CSE 555 Homework Three Sample Solutions

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1 CSE 555 Homework Three Sample Solutions Question 1 Let Σ be a fixed alphabet. A PDA M with n states and stack alphabet of size σ may recognize the empty language; if its language is non-empty then denote by µ(m) the length of the shortest string that it accepts. Now define m(n, σ) = max{µ(m) : M is a PDA with n states and stack alphabet of size σ having L(M) }. Prove or disprove: There is an algorithm that, given n and σ, finds a PDA M with n states and stack alphabet of size σ having µ(m) = m(n, σ). Solution: We prove it. Note that E P DA is decidable, and there are only finitely many PDAs that have n states and stack alphabet of size σ: D = On input n, σ: 1. Let L be the list of all PDAs with n states and σ characters in the stack alphabet. 2. ρ M null. 4. For every PDA P L: (a) Run E P DA on input P (b) If E P DA accepts, proceed to the next PDA. (c) Otherwise, enumerate strings s in lexicographic order (in terms of the PDA s input alphabet): i. Run A P DA on P, s. ii. If A P DA rejects, continue on to the next string. iii. Else (A P DA accepts) A. if ρ < s, set M P, ρ s, B. go back to start on the next PDA. 5. Output M. Question 2 Let Σ be a fixed alphabet. A TM M with n states and tape alphabet of size σ may recognize the empty language; if its language is non-empty then denote by µ(m) the length of the shortest string that it accepts. Now define m(n, σ) = max{µ(m) : M is a TM with n states and tape alphabet of size σ having L(M) }. Prove or disprove: There is an algorithm that, given n and σ, finds a TM M with n states and tape alphabet of size σ having µ(m) = m(n, σ). Solution: We disprove it. We can just make another TM, using the recursion theorem, that is guaranteed to have a larger value of m(n, σ) than the current TM. D = On input w: 1. Obtain own description, D, by the recursion theorem. 2. Let n be the number of states in D, σ the number of tape alphabet characters. 3. Calculate N, a TM with n states and tape alphabet size σ for which the shortest string accepted has length m(n, σ). 4. If w is even, reject. 5. Write w = xay with x = y and a = 1; if x y reject. 6. Simulate N on x and if it accepts, accept. (If it rejects or runs forever, we will not accept.) Then the length of the shortest string accepted by D is 2m(n, σ) + 1, a contradiction. Question 3 We know that T h(n, +) is decidable but T h(n, +, ) is undecidable. We want to handle a limited form of multiplication in a decidable theory. To do this, fix a positive integer value n, and permit multiplication by integers k {1,..., n}. Specifically, let k be the relation {(x, kx) : x N}. Show that T h(n, +, 1,..., n ) is decidable. (Hint: Show how 1

2 to define k using + for every k.) Solution sketch: define this relation as M k (a, b) (i.e., k a = b). Note that M 2 (a, b) = P LUS(a, a, b). Now recursively define M k : M k (a, b) = zp LUS(a, z, b) M k 1 (a, z), where z is a new variable not defined previously. By an induction argument, once we defined M k 1 in terms of P LUS, we can define M k in terms of P LUS. Therefore, T h(n, +, 1,..., n ) is decidable because we can make a DFA for the entire formula ψ, because we know how to make a DFA for P LUS. Now we have a formula in T h(n, +). Therefore, T h(n, +, 1,..., n ) is decidable. Question 4a Using your method for question 3 and the algorithm for deciding T h(n, +), determine the truth or falsity of each of the following statements. Show all steps (first write each as an equivalent statement in T h(n, +) using your method from question 3; you may have to introduce new variables and quantifiers to do this). a b[2a + 2b = 2(a + b)]. Solution: Rewrite the formula using substitutions using our method above: which is the same as: a b x 1 x 2 s s 2 [M 2 (a, x 1 ) M 2 (b, x 2 ) P LUS(a, b, s) M 2 (s, s 2 ) P LUS(x 1, x 2, s 2 )] a b x 1 x 2 s s 2 [P LUS(a, a, x 1 ) P LUS(b, b, x 2 ) P LUS(a, b, s) P LUS(s, s, s 2 ) P LUS(x 1, x 2, s 2 )] The representation of the binary vectors has the form: a b x 1 x 2 s s 2 with each of the 6 entries in {0, 1}. For simplicity, we will interpret a binary number as abx 1 x 2 ss 2, with a being the most significant bit, and s 2 the least significant. For example, if the string is , then a = 1, b = 0,, s 2 = 1. Let the DFAs in the description for each P LUS relation be named as D 1,, D 5. We won t show all the steps in creating each of the DFAs (since there are 192 transitions for each); however, we produce the DFA for L(D 1 ) L(D 5 ) in Tables 1 and 2 (after performing DFA minimization). The tables were produced by the usual DFA intersection algorithm with minimization, written in Python, and not by hand. An entry of indicates a dead state, and all other entries correspond to the state name. The left column corresponds to the length-6 binary vector described above. The DFA now, as described in the T h(n, +) algorithm, is A 6. Now we handle s 2 to get A 5. Since there are no transitions that lead to multiple non-dead states at the same time from any state, we can just collapse rows, and add a new start state with ɛ-transitions to states 0 and. Note that since we a dead state never has transitions out of itself, we can skip the step of having a new start state, and just use the existing NFA from collapsing the rows. But as we can see, we have no nondeterminism. Therefore, this is still a DFA, as is shown in Table 3. We can perform basic DFA minimization by removing states 4, 21, and 121 (since they are equivalent to 0, 19, and 117, respectively). See Table 4 for details: an xxx entry references one of the removed states, and all other entries are reflected to refer to the other equivalent state for each. Now we handle s to get A 4. We add a new state with ɛ-transitions to states 0 and 19, and the resulting NFA is shown in Table 5, and the minimized equivalent DFA is shown in Table 6. Now we handle x 2 to get A 3, as before. The resulting minimized DFA is shown in Table 7. Now we handle x 1 to get A 2. The resulting minimized DFA is shown in Table 8. Now we handle b to get A 1. We first complement the DFA in Table 8, apply as before, then complement back. The resulting minimized DFA is shown in Table 9. Now we handle a to get A 0. Carrying through the steps has A 0 be a single state that is accepting. Therefore, since A 0 accepts ɛ, this this statement is true and proved. Question 4b Using your method for question 3 and the algorithm for deciding T h(n, +), determine the truth or falsity of each of the following statements. Show all steps (first write each as an equivalent statement in T h(n, +) using your method from question 3; you may have to introduce new variables and quantifiers to do this). c a b[2a + 3b = c]. Solution: we can rewrite the formula using substitutions using our method above: 2

3 Table 1: Question 3a: Intersection DFA (64-character alphabet), Part

4 Table 2: Question 3a: Intersection DFA (64-character alphabet), Part

5 Table 3: Question 3a: Intersection DFA (32-character alphabet)

6 Table 4: Question 3a: Intersection DFA (32-character alphabet), Minimized 0 xxx 19 xxx xxx xxx 19 xxx xxx 19 xxx xxx xxx xxx 45 xxx xxx 32 xxx xxx xxx xxx 91 xxx xxx 85 xxx xxx xxx xxx 128 xxx xxx 117 xxx xxx xxx 117 Table 5: Question 3a: 16-character alphabet NFA 0 xxx 19 xxx xxx 128 s ɛ 0, {0,19} xxx {19} xxx 0001 {0,19} {0,19} {19} 0010 {0,19} {0,19} 0011 {0,19} xxx {19} 0100 {32,45} xxx {32,45} xxx 0101 {32,45} {32,37} {32,45} 0110 {32,45} {32,37} 0111 {32,37} xxx {32,45} 1000 {85,91} xxx {85,91} xxx 1001 {85,91} {85,91} {85,91} 1010 {85,91} {85,91} 1011 {85,91} xxx {85,91} 1100 {117} xxx {117,128} xxx 1101 {117} {117,128} {117,128} 1110 {117} {117,128} 1111 {117,128} xxx {117,128} 6

7 Table 6: Question 3a: 16-character alphabet DFA, Minimized Table 7: Question 3a: 8-character alphabet DFA, Minimized. The only final state is 1, and is also the start state Table 8: Question 3a: 4-character alphabet DFA, Minimized. The only final state is 0, and is also the start state Table 9: Question 3a: 2-character alphabet DFA, Minimized. The only final state is 0, and is also the start state

8 c a b a 1 b 1 [M 2 (a, a 1 ) M 3 (b, b 1 ) P LUS(a 1, b 1, c)] which is the same as: c a b a 1 b 1 z[p LUS(a, a, a 1 ) P LUS(b, z, b 1 ) P LUS(b, b, z) P LUS(a 1, b 1, c)] We won t show all the details, but the process is nearly identical to the (a) part, and turns out to be false. The proof requires constructing the DFAs and working with the quantifiers, but intuitively, it is false by setting c = 1; there do not exist a, b such that 2a + 3b = c. Question 5 MIN A TM TM = { M : M is a TM with an oracle for A TM for which the size of M is no larger than the size of M when M is a TM with an oracle for A TM for which L(M) = L(M )}. Prove that MIN A TM TM Solution: Suppose MIN A TM is not Turing recognizable. TM were recognizable; therefore, it is enumerable by an enumerator E. Construct the following TM which has an oracle for A TM : C A TM = On input w: 1. Obtain own description, C A TM, as given by the Oracle recursion theorem. 2. Run E until a TM with an oracle for A TM, D A TM, with longer description than C A TM, is found. 3. Run D A TM on w. The argument for contradiction is the same as for MIN TM, except we didn t establish the recursion theorem for oracle TMs. A solution to this is in the HW #3 solutions of Spring 2014 s offering of CSE555: ccolbou/src/555hw3s14sol.pdf 8