match
C) Z M 0 and 1) .. R7
N and N) Z M 0 and 1) .. R9
0 and N) Z M 0 and 1) .. Match
N) Z M 1) … R9
1) Z M 1) if (0 and 0) id = 1*1 $ Match
Z M if (0 and 0) id = 1*1 $ R2
L M if (0 and 0) id = 1*1 $ R3
if C Z M M if (0 and 0) id = 1*1 $ Match
C Z M M (0 and 0) id = 1*1 $ R6
(C) Z M M (0 and 0) id = 1*1 $ Match
C) Z M M 0 and 0) id = 1*1 $ R7
N and N) Z M M 0 and 0) id = 1*1 $ R9
0 and N) Z M M 0 and 0) id = 1*1 $ Match
N) Z M M 0) id = 1*1 $ R9
0) Z M M 0) id = 1*1 $ Match
Z M M id = 1*1 $ R1
E M M id = 1*1 $ R8
id = N * N M M id = 1*1 $ Match
N * N M M 1*1 $ R10
1 * N M M 1*1 $ Match
N M M 1 $ R10
1 M M 1 $ Match
M M $ R4
M $ R4

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Empty $ Accept
d. Show the 10 steps for bottom-up parsing of the sample above. [10 marks]
Solution:
Stack Input Action
$ if (0 and 1) if (0 and 0) id = 1*1
$
Shift
$if (0 and 1) … Shift
$if( 0 and 1) …. Shift
$if(0 and 1) …. Reduce R9
$if(N and 1) …. Shift
$if(N and 1) …. Shift
$if(N and 1 ) if (0 and 0) id = 1*1 $ Reduce R10
$if(N and N ) if (0 and 0) id = 1*1 $ Reduce R7
$if(C ) if (0 and 0) id = 1*1 $ Shift
$if(C) if (0 and 0) id = 1*1 $ Reduce R6
$if C if (0 and 0) id = 1*1 $ Shift
$if C if (0 and 0) id = 1*1 $ Shift
$if C if( 0 and 0) id = 1*1 $ Shift
$if C if(0 and 0) id = 1*1 $ Reduce R9
$if C if(N and 0) id = 1*1 $ Shift
$if C if(N and 0) id = 1*1 $ Shift
$if C if(N and 0 ) id = 1*1 $ Reduce R9
$if C if(N and N ) id = 1*1 $ Reduce R7
$if C if(C ) id = 1*1 $ Shift
$if C if(C) id = 1*1 $ Shift
$if C if(C) id = 1*1 $ Reduce R6
$if C if C id = 1*1 $ Shift
$if C if C id 1*1 $ Shift
$if C if C id = 1*1 $ Shift
$if C if C id = 1 *1 $ Reduce R10
$if C if C id = N *1 $ Shift
$if C if C id = N * 1 $ Shift
$if C if C id = N * 1 $ Reduce R10
$if C if C id = N * N $ Reduce R8
$if C if C E $ Reduce R1
$if C if C Z $ Reduce R4
$if C if C Z M $ Reduce R3
$if C L $ Reduce R2
$if C Z $ Reduce R4
$if C Z M $ Reduce R3
$ L $ Reduce R2
$Z $ Accept