SFWR ENG 2FA3. Solution to the Assignment #3

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1[20]. Exercise 12.42 (page 247 of the Gries-Schneider textbook).

Recall the checklist for proving loop correct:

(a) P is true before execution of the loop.

(b) P is a loop invariant: {P∧B} S{P}.

(c) Execution of the loop terminates.

(d) R holds upon termination: P∧¬B =⇒ R.

(a)[5] Recall F0 = 0,F1 = 1 and Fn = Fn−1 +Fn−2 for n > 1, and additionallyF−1 = 1,F1 = F0 +F−1.

{Q : n≥ 0}k,b,c := 0,1,0;{invariant P : 0≤ k ≤ n∧b = Fk−1∧ c = Fk}do k 6= n→ k,b,c := k+1,c,b+ c od{R : c = Fn}

Hence:

(a) Since k,b,c := 0,1,0; sets k = 0,b = 1,c = 0 then P is clearly true before exe-cution of the loop.

(b) It suffices to show that P[k,b,c := k+1,c,b+ c]⇐ P∧B.

(0≤ k ≤ n∧b = Fk−1∧ c = Fk)[k,b,c := k+1,c,b+ c]

⇐⇒ 〈 rules for concurrent substitution 〉0≤ k+1≤ n∧ c = Fk+1−1∧b+ c = Fk+1

⇐⇒ 〈 arithmetic 〉−1≤ k ≤ n−1∧ c = Fk ∧b+ c = Fk+1

⇐⇒ 〈 Fk+1 = Fk +Fk−1 and arithmetic 〉−1≤ k < n∧ c = Fk ∧b = Fk−1

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⇐ 〈 arithmetic and logic 〉0≤ k ≤ n∧b = Fk−1∧ c = Fk ∧ k 6= n ⇐⇒ P∧B

Hence P∧B =⇒ P[k,b,c := k+1,c,b+c], so P is a loop invariant for this loop.

(c) Execution of the loop clearly terminates as each time k is increased by 1 and theloop stops when k = n.

(d) We have P∧¬B ⇐⇒ P∧ (k = n) ⇐⇒(0≤ k ≤ n∧b = Fk−1∧ c = Fk)∧ k = n ⇐⇒b = Fn−1∧ c = Fn =⇒ (c = Fn) ⇐⇒ R.

Hence R holds upon termination.

(b)[5] Prove the correctness of the below algorithm

{Q : n≥ 0}x,k := 0,0;{invariant P : 0≤ k ≤ n∧ x = ∑(i | 0≤ i < k : b[i])}do k 6= n→ x,k := x+b[k],k+1 od{R : x = ∑(i | 0≤ i < n : b[i])}

Hence:

(a) Since x,k := 0,0 sets x = 0,k = 0 then P is clearly true before execution of theloop.

(b) It suffices to show that P[x,k := x+b[k],k+1]⇐ P∧ (k 6= n).

(0≤ k ≤ n∧ x = ∑(i | 0≤ i < k : b[i]))[x,k := x+b[k],k+1]

⇐⇒ 〈 rules for concurrent substitution in quantifiers 〉0≤ k+1≤ n∧ x+b[k] = ∑(i | 0≤ i < k+1 : b[i])

⇐⇒ 〈 arithmetic 〉−1≤ k ≤ n−1∧ x = ∑(i | 0≤ i < k+1 : b[i])−b[k]

⇐⇒ 〈 domain split for quantifiers 〉−1≤ k ≤ n−1∧ x = ∑(i | 0≤ i < k : b[i])+b[k]−b[k]

⇐ 〈 arithmetic and logic 〉(0≤ k ≤ n∧ x = ∑(i | 0≤ i < k : b[i]))∧ (k 6= n)

Hence P∧B =⇒ P[x,k := x+b[k],k+1], so P is a loop invariant for this loop.

(c) Execution of the loop clearly terminates as each time k is increased by 1 and theloop stops when k = n.

(d) We have P∧¬B ⇐⇒ P∧ (k = n) ⇐⇒(0≤ k ≤ n∧ x = ∑(i | 0≤ i < k : b[i]))∧ (k = n) ⇐⇒x = ∑(i | 0≤ i < n : b[i]) ⇐⇒ R.

Hence R holds upon termination.

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(c)[5] Prove the correctness of the below algorithm

{Q : n≥ 0}x,k := 0,n;{invariant P : 0≤ k ≤ n∧ x = ∑(i | k ≤ i < n : b[i])}do k 6= 0→ x,k := x+b[k−1],k−1 od{R : x = ∑(i | 0≤ i < n : b[i])}

Hence:

(a) Since x,k := 0,0 sets x = 0,k = 0 then P is clearly true before execution of theloop.

(b) It suffices to show that P[x,k := x+b[k−1],k−1]⇐ P∧ (k 6= n).

(0≤ k ≤ n∧ x = ∑(i | k ≤ i < n : b[i]))[x,k := x+b[k−1],k−1]

⇐⇒ 〈 rules for concurrent substitution in quantifiers 〉0≤ k−1≤ n∧ x+b[k−1] = ∑(i | k−1≤ i < n : b[i])

⇐⇒ 〈 arithmetic 〉1≤ k ≤ n+1∧ x = ∑(i | k−1≤ i < n : b[i])−b[k−1]

⇐⇒ 〈 domain split for quantifiers 〉1≤ k ≤ n+1∧ x = ∑(i | k−1 < i < n : b[i])−b[k−1]+b[k−1]

⇐ 〈 arithmetic and logic 〉(0≤ k ≤ n∧ x = ∑(i | k ≤ i < n : b[i]))∧ (k 6= 0)

Hence P∧B =⇒ P[x,k := x− b[k− 1],k− 1], so P is a loop invariant for thisloop.

(c) Execution of the loop clearly terminates as each time k is decreased by 1 and theloop stops when k = 0.

(d) We have P∧¬B ⇐⇒ P∧ (k = n) ⇐⇒()∧ (k = 0) ⇐⇒x = ∑(i | 0≤ i < n : b[i]) ⇐⇒ R.

Hence R holds upon termination.

(d)[5] Prove the correctness of the below algorithm

{Q : 0≤ X ∧0≤ Y}x,y := X ,Y ;{invariant P : x gcd y = X gcd Y ∧0≤ x∧0≤ y}do y 6= 0→ x,y := y,x mod y od{R : X gcd Y = x}

Hence:

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(a) Since x,y := X ,Y sets x = X ,y = Y then P is clearly true before execution of theloop.

(b) It suffices to show that P[x,y := y,x mod y]⇐ P∧ (y 6= 0).

(x gcd y = X gcd Y ∧0≤ x∧0≤ y)[x,y := y,x mod y]

⇐⇒ 〈 rules for concurrent substitution 〉(y gcd (x mod y) = X gcd Y ∧0≤ y∧0≤ x mod y

⇐⇒ 〈 since x gcd y = y gcd (x mod y), see page 248 ofGries-Schneider textbook 〉

(x mod y = X gcd Y ∧0≤ y∧0≤ x mod y

⇐ 〈 since x mod y < x 〉x gcd y = X gcd Y ∧0≤ x∧0≤ y

⇐ 〈 as α∧β⇒ α 〉(x gcd y = X gcd Y ∧0≤ x∧0≤ y)∧ (y 6= 0)

Hence P∧B =⇒ P[x,y := y,x mod y], so P is a loop invariant for this loop.

(c) Execution of the loop clearly terminates as y < x mod y so each time y is in-creased by at least 1 and the loop stops when y = 0.

(d) We have P∧¬B ⇐⇒ P∧ (y = 0) ⇐⇒(x gcd y = X gcd Y ∧0≤ x∧0≤ y)∧ (y = 0) ⇐⇒X gcd Y = x gcd y = x gcd 0 = x ⇐⇒ R.

Hence R holds upon termination.

2.[7] Let M be the following deterministic finite automaton: Q = {q0,q1,q2},Σ = {a,b}, q0 is the initial state, F = {q2}, and δ : Q×Σ→ Q

δ a bq0 q0 q1q1 q2 q1q2 q2 q0

(a)[2] Give the state diagram (graphical representation) of M

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C q0

a

b

ba

a

b

q1

q2

(b)[4] Trace the computations (draw paths consisting of states and actions) that pro-cess the sequences:

i[1] abaaq0→ a→ q0→ b→ q1→ a→ q2→ a→ q2

ii[1] bbbabbq0→ b→ q1→ b→ q1→ b→ q1→ a→ q2→ b→ q0→ b→ q1

iii[1] bababaq0→ b→ q1→ a→ q2→ b→ q0→ a→ q0→ b→ q1→ a→ q2

iv[1] bbbaaq0→ b→ q1→ b→ q1→ b→ q1→ a→ q2→ a→ q2

c.[2] Which of the sequences from (b) are accepted by M?

abaa,bababa,bbbaa

3.[5] Let M be the following deterministic finite automaton: Q = {q0,q1,q2},Σ = {a,b}, q0 is the initial state, F = {q0}, and δ : Q×Σ→ Q

δ a bq0 q1 q0q1 q1 q2q2 q1 q0

(a)[2] Give the state diagram (graphical representation) of M

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C q0

b

a

ab

b

q1

q2

a

(b)[1] Trace the computations (draw paths consisting of states and actions) that pro-cess the sequence babaab:

q0→ b→ q0→ a→ q1→ b→ q2→ a→ q1→ a→ q1→ b→ q2

(c)[2] Give an example of two sequences that are accepted by M and another twosequence that are not accepted by M.

abb - accepted, ba - rejected

3a.[15] Consider a typical vending machine that dispenses a person’s choice ofcandy after it has received a total of 30 cents in nickels (5 cents), dimes (10 cents),and quarters (25 cents). In this case, the automaton’s alphabet consists of three differ-ent sizes of coins, a state of the machine is the total amount of money that has beenreceived since the last candy was dispensed, the initial state is that of not having re-ceived any coins since the last candy was dispensed, and a final state is that of havingreceived at least 30 cents.

(a)[10] Assuming that the machine is not designed to make change and will thereforemerely consume any overpayment, define a finite deterministic automaton thatmodels the vending machine.

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C 0 5

10

15

20

25 $30

nickel

dime

quarter

quarternickel

dime

quarternickeldime

dime

quarternickel

quarter

nickel quarter

dime

quarter

dimenickel

quarter

dime

nickel

(b)[5.] Give a finite deterministic automaton for a machine that returns change.

Similarly as (a) above. The state ‘≥ 30’ has to be replaced by the states ‘30 andno change’, ‘35-change 5’, ‘40-change 10’, ‘45-change 15’, ‘50-change 20’ and‘55-change 25’. For instance from the state ‘20’, nickel goes to ‘30’, dime to‘34’, quarter to ‘25’, etc.

4.[15] Design deterministic finite state automata for the following sets:

(c)[5] the set of strings x ∈ {0,1}∗ such that #0(x) is even and #1(x) is a multiple ofthree.

Note that for each natural number x, x mod 2 is either 0 or 1, while x mod 3 is 0,1 or 2.The automaton will have 6 states, denoted as (k, l), where k = 0,1 and l = 0,1,2. Thestate (k, l) is interpreted as a string that starts from the initial state and ends in (k, l)has k = #0(x) mod 2 and l = #1(x) mod 3.

The automaton is given below:

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(d)[10] the set of strings over the alphabet {a,b} containing at least three occur-rences of three consecutive b’s, overlapping permitted (e.f., the string bbbb should beaccepted).

First note that the following automaton accepts all strings over the alphabet {a,b}containing at least one occurrence of three consecutive b’s, overlapping permitted.

Hence the full solutions looks as follows:

Note that the solutions with regular expressions or nondeterministic automata are moreintuitive:

(a+b)∗bbb(a+b)∗bbb(a+b)∗bbb(a+b)∗

and

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5[16]. Consider the following two deterministic finite state automata.

Use the product construction to produce deterministic automata accepting:

(a)[8] the intersection of the two sets accepted by these automata.

(b)[8] the union of the two sets accepted by these automata.

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6.[10] Let M = (Q,Σ,δ,s,F) be an arbitrary DFA. Prove by induction on |y| that forall strings x,y ∈ Σ∗ and q ∈ Q,

δ̂(q,xy) = δ̂(δ̂(q,x),y),

where δ̂ is the extended version of δ.Proof. It is clearly true for y = ε as for every q we have δ̂(q,ε) = q, soδ̂(δ̂(q,x),ε) = δ̂(q,x) = δ̂(q,xε).

Assume it holds for y and consider ya where a ∈ Σ.

δ̂(q,xya)= 〈 definition of δ̂ 〉δ(δ̂(q,xy),a)= 〈 inductive assumption for y 〉δ(δ̂(δ̂(q,x),y),a)= 〈 definition of δ̂ 〉δ̂(δ̂(q,x),ya)

So we are done.

7.[8] Convert the following nondeterministic automaton into a deterministic one (Ex-ercise 1 from page 302 of the Kozen textbook).

Solution:

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8.[10] Show that for any A∈ Σ∗, if A is regular then so is rev A (Exercise 2 from page302 of the Kozen textbook).

The solution is illustrated by the following example. If the first automaton acceptsA then all automata after step 2 accept rev A.

In principle, we first reverse arrows, next make initial state final and all final state ini-tial. Now we have to solve the problem of multiple initial states. The simplest solutionis to use ε-moves, but we can live without them too.

Formal solutions is the following. Let M = (Q,Σ,δ,s0,F) be a deterministic au-tomaton such that L(M) = A.

Define a non-deterministic automaton Mrev = (Qrev,Σ,∆rev,s00,{s0}), where s0

0 /∈Q,Qrev = Q∪{s0

0}, and ∆rev : Qrev×Σ→ 2Qrevis defined as follows:

∆rev(p,a) = {q | δ(q,a) = p},

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for all p ∈ Q and a ∈ Σ, and

∆rev(s0

0,a) = {q | ∃(p | p ∈ F : δ(q,a) = p},

for all a ∈ Σ.Clearly L(Mrev) = L(M) = A.This suffices for the proof as for every regular set A there exists a deterministic

automaton M such that L(M) = A. For the initial deterministic automaton this formalconstruction will give an automaton after step 3.

An extension of the formal construction for the initial non-deterministic automaton(not needed for the proof per se), is the following.

Let M = (Q,Σ,∆,s0,F) be a non-deterministic automaton such that L(M) = A.Define a non-deterministic automaton Mrev = (Qrev,Σ,∆rev,s0

0,{s0}), where s00 /∈Q,

Qrev = Q∪{s00}, and ∆rev : Qrev×Σ→ 2Qrev

is defined as follows:

∆rev(p,a) = {q | p ∈ ∆(q,a)},

for all p ∈ Q and a ∈ Σ, and

∆rev(s0

0,a) = {q | ∃(p | p ∈ F : p ∈ ∆(q,a)},

for all a ∈ Σ.

9.[10] Give regular expressions for each of the following subsets of {a,b}∗:

(b)[5] {x | x contains an odd number of b’s}.

a∗(ba∗ba∗)∗ba∗

(c)[5] {x | x contains an even number of a’s or an odd number of b’s}.

b∗(ab∗ab∗)∗+a∗(ba∗ba∗)∗ba∗

10.[15] Give deterministic finite automata accepting the sets of strings matching thefollowing regular expressions.

(a)[10] (000∗+111∗)∗

It helps to decipher this expression. Let A = φ((000∗+ 111∗)∗), where φ is inter-pretation. Then x∈ A if and only if x = ε or x = x1 . . .xn such that each xi either belongsto 000∗ or to 111∗ (more precisely either to φ(000∗) or φ(111∗), but we often identifya regular expression with the language it defines). In other words, each 0 must be fol-lowed by one ore more 0’s and each 1 must be followed by one ore more 1’s, whichleads to the following simple solution:

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(b)[5] (01+10)(01+10)(01+10)

11.[11] Show that the following sets are not regular.

(a)[5] {anbm | n = 2m}

Recall the Pumping Lemma:

Lemma 1 (Most popular version)If A is a regular set, then there is a number p such that for every s ∈ A, if |s| ≥ p, thens = xyz and

1. for each i≥ 0, xyiz ∈ A

2. |y|> 0

3. |xy| ≤ p

We will use this lemma to show that A = {anbm | n = 2m} is not regular.

• We take p be the number from pumping lemma and consider s = apb2p. Hence|s|= 3p > p.

• Let x,y,z be such that s = xyz,|y|> 0, i.e. y 6= ε,and |xy| ≤ p.

• This means that y = ak where 1≤ k ≤ p.

• Now we have xy0z = ap−kb2p, and p− k 6= p, so xy0z /∈ A!

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• Hence the Pumping Lemma is not satisfied, i.e. A is not regular.

(d)[6] The set PAREN of balanced strings of parentheses ( ).

Recall that if A and B are regular than A∩B is regular too.The set B = (∗)∗ is clearly regular (it is just a∗b∗ with a = ( and b =)).Hence if PAREN is regular then PAREN∩B is also regular.But PAREN∩B = {(n)n | n > 0}, and the set {(n)n | n > 0} is NOT regular (it is

just {anbn | n > 0} with a = ( and b =)).Therefor PAREN is not regular.

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