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\author{Filip Crljeni\'{c}}
\begin{document}
\begin{center}
Filip Crljeni\'{c}\\
Prvi zadatak 15 bodova.\\
\end{center}
Za jezik:
$$ L = \bigl\{ \tt{a}^m\tt{b}^n\tt{c}^p \big|\ m+n+p\equiv 1\pmod{3} \bigr\} $$
  pronadite sve reziduale te konstruirajte minimalni DKA koji ga prepoznaje. Konstruirajte automat koji prepoznaje $L^{c}$.
  
  Rje\v{s}enje:
  \\
  $$a^{-1}L=\bigl\{ \tt{a}^m\tt{b}^n\tt{c}^p \big|\ m+n+p\equiv 0\pmod{3} \bigr\}=L_{1}$$
  $$b^{-1}L=\bigl\{ \tt{b}^n\tt{c}^p \big|\ n+p\equiv 0\pmod{3} \bigr\}=L_{2}$$
  $$c^{-1}L=\bigl\{ \tt{c}^p \big|\ p\equiv 0\pmod{3} \bigr\}=L_{3}$$
  \\
  $$a^{-1}L_{1}=\bigl\{ \tt{a}^m\tt{b}^n\tt{c}^p \big|\ m+n+p\equiv 2\pmod{3} \bigr\}=L_{4}$$ 
	$$b^{-1}L_{1}=\bigl\{ \tt{b}^n\tt{c}^p \big|\ n+p\equiv 2\pmod{3} \bigr\}=L_{5}$$
	$$c^{-1}L_{1}=\bigl\{ \tt{c}^p \big|\ p\equiv 2\pmod{3} \bigr\}=L_{6}$$  	
	\\
	$$a^{-1}L_{2}=\emptyset$$
	$$b^{-1}L_{2}=\bigl\{ \tt{b}^n\tt{c}^p \big|\ n+p\equiv 2\pmod{3} \bigr\}=L_{5}$$
	$$c^{-1}L_{2}=\bigl\{ \tt{c}^p \big|\ p\equiv 2\pmod{3} \bigr\}=L_{6}$$  
	\\
	$$a^{-1}L_{3}=\emptyset$$
	$$b^{-1}L_{3}=\emptyset$$
	$$c^{-1}L_{3}=\bigl\{ \tt{c}^p \big|\ p\equiv 2\pmod{3} \bigr\}=L_{6}$$
	\\
	$$a^{-1}L_{4}=L$$
	$$b^{-1}L_{4}=\bigl\{ \tt{b}^n\tt{c}^p \big|\ n+p\equiv 1\pmod{3} \bigr\}=L_{7}$$
	$$c^{-1}L_{4}=\bigl\{ \tt{c}^p \big|\ p\equiv 1\pmod{3} \bigr\}=L_{8}$$
	\\
	$$a^{-1}L_{5}=\emptyset$$
	$$b^{-1}L_{5}=L_{7}$$
	$$c^{-1}L_{5}=L_{8}$$
	\\
	$$a^{-1}L_{6}=\emptyset$$
	$$b^{-1}L_{6}=\emptyset$$
	$$c^{-1}L_{6}=L_{8}$$
	\\
	$$a^{-1}L_{7}=\emptyset$$
	$$b^{-1}L_{7}=L_{2}$$
	$$c^{-1}L_{7}=L_{3}$$
	\\
	$$a^{-1}L_{8}=\emptyset$$
	$$b^{-1}L_{8}=\emptyset$$
	$$c^{-1}L_{8}=L_{3}$$
	\\
	Min DKA:
	\\
		\includegraphics[scale=0.75]{zad1.1}
	\\
	
	DKA za $L^{c}$:
	\\
	\includegraphics[scale=0.75]{zad1c.1}
\\
\newpage
\begin{center}
Drugi zadatak 15 bodova\\
\end{center}

Dani NDKA prebacite u ekvivalentni DKA
\\
\includegraphics[scale=0.75]{zad2.1}
\\
Rje\v{s}enje.
\\
$Q=\rho\left(\left\{a,b,c\right\}\right),q_{0}=E\left(\left\{1,2\right\}\right)=\left\{1234\right\}$\\
$F=\left\{\left\{3\right\},\left\{13\right\},\left\{23\right\},\left\{34\right\},\left\{123\right\},\left\{134\right\},\left\{234\right\},\left\{1234\right\}\right\}$\\
\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c}
 & 1 & 2 & 3 & 4 & 12 & 13 & 14 & 23 & 24 & 34 & 123 & 124 & 134 & 234 & 1234 \\
 \hline
a & 34 & 4 & 13 & 4 & 34 & 134 & 34 & 134 & 4 & 134 & 134 & 34 & 134 & 134 & 134\\
\hline
b & 13 & 4 & 24 & $\emptyset$ & 134 & 123 & 13 & 24 & $\emptyset$ & 24 & 1234 & 13 & 1234 & 24 & 1234\\
\hline
\end{tabular}
\\
\newline
DKA
\\
\includegraphics[scale=0.75]{zad2b.1}
\\
Nedosti\v{z}na stanja su 1,2,3,4,12,13,14,23,24,34,123,124,234.

\end{document}
