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Let $\mathit{U}\mathbf{=}\mathbf{\backslash }\mathit{l}\mathit{e}\mathit{f}\mathit{t}\mathbf{\backslash }\mathbf{\{}\mathit{l}\mathit{e}\mathit{f}\mathit{t}\mathbf{\backslash }\mathit{l}\mathit{a}\mathit{n}\mathit{g}\mathit{l}\mathit{e}\mathbf{}\mathit{M}\mathbf{,}\mathit{x}\mathbf{,}\mathbf{\backslash }\mathbf{\#}\mathbf{^}\mathbf{\{}\mathit{t}\mathbf{\}}\mathbf{\backslash }\mathit{r}\mathit{i}\mathit{g}\mathit{h}\mathit{t}\mathbf{\backslash }\mathit{r}\mathit{a}\mathit{n}\mathit{g}\mathit{l}\mathit{e}\mathbf{\backslash }\mathit{m}\mathit{i}\mathit{d}\mathbf{\backslash }\mathit{r}\mathit{i}\mathit{g}\mathit{h}\mathit{t}$$\mathit{N}\mathit{T}\mathit{M}\mathbf{}\mathit{M}$accepts $\mathit{x}$within steps on at least one branch}. Note thatlocalid="1663241578125" $\mathit{M}$isn鈥檛 required to halt on all branches. Show that$\mathit{U}$is$\mathit{N}\mathit{P}$-complete.

Show that if $\mathit{P}\mathbf{=}\mathit{N}\mathit{P}$, a polynomial time algorithm exists that produces a satisfying assignment when given a satisfiable Boolean formula. (Note: The algorithm you are asked to provide computes a function; but $\mathit{N}\mathit{P}$contains languages, not functions. The $\mathit{P}\mathbf{=}\mathit{N}\mathit{P}$assumption implies that $\mathit{S}\mathit{A}\mathit{T}$is in $\mathit{P}$, so testing satisfiability is solvable in polynomial time. But the assumption doesn鈥檛 say how this test is done, and the test may not reveal satisfying assignments. You must show that you can find them anyway. Hint: Use the satisfiability tester repeatedly to find the assignment bit-by-bit.)

Show that if $\mathit{P}\mathbf{=}\mathit{N}\mathit{P}$ , you can factor integers in polynomial time. (See the note in Problem 7.38.)

Show that if $A鈮�{}_{T}B$and $B鈮�{}_{T}C$, then$A鈮�{}_{T}C$ .

Which of the following pairs of numbers are relatively prime? Show the calculations that led to your conclusions

$$\begin{gathered} \mathbf{a}.\mathbf{1274}\mathbf{and}\mathbf{10505} \\ \mathbf{b}\mathbf{.}\mathbf{7289}\mathbf{and}\mathbf{8029} \end{gathered}$$

LetAbe the set$\mathbf{\{}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{\}}$and$\mathbf{B}$be the set$\mathbf{\{}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\}}$.

1. IsAa subset ofB?
2. IsBa subset ofA?
3. What is$\mathbf{A}\mathbf{鈭�}\mathbf{B}$?
4. What is$\mathbf{A}\mathbf{鈭�}\mathbf{B}$?
5. What is$\mathbf{A}\mathbf{脳}\mathbf{B}$?
6. What is the power set ofB ?

Modify the proof of Theorem 3.16 to obtain Corollary 3.19, showing that a language is decidable if some nondeterministic Turing machine decides it. (You may assume the following theorem about trees. If every node in a tree has finitely many children and every branch of the tree has finitely many nodes, the tree itself has finitely many nodes.)

Question:Consider the algorithm MINIMIZE, which takes a DFA as input and outputs DFA .

MINIMIZE = 鈥淥n input , where $\mathbf{\text{M=}}\left(Q,危,未,q0,A\right)$ 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 $\mathbf{a}\mathbf{鈭�}\mathbf{危}$ :

6. Add the edge (q,r) to G if $\left(\mathbf{未}\left(\mathbf{q},\mathbf{a}\right),\mathbf{未}\left(\mathbf{r},\mathbf{a}\right)\right)$ is an edge of G .

7. For each state $\mathbf{q}\mathbf{,}\mathbf{let}\left[q\right]$ be the collection of $\mathbf{states}\left[\mathbf{q}\right]=\{\mathbf{r}鈭�\mathbf{Q}|\mathbf{no}$edge joins q and r in G }.

8.Form a new DFA $\mathbf{M}\text{'}=\left(\mathbf{Q}\text{'},\mathbf{危},\mathbf{未}\text{'},\mathbf{q}{\text{'}}_0,\mathbf{A}\text{'}\right)$where

$$\begin{gathered} \mathbf{\text{Q'={}}\left[q\right]\mathbf{|}\mathit{q}\mathbf{鈭�}\mathit{Q}\mathbf{\}}\left(if\left[q\right]=\left[r\right],onlyoneofthemisinQ\text{'}\right)\mathbf{,}\mathit{未}\mathbf{\text{'}}\left(\left[q\right],a\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\left[未\left(q,a\right)\right]\mathit{f}\mathit{o}\mathit{r}\mathit{e}\mathit{v}\mathit{e}\mathit{r}\mathit{y}\mathit{q}\mathbf{鈭�}\mathit{Q}\mathit{a}\mathit{n}\mathit{d}\mathit{a}\mathbf{鈭�}\mathit{危}\mathbf{,}\mathit{q}\mathbf{00}\mathbf{=}\left[q0\right]\mathbf{,}\mathit{a}\mathit{n}\mathit{d}\mathit{A}\mathbf{0}\mathbf{=}\mathbf{\left\{}\left[q\right]\mathbf{\right|}\mathit{q}\mathbf{鈭�}\mathit{A}\mathbf{\}} \end{gathered}$$

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 $\underset{}{鈭�}=\left\{0,1\right\}$. Let B be the collection of strings that contain at least one 1 in their second half. In other words,

a. Give a PDA that recognizes B

b. Give a CFG that generates B .

Let. $\mathbf{鈭�}\mathbf{=}\mathbf{\{}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{\}}$Let${\mathit{C}}_{\mathbf{1}}$ be the language of all strings that contain a 1 in their middle third.

Let ${\mathit{C}}_{\mathbf{2}}$be the language of all strings that contain two 1s in their middle third. So ${\mathit{C}}_{\mathbf{1}}\mathbf{=}\mathbf{\left\{}\mathbf{x}\mathbf{y}\mathbf{z}\mathbf{|}\mathbf{x}\mathbf{,}\mathbf{z}\mathbf{鈭�}\mathbf{鈭�}\mathbf{*}\mathbf{\text{andy}}\mathbf{鈭�}\mathbf{鈭�}\mathbf{\text{*1}}\mathbf{鈭�}\mathbf{\text{*,where}}\mathbf{\left|}\mathbf{x}\mathbf{\right|}\mathbf{=}\mathbf{\left|}\mathbf{z}\mathbf{\right|}\mathbf{鈮�}\mathbf{\left|}\mathbf{y}\mathbf{\right|}\mathbf{\right\}}$and ${\mathit{C}}_{\mathbf{2}}\mathbf{=}\mathbf{\left\{}\mathbf{x}\mathbf{y}\mathbf{z}\mathbf{|}\mathbf{x}\mathbf{,}\mathbf{z}\mathbf{鈭�}\mathbf{鈭�}\mathbf{*}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{\text{鈥�}}\mathbf{y}\mathbf{鈭�}\mathbf{鈭�}\mathbf{*}\mathbf{1}\mathbf{鈭�}\mathbf{*}\mathbf{1}\mathbf{鈭�}\mathbf{*}\mathbf{,}\mathbf{\text{鈥娾赌�}}\mathbf{w}\mathbf{h}\mathbf{e}\mathbf{r}\mathbf{e}\mathbf{\text{鈥�}}\mathbf{\left|}\mathbf{x}\mathbf{\right|}\mathbf{=}\mathbf{\left|}\mathbf{z}\mathbf{\right|}\mathbf{鈮�}\mathbf{\left|}\mathbf{y}\mathbf{\right|}\mathbf{\right\}}$.

b. Show that${\mathit{C}}_{\mathbf{2}}$ is not a CFL