Saturday
CSCI 5100
Amber Huffman, a Principal Engineer at Google, has spent her career defining standards for the hardware industry. Now with the rapid growth of AI the demand and uses for Open Source Hardware have never been higher. Learn what Open Source Hardware is and why it is important to all of us.
A 5-tuple \(M = (Q, \Sigma, \delta, q_0, F)\):
\[ \forall L(M_1), L(M_2): \exists M_3 : L(M_3) = L(M_1) \cup L(M_2) \]
Proof
\[ \begin{aligned} L(&Q_1, \Sigma, \delta_1, q_1, F_1) \cup L(Q_2, \Sigma, \delta_2, q_2, F_2) = \\ L(&Q_1 \times Q_2, \Sigma, \\ &\delta((q,q')a),\rightarrow (\delta_1(q,a), \delta_2(q',a)), \\ &(q_1, q_2), \\ &\{ (F_1 \times Q_2) \cup (Q_1 \times F_2) \})\blacksquare \end{aligned} \]
\[ \forall M_{NFA}: \exists M_{DFA} : L(M_{NFA})) = L(M_{DFA})) \]
Proof
\[ \begin{aligned} M_{NFA}(&Q, \Sigma, \delta, q_1, F_1) = \\ M_{DFA}(&\mathcal{P} (Q), \Sigma, \delta, \{q_1\}, \\ & \{ R \in \mathcal{P} (Q) | R \cap F \neq \varnothing \})\blacksquare \end{aligned} \]
\[ \forall L(M_1), L(M_2): \exists M_3 : L(M_3) = L(M_1)L(M_2) \]
Proof
\[ \begin{aligned} L(&Q_1, \Sigma, \delta_1, q_1, F_1)L(Q_2, \Sigma, \delta_2, q_2, F_2) = \\ L(&Q_1 \sqcup Q_2, \\ &\Sigma, \\ &\delta_1 \sqcup \delta_2 \cup \{ (f, \varnothing) \rightarrow \{q_2\} | f \in F_1 \} \\ &q_1, \\ &F_2)\blacksquare \end{aligned} \]
\[ \forall L(M_1): \exists M_3 : L(M_2) = L(M_1)* \]
Proof
\[ \begin{aligned} L(&Q_1, \Sigma, \delta_1, q_1, F_1)* = \\ L(&Q_1 \cup \{q_0\}, \\ &\Sigma, \\ &\delta_1 \cup \{ (f, \varnothing) \rightarrow \{q_1\} | f \in F_1 \cup \{q_0\} \} \\ &q_0, \\ &f \in F_1 \cup \{q_0\})\blacksquare \end{aligned} \]
\[ \begin{aligned} \exists M_1, M_2, M_3 :\\ L(M_1) &= a \in \Sigma \\ L(M_2) &= \Sigma^0 \\ L(M_2) &= \varnothing \end{aligned} \]
Proof
\[ \begin{aligned} M_1 &= (\{q_0,q_1\}, &\Sigma, &\{ (q_0, a) \rightarrow \{q_1\} \}, &q_0, &\{q_1\}) \\ M_2 &= (\{q_0\}, &\Sigma, &\{ (q_0, a) \rightarrow \{q_1\} \}, &q_0, &\{q_1\}) \\ M_3 &= (\{q_0\}, &\Sigma, &\varnothing, &q_0, &\{q_1\}) \\ \end{aligned} \]
\[ \forall R : \exists M : R = L(M) \]
Proof
\[ \forall |G| = 2 := (\{q_0,q_1\},\Sigma,\delta,q_0,\{q_1\}) : \exists R : R = L(G) \]
Proof
By definition, \(G = (\{q_0,q_1\},\Sigma,\delta,q_0,\{q_1\})\)
By definition, \(\delta = \{ (q_0, R') \rightarrow q_1 \}\)
Let \(R = R'\)
\(\blacksquare\)
\[ \forall |G| = (k > 2) : \exists |G| = (k - 1) : L(G) = L(G') \]
Proof
By definition, \(G = (Q,\Sigma,\delta,q_0,\{q_n\})\)
Select arbitrary \(q_i \in Q \setminus \{q_0, q_n\}\)
Take \(G' = (Q \setminus \{q_i\},\Sigma,\delta_{-i},q_0,\{q_n\})\)
\[ \forall M : \exists R : R = L(M) \]
Proof
Proof
\[ \begin{align*} G_2& \\ &\left. \begin{aligned} &E \rightarrow E+T \quad | \quad T\\ &T \rightarrow T\times F \quad | \quad F \\ &F \rightarrow ( E ) \quad | \quad a \end{aligned} \right. \end{align*} \]
\[ \begin{align*} G_3& \\ &\left. \begin{aligned} &E \rightarrow E+E \quad | \quad E \times E \quad | \quad ( E ) \quad | \quad a\\ \end{aligned} \right. \end{align*} \]