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| **7.1.1.** (Solved during lecture) Define the reversal of a language $ L $ as $ rev(L) = \{ w \in \Sigma^* | rev(w) \in L \}$, where $ rev(c_1c_2 \dots c_n) = c_nc_{n - 1} \dots c_1$ , with $ c_i \in \Sigma, 1 \leq i \leq n $. Show that reversal is a closure property. | **7.1.1.** (Solved during lecture) Define the reversal of a language $ L $ as $ rev(L) = \{ w \in \Sigma^* | rev(w) \in L \}$, where $ rev(c_1c_2 \dots c_n) = c_nc_{n - 1} \dots c_1$ , with $ c_i \in \Sigma, 1 \leq i \leq n $. Show that reversal is a closure property. | ||
| + | **7.1.2.** Suppose $math[A] is a DFA. Write another DFA which accepts only those words $math[w] such that both $math[w \in L(A)] and $math[reverse(w) \in L(A)]. | ||
| - | **7.1.2.** Identify **different strategies** to verify that two regular expressions generate the same language. State them as algorithmic procedures. | + | <hidden Solution 7.1.2.> |
| - | + | ||
| - | <hidden> All strategies require building DFAs from the regexes. | + | |
| - | - You can use an existing minimisation algorithm to find the minimal DFAs. Then, label each state from each DFA from 0 to |K|. Fix an ordering of the symbols of the alphabet. Sort the transitions by symbol. Make a textual representation of each DFA which includes the number of states and the sorted transition function. If the textual representations of the two DFAs are identical then they accept the same language. | + | |
| - | - Use the indistinguishability algorithm to check if the two initial states of the two DFAs are indistinguishable. | + | |
| - | - Check $math[L(A_1) \subseteq L(A_2)] and $math[L(A_2) \subseteq L(A_1)]. The condition $math[L(A) \subseteq L(B)] can be restated as $math[L(A) \cap complement(L(B)) = \emptyset]. Build the complement DFA for B, and use the product construction with A. Check if the resulting DFA has any final state accessible from the initial one. | + | |
| - | </hidden> | + | |
| - | + | ||
| - | + | ||
| - | **7.1.2.** Write a regex for the complement of $math[L((10 \cup 0)^*(1 \cup \epsilon))]. | + | |
| - | + | ||
| - | + | ||
| - | <hidden> | + | |
| - | Build the DFA for that regex and then build the complement of that DFA. Transform it into a regex. | + | |
| - | </hidden> | + | |
| - | + | ||
| - | + | ||
| - | **7.1.3.** Suppose $math[A] is a DFA. Write another DFA which accepts only those words $math[w] such that both $math[w \in L(A)] and $math[reverse(w) \in L(A)]. | + | |
| - | + | ||
| - | + | ||
| - | <hidden> | + | |
| Build the DFA $math[A^R] which accepts $math[reverse(L(A))]. Build the product construction between $math[A] and $math[A^R]. | Build the DFA $math[A^R] which accepts $math[reverse(L(A))]. Build the product construction between $math[A] and $math[A^R]. | ||
| </hidden> | </hidden> | ||
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| - | <hidden> | + | <hidden Solution 7.2.1.> |
| $math[L/a = L((aaa \cup ba)^*(aa \cup b))] | $math[L/a = L((aaa \cup ba)^*(aa \cup b))] | ||
| </hidden> | </hidden> | ||
| Line 44: | Line 25: | ||
| - | <hidden> | + | <hidden Solution 7.2.2.> |
| $math[L/a = L(a^*)] | $math[L/a = L(a^*)] | ||
| </hidden> | </hidden> | ||
| Line 52: | Line 33: | ||
| - | <hidden> | + | <hidden Solution 7.2.3.> |
| Let $math[A] be a DFA which accepts $math[L]. We build a new DFA $math[A/c] which accepts $math[L/c]. This new DFA has the same number of states, initial state, same transitions and same alphabet. The set of final states is different though: $math[F/c = \{q' \in K | \exists q \in F. \delta(q',c) = q\}]. In other words, the final states of $math[A/c] are those states which are $math[c-]predecessors of some final state of $math[A]. Note that some $math[q'] may be equal to $math[q], and it may also be the case that no such predecessors exist. | Let $math[A] be a DFA which accepts $math[L]. We build a new DFA $math[A/c] which accepts $math[L/c]. This new DFA has the same number of states, initial state, same transitions and same alphabet. The set of final states is different though: $math[F/c = \{q' \in K | \exists q \in F. \delta(q',c) = q\}]. In other words, the final states of $math[A/c] are those states which are $math[c-]predecessors of some final state of $math[A]. Note that some $math[q'] may be equal to $math[q], and it may also be the case that no such predecessors exist. | ||
| </hidden> | </hidden> | ||
| Line 60: | Line 41: | ||
| - | <hidden> | + | <hidden Solution 7.2.4.> |
| The easiest way is to rely on other closure properties: $math[c/L = reverse(reverse(L)/c)] | The easiest way is to rely on other closure properties: $math[c/L = reverse(reverse(L)/c)] | ||
| </hidden> | </hidden> | ||
| Line 70: | Line 51: | ||
| - | <hidden> | + | <hidden Solution 7.3.1.> |
| Suppose $math[A] is the accepting DFA of L. We build a new NFA from A by adding $math[\epsilon-]transitions from the initial state to all other states of A (obs: if the initial state of A had transitions that entered it, then, in the new NFA, create another initial state that has an $math[\epsilon-]transition to the former initial state). Thus, we can nondeterministically //skip// any prefix $math[x] of any accepted word from $math[L]. This new DFA accepts $math[\text{suffix(L)}] | Suppose $math[A] is the accepting DFA of L. We build a new NFA from A by adding $math[\epsilon-]transitions from the initial state to all other states of A (obs: if the initial state of A had transitions that entered it, then, in the new NFA, create another initial state that has an $math[\epsilon-]transition to the former initial state). Thus, we can nondeterministically //skip// any prefix $math[x] of any accepted word from $math[L]. This new DFA accepts $math[\text{suffix(L)}] | ||
| </hidden> | </hidden> | ||
| Line 83: | Line 64: | ||
| - | <hidden> | + | <hidden Solution 7.4.1.> |
| $math[min(L)] is the set of words from $math[L] that have no proper ($math[\neq \epsilon]) prefix in $math[L]. | $math[min(L)] is the set of words from $math[L] that have no proper ($math[\neq \epsilon]) prefix in $math[L]. | ||
| </hidden> | </hidden> | ||
| Line 91: | Line 72: | ||
| - | <hidden> | + | <hidden Solution 7.4.2.> |
| $math[min(L(a^*)) = \{\epsilon\}] | $math[min(L(a^*)) = \{\epsilon\}] | ||
| </hidden> | </hidden> | ||
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| - | <hidden> | + | <hidden Solution 7.4.3.> |
| No word from $math[L(a^*b)] has a prefix in this same language, hence the answer is $math[L(a^*b)]. | No word from $math[L(a^*b)] has a prefix in this same language, hence the answer is $math[L(a^*b)]. | ||
| </hidden> | </hidden> | ||
| Line 107: | Line 88: | ||
| - | <hidden> | + | <hidden Solution 7.4.4.> |
| Let $math[A] be the accepting DFA of L. If a word $math[w=xy \in L] has a proper prefix $math[x] in $math[L], then $math[(q_0,xy) \vdash^* (q_1,x) \vdash^* (q_2,\epsilon)] where $math[q_1,q_2] are final states. We need to //break-off// those paths leading from one final state (such as $math[q_1]) to another. | Let $math[A] be the accepting DFA of L. If a word $math[w=xy \in L] has a proper prefix $math[x] in $math[L], then $math[(q_0,xy) \vdash^* (q_1,x) \vdash^* (q_2,\epsilon)] where $math[q_1,q_2] are final states. We need to //break-off// those paths leading from one final state (such as $math[q_1]) to another. | ||
| Line 118: | Line 99: | ||
| - | <hidden> | + | <hidden Solution 7.4.5.> |
| The language $math[R] contains all those words formed by taking any suffix $math[x] of a word from $math[L] an appending its respective prefix $math[y] to it. | The language $math[R] contains all those words formed by taking any suffix $math[x] of a word from $math[L] an appending its respective prefix $math[y] to it. | ||
| We start from the accepting DFA $math[A] of $math[L]. We create a new initial state and a new final one. All states of A are no longer initial or final in our construction. We create $math[n] copies of $math[A], denoted as $math[A_q], one for each state $math[q\in K] of $math[A]. | We start from the accepting DFA $math[A] of $math[L]. We create a new initial state and a new final one. All states of A are no longer initial or final in our construction. We create $math[n] copies of $math[A], denoted as $math[A_q], one for each state $math[q\in K] of $math[A]. | ||
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| ===== Extras ===== | ===== Extras ===== | ||
| - | **7.5.1.** Invent a language transformation or an operation between languages. See if you can show that it is a closure property. | + | **7.5.1.** Identify **different strategies** to verify that two regular expressions generate the same language. State them as algorithmic procedures. |
| + | |||
| + | <hidden Solution 7.5.1.> | ||
| + | All strategies require building DFAs from the regexes. | ||
| + | - You can use an existing minimisation algorithm to find the minimal DFAs. Then, label each state from each DFA from 0 to |K|. Fix an ordering of the symbols of the alphabet. Sort the transitions by symbol. Make a textual representation of each DFA which includes the number of states and the sorted transition function. If the textual representations of the two DFAs are identical then they accept the same language. | ||
| + | - Use the indistinguishability algorithm to check if the two initial states of the two DFAs are indistinguishable. | ||
| + | - Check $math[L(A_1) \subseteq L(A_2)] and $math[L(A_2) \subseteq L(A_1)]. The condition $math[L(A) \subseteq L(B)] can be restated as $math[L(A) \cap complement(L(B)) = \emptyset]. Build the complement DFA for B, and use the product construction with A. Check if the resulting DFA has any final state accessible from the initial one. | ||
| + | </hidden> | ||
| + | |||
| + | |||
| + | **7.5.2.** Invent a language transformation or an operation between languages. See if you can show that it is a closure property. | ||