CpSc 421 Homework 1 Solutions
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1 CpSc 421 Homework 1 Solutions 1. (15 points) Let Σ = {a, b, c}. Figure 7 depicts two finite state machines that read Let L a and L b denote the languages recognized by DFA (a) and DFA (b) respectively. (a) (6 points) For each of L a and L b list three strings in Σ that are in the language and three strings in Σ that are not in the language. Solution: {abc, aabbcc, aaaabccc} L a {ɛ, acabc, c} L a = {ɛ, aab, aaacaab} L b {aaab, aaacc, a 7 } L b = (b) (3 points) Write a short, English description of language, L a. Solution: Language L a is the set of all strings starting with one or more a followed by one or more b followed by one or more c. (c) (6 points) Write a short, English description of language, L b. Solution: Language L b is the set of all strings s such that (1) the #a(s) 2#b(s) 3#c(s) = 0; (2) For all prefixes x of s, 0 #a(x) 2#b(x) 3#c(x) (15 points) (inspired by Sipser exercise 1.6 (p. 84): Give state diagrams of DFAs recognizing the following languages. For each language, the alphabet is {0, 1}. (a) {w w begins and ends with the same symbol}. (b) {w w contains three consecutive 1s}. (c) {w w contains neither the substring 010 nor the substring 101}. Solution: See below. 3. (20 points) (inspired by Sipser exercise 1.7 (p. 84): Give state diagrams of NFAs with the specified number of states recognizing each of the following languages. For each language, the alphabet is Σ with Σ = {0, 1}. Solution: See below. (a) The set of strings w that end with the substring 010: {w x Σ. w = x010} Use four states. (b) The set of strings w that contain the substring 010: {w x, y Σ. w = x010y} Use four states. (c) The set of strings w that can be written as x 1 x 2 x k for some k 0, with each x i is an element of {01, 10, 001, 0011} Use eleven states. (d) The set of strings whose length is a multiple of three plus a multiple of five: {w m, n N. w = 3m + 5n} Use eight states.
2 Figure 1: Solution for question 2a Figure 2: Solution for question 2b
3 Figure 3: Solution for question 2c Figure 4: Solution for question 3a Figure 5: Solution for question 3b Figure 6: Solution for question 3c
4 Figure 7: Solution for question 3d 4. (20 points): Let Σ = {0, 1}. Let L k be the set of all strings whose k th from end symbol is a 1: L k = {w Σ x Σ, y Σ k 1. w = x1y} In the September 12 notes, we claimed that any DFA that recognizes L k must have at least 2 k states. Prove this claim. Solution: Suppose the claim does not hold, i.e. there exists a DFA A k = (Q k, Σ, δ, q 0, F ) recognizing L k with Q k < 2 k. Since there are 2 k strings of length k over Σ = {0, 1}, there exists two strings s 1, s 2 Σ k such that s 1 s 2 and A k is in the same state after processing both s 1 and s 2. Without loss of generarilty, assume that s 1 and s 2 differ in symbol i, i.e. s i 1 = 0 and s i 2 = 1 with 0 i k 1. Then A k is in the same state after processing strings s 1 0 i and s 2 0 i, but s 1 0 i / L k and s 2 0 i L k, a contradiction. Therefore, Q k 2 k. 5. (20 points, extra credit): In this exercise, you will prove the equivalence of NFA acceptances as defined in Sipser and the formulation that I gave in class. This problem statement is long because I first summarize both Sipser s and my definitions of acceptance. If you are comfortable with both, you can skip to the end of the problem statement where I ask you to prove the equivalence of the two formulations. Let N mrg = (Q, Σ,, q 0, F ) be an NFA as defined in the September 12 notes. Q is a finite set of states; Σ is a finite alphabet; Q Σ ɛ Q is the transition relation; q 0 Q is the initial state; and F Q is the set of accepting states. where Σ ɛ = Σ {ɛ}. Here are the formulas that we used to define NFA acceptance (see the slides for explanations). close ɛ (q) is the subset of Q where p close ɛ (q) iff p = q q close ɛ (q). (q, ɛ, p) We extended close ɛ to sets as close ɛ (G) = close ɛ (q) q G
5 Then we defined: step(q, c) = close ɛ ({q (q, c, q ) }), c Σ step(g, c) = step(q, c), G Q, c Σ q G (G, ɛ) = close ɛ (G), G Q (G, x c) = step( (G, x), c), G Q, x Σ, c Σ Finally, we said that N mrg accepts s iff ({q 0 }, s) F Now, for Sipser s version. Given N mrg = (Q, Σ,, q 0, F ) as described above, let N ms = (Q, Σ, δ, q 0, F ) with δ : Q Σ ɛ 2 Q where 2 Q is the power set of Q and δ(q, c) = {q (q, c, q ) } Sipser says that NFA N ms accepts string s iff we can find y 1, y 2,..., y m Σ ɛ and r 0, r 2,... r m Q such that s = y 1 y 2 y m ; r 0 = q 0 ; for i in 0... m 1, r i+1 δ(r i, y i + 1); r m F. Prove that N ms accepts a string (i.e. acceptance by Sipser s condition), iff N mrg accepts the string (i.e. acceptance by the condition given in class).
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