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Chapter 0: Q22P (page 1)
Show that A is Turing-recognizable
It is proved that A is Turing Recognizable.
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Question:Consider the algorithm MINIMIZE, which takes a DFA as input and outputs DFA .
MINIMIZE = 鈥淥n input , where is a DFA:
1.Remove all states of G that are unreachable from the start state.
2. Construct the following undirected graph G whose nodes are the states of .
3. Place an edge in G connecting every accept state with every non accept state. Add additional edges as follows.
4. Repeat until no new edges are added to G :
5. For every pair of distinct states q and r of and every :
6. Add the edge (q,r) to G if is an edge of G .
7. For each state be the collection of edge joins q and r in G }.
8.Form a new DFA where
9. Output ( M')鈥�
a. Show that M and M' are equivalent.
b. Show that M0 is minimal鈥攖hat is, no DFA with fewer states recognizes the same language. You may use the result of Problem 1.52 without proof.
c. Show that MINIMIZE operates in polynomial time.
Let is a satisfiable cnf-formula where each variable appears in at most k places}.
a. Show that .
b. Show that-complete.
Show that
Myhill鈥揘erode theorem. Refer to Problem . Let L be a language and let X be a set of strings. Say that X is pairwise distinguishable by L if every two distinct strings in X are distinguishable by L. Define the index of L to be the maximum number of elements in any set that is pair wise distinguishable by L . The index of L may be finite or infinite.
a. Show that if L is recognized by a DFA with k states, L has index at most k.
b. Show that if the index of L is a finite number K , it is recognized by a DFA with k states.
c. Conclude that L is regular iff it has finite index. Moreover, its index is the size of the smallest DFA recognizing it.
Question: Describe the error in the following 鈥減roof鈥� that is not a regular language. (An error must exist because is regular.) The proof is by contradiction. Assume that is regular. Let p be the pumping length for localid="1662103472623" given by the pumping lemma. Choose s to be the string 0p1p . You know that s is a member of 0*1*, but Example 1.73 shows that s cannot be pumped. Thus you have a contradiction. So is not regular.
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