Exam (elaborations)
COS3701 EXAM PACK 2025
- Course
- COS3701
- Institution
- University Of South Africa
COS3701 EXAM PACK 2025
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COS3701
Jan/Feb 2023
UNIVERSITY EXAMINATIONS
Jan/Feb 2023
COS3701
Theoretical Computer Science III
80 Marks
Duration 2 Hours
Welcome to the COS3701 exam.
Examiner name: Ms DR Mokwana
Internal moderator name: Prof F Bankole
This paper consists of 5 pages.
Instructions:
• Upload you answer script in a single PDF file format not password
protected
• No emailed scripts will be marked
• Preview your submission to ensure legibility and correct script file
has been uploaded
• Students who have not utilised IRIS invigilation app will be
subjected to disciplinary process
• Students suspected of dishonesty conduct during examination will
be subjected to disciplinary process
• Write neatly and legibly
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COS3701
Jan/Feb 2023
• The mark for each question is in brackets next to the question
Question 1 [16]
(a) Determine a regular expression for the language L over the alphabet
{a, b} that consists of all words that have at least one b but contain
exactly one aa substring (and no other as).
Example of words in the language are aab, bbbaabbb,
bbbbaabbbbbbbb etc.
Examples of words that are not in the language are a, aba, bbab,
aaabbbb, baabbbbaabb
etc. (2)
Design a deterministic finite automaton (DFA) that will recognise all
(b)
of the words in L
as defined above. (4)
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COS3701
Jan/Feb 2023
(c) Use Theorem 21 to develop a context-free grammar (CFG) for (4)
the language L.
(d) Convert the following CFG to Chomsky Normal Form
(CNF):
S → aX Y | Y b
X→XZYZ|
a Y → bY |
Λ
Z→a|Λ (6)
Question 2 [10]
Build a deterministic pushdown automata (DPDA) that
accepts the language
L = {(ab)n(aa)m(ba)n−1 | n ≥ 1, m ≥ 1} over the alphabet Σ =
{a, b}.
Question 3 [12]
The pumping lemma with length for context-free languages (CFLs) can
be stated as follows: Let L be a CFL generated by a CFG in CNF with p
live productions. Then any word w in L with length > 2p can be broken
into five parts:
w = uvxyz such that length(vxy) ≤ 2p length(x) > 0 length(v) +
length(y) > 0 and such that all the words uvnxynz with n ∈ {2, 3,
4, . . .} are also in the language L.
Use the pumping lemma with length to prove that the language
L = {(a)n(b)2n+2(a)n−1 | n ≥ 1}
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