# CS代考 FIT2014 Theory of Computation FINAL EXAM SAMPLE SOLUTIONS – cscodehelp代写

Faculty of Information Technology Monash University
FIT2014 Theory of Computation FINAL EXAM SAMPLE SOLUTIONS
2nd Semester 2015
Instructions:
3 hours writing time.
No books, calculators or devices. Total marks on the exam = 120.
Working Space
Question 1 (3 marks)
You are choosing a main course at a restaurant, from a menu containing three items: bibim- bap, haggis and manakish.
Suppose we have propositions B, H and M, with the following meanings.
B: You choose bibimbap. H: You choose haggis. M: You choose manakish.
Use B, H and M to write a proposition, in Conjunctive Normal Form, that is True precisely when you choose either the bibimbap or both the other items.
Allowing either or both of these two options: (B ∨ H) ∧ (B ∨ M)
Allowing exactly one of these two options: (B ∨ H) ∧ (B ∨ M) ∧ (B ∨ H ∨ M)
Question 2 (5 marks)
Suppose you have predicates belongsToP and belongsToNP with the following meanings, where variable X represents an arbitrary language:
belongsToP(X): the language X belongs to the class P. belongsToNP(X): the language X belongs to the class NP.
(a) In the space below, write a statement in predicate logic with the meaning: P is a proper subset of NP (i.e., P ⊆ NP and P ̸= NP).
To do this, you may only use: the above two predicates; quantifiers; logical connectives; equality. (In particular, you may not use ⊆ or ⊂ or ̸=, etc, and in fact they would not help.)
(∀X : belongsToP(X) ⇒ belongsToNP(X))∧(∃X : belongsToNP(X)∧¬belongsToP(X))
(∀X belongsToP(X) ⇒ belongsToNP(X)) ∧ (¬∀X ¬belongsToNP(X) ∨ belongsToP(X)) (∀X belongsToP(X) ⇒ belongsToNP(X)) ∧ (¬∀X belongsToNP(X) ⇒ belongsToP(X))
(b) For the statement “P is a proper subset of NP”, which one of the following holds? Circle (A), (B), (C) or (D) to indicate your answer.
(A) The statement is True.
(B) The statement is False.
(C) It is not known whether it’s True or False, but it’s generally believed to be True. 􏰂􏰁
(D) It is not known whether it’s True or False, but it’s generally believed to be False.
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Question 3 (7 marks)
A knob is used to select one of three positions A, B, C, as shown in the following diagram.
The position of the knob is recorded every second for some nonzero period of time.
The knob can never be turned from A to C, or from C to A, without spending at least one second in position B. The initial position of the knob can be any of A, B, C.
Consider the language of all possible strings of knob positions recorded in this way.
(a) Give a regular expression for this language.
If empty string forbidden:
AA∗ ∪ CC∗ ∪ (A∗ ∪ C∗)B(A∗ ∪ C∗)(B(A∗ ∪ C∗))∗
Many alternatives are possible, e.g.: ((AA∗ ∪ CC∗)(B(A∗ ∪ C∗))∗) ∪ B(A∗ ∪ C∗)(B(A∗ ∪ C∗))∗, . . . If empty string allowed:
Give a Finite Automaton for this language.
((A∗ ∪ C∗)B)∗(A∗ ∪ C∗)
Many alternatives are possible, e.g.: ((A∗B)∗(C∗B)∗)∗(A∗ ∪ C∗), (A∗ ∪ C∗)(BA∗ ∪ BC∗)∗, . . .
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Question 4 (5 marks)
Consider the following Generalised Nondeterministic Finite Automaton (GNFA). Construct an equivalent GNFA with the top state, q, removed.
Show your answer by writing in the four missing transition labels, on the dotted lines, in the second diagram below.
bab U aba*ba 2
(a U b)a*ba 3
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Question 5 (6 marks)
Consider the following Finite Automaton.
3 5 ACCEPT
ab aba aabb
(a) Which, if any, of the following strings are accepted by this FA? Circle the ones that are accepted.
(b) Consider the string aabab. Show how to split it up into three parts xyz such that, for all i ≥ 0, the string xyiz is also accepted by the FA.
Do this by drawing vertical lines between the parts of the string given on the next line:
This question is about the ideas behind the Pumping Lemma for Regular Languages.
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Question 6 (4 marks)
The following table describes part of a Finite Automaton with three states. The second row, for state 2, has not been filled in.
What could the state-2 row be, if you are given that the number of states of this FA is not minimum?
Write all possible state-2 rows in the second table below. Use as many rows as you need, one for each possible state-2 row. You may not need all the rows available.
state Start 1 2 Final 3
state 213 223 2
By allowing State 2 to be a Final State, some extra possibilities arise, shown in the third and fourth rows below.
state a b 213 223
Final 2 Final 2
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Question 7 (6 marks) This question uses the following Definitions and Facts.
Definitions:
• If L is any language, the reversal of L is the set of all reversals of strings in L. In other words, you obtain the reversal of L by taking every string in L and writing it backwards. So, for example, the string abb becomes bba.
• If L is any language over the alphabet {a,b}, then the interchange of a and b forms a new language as follows: take every string in L, and replace every a by b and every b by a, simultaneously. So, for example, the string abb becomes baa.
• The class of regular languages is closed under reversal.
• The class of regular languages is closed under interchange of a and b. • ThelanguageAB:={anbn : n∈N}isnotregular.
Using these facts, and any other closure properties of regular languages you like, prove by contradiction that the language
{ambn :m≤n}
is not regular.
Assume, by way of contradiction, that the language { ambn : m ≤ n } is regular. Then its reversal, namely { bnam : m ≤ n }, is regular, since the class of regular languages is closed under reversal (using the first fact). Then we interchange a and b, giving the language { anbm : m ≤ n }, which must be regular, since the class of regular languages is closed under interchange of a and b (using the second fact). Take the intersection of the language we started with, and this last language. The former’s b-part is at least as long as its a-part, while the latter’s a-part is at least as long as its b-part. Their intersection consists of strings where the two parts have the same size:
{ambn : m≤n}∩{anbm : m≤n}={anbn : n∈N}=AB,
which must be regular, since the class of regular languages is closed under intersection (as discussed in lectures). But we know that AB is not regular (third fact). So we have a contradiction. So the original assumption, that the language given in the question is regular, is wrong. So that language is not regular.
Question 8 (3 marks)
Explain how to derive, for any Finite Automaton, a regular grammar for the language recognised by the FA.
For each state, create a new non-terminal symbol to represent that state. The initial state is represented by the start symbol S.
For each transition, create a production rule as follows. If the transition has label x and goes from state P to state Q, the production rule is P → xQ. For each final state, create a production rule whose right-hand side is the empty string. So, for a final state represented by non-terminal symbol P , we create the rule P → ε.
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Question 9
Consider the
(a) Give a Each step in
(12 marks)
(b) Give a
S→X (1) X → sXh (2) X→ε (3)
derivation for the string ssshhh.
your derivation must be labelled, on its right, by the number of the rule used.
⇒ ssXhh (2) ⇒ sssXhhh (2) ⇒ ssshhh (3)
parse tree for the same string, ssshhh.
following Context-Free Grammar:
(c) Prove by induction on n, that for all n ≥ 0, the string snhn has a derivation in this grammar of n + 2 steps.
Inductive basis: when n = 0, the string is empty. The empty string can be derived from this grammar in two steps: S ⇒ X ⇒ ε, which is n+2 steps with n = 0. So the statement is true for n = 0.
Now suppose n ≥ 1, and assume that the statement is true for n−1, i.e., any string sn−1hn−1 has a derivation of length (n − 1) + 2 = n + 1 steps. (This is our Inductive Hypothesis.)
This derivation looks like
S ⇒ X ⇒ ··· ⇒ sn−1hn−1 . 􏰇 􏰆􏰅 􏰈
n + 1 steps
Observe here that the first rule must be rule (1), since that is the only one that uses S.
Now consider the string snhn. Since n ≥ 1, we can write this as: snhn = ssn−1hn−1h
Take the derivation S ⇒ X ⇒ ··· ⇒ sn−1hn−1 given above, and do the following. First, omit S and the first production, at the very start. Then we have a derivation from X to sn−1hn−1 in n steps. Then, for each string in this derivation, put an extra s at the start and an extra h at the end. This gives us the partial derivation:
sXh ⇒ ··· ⇒ ssn−1hn−1h 􏰇 􏰆􏰅 􏰈
(The fact that it’s still a valid derivation follows from the context-free property.)
We now prepend this derivation with the two-step derivation of sXh from S, which is S ⇒ X ⇒ sXh, to give the n + 2-step derivation of snhn:
S ⇒X ⇒sXh⇒···⇒ssn−1hn−1h =snhn. 􏰇 􏰆􏰅 􏰈
n + 2 steps
This completes the inductive step.
The statement is therefore true for all n ≥ 0, by the Principle of Mathematical Induction.
Question 10 (6 marks)
This question uses the same Context-Free Grammar as the previous question. Here it is again for convenience:
S→X (1) X → sXh (2) X→ε (3)
Complete the following diagram to give a Pushdown Automaton for the language generated by this grammar.
s,s→ε isoptional h, h → ε
ε, ε → S 1234
Instead of the triangle below state 3, you could have a 2-cycle with two transitions
s, X → h then ε, ε → X . (This effectively combines the first two arcs of the triangle.)
Official use only
Question 11 (5 marks)
State (without proof) the Pumping Lemma for Context-Free Languages, and briefly describe its main purpose.
Let L be a Context-Free Language with k non-terminal symbols in Chomsky Normal Form. For any string w of length ≥ 2k−1, there is a partition of w into five substrings, w = uvxyz, such that:
• at least one of v, y is nonempty,
• |vxy| ≤ 2k, and
• for all i ≥ 0, the string uvixyiz belongs to L.
Its main purpose is to help show that certain languages are not context-free.
Official use only
Question 12
Consider the following Turing machine.
Trace the execution of this Turing machine, writing your answer in the spaces provided
on the next page.
The lines show the configuration of the Turing machine at the start of each step. For each line, fill in the state and the contents of the tape. On the tape, you should indicate the currently-scanned character by underlining it, and you should show the first blank character as ∆ (but there is no need to show subsequent blank characters).
You should not need all the lines provided.
To get you started, the first line has been filled in already.
At start of step 1:
At start of step 2:
At start of step 3:
At start of step 4:
At start of step 5:
At start of step 6:
At start of step 7:
At start of step 8:
At start of step 9:
At start of step 10:
At start of step 11:
At start of step 12:
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Working Space
Question 13 (4 marks)
For each of the following decision problems, indicate whether or not it is decidable.
You may assume that, when Turing machines are encoded as strings, this is done using the Code-Word Language (CWL).
Decision Problem
Input: a Turing machine M.
Question: Does there exist a string w that is accepted by M in at most 7 steps?
Input: a Turing machine M.
Question: Does there exist a string w that is accepted by M?
Input: a string w.
Question: Does there exist a Turing machine that accepts w?
Input: a Turing machine M, and a string w. Question: Is w the encoding, in CWL, of M?
(tick one box in each row)
Undecidable
Undecidable
Undecidable
Undecidable
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Question 14 (10 marks)
The Venn diagram on the right shows several classes of languages. For each language (a)–(j) in the list below, indicate which classes it belongs to, and which it doesn’t belong to, by placing its corresponding letter in the correct region of the diagram.
If a language does not belong to any of these classes, then place its letter above the top of the diagram.
You may assume that, when Turing machines are encoded as strings, this is done using Code-Word Language (CWL), with input alphabet {a,b} and tape alphabet {a,b,#,∆}.
The set of all palindromes of even length.
The set of all positive integers, in binary, whose number of bits is odd.
The set of all strings of correctly matched parentheses.
The set of all encodings of Turing machines that have at least three states.
The set of all encodings of Turing machines that have at most three states. The set of all encodings of Turing machines that halt for some input.
The set of all encodings of Turing machines that loop forever for all inputs.
The set of all satisfiable Boolean expressions.
The set of all satisfiable Boolean expressions in Conjunctive Normal Form in which
every variable appears at most once (so each variable has only one literal).
(j) The set of all nondecreasing strings of digits, i.e., strings over the digits 0,1,. . . ,9 such that no digit is ever followed by a lower digit. (So 03348 is ok, but 03328 is not ok.)
recursively enumerable (r.e.)
Context-Free
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Question 15 (6 marks)
If L is a language and s is a string, then L∆s denotes the language formed by removing s from L if it is there already, and adding s to L otherwise. In other words,
􏰄L{s}, ifs∈L, L∆s:= L∪{s}, ifs̸∈L.
Prove that L is decidable if and only if L∆s is decidable.
Suppose L is decidable. Let D be a decider for L. We use it to build the following
decider for L∆s.
Input: string x.
If x = s then return the opposite answer to D(s), else return D(x).
This works because the only difference between the two languages is for the string s, on which they disagree; but any other string is either in both the languages or in neither of them.
Therefore L∆s is decidable.
Exactly the same argument shows that, if L∆s is decidable, then so is L. (In fact, if we now make D a decider for L∆s, then the above decider becomes a decider for L.) Therefore L is decidable if and only if L∆s is decidable.
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Question 16 (5 marks)
Below is a proof that the Halting Problem is undecidable. But some parts are missing; these are shown as blank spaces, underlined.
Your task is to fill in the underlined spaces to complete the proof. In some cases, options are given in square brackets.
Proof. Assume that the Halting Problem is actually decidable.
Then the Halting Problem has a decider D.
Using D, we can construct a Turing machine E that does the following:
1. Input: encoding of a Turing machine M.
2. Run decider D to determine whether or not M halts when given itself as input.
3. IF the answer from D is Yes, then
loop forever
4. IF the answer from D is No, then stop
Now consider what happens if E is given, as input, an encoding of itself.
If E halts on input E, then running D in line 2 gives the answer Yes
So, using statement 3 [3 or 4], we see that D actually loops forever . On the other hand,
if E does not halt on input E, then running D in line 2 gives the answer No
So, using statement 4 [3 or 4], we see that D actually stops
So the Halting Problem must be undecidable.
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Question 17 (6 marks)
Suppose you have an enumerator M1 for a language L and another enumerator M2 for its complement L.
(a) Explain how to construct a decider for L that uses the enumerators M1 and M2. Decider for L:
Input: string x.
Run M1 and M2 in parallel. (This can be done by a loop which keeps track of the entire configuration of both machines and, in each iteration, simulates one step of each machine.) Whenever either machine outputs a string, we check if that string equals x.
If we see x as an output string from M1, then we know x ∈ L, and we stop and output Yes. If we see x as an output string from M2, then we know x ∈ L, and we stop and output No.
We must eventually see x as an output string from exactly one of the two machines (not both, not neither), since one machine enumerates L and the other enumerates its comple- ment, namely L. (Every string belongs either to L or to L.) So the above computation must always halt eventually.
Other algorithms are possible. E.g., for each i ∈ N, simulate the first i steps of M1 and then the first i steps of M2. If either produces x, then output Yes or No according as x was produced by M1 or M2. If neither produces x, continue.
(b) What does this tell you about recursively enumerable (r.e.) languages whose complements are also recursively enumerable? (Only one short sentence is required here.)
Such languages must be decidable.
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Working Space
Question 18 (13 marks)
A perfect matching in a graph G is a subset X of the edge set of G that meets each vertex exactly once. In other words, no two edges in X share a vertex, and each vertex of G is incident with exactly one edge in X.
The PERFECT MATCHING decision problem is as follows.
PERFECT MATCHING
Input: Graph G.
Question: Does G have a perfect matching?
For example, in the following graph, the edge set {a,e} is a perfect matching. But {a,b,e} is not a perfect matching (since, for example, a and b share a vertex), and {a} is not a perfect matching (since some vertices are not incident with the edge in this set).
d Let W be the above graph.
(a)ConstructaBooleanexpressionEW inConjunctiveNormalFormsuchthatthesatisfying truth assignments for EW correspond to perfect matchings in the above graph W.
(a∨b)∧(¬a∨¬b)∧ (a∨c∨d)∧(¬a∨¬c)∧(¬a∨¬d)∧(¬c∨¬d)∧ (b∨c∨e)∧(¬b∨¬c)∧(¬b∨¬e)∧(¬c∨¬e)∧ (d∨e)∧(¬d∨¬e).
(b) Give a polynomial-time reduction from PERFECT MATCHING to SATISFIABILITY.
Input: graph G. For each vertex v: {
Let the edges incident with v be e1,e2,…,ed.
We need to create clauses which, when AND-ed together, say: Exactly one of these edges is in the matching.
Create a new clause, e1 ∨···∨ed.
This says: at least one of e1, . . . , ed is in the matching.
For each pair ei,ej of edges incident with v: Create a new clause, ¬ei ∨ ¬ej .
This says: at least one of ei,ej is not in the matching.
Combine all these clauses using ∧, to create the conjunction of all of them. Output the resulting expression.
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Question 19 (8 marks)
Prove that the GRAPH 4-COLOURABILITY problem is NP-complete, by reduction from GRAPH 3-COLOURABILITY. You may assume that GRAPH 3-COLOURABILITY is NP-complete.
Definitions:
For any positive integer k, a k-colouring in a graph G is an assignment of “colours” from the set {1,2,…,k} to the vertices of G such that (a) each vertex gets exactly one colour from the set, and (b) adjacent vertices get different colours.
GRAPH 3-COLOURABILITY
Input: Graph G.
Question: Does G have a 3-colouring?
GRAPH 4-COLOURABILITY
Input: Graph G.
Question: Does G have a 4-colouring?
GRAPH 4-COLOURABILITY is in NP, since it has a polynomial-time verifier: the cer- tificate is a 4-colouring, and this can be checked in polynomial time: for each edge, check that its endpoints are differently coloured, and check that the total number of colours used is at most 4.