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Diffstat (limited to 'Homework/cs5000')
74 files changed, 5098 insertions, 0 deletions
diff --git a/Homework/cs5000/.DS_Store b/Homework/cs5000/.DS_Store Binary files differnew file mode 100644 index 0000000..eda95e0 --- /dev/null +++ b/Homework/cs5000/.DS_Store diff --git a/Homework/cs5000/computability_complexity_languages.pdf b/Homework/cs5000/computability_complexity_languages.pdf Binary files differnew file mode 100644 index 0000000..b88520e --- /dev/null +++ b/Homework/cs5000/computability_complexity_languages.pdf diff --git a/Homework/cs5000/hw01/BrooksWeber_DFA.pdf b/Homework/cs5000/hw01/BrooksWeber_DFA.pdf Binary files differnew file mode 100644 index 0000000..91a09fa --- /dev/null +++ b/Homework/cs5000/hw01/BrooksWeber_DFA.pdf diff --git a/Homework/cs5000/hw01/BrooksWeber_NFA.pdf b/Homework/cs5000/hw01/BrooksWeber_NFA.pdf Binary files differnew file mode 100644 index 0000000..a3fba1f --- /dev/null +++ b/Homework/cs5000/hw01/BrooksWeber_NFA.pdf diff --git a/Homework/cs5000/hw01/CS5000_F23_HW01.pdf b/Homework/cs5000/hw01/CS5000_F23_HW01.pdf Binary files differnew file mode 100644 index 0000000..d6413e0 --- /dev/null +++ b/Homework/cs5000/hw01/CS5000_F23_HW01.pdf diff --git a/Homework/cs5000/hw01/CS5000_F23_HW1.org b/Homework/cs5000/hw01/CS5000_F23_HW1.org new file mode 100644 index 0000000..d5ac1c5 --- /dev/null +++ b/Homework/cs5000/hw01/CS5000_F23_HW1.org @@ -0,0 +1,60 @@ +#+TITLE: HW 01 +#+AUTHOR: Elizabeth Hunt +#+STARTUP: entitiespretty fold inlineimages +#+LATEX_HEADER: \notindent \notag \usepackage{amsmath} \usepackage[a4paper,margin=1in,landscape]{geometry} +#+LATEX: \setlength\parindent{0pt} +#+OPTIONS: toc:nil + +* Question One +#+attr_latex: :width 400px +[[./img/line_robor.png]] + +$Q$ = { readsensors, turnleft, moveforward, turnright } + +$\Sigma$ = { white, blackonleft, leftturncomplete, blackonmiddle, movecomplete, +blackonright, rightturncomplete } + +$\delta(p_0, a) = p_0$ +$\delta(p_0, b) = p_1$ +$\delta(p_0, d) = p_2$ +$\delta(p_0, f) = p_3$ +$\delta(p_1, c) = p_0$ +$\delta(p_2, e) = p_0$ +$\delta(p_3, g) = p_0$ + +$F = \emptyset$ + +* Question Two +#+attr_latex: :width 400px +[[./img/blob_robor.png]] + + +$Q$ = { readsensors, turnleft, moveforward, turnright } + +$\Sigma$ = { white, obstacleleft, leftturncomplete, obstacleboth, movecomplete, +obstacleright, rightturncomplete } + +$\delta(p_0, a) = p_0$ +$\delta(p_0, b) = p_1$ +$\delta(p_0, d) = p_2$ +$\delta(p_0, f) = p_3$ +$\delta(p_1, c) = p_0$ +$\delta(p_2, e) = p_0$ +$\delta(p_3, g) = p_0$ + +$F = \emptyset$ + +* Question Three +#+attr_latex: :width 400px +[[./img/sus_man.png]] + +Thus a robot could take ~puton(C, T)~, ~puton(B, C)~, and finally ~puton(A, B)~. + +* Question Four +| State | 0 | 1 | 2 | +| q_0 | { q_0, q_1, q_2} | | | +| q_1 | | {q_1, q_2} | | +| q_2 | | | {q_2} | + +#+attr_latex: :width 200px +[[./img/no_epsilon.png]] diff --git a/Homework/cs5000/hw01/CS5000_F23_HW1.pdf b/Homework/cs5000/hw01/CS5000_F23_HW1.pdf Binary files differnew file mode 100644 index 0000000..39dfc1c --- /dev/null +++ b/Homework/cs5000/hw01/CS5000_F23_HW1.pdf diff --git a/Homework/cs5000/hw01/CS5000_F23_HW1.tex b/Homework/cs5000/hw01/CS5000_F23_HW1.tex new file mode 100644 index 0000000..42eac8b --- /dev/null +++ b/Homework/cs5000/hw01/CS5000_F23_HW1.tex @@ -0,0 +1,96 @@ +% Created 2023-09-15 Fri 20:36 +% Intended LaTeX compiler: pdflatex +\documentclass[11pt]{article} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +\usepackage{graphicx} +\usepackage{longtable} +\usepackage{wrapfig} +\usepackage{rotating} +\usepackage[normalem]{ulem} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{capt-of} +\usepackage{hyperref} +\notindent \notag \usepackage{amsmath} \usepackage[a4paper,margin=1in,landscape]{geometry} +\author{Elizabeth Hunt} +\date{\today} +\title{HW 01} +\hypersetup{ + pdfauthor={Elizabeth Hunt}, + pdftitle={HW 01}, + pdfkeywords={}, + pdfsubject={}, + pdfcreator={Emacs 28.2 (Org mode 9.7-pre)}, + pdflang={English}} +\begin{document} + +\maketitle +\setlength\parindent{0pt} + +\section{Question One} +\label{sec:org0490f98} +\begin{center} +\includegraphics[width=400px]{./img/line_robor.png} +\end{center} + +\(Q\) = \{ readsensors, turnleft, moveforward, turnright \} + +\(\Sigma\) = \{ white, blackonleft, leftturncomplete, blackonmiddle, movecomplete, +blackonright, rightturncomplete \} + +\(\delta(p_0, a) = p_0\) +\(\delta(p_0, b) = p_1\) +\(\delta(p_0, d) = p_2\) +\(\delta(p_0, f) = p_3\) +\(\delta(p_1, c) = p_0\) +\(\delta(p_2, e) = p_0\) +\(\delta(p_3, g) = p_0\) + +\(F = \emptyset\) + +\section{Question Two} +\label{sec:orgdb1d9e7} +\begin{center} +\includegraphics[width=400px]{./img/blob_robor.png} +\end{center} + + +\(Q\) = \{ readsensors, turnleft, moveforward, turnright \} + +\(\Sigma\) = \{ white, obstacleleft, leftturncomplete, obstacleboth, movecomplete, +obstacleright, rightturncomplete \} + +\(\delta(p_0, a) = p_0\) +\(\delta(p_0, b) = p_1\) +\(\delta(p_0, d) = p_2\) +\(\delta(p_0, f) = p_3\) +\(\delta(p_1, c) = p_0\) +\(\delta(p_2, e) = p_0\) +\(\delta(p_3, g) = p_0\) + +\(F = \emptyset\) + +\section{Question Three} +\label{sec:orgf7c103c} +\begin{center} +\includegraphics[width=400px]{./img/sus_man.png} +\end{center} + +Thus a robot could take \texttt{puton(C, T)}, \texttt{puton(B, C)}, and finally \texttt{puton(A, B)}. + +\section{Question Four} +\label{sec:org54c6f53} +\begin{center} +\begin{tabular}{llrr} +State & 0 & 1 & 2\\[0pt] +q\textsubscript{0} & \{ q\textsubscript{0}, q\textsubscript{1}, q\textsubscript{2}\} & & \\[0pt] +q\textsubscript{1} & & \{q\textsubscript{1}, q\textsubscript{2}\} & \\[0pt] +q\textsubscript{2} & & & \{q\textsubscript{2}\}\\[0pt] +\end{tabular} +\end{center} + +\begin{center} +\includegraphics[width=200px]{./img/no_epsilon.png} +\end{center} +\end{document}
\ No newline at end of file diff --git a/Homework/cs5000/hw01/RobotLineAndBlobFollowingProblems.pdf b/Homework/cs5000/hw01/RobotLineAndBlobFollowingProblems.pdf Binary files differnew file mode 100644 index 0000000..8f0d88e --- /dev/null +++ b/Homework/cs5000/hw01/RobotLineAndBlobFollowingProblems.pdf diff --git a/Homework/cs5000/hw01/img/blob_robor.png b/Homework/cs5000/hw01/img/blob_robor.png Binary files differnew file mode 100644 index 0000000..a579f5c --- /dev/null +++ b/Homework/cs5000/hw01/img/blob_robor.png diff --git a/Homework/cs5000/hw01/img/line_robor.png b/Homework/cs5000/hw01/img/line_robor.png Binary files differnew file mode 100644 index 0000000..cbaddda --- /dev/null +++ b/Homework/cs5000/hw01/img/line_robor.png diff --git a/Homework/cs5000/hw01/img/no_epsilon.png b/Homework/cs5000/hw01/img/no_epsilon.png Binary files differnew file mode 100644 index 0000000..9d2d0ff --- /dev/null +++ b/Homework/cs5000/hw01/img/no_epsilon.png diff --git a/Homework/cs5000/hw01/img/sus_man.png b/Homework/cs5000/hw01/img/sus_man.png Binary files differnew file mode 100644 index 0000000..d2a617e --- /dev/null +++ b/Homework/cs5000/hw01/img/sus_man.png diff --git a/Homework/cs5000/hw02/CS5000_F23_EH_HW02.org b/Homework/cs5000/hw02/CS5000_F23_EH_HW02.org new file mode 100644 index 0000000..5fd935c --- /dev/null +++ b/Homework/cs5000/hw02/CS5000_F23_EH_HW02.org @@ -0,0 +1,65 @@ +#+TITLE: HW 02 +#+AUTHOR: Elizabeth Hunt +#+STARTUP: entitiespretty fold inlineimages +#+LATEX_HEADER: \notindent \notag \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} +#+LATEX: \setlength\parindent{0pt} +#+OPTIONS: toc:nil + +* Problem 1 +** Part 1 +$L(M) = \{ x | x \in \{ a \}^{*} \} \cup \{ x | x \in \{ b \}^{*} \} \cup \{ x | x \in \{ c \}^{*} \}$ + +** Part 2 + +#+attr_latex: :width 300px +[[./img/p01_02.png]] + +$Q = {q_0, q_1, q_2, q_3}$ + +$\Sigma = {a, b, c}$ + +$F = {q_1, q_2, q_3}$ + +$\delta(q_0, a) = {q_1}, \delta(q_1, a) = {q_1}, \delta(q_0, b) = {q_2}, \delta(q_2, b) = {q_2}, \delta(q_0, c) = {q_3}, \delta(q_3, c) = c$ + +* Problem 2 +** Work +| subset | 0 | 1 | +| {q_0} | {q_0, q_1} | {q_1} | +| {q_1} | {q_2} | {q_2} | +| {q_2} | {q_2} | {q_2} | +| {q_0, q_1} | {q_0, q_1} \cup {q_2} | {q_1} \cup {q_2} | +| {q_0, q_1, q_2} | {q_0, q_1} \cup {q_2} \cup {q_2} | {q_1} \cup {q_2} \cup {q_2} | +| {q_1, q_2} | {q_2} \cup {q_2} | {q_2} \cup {q_2} | + + +** Solution +#+attr_latex: :width 300px +[[./img/p02.png]] + +$Q_D = \{q_0, q_1, q_2, q_3, q_4, q_5\}$ + +$\Sigma = \{0, 1\}$ + +$F_D = \{q_1, q_2, q_4, q_5\}$ + +$\delta_D(q_0, 0) = q_1$ + +$\delta_D(q_0, 1) = q_2$ + +$\delta_D(q_1, 0) = q_5$ + +$\delta_D(q_1, q) = q_4$ + +$\delta_D(q_2, 0) = \delta_D(q_2, 1) = q_3$ + +$\delta_D(q_3, 0) = \delta_D(q_3, 1) = q_3$ + +$\delta_D(q_4, 0) = \delta_D(q_4, 1) = q_4$ + +$\delta_D(q_5, 0) = q_5$ + +$\delta_D(q_5, 1} = q_4$ + +* Problem 3 +See attached python diff --git a/Homework/cs5000/hw02/CS5000_F23_EH_HW02.pdf b/Homework/cs5000/hw02/CS5000_F23_EH_HW02.pdf Binary files differnew file mode 100644 index 0000000..5344c34 --- /dev/null +++ b/Homework/cs5000/hw02/CS5000_F23_EH_HW02.pdf diff --git a/Homework/cs5000/hw02/CS5000_F23_EH_HW02.tex b/Homework/cs5000/hw02/CS5000_F23_EH_HW02.tex new file mode 100644 index 0000000..f578070 --- /dev/null +++ b/Homework/cs5000/hw02/CS5000_F23_EH_HW02.tex @@ -0,0 +1,102 @@ +% Created 2023-10-07 Sat 14:51 +% Intended LaTeX compiler: pdflatex +\documentclass[11pt]{article} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +\usepackage{graphicx} +\usepackage{longtable} +\usepackage{wrapfig} +\usepackage{rotating} +\usepackage[normalem]{ulem} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{capt-of} +\usepackage{hyperref} +\notindent \notag \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} +\author{Elizabeth Hunt} +\date{\today} +\title{HW 02} +\hypersetup{ + pdfauthor={Elizabeth Hunt}, + pdftitle={HW 02}, + pdfkeywords={}, + pdfsubject={}, + pdfcreator={Emacs 28.2 (Org mode 9.7-pre)}, + pdflang={English}} +\begin{document} + +\maketitle +\setlength\parindent{0pt} + +\section{Problem 1} +\label{sec:org41ad199} +\subsection{Part 1} +\label{sec:orgabe730c} +\(L(M) = \{ x | x \in \{ a \}^{*} \} \cup \{ x | x \in \{ b \}^{*} \} \cup \{ x | x \in \{ c \}^{*} \}\) + +\subsection{Part 2} +\label{sec:orgb60fdea} + +\begin{center} +\includegraphics[width=300px]{./img/p01_02.png} +\end{center} + +\(Q = {q_0, q_1, q_2, q_3}\) + +\(\Sigma = {a, b, c}\) + +\(F = {q_1, q_2, q_3}\) + +\(\delta(q_0, a) = {q_1}, \delta(q_1, a) = {q_1}, \delta(q_0, b) = {q_2}, \delta(q_2, b) = {q_2}, \delta(q_0, c) = {q_3}, \delta(q_3, c) = c\) + +\section{Problem 2} +\label{sec:org9cf81d0} +\subsection{Work} +\label{sec:org2daff9f} +\begin{center} +\begin{tabular}{lll} +subset & 0 & 1\\[0pt] +\{q\textsubscript{0}\} & \{q\textsubscript{0}, q\textsubscript{1}\} & \{q\textsubscript{1}\}\\[0pt] +\{q\textsubscript{1}\} & \{q\textsubscript{2}\} & \{q\textsubscript{2}\}\\[0pt] +\{q\textsubscript{2}\} & \{q\textsubscript{2}\} & \{q\textsubscript{2}\}\\[0pt] +\{q\textsubscript{0}, q\textsubscript{1}\} & \{q\textsubscript{0}, q\textsubscript{1}\} \(\cup\) \{q\textsubscript{2}\} & \{q\textsubscript{1}\} \(\cup\) \{q\textsubscript{2}\}\\[0pt] +\{q\textsubscript{0}, q\textsubscript{1}, q\textsubscript{2}\} & \{q\textsubscript{0}, q\textsubscript{1}\} \(\cup\) \{q\textsubscript{2}\} \(\cup\) \{q\textsubscript{2}\} & \{q\textsubscript{1}\} \(\cup\) \{q\textsubscript{2}\} \(\cup\) \{q\textsubscript{2}\}\\[0pt] +\{q\textsubscript{1}, q\textsubscript{2}\} & \{q\textsubscript{2}\} \(\cup\) \{q\textsubscript{2}\} & \{q\textsubscript{2}\} \(\cup\) \{q\textsubscript{2}\}\\[0pt] +\end{tabular} +\end{center} + + +\subsection{Solution} +\label{sec:org7701586} +\begin{center} +\includegraphics[width=300px]{./img/p02.png} +\end{center} + +\(Q_D = \{q_0, q_1, q_2, q_3, q_4, q_5\}\) + +\(\Sigma = \{0, 1\}\) + +\(F_D = \{q_1, q_2, q_4, q_5\}\) + +\(\delta_D(q_0, 0) = q_1\) + +\(\delta_D(q_0, 1) = q_2\) + +\(\delta_D(q_1, 0) = q_5\) + +\(\delta_D(q_1, q) = q_4\) + +\(\delta_D(q_2, 0) = \delta_D(q_2, 1) = q_3\) + +\(\delta_D(q_3, 0) = \delta_D(q_3, 1) = q_3\) + +\(\delta_D(q_4, 0) = \delta_D(q_4, 1) = q_4\) + +\(\delta_D(q_5, 0) = q_5\) + +\(\delta_D(q_5, 1} = q_4\) + +\section{Problem 3} +\label{sec:orge1c9402} +See attached python +\end{document}
\ No newline at end of file diff --git a/Homework/cs5000/hw02/CS5000_F23_HW02.pdf b/Homework/cs5000/hw02/CS5000_F23_HW02.pdf Binary files differnew file mode 100644 index 0000000..60b4d50 --- /dev/null +++ b/Homework/cs5000/hw02/CS5000_F23_HW02.pdf diff --git a/Homework/cs5000/hw02/cs5000_f23_hw02_uts.py b/Homework/cs5000/hw02/cs5000_f23_hw02_uts.py new file mode 100644 index 0000000..a45fba6 --- /dev/null +++ b/Homework/cs5000/hw02/cs5000_f23_hw02_uts.py @@ -0,0 +1,45 @@ +#################################################### +# CS 5000: F23: Assignment 2: Unit Tests +# bugs to vladimir kulyukin in canvas +##################################################### + +import numpy as np +from Mod3DFA import Mod3DFA +from Mod7DFA import Mod7DFA +import unittest + + +class cs5000_f23_hw02_uts(unittest.TestCase): + def test_assgn_02_ut_01(self): + print("\n***** Testing Mod3DFA: Hw02: UT01 ************") + lower, upper = 0, 10001 + dfa = Mod3DFA() + for i in range(lower, upper): + bstr = "{0:b}".format(i) + if i % 3 == 0: + assert dfa.processString(bstr) == True + # print('assertion passed on {}'.format(i)) + else: + assert dfa.processString(bstr) == False + # print('assertion passed on {}'.format(i)) + print("Mod7DFA: HW02: UT01 passed...") + + def test_assgn_02_ut_02(self): + print("\n***** Mod7DFA: HW03: UT02 ************") + lower, upper = 0, 1000001 + dfa = Mod7DFA() + for i in range(lower, upper): + bstr = "{0:b}".format(i) + if i % 7 == 0: + assert dfa.processString(bstr) == True + # print('assertion passed on {}'.format(i)) + else: + assert dfa.processString(bstr) == False + # print('assertion passed on {}'.format(i)) + print("Mod7DFA: HW02: UT02: passed...") + + +### ================ Unit Tests ==================== + +if __name__ == "__main__": + unittest.main() diff --git a/Homework/cs5000/hw02/hw02.zip b/Homework/cs5000/hw02/hw02.zip Binary files differnew file mode 100644 index 0000000..604aa07 --- /dev/null +++ b/Homework/cs5000/hw02/hw02.zip diff --git a/Homework/cs5000/hw02/hw02/Mod3DFA.py b/Homework/cs5000/hw02/hw02/Mod3DFA.py new file mode 100644 index 0000000..93a7c5c --- /dev/null +++ b/Homework/cs5000/hw02/hw02/Mod3DFA.py @@ -0,0 +1,50 @@ +#########################################
+## Mod3DFA.py
+## bugs to vladimir kulyukin in canvas
+#########################################
+
+class Mod3DFA:
+
+ def __q0(self, s, pos):
+ if len(s) == pos:
+ return True
+ elif s[pos] == '0':
+ return self.__q0(s, pos+1)
+ elif s[pos] == '1':
+ return self.__q1(s, pos+1)
+ else:
+ return self.__q3(s, pos+1)
+
+ def __q1(self, s, pos):
+ if len(s) == pos:
+ return False
+ elif s[pos] == '0':
+ return self.__q2(s, pos+1)
+ elif s[pos] == '1':
+ return self.__q0(s, pos+1)
+ else:
+ return self.__q3(s, pos+1)
+
+ def __q2(self, s, pos):
+ if len(s) == pos:
+ return False
+ elif s[pos] == '0':
+ return self.__q1(s, pos+1)
+ elif s[pos] == '1':
+ return self.__q2(s, pos+1)
+ else:
+ return self.__q3(s, pos+1)
+
+ def __q3(self, s, pos):
+ if (len(s) == pos):
+ return False
+ else:
+ return self.__q3(s, pos+1)
+
+ def processString(self, s):
+ if s[0] == '0':
+ return self.__q0(s, 1)
+ elif s[0] == '1':
+ return self.__q1(s, 1)
+ else:
+ return self.__q3(s, pos+1)
diff --git a/Homework/cs5000/hw02/hw02/Mod7DFA.py b/Homework/cs5000/hw02/hw02/Mod7DFA.py new file mode 100644 index 0000000..32cea44 --- /dev/null +++ b/Homework/cs5000/hw02/hw02/Mod7DFA.py @@ -0,0 +1,50 @@ +#########################################
+## Mod7DFA.py
+## Elizabeth Hunt
+## A02364151
+#########################################
+
+
+class DFA:
+ def __init__(
+ self,
+ alphabet: set[str],
+ initial_state: str,
+ error_state: str,
+ final_states: set[str],
+ delta: dict[dict[str, str]],
+ ):
+ self.alphabet = alphabet
+ self.initial_state = initial_state
+ self.error_state = error_state
+ self.final_states = final_states
+ self.delta = delta
+
+ def processString(self, s):
+ state = self.initial_state
+ for i in s:
+ if i not in self.alphabet or state == self.error_state:
+ state = self.error_state
+ continue
+ state = self.delta[state][i]
+ return state in self.final_states
+
+
+# ramblings: https://excalidraw.com/#json=YUK684eVJcffkGuAAG3rr,K__bSFw7kGK3dLQggSh_qA
+class Mod7DFA(DFA):
+ def __init__(self):
+ super().__init__(
+ set(["1", "0"]),
+ "7m + 0",
+ "err",
+ set(["7m + 0"]),
+ {
+ "7m + 0": {"0": "7m + 0", "1": "7m + 1"},
+ "7m + 1": {"0": "7m + 2", "1": "7m + 3"},
+ "7m + 2": {"0": "7m + 4", "1": "7m + 5"},
+ "7m + 3": {"0": "7m + 6", "1": "7m + 0"},
+ "7m + 4": {"0": "7m + 1", "1": "7m + 2"},
+ "7m + 5": {"0": "7m + 3", "1": "7m + 4"},
+ "7m + 6": {"0": "7m + 5", "1": "7m + 6"},
+ },
+ )
diff --git a/Homework/cs5000/hw02/hw02/__pycache__/Mod3DFA.cpython-311.pyc b/Homework/cs5000/hw02/hw02/__pycache__/Mod3DFA.cpython-311.pyc Binary files differnew file mode 100644 index 0000000..7b6a27a --- /dev/null +++ b/Homework/cs5000/hw02/hw02/__pycache__/Mod3DFA.cpython-311.pyc diff --git a/Homework/cs5000/hw02/hw02/__pycache__/Mod7DFA.cpython-311.pyc b/Homework/cs5000/hw02/hw02/__pycache__/Mod7DFA.cpython-311.pyc Binary files differnew file mode 100644 index 0000000..9faf32a --- /dev/null +++ b/Homework/cs5000/hw02/hw02/__pycache__/Mod7DFA.cpython-311.pyc diff --git a/Homework/cs5000/hw02/hw02/cs5000_f23_hw02_uts.py b/Homework/cs5000/hw02/hw02/cs5000_f23_hw02_uts.py new file mode 100644 index 0000000..e2093d4 --- /dev/null +++ b/Homework/cs5000/hw02/hw02/cs5000_f23_hw02_uts.py @@ -0,0 +1,45 @@ +#################################################### +# CS 5000: F23: Assignment 2: Unit Tests +# bugs to vladimir kulyukin in canvas +##################################################### + +import numpy as np +from Mod3DFA import Mod3DFA +from Mod7DFA import Mod7DFA +import unittest + + +class cs5000_f23_hw02_uts(unittest.TestCase): + def test_assgn_02_ut_01(self): + print("\n***** Testing Mod3DFA: Hw02: UT01 ************") + lower, upper = 0, 1000001 + dfa = Mod3DFA() + for i in range(lower, upper): + bstr = "{0:b}".format(i) + if i % 3 == 0: + assert dfa.processString(bstr) == True + # print('assertion passed on {}'.format(i)) + else: + assert dfa.processString(bstr) == False + # print('assertion passed on {}'.format(i)) + print("Mod7DFA: HW02: UT01 passed...") + + def test_assgn_02_ut_02(self): + print("\n***** Mod7DFA: HW02: UT02 ************") + lower, upper = 0, 1000001 + dfa = Mod7DFA() + for i in range(lower, upper): + bstr = "{0:b}".format(i) + if i % 7 == 0: + assert dfa.processString(bstr) == True + # print('assertion passed on {}'.format(i)) + else: + assert dfa.processString(bstr) == False + # print('assertion passed on {}'.format(i)) + print("Mod7DFA: HW02: UT02: passed...") + + +### ================ Unit Tests ==================== + +if __name__ == "__main__": + unittest.main() diff --git a/Homework/cs5000/hw02/hw02/hw02.pdf b/Homework/cs5000/hw02/hw02/hw02.pdf Binary files differnew file mode 100644 index 0000000..c33b60f --- /dev/null +++ b/Homework/cs5000/hw02/hw02/hw02.pdf diff --git a/Homework/cs5000/hw02/img/p01_02.png b/Homework/cs5000/hw02/img/p01_02.png Binary files differnew file mode 100644 index 0000000..ce2a732 --- /dev/null +++ b/Homework/cs5000/hw02/img/p01_02.png diff --git a/Homework/cs5000/hw02/img/p02.png b/Homework/cs5000/hw02/img/p02.png Binary files differnew file mode 100644 index 0000000..f649e12 --- /dev/null +++ b/Homework/cs5000/hw02/img/p02.png diff --git a/Homework/cs5000/hw03/CS5000_F23_hw03.org b/Homework/cs5000/hw03/CS5000_F23_hw03.org new file mode 100644 index 0000000..a0f08de --- /dev/null +++ b/Homework/cs5000/hw03/CS5000_F23_hw03.org @@ -0,0 +1,43 @@ +#+TITLE: HW 02 +#+AUTHOR: Elizabeth Hunt +#+STARTUP: entitiespretty fold inlineimages +#+LATEX_HEADER: \notindent \notag \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} +#+LATEX: \setlength\parindent{0pt} +#+OPTIONS: toc:nil + +* Question One +** Partition Refinement +{ {q_0, q_1, q_3}, {q_2, q_4} } + +S_1 = {(q_0, q_1), (q_0, q_3), (q_1, q_3)} +S_2 = {(q_2, q_4)} + +\delta(q_0, 1) = q_3 \in S_1 +\delta(q_1, 1) = q_4 \in S_2 + +(q_0, q_1) need to be split + +\delta(q_0, 0) = q_1 \in S_1 +\delta(q_3, 0) = q_2 \in S_2 + +(q_1, q_2) need to be split + +\forall x \in \Sigma, \delta(q_1, x) = \delta(q_3, x) +so {q_1, q_3} does not need to be split + +In S_2, \delta(q_2, 0) \in S_1 and \delta(q_4, 0) \in S_2, thus need to be split + +Finally, the refined partitions are {{q_0}, {q_1, q_3}, {q_2}, {q_4}} + +** Minimization +| a \in \Sigma | {q_0} | {q_1, q_3} | {q_2} | {q_4} | +| 0 | {q_1, q_3} | {q_2} | {q_1, q_3} | {q_4} | +| 1 | {q_1, q_3} | {q_4} | {q_4} | {q_4} | + +with d_0 = {q_0}, d_1 = {q_1, q_3}, d_2 = {q_2} and d_3 = {q_4} + +#+attr_latex: :width 350px +[[./img/min_dfa.png]] + +* Question Two +See attached python diff --git a/Homework/cs5000/hw03/CS5000_F23_hw03.tex b/Homework/cs5000/hw03/CS5000_F23_hw03.tex new file mode 100644 index 0000000..634470a --- /dev/null +++ b/Homework/cs5000/hw03/CS5000_F23_hw03.tex @@ -0,0 +1,76 @@ +% Created 2023-09-20 Wed 15:58 +% Intended LaTeX compiler: pdflatex +\documentclass[11pt]{article} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +\usepackage{graphicx} +\usepackage{longtable} +\usepackage{wrapfig} +\usepackage{rotating} +\usepackage[normalem]{ulem} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{capt-of} +\usepackage{hyperref} +\notindent \notag \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} +\author{Elizabeth Hunt} +\date{\today} +\title{HW 02} +\hypersetup{ + pdfauthor={Elizabeth Hunt}, + pdftitle={HW 02}, + pdfkeywords={}, + pdfsubject={}, + pdfcreator={Emacs 28.2 (Org mode 9.7-pre)}, + pdflang={English}} +\begin{document} + +\maketitle +\setlength\parindent{0pt} + +\section{Question One} +\label{sec:org859695d} +\subsection{Partition Refinement} +\label{sec:org55dda6e} +\{ \{q\textsubscript{0}, q\textsubscript{1}, q\textsubscript{3}\}, \{q\textsubscript{2}, q\textsubscript{4}\} \} + +S\textsubscript{1} = \{(q\textsubscript{0}, q\textsubscript{1}), (q\textsubscript{0}, q\textsubscript{3}), (q\textsubscript{1}, q\textsubscript{3})\} +S\textsubscript{2} = \{(q\textsubscript{2}, q\textsubscript{4})\} + +\(\delta\)(q\textsubscript{0}, 1) = q\textsubscript{3} \(\in\) S\textsubscript{1} +\(\delta\)(q\textsubscript{1}, 1) = q\textsubscript{4} \(\in\) S\textsubscript{2} + +(q\textsubscript{0}, q\textsubscript{1}) need to be split + +\(\delta\)(q\textsubscript{0}, 0) = q\textsubscript{1} \(\in\) S\textsubscript{1} +\(\delta\)(q\textsubscript{3}, 0) = q\textsubscript{2} \(\in\) S\textsubscript{2} + +(q\textsubscript{1}, q\textsubscript{2}) need to be split + +\(\forall\) x \(\in\) \(\Sigma\), \(\delta\)(q\textsubscript{1}, x) = \(\delta\)(q\textsubscript{3}, x) +so \{q\textsubscript{1}, q\textsubscript{3}\} does not need to be split + +In S\textsubscript{2}, \(\delta\)(q\textsubscript{2}, 0) \(\in\) S\textsubscript{1} and \(\delta\)(q\textsubscript{4}, 0) \(\in\) S\textsubscript{2}, thus need to be split + +Finally, the refined partitions are \{\{q\textsubscript{0}\}, \{q\textsubscript{1}, q\textsubscript{3}\}, \{q\textsubscript{2}\}, \{q\textsubscript{4}\}\} + +\subsection{Minimization} +\label{sec:org870d4d3} +\begin{center} +\begin{tabular}{rllll} +a \(\in\) \(\Sigma\) & \{q\textsubscript{0}\} & \{q\textsubscript{1}, q\textsubscript{3}\} & \{q\textsubscript{2}\} & \{q\textsubscript{4}\}\\[0pt] +0 & \{q\textsubscript{1}, q\textsubscript{3}\} & \{q\textsubscript{2}\} & \{q\textsubscript{1}, q\textsubscript{3}\} & \{q\textsubscript{4}\}\\[0pt] +1 & \{q\textsubscript{1}, q\textsubscript{3}\} & \{q\textsubscript{4}\} & \{q\textsubscript{4}\} & \{q\textsubscript{4}\}\\[0pt] +\end{tabular} +\end{center} + +with d\textsubscript{0} = \{q\textsubscript{0}\}, d\textsubscript{1} = \{q\textsubscript{1}, q\textsubscript{3}\}, d\textsubscript{2} = \{q\textsubscript{2}\} and d\textsubscript{3} = \{q\textsubscript{4}\} + +\begin{center} +\includegraphics[width=350px]{./img/min_dfa.png} +\end{center} + +\section{Question Two} +\label{sec:orga1d2018} +See attached python +\end{document}
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b/Homework/cs5000/hw03/hw03/__pycache__/cs5000_f23_hw03.cpython-311.pyc diff --git a/Homework/cs5000/hw03/hw03/cs5000_f23_hw03.py b/Homework/cs5000/hw03/hw03/cs5000_f23_hw03.py new file mode 100644 index 0000000..f32a7c1 --- /dev/null +++ b/Homework/cs5000/hw03/hw03/cs5000_f23_hw03.py @@ -0,0 +1,110 @@ +############################################### +# cs5000_f23_hw03.py +# Elizabeth Hunt +# A02364151 +############################################### +import functools +from collections.abc import Iterable + + +def power_set(s: list) -> list[list]: + if len(s) == 0: + return [[]] + + head, tail = [s[0], s[1:]] + tail_set = power_set(tail) + + res = [] + for i in tail_set: + res.append(i + [head]) + res.append(i) + + return res + + +def deterministic_states_key(s: Iterable[str]) -> str: + return "".join(sorted(s)) + + +def reachable( + begin: str, + states: set[str], + alphabet: set[str], + delta: dict[tuple[str, str], set[str]], +) -> set[str]: + visited = set() + + def dfs(state: str): + visited.add(state) + + for a in alphabet: + transition = (state, a) + if transition in delta and delta[transition] not in visited: + new_state = delta[transition] + dfs(new_state) + + dfs(begin) + return visited + + +def nfa_to_dfa( + nfa: tuple[set[str], set[str], dict[tuple[str, str], set[str]], set[str]] +) -> tuple[set[str], set[str], dict[tuple[str, str], set[str]], set[str]]: + (states, alphabet, delta, start, accepting) = nfa + + p_states = power_set(list(states)) + p_inv = dict((deterministic_states_key(s), str(i)) for i, s in enumerate(p_states)) + start_d = next( + str(i) for i, s in enumerate(p_states) if len(s) == 1 and s[0] == start + ) + f_d = set() + delta_d = {} + + for i, q in enumerate(p_states): + for a in alphabet: + nfa_to_states = functools.reduce( + lambda acc, x: acc.union(delta[(x, a)] if (x, a) in delta else set()), + q, + set(), + ) + key = deterministic_states_key(nfa_to_states) + + if key in p_inv: + delta_d[(str(i), a)] = p_inv[key] + + for i, q in enumerate(p_states): + if len(set(q).intersection(accepting)) > 0: + f_d.add(str(i)) + + reachable_states = reachable(start_d, list(range(len(p_states))), alphabet, delta_d) + pruned_delta_d = dict( + list( + filter( + lambda x: x[0][0] in reachable_states and x[1] in reachable_states, + delta_d.items(), + ) + ) + ) + + return ( + reachable_states, + alphabet, + pruned_delta_d, + start_d, + f_d.intersection(reachable_states), + ) + + +def display_dfa( + dfa: tuple[set[str], set[str], dict[tuple[str, str], set[str]], set[str]] +): + (states, alphabet, delta, start, accepting) = dfa + + print(f"STATES: {states}") + print(f"SIGMA: {alphabet}") + print(f"START STATE: {start}") + print("DELTA:") + for i, entry in enumerate(delta.items()): + (key, value) = entry + print(f"{i}) d({key}) = {value}") + print(f"FINAL STATES: {accepting}") diff --git a/Homework/cs5000/hw03/hw03/cs5000_f23_hw03_uts.py b/Homework/cs5000/hw03/hw03/cs5000_f23_hw03_uts.py new file mode 100644 index 0000000..173d3cf --- /dev/null +++ b/Homework/cs5000/hw03/hw03/cs5000_f23_hw03_uts.py @@ -0,0 +1,63 @@ +#################################################### +# CS5000: F23: Assignment 3: Unit Tests +# Description: Two unit tests for the subset construction +# algorithm. +# bugs to vladimir kulyukin in canvas +##################################################### + +from cs5000_f23_hw03 import nfa_to_dfa, display_dfa +import unittest + +class CS5000F23Assign03UnitTests(unittest.TestCase): + + def test_assgn_03_ut_01(self): + print('\n***** Assign 03: Subset Construction UT 01 *****') + NFA_DELTA_01 = {} + NFA_DELTA_01[('q0', '0')] = set(['q0']) + NFA_DELTA_01[('q0', '1')] = set(['q0', 'q1']) + NFA_DELTA_01[('q1', '0')] = set(['q2']) + NFA_DELTA_01[('q1', '1')] = set(['q2']) + NFA_01 = (set(['q0','q1', 'q2']), + set(['0', '1']), + NFA_DELTA_01, + 'q0', + set(['q2'])) + + dfa_01 = nfa_to_dfa(NFA_01) + display_dfa(dfa_01) + + def test_assgn_03_ut_02(self): + print('\n***** Assign 03: Subset Construction UT 02 *****') + NFA_DELTA_02 = {} + NFA_DELTA_02[('q0', '0')] = set(['q0', 'q1']) + NFA_DELTA_02[('q0', '1')] = set(['q1']) + NFA_DELTA_02[('q1', '0')] = set(['q2']) + NFA_DELTA_02[('q1', '1')] = set(['q2']) + NFA_DELTA_02[('q2', '0')] = set(['q2']) + NFA_DELTA_02[('q2', '1')] = set(['q2']) + NFA_02 = (set(['q0', 'q1', 'q2']), + set(['0', '1']), + NFA_DELTA_02, + 'q0', + set(['q1'])) + dfa_02 = nfa_to_dfa(NFA_02) + display_dfa(dfa_02) + +### ================ Unit Tests ==================== + +if __name__ == '__main__': + unittest.main() + + + + + + + + + + + + + + diff --git a/Homework/cs5000/hw03/img/min_dfa.png b/Homework/cs5000/hw03/img/min_dfa.png Binary files differnew file mode 100644 index 0000000..3d4d529 --- /dev/null +++ b/Homework/cs5000/hw03/img/min_dfa.png diff --git a/Homework/cs5000/hw04.zip b/Homework/cs5000/hw04.zip Binary files differnew file mode 100644 index 0000000..0252337 --- /dev/null +++ b/Homework/cs5000/hw04.zip diff --git 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Then we assume that +$G_k$ is irregular. If so, we can show $G_{k+1}$ is irregular because we can only construct a FA to recognize +$G_{k+1}$ if and only if we can concatenate a FA recognizing $G_k$ in an epsilon transition with another FA recognizing $G_1$; which is +not existant. By induction, any such $G_i | i \in \mathds{N}$ is irregular. + +Each $G$ is also a proper sublanguage since for each $i \in \mathds{N}$ we can construct $(01)^{i+1}$ +which is not in $G_i$ but in $L$, so $\nexist x \in G | x = L$. For extra clarity we know every string in $G_i$ is +also in $L$ since $L$ is really the Kleene Closure. + +Thus there are at least $\aleph_0$ infinitely many such non-regular proper sublanguages of the regular language $L$. + +* Question Two +** One (adapted from slide notes in Lecture 5) +Consider a minimal DFA $M$ that recognizes $L$; $L = L(M)$ with $k$ states. + +Then consider the string $a^k b^k c^k$; to first recognize $a^k$ we go through $k+1$ states, so we can +find a loop in the path taken via $\delta$ such that there exists $q = \delta^{\star}(q, a^i) | i > 0$. + +If we pump this loop zero times, then for the string $a^j b^k c^k$, $j < k$; for one or more times, +$j > k$; thus $j < k$ or $j > k$ but $j \neq k$, a contradiction from the original definition. + +** Two + +Any string in this language is an even number of $a$'s, recognized by this FA (thus, is a regular language). + ++ $F = \{q_0\}$ ++ $\Sigma = \{a\}$ ++ $S = q_0$ ++ $\delta(q_0, a) = q_1, \delta(q_1, a) = q_0$ ++ $Q = \{q_0, q_1\}$ + +** Three +Consider a minimal DFA $M$ that recognizes $L$; $L = L(M)$ with $k$ states. By the pumping lemma +each string $x \in L$ such that $|x| \geq k$ is of the form $uvw$ with $|uv| \leq k$, $|v| \geq 1$ and +$uv^i w \in L \forall i \geq 0$. + +For any such $k$ we create the string $a^k c a^k \in L$, and because $|uv| \leq k$ then $uv$ matches at most +$a^k$. So, $u = a^m, v = a^n$ with $m + n \leq k$, and thus $w = a^{k - (m+n)} c a^k$. Additionally, since $|v| \geq 1$, +$n \ge 1$. + +By pumping $v$ zero times we then have $a^m a^{k-(m+n)}c a^k = a^{k-n} c a^k \notin L$ as $n \geq 1$ so $L$ must be irregular. + +** Four +Consider a minimal DFA $M$ that recognizes $L$; $L = L(M)$ with $k$ states. By the pumping lemma +each string $x \in L$ such that $|x| \geq k$ is of the form $uvw$ with $|uv| \leq k$, $|v| \geq 1$ and +$uv^i w \in L \forall i \geq 0$. + +For any such $k$ we create the string $a^k c^k b^{2k} \in L$ and because $|uv| \leq k$ then $uv$ matches at most +$a^k$. So, $u = a^m, v = a^n$ with $m + n \leq k$, and since $|v| \geq 1$, $n \ge 1$ thus $w = a^{k - (m+n)} c^k b^{2k}$. +Additionally, since $|v| \geq 1$, $n \ge 1$. + +By pumping $v$ zero times we then obtain $a^m a^{k-(m+n)} c^k b^2k = a^{k-n} c^k b^{2k}$ but then $k-n + k \neq 2k$ as +$n \geq 1$, a contradiction; $L$ is irregular. + +** Five +Consider a minimal DFA $M$ that recognizes $L$; $L = L(M)$ with $k$ states. By the pumping lemma +each string $x \in L$ such that $|x| \geq k$ is of the form $uvw$ with $|uv| \leq k$, $|v| \geq 1$ and +$uv^i w \in L \forall i \geq 0$. + +For any such $k$ we create the string $0^k 1^{k-1} \in L$, and because $|uv| \leq k$ then $uv$ matches at most +$0^k$. So, $u = 0^m, v = 0^n$ with $m + n \leq k$, thus $w = 0^{k - (m+n)} 1^{k-1}$. Additionally, since $|v| \geq 1$, $n \ge 1$. + +By pumping $v$ zero times we obtain the string $0^m 0^{k-(m+n)} 1^{k-1} = 0^{k-n} 1^{k-1}$ and $k-n$ cannot be greater than +$k-1$, a contradiction; $L$ is irregular. + +* Question Three + +(pictorial draft) DFA + +#+attr_latex: :width 200px +[[./img/problem_3_dfa.png]] + +And: + ++ $\Sigma = \{a, b\}$ ++ $Q = \{q_0, q_1, q_2, q_3\}$ ++ $F = \{q_2\}$ ++ $S = q_0$ ++ $\delta(q_0, a) = q_1$, $\delta(q_0, b) = \emptyset$, $\delta(q_1, a) = \emptyset$, $\delta(q_1, b) = q_2$, $\delta(q_2, a) = q_3$, $\delta(q_2, a = q_1)$, + $\delta(q_3, a) = q_2$, $\delta(q_3, b) = \emptyset$ + +We can build a FA that recognizes strings in $L(G)$, so it is regular. + +* Question Four +$G = (\{S\}, \{0,1,\cdots,9\}, S, \{S \rightarrow 0S | 1S | 2S | 3S | 4S | 5S | 6S | 7S | 8S | 9S | \epsilon\})$ diff --git a/Homework/cs5000/hw04/hw04.pdf b/Homework/cs5000/hw04/hw04.pdf Binary files differnew file mode 100644 index 0000000..3e97c6f --- /dev/null +++ b/Homework/cs5000/hw04/hw04.pdf diff --git a/Homework/cs5000/hw04/hw04.tex b/Homework/cs5000/hw04/hw04.tex new file mode 100644 index 0000000..9aa70a1 --- /dev/null +++ b/Homework/cs5000/hw04/hw04.tex @@ -0,0 +1,146 @@ +% Created 2023-09-27 Wed 14:13 +% Intended LaTeX compiler: pdflatex +\documentclass[11pt]{article} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +\usepackage{graphicx} +\usepackage{longtable} +\usepackage{wrapfig} +\usepackage{rotating} +\usepackage[normalem]{ulem} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{capt-of} +\usepackage{hyperref} +\notindent \notag \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} +\author{Elizabeth Hunt} +\date{\today} +\title{HW 04} +\hypersetup{ + pdfauthor={Elizabeth Hunt}, + pdftitle={HW 04}, + pdfkeywords={}, + pdfsubject={}, + pdfcreator={Emacs 28.2 (Org mode 9.7-pre)}, + pdflang={English}} +\begin{document} + +\maketitle +\setlength\parindent{20pt} + +\section{Question One} +\label{sec:org1166362} +Consider the regular language \(L\) over \(\Sigma = \{0,1,2\}\) such that \(L = \{x^{\star} | x \in \Sigma\}\) +, then \(L\) is a language; its members being all possible combinations of any length of all +\(x \in \Sigma\). + +\(L\) is a \uline{regular} language since there is a FA to describe it: + +\begin{itemize} +\item \(F = \{q_0\}\) +\item \(\Sigma = \{0,1,2\}\) +\item \(S = q_0\) +\item \(\delta(q_0, 0) = q_0\), \(\delta(q_0, 1) = q_0\), \(\delta(q_0, 2) = q_0\) +\item \(Q = {q_0}\) +\end{itemize} + +Let a set of languages \(G\) exist such that \(G_1 = \{0^i 1^i | i \geq 0\}\), \(G_2 = \{(0^i 1^i)(0^i 1^i) | i \geq 0\}\) +\(\cdots\) \(G_n = \{(0^i 1^i)^n | i \geq 0\}\). \(G_1\) is irregular by the proof found in Lecture 5. Then we assume that +\(G_k\) is irregular. If so, we can show \(G_{k+1}\) is irregular because we can only construct a FA to recognize +\(G_{k+1}\) if and only if we can concatenate a FA recognizing \(G_k\) in an epsilon transition with another FA recognizing \(G_1\); which is +not existant. By induction, any such \(G_i | i \in \mathds{N}\) is irregular. + +Each \(G\) is also a proper sublanguage since for each \(i \in \mathds{N}\) we can construct \((01)^{i+1}\) +which is not in \(G_i\) but in \(L\), so \(\nexist x \in G | x = L\). For extra clarity we know every string in \(G_i\) is +also in \(L\) since \(L\) is really the Kleene Closure. + +Thus there are at least \(\aleph_0\) infinitely many such non-regular proper sublanguages of the regular language \(L\). + +\section{Question Two} +\label{sec:orgf857515} +\subsection{One (adapted from slide notes in Lecture 5)} +\label{sec:org606e4e3} +Consider a minimal DFA \(M\) that recognizes \(L\); \(L = L(M)\) with \(k\) states. + +Then consider the string \(a^k b^k c^k\); to first recognize \(a^k\) we go through \(k+1\) states, so we can +find a loop in the path taken via \(\delta\) such that there exists \(q = \delta^{\star}(q, a^i) | i > 0\). + +If we pump this loop zero times, then for the string \(a^j b^k c^k\), \(j < k\); for one or more times, +\(j > k\); thus \(j < k\) or \(j > k\) but \(j \neq k\), a contradiction from the original definition. + +\subsection{Two} +\label{sec:org6221b61} + +Any string in this language is an even number of \(a\)'s, recognized by this FA (thus, is a regular language). + +\begin{itemize} +\item \(F = \{q_0\}\) +\item \(\Sigma = \{a\}\) +\item \(S = q_0\) +\item \(\delta(q_0, a) = q_1, \delta(q_1, a) = q_0\) +\item \(Q = \{q_0, q_1\}\) +\end{itemize} + +\subsection{Three} +\label{sec:org4423d9f} +Consider a minimal DFA \(M\) that recognizes \(L\); \(L = L(M)\) with \(k\) states. By the pumping lemma +each string \(x \in L\) such that \(|x| \geq k\) is of the form \(uvw\) with \(|uv| \leq k\), \(|v| \geq 1\) and +\(uv^i w \in L \forall i \geq 0\). + +For any such \(k\) we create the string \(a^k c a^k \in L\), and because \(|uv| \leq k\) then \(uv\) matches at most +\(a^k\). So, \(u = a^m, v = a^n\) with \(m + n \leq k\), and thus \(w = a^{k - (m+n)} c a^k\). Additionally, since \(|v| \geq 1\), +\(n \ge 1\). + +By pumping \(v\) zero times we then have \(a^m a^{k-(m+n)}c a^k = a^{k-n} c a^k \notin L\) as \(n \geq 1\) so \(L\) must be irregular. + +\subsection{Four} +\label{sec:org719bada} +Consider a minimal DFA \(M\) that recognizes \(L\); \(L = L(M)\) with \(k\) states. By the pumping lemma +each string \(x \in L\) such that \(|x| \geq k\) is of the form \(uvw\) with \(|uv| \leq k\), \(|v| \geq 1\) and +\(uv^i w \in L \forall i \geq 0\). + +For any such \(k\) we create the string \(a^k c^k b^{2k} \in L\) and because \(|uv| \leq k\) then \(uv\) matches at most +\(a^k\). So, \(u = a^m, v = a^n\) with \(m + n \leq k\), and since \(|v| \geq 1\), \(n \ge 1\) thus \(w = a^{k - (m+n)} c^k b^{2k}\). +Additionally, since \(|v| \geq 1\), \(n \ge 1\). + +By pumping \(v\) zero times we then obtain \(a^m a^{k-(m+n)} c^k b^2k = a^{k-n} c^k b^{2k}\) but then \(k-n + k \neq 2k\) as +\(n \geq 1\), a contradiction; \(L\) is irregular. + +\subsection{Five} +\label{sec:org9f5d6ff} +Consider a minimal DFA \(M\) that recognizes \(L\); \(L = L(M)\) with \(k\) states. By the pumping lemma +each string \(x \in L\) such that \(|x| \geq k\) is of the form \(uvw\) with \(|uv| \leq k\), \(|v| \geq 1\) and +\(uv^i w \in L \forall i \geq 0\). + +For any such \(k\) we create the string \(0^k 1^{k-1} \in L\), and because \(|uv| \leq k\) then \(uv\) matches at most +\(0^k\). So, \(u = 0^m, v = 0^n\) with \(m + n \leq k\), thus \(w = 0^{k - (m+n)} 1^{k-1}\). Additionally, since \(|v| \geq 1\), \(n \ge 1\). + +By pumping \(v\) zero times we obtain the string \(0^m 0^{k-(m+n)} 1^{k-1} = 0^{k-n} 1^{k-1}\) and \(k-n\) cannot be greater than +\(k-1\), a contradiction; \(L\) is irregular. + +\section{Question Three} +\label{sec:org9247330} + +(pictorial draft) DFA + +\begin{center} +\includegraphics[width=200px]{./img/problem_3_dfa.png} +\end{center} + +And: + +\begin{itemize} +\item \(\Sigma = \{a, b\}\) +\item \(Q = \{q_0, q_1, q_2, q_3\}\) +\item \(F = \{q_2\}\) +\item \(S = q_0\) +\item \(\delta(q_0, a) = q_1\), \(\delta(q_0, b) = \emptyset\), \(\delta(q_1, a) = \emptyset\), \(\delta(q_1, b) = q_2\), \(\delta(q_2, a) = q_3\), \(\delta(q_2, a = q_1)\), +\(\delta(q_3, a) = q_2\), \(\delta(q_3, b) = \emptyset\) +\end{itemize} + +We can build a FA that recognizes strings in \(L(G)\), so it is regular. + +\section{Question Four} +\label{sec:org7d011c1} +\(G = (\{S\}, \{0,1,\cdots,9\}, S, \{S \rightarrow 0S | 1S | 2S | 3S | 4S | 5S | 6S | 7S | 8S | 9S | \epsilon\})\) +\end{document}
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+# module: cnfg.py
+# description: Chomsky Normal Form Grammar
+# bugs to vladimir kulyukin in canvas
+###############################################
+
+class CNFG:
+
+ def __init__(self, startSymbol = "S", productions = dict()):
+ self._productions = productions
+ self._startSymbol = startSymbol
+
+ def __str__(self):
+ productions = list()
+ for lhs in self._productions:
+ for rhs in self._productions[lhs]:
+ productions.append(lhs + " -> " + str(rhs))
+ return "\n".join(productions)
+
+ def add_production(self, lhs, rhs1, rhs2 = None):
+ if lhs in self._productions:
+ rhs = self._productions.get(lhs)
+ rhs.append(CNFProductionRHS(rhs1, rhs2))
+ else:
+ rhs = list()
+ rhs.append(CNFProductionRHS(rhs1, rhs2))
+ self._productions[lhs] = rhs
+
+ def fetch_lhs(self, rhs1, rhs2 = None):
+ lhs_list = set()
+ for lhs in self._productions:
+ for prod_rhs in self._productions[lhs]:
+ if rhs2 is None:
+ if prod_rhs.is_rhs1() and prod_rhs.is_rhs1_equal(rhs1):
+ lhs_list.add(lhs)
+ else:
+ if prod_rhs.is_rhs2() and prod_rhs.is_rhs2_equal(rhs1, rhs2):
+ lhs_list.add(lhs)
+
+ return lhs_list
+
+ def clear_productions(self):
+ self._productions.clear()
+
+ def get_start_symbol(self):
+ return self._startSymbol
+
+ def display(self):
+ print(str(self))
+
+
+class CNFProductionRHS:
+ def __init__(self, rhs1, rhs2 = None):
+ if rhs2 is None:
+ self._rhs_list = list(rhs1)
+ else:
+ self._rhs_list = list([rhs1, rhs2])
+
+ def __str__(self):
+ return "".join(self._rhs_list)
+
+ def is_rhs1(self):
+ return len(self._rhs_list) == 1
+
+ def is_rhs2(self):
+ return len(self._rhs_list) == 2
+
+ def is_rhs1_equal(self, rhs):
+ return self._rhs_list[0] == rhs
+
+ def is_rhs2_equal(self, rhs1, rhs2):
+ return self._rhs_list[0] == rhs1 and self._rhs_list[1] == rhs2
+
+ def get_productions(self):
+ if self.is_rhs1():
+ return self._rhs_list[0]
+ elif self.is_rhs2():
+ return self._rhs_list[0], self._rhs_list[1]
+ else:
+ return None
+
+
diff --git a/Homework/cs5000/hw05/PyCYK/cyk.py b/Homework/cs5000/hw05/PyCYK/cyk.py new file mode 100644 index 0000000..e02e34a --- /dev/null +++ b/Homework/cs5000/hw05/PyCYK/cyk.py @@ -0,0 +1,43 @@ +###############################################
+# module: cyk.py
+# Elizabeth Hunt
+# A02364151
+###############################################
+
+
+def format_table(derives: list[list[set]], max_row: int) -> str:
+ table = ""
+ n = len(derives)
+ for row in range(max_row):
+ curr_row = f"{row+1}: "
+ for col in range(n - row):
+ set_of_nonterminals = " ".join(derives[row][col])
+ curr_row += f"|{set_of_nonterminals}|"
+ table += curr_row + "\n"
+ return table
+
+
+class CYK(object):
+ @staticmethod
+ def is_in_cfl(test_string, cnfg, table_display_flag=False):
+ n = len(test_string)
+ derives = [[set() for _ in range(n)] for _ in range(n)]
+
+ for i, terminal in enumerate(test_string):
+ derives[0][i] |= cnfg.fetch_lhs(terminal)
+ if table_display_flag:
+ print(format_table(derives, 1))
+
+ for length in range(2, n + 1):
+ for start in range(n - length + 1):
+ for split_idx in range(1, length):
+ for first in derives[split_idx - 1][start]:
+ for second in derives[length - split_idx - 1][
+ start + split_idx
+ ]:
+ derives[length - 1][start] |= cnfg.fetch_lhs(first, second)
+
+ if table_display_flag:
+ print(format_table(derives, length))
+
+ return "S" in derives[n - 1][0]
diff --git a/Homework/cs5000/hw05/PyCYK/cyktest.py b/Homework/cs5000/hw05/PyCYK/cyktest.py new file mode 100644 index 0000000..0fa5e59 --- /dev/null +++ b/Homework/cs5000/hw05/PyCYK/cyktest.py @@ -0,0 +1,272 @@ +###############################################
+# module: cyktest.py
+# description: tests CYK in cyk.py
+# bugs to vladimir kulyukin via canvas
+###############################################
+
+import unittest
+import cyk
+from cyk import CYK
+from cnfg import CNFG
+
+class TestCykAlgorithm(unittest.TestCase):
+ _grammar = CNFG()
+
+ def _use_grammar1(self):
+ ## 1. S -> AB
+ ## 2. S --> BC
+ ## 3. A --> BA
+ ## 4. A --> a
+ ## 5. B --> CC
+ ## 6. B --> b
+ ## 7. C --> AB
+ ## 8. C --> a
+ self._grammar.clear_productions()
+ self._grammar = CNFG()
+ self._grammar.add_production("S", "A", "B")
+ self._grammar.add_production("S", "B", "C")
+ self._grammar.add_production("A", "B", "A")
+ self._grammar.add_production("A", "a")
+ self._grammar.add_production("B", "C", "C")
+ self._grammar.add_production("B", "b")
+ self._grammar.add_production("C", "A", "B")
+ self._grammar.add_production("C", "a")
+
+ def _use_grammar2(self):
+ ## 1. S -> AB
+ ## 2. S --> BB
+ ## 3. A --> CC
+ ## 4. A --> AB
+ ## 5. A --> a
+ ## 6. B --> BB
+ ## 7. B --> CA
+ ## 8. B --> b
+ ## 9. C --> BA
+ ## 10. C --> AA
+ ## 11. C --> b
+ self._grammar.clear_productions()
+ self._grammar = CNFG();
+ self._grammar.add_production("S", "A", "B")
+ self._grammar.add_production("S", "B", "B")
+ self._grammar.add_production("A", "C", "C")
+ self._grammar.add_production("A", "A", "B")
+ self._grammar.add_production("A", "a")
+ self._grammar.add_production("B", "B", "B")
+ self._grammar.add_production("B", "C", "A")
+ self._grammar.add_production("B", "b")
+ self._grammar.add_production("C", "B", "A")
+ self._grammar.add_production("C", "A", "A")
+ self._grammar.add_production("C", "b")
+
+ def _use_grammar3(self):
+ ## 1. S -> AD1
+ ## 2. D1 --> BC
+ ## 3. C --> BD2
+ ## 4. D2 --> AB
+ ## 5. C --> c
+ ## 6. B --> BB
+ ## 7. B --> b
+ ## 8. A --> a
+ self._grammar._productions.clear()
+ self._grammar = CNFG();
+ self._grammar.add_production("S", "A", "D1")
+ self._grammar.add_production("D1", "B", "C")
+ self._grammar.add_production("C", "B", "D2")
+ self._grammar.add_production("D2", "A", "B")
+ self._grammar.add_production("C", "c")
+ self._grammar.add_production("B", "B", "B")
+ self._grammar.add_production("B", "b")
+ self._grammar.add_production("A", "a")
+
+ def test_ut0(self):
+ print("===== Test 00 =====")
+ grammar = CNFG()
+ grammar.clear_productions()
+ grammar.add_production("S", "A", "B")
+ grammar.add_production("S", "B", "C")
+ grammar.add_production("A", "B", "A")
+ grammar.add_production("A", "a")
+ grammar.add_production("B", "C", "C")
+ grammar.add_production("B", "b")
+ grammar.add_production("C", "A", "B")
+ grammar.add_production("C", "a")
+ print("LHS for RHS AB is {}".format(grammar.fetch_lhs("A", "B")))
+
+ ### ===================== Grammar 1 Tests #############################
+
+ def test1a(self):
+ print("===== Test 1a =====")
+ self._use_grammar1()
+ #self._grammar.display()
+ input_str = 'ab'
+ print('Input string: ' + input_str)
+ result = (CYK.is_in_cfl(input_str, self._grammar))
+ print('Result = ' + str(result))
+
+ def test1(self):
+ print("===== Test 1 =====")
+ self._use_grammar1()
+ self._grammar.display()
+ input_str = 'baaba'
+ print('Input string: ' + input_str)
+ result = (CYK.is_in_cfl(input_str, self._grammar))
+ print('Result = ' + str(result))
+
+ def test2(self):
+ print("===== Test 2 =====")
+ self._use_grammar1()
+ #self._grammar.display()
+ input_str = 'baaa'
+ print('Input string: ' + input_str)
+ result = (CYK.is_in_cfl(input_str, self._grammar))
+ print('Result = ' + str(result))
+
+ def test3(self):
+ print("===== Test 3 =====")
+ self._use_grammar1()
+ #self._grammar.display()
+ input_str = 'baba'
+ print('Input string: ' + input_str)
+ result = (CYK.is_in_cfl(input_str, self._grammar))
+ print('Result = ' + str(result))
+
+ def test4(self):
+ print("===== Test 4 =====")
+ self._use_grammar1()
+ #self._grammar.display()
+ input_str = 'baaabab'
+ print('Input string: ' + input_str)
+ result = (CYK.is_in_cfl(input_str, self._grammar))
+ print('Result = ' + str(result))
+
+ def test4a(self):
+ print('===== Test 4a =====')
+ self._use_grammar1()
+ input_str = 'baab'
+ print('Input string: {}'.format(input_str))
+ result = (CYK.is_in_cfl(input_str, self._grammar, True))
+ print('Result = ' + str(result))
+
+ def test4b(self):
+ print('===== Test 4b =====')
+ self._use_grammar1()
+ input_str = 'aaab'
+ print('Input string: {}'.format(input_str))
+ result = (CYK.is_in_cfl(input_str, self._grammar, True))
+ print('Result = ' + str(result))
+
+ ### ===================== Grammar 2 Tests #############################
+
+ def test5(self):
+ print("===== Test 5 =====")
+ self._use_grammar2()
+ #self._grammar.display()
+ input_str = 'aabb'
+ print('Input string: ' + input_str)
+ result = (CYK.is_in_cfl(input_str, self._grammar))
+ print('Result = ' + str(result))
+
+ def test5a(self):
+ print("===== Test 5a =====")
+ self._use_grammar2()
+ #self._grammar.display()
+ input_str = 'bbb'
+ print('Input string: ' + input_str)
+ result = (CYK.is_in_cfl(input_str, self._grammar))
+ print('Result = ' + str(result))
+
+ def test5b(self):
+ print("===== Test 5b =====")
+ self._use_grammar2()
+ #self._grammar.display()
+ input_str = 'bbb'
+ print('Input string: ' + input_str)
+ result = (CYK.is_in_cfl(input_str, self._grammar))
+ print('Result = ' + str(result))
+
+ def test5c(self):
+ print("===== Test 5c =====")
+ self._use_grammar2()
+ #self._grammar.display()
+ input_str = 'cccccb'
+ print('Input string: ' + input_str)
+ result = (CYK.is_in_cfl(input_str, self._grammar))
+ print('Result = ' + str(result))
+
+ def test5d(self):
+ print("===== Test 5d =====")
+ self._use_grammar2()
+ #self._grammar.display()
+ input_str = 'bababaaa'
+ print('Input string: ' + input_str)
+ result = (CYK.is_in_cfl(input_str, self._grammar))
+ print('Result = ' + str(result))
+
+ ### ===================== Grammar 3 Tests #############################
+
+ def test6(self):
+ print("===== Test 6 =====")
+ self._use_grammar3()
+ input_str = 'abc'
+ print('Input string: ' + input_str)
+ result = (CYK.is_in_cfl(input_str, self._grammar))
+ print('Result = ' + str(result))
+
+ def test7(self):
+ print("===== Test 7 =====")
+ self._use_grammar3()
+ input_str = 'abbbabb'
+ print('Input string: ' + input_str)
+ result = (CYK.is_in_cfl(input_str, self._grammar))
+ print('Result = ' + str(result))
+
+ def test7b(self):
+ print("===== Test 7b =====")
+ self._use_grammar3()
+ input_str = 'abbc'
+ print('Input string: ' + input_str)
+ result = (CYK.is_in_cfl(input_str, self._grammar))
+ print('Result = ' + str(result))
+
+ def test8(self):
+ print("===== Test 8 =====")
+ self._use_grammar3()
+ input_str = 'bbc'
+ print('Input string: ' + input_str)
+ result = (CYK.is_in_cfl(input_str, self._grammar))
+ print('Result = ' + str(result))
+
+ def test9(self):
+ print("===== Test 9 =====")
+ self._use_grammar3()
+ input_str = 'aaabb'
+ print('Input string: ' + input_str)
+ result = (CYK.is_in_cfl(input_str, self._grammar))
+ print('Result = ' + str(result))
+
+ def test10(self):
+ print("===== Test 10 =====")
+ self._use_grammar3()
+ input_str = 'ab'
+ print('Input string: ' + input_str)
+ result = (CYK.is_in_cfl(input_str, self._grammar))
+ print('Result = ' + str(result))
+
+if __name__ == '__main__':
+ unittest.main()
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/Homework/cs5000/hw06/hw06.org b/Homework/cs5000/hw06/hw06.org new file mode 100644 index 0000000..c48e3f5 --- /dev/null +++ b/Homework/cs5000/hw06/hw06.org @@ -0,0 +1,174 @@ +#+TITLE: HW 06 +#+AUTHOR: Elizabeth Hunt (A02364151) +#+STARTUP: entitiespretty fold inlineimages +#+LATEX_HEADER: \notindent \notag \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} +#+LATEX: \setlength\parindent{20pt} +#+OPTIONS: toc:nil + +* Problem One +** 1 - { a^i b^j c^k | i > j > k } +We assume that there exists a Context Free Language, $L$ such that $L = { a^i b^j c^k | i > j > k }$. +Then by the Pumping Lemma for Context Free Languages, for each $z \in L$ of sufficient length $n$ we can +split $z$ into the sequence of substrings $uvwxy$ with $|vx| \geq 1$ and $|vwx| \le n$. Any such "pumped" +string $uv^i w x^i y \forall i \ge 0$ is also in $L$. + +Consider the string $a^{n+2} b^{n+1} c^{}^n \in L$ which is always longer than $n$ (the integer of the Pumping Lemma) +so it is of sufficient length. Splitting this up according to the aforementioned Lemma, we know +$|vwx| \le n$ and thus the string $vx$ contains either all of one letter or two different letters. + +If $vx$ contains only $a$'s or $b$'s then we can "pump" $vx$ 0 times to produce $z^* = u v^0 w x^0 y$ and +since $|vx| \geq 1$ then we lose $p \geq 1$ $a$'s or $b$'s. Thus, in the case of $a$, $(n+2)-p \leq n+1$ and +$z^*$ is not in $L$, and in the case of $b$, $(n+1)-p \leq n$, with the same conclusion. + +If $vx$ contains only $c$'s then we can "pump" $vx$ $n$ times to produce $z^* = u v^n w x^n y$ at least +$2n$ the number of $c's$ (which is always $\ge n$), and there would be equal or more $c$'s to +$b$'s and thus $z^* \notin L$. + +If $vx$ contains both $a$'s and $b$'s then we can "pump" $vx$ 0 times similar to the case of only $a$'s +and $b$'s to produce a string with an equal or less number of $b$'s than $c$'s. + +If $vx$ contains both $b$'s and $c$'s then we can "pump" $vx$ $3n$ times to come up with a string that +will always have $b$'s and $c$'s greater or eqaul to the number of $a$'s. + +** 2 - { a^i b^j | i = j^2 } + +We assume that there exists a Context Free Language, $L$ such that $L = { a^i b^j | i = j^2 }$. + +Then by the Pumping Lemma for Context Free Languages, for each $z \in L$ of sufficient length $n$ we can +split $z$ into the sequence of substrings $uvwxy$ with $|vx| \geq 1$ and $|vwx| \le n$. Any such "pumped" +string $uv^i w x^i y \forall i \ge 0$ is also in $L$. + +Consider the string $a^{n^2}^{} b^{}^n \in L$ which is always longer than $n$ (the integer of the Pumping Lemma) +so it is of sufficient length. Splitting this up according to the aforementioned Lemma, we know +$|vwx| \le n$ and thus the string $vx$ contains either all of one letter or two different letters. + +1. If $vx$ contains only $a$'s then we can "pump" $vx$ 0 times to produce $z^* = u v^0 w x^0 y$ and + since $|vx| \geq 1$ then we lose $p \geq 1$ $a$'s and thus $z^* = a^{n^2 - p} b^n \notin L$ as $n^2 - p \ne n^2$. + +2. If $vx$ contains only $b$'s then we can "pump" $vx$ 2 times to produce $z^* = u v^2 w x^2 y$ and + since $|vx| \geq 1$ then we add $p \geq 2$ $b$'s and thus $z^* = a^{n^2} b^{n+p} \notin L$ as $n^2 \ne (n + p)^2$. + +3. If $vx$ contains both $b$'s and $a$'s then we can prove that we can construct a number of times to pump + such that the relationship for $i = j^2$ cannot be satisfied: + + $v$ must contain only $a$'s and $x$ must contain only $b$'s; else we could pump and obtain a string + in which the order of $a$'s and $b$'s is disrupted. + + Let $y = |v|$ and $u = |x|$ with $1 \leq y + u \le n$ since $|vwx| \le n$ and $|vx| \geq 1$. + Then we can pump $p \geq 1$ times to create the string + $z^* = a^{(n^2 - y) + py} b^{(n - u) + pu} = a^{n^2 + p(y-1)} b^{n + p(u-1)}$. + + Then for any such $n, u, y$ we can find a $p$ such that + $(n + p(u - 1))(n + p(u - 1)) = n^2 + 2pn(u-1) + p^2(u-1)^2 \neq n^2 + p(y-1)$ + since $u$ and $y$ are positive. + +** 3 - { a^i | i is prime } + +We assume that there exists a Context Free Language, $L$ such that $L = { a^i b^j | i = j^2 }$. + +Then by the Pumping Lemma for Context Free Languages, for each $z \in L$ of sufficient length $n$ we can +split $z$ into the sequence of substrings $uvwxy$ with $|vx| \geq 1$ and $|vwx| \le n$. Any such "pumped" +string $uv^i w x^i y \forall i \ge 0$ is also in $L$. + +Consider the string $z = a^{p(n)} \in L$ where $p(n)$ is a function returning the first prime number greater +than $n$; this string is always longer than $n$ (the integer of the Pumping Lemma) so +it is of sufficient length. Splitting this up according to the aforementioned Lemma, we know +$|vwx| \le n$ and thus the string $vx$ is made up of one or more $a$'s. + +Given $t = |v|$ and $u = |x|$, and $1 \le t + u \le n$; we can simply pump $vx$ $p(n)$ times to obtain +the string $z^* = uv^{tp(n)} w x^{up(n)}y$ which is $a^{p(n) + tp(n) + up(n)}$. And $p(n) + tp(n) + up(n)$ is not prime +since it can be factored into $p(n)(1 + t + u)$; as either $t$ or $u$ is greater than one, then +$z^* \notin L$. + +* Problem Two + +[[https://bit.ly/cs5000-simponic-hw06-p02]] + + +* Problem Three + +[[https://bit.ly/cs5000-simponic-hw06-p03]] + +In english(ish): + +1. If the current symbol is $B$, move right and goto 2 + +2. If the current symbol is $a, b, c$, move right and goto 2. If the current symbol is $B$, move Left and goto 3. + If the current symbol is * we move left and goto 6. + +3. We're at the last $c$ in the string. If the current symbol is $c$ then we print $ and go left until + we see an $a$. Then, move Left and go to 4. + +4. If the current symbol is $b$ then we print * and go right until we see $. At which we move left and + goto 3. If the symbol is * then we move right and goto 5. + +5. If the current symbol is b, c, *, we move right and goto 5. If it's $ we move left and goto 2 to do the next + iteration. + +6. We move left until we hit $B$ at which we move right and goto 7. + +7. We're now at the beginning of the $a$'s after replacing all $c$'s with $ and $b$'s with *. If we read + an $a$ we print $+$ and goto 8. If we see a $-$ (which we replace $*$'s with), we accept. + +8. If the current symbol is $+, a, *$, we move right and goto 8. If we see $ or $-$ we move left and + goto 9. + +9. If the current symbol is $*$ then we print $-$ and goto 9. If we see $-$ we move left and goto 10. + +10. If the current symbol is $*, a$ then we move left and goto 10. If we see $+$ we move right and goto + 7 to do another iteration. + + +* Problem Four +We can write the control tuples for the two-tape turing machine in the form: $(q_{\text{from}} , s_1, s_2, a_1, a_2, q_{\text{to}})$ +where s_n is the expected symbol on tape $n$, $a_n$ is the action (R / L / new symbol) to perform on tape $n$, $q_{\text{from}}$ +is the current state, and $q_{\text{to}}$ is the next state. + +At a high level we can translate these control tuples to a single tape by separating the single tape into two sections; the +first section simulates the first tape, and the other the second tape. These can be split by a single symbol ~+~ not in the alphabet (with the +first cell also being ~+~, representing the beginning of the input on the first tape). Additionally, the position of the tape +head of the turing machine being simulated in its section can be delimited by some symbol not in the input alphabet in each section; call this +symbol ~*~. + +Thus, to build the initial tape we can construct the tape ~+ * <input on tape 1> + * <input on tape 2>~ (the head being at the beginning). +Then, for each tuple in the two-tape turing machine $(q_{\text{from}} , s_1, s_2, a_1, a_2, q_{\text{to}})$ we construct a set of new states that will: + +1. Search until we find ~*~ (we're at the head of the first "tape"), at which we go right one, and if we read s_1 goto 2. +2. Search until we find ~*~ (we're at the head of the second "tape"), at which we go right one, and if we read s_2 goto 3. +3. We perform a_2. + + If $s_2$ is the ~+~ symbol, we perform a new set of states to analyze the next symbol after; if it's non-~+~ we shift all symbols proceeding + to the right (this is a catch for 4 in the case we need to overwrite the ~+~ cell delimiting the two sections) and print a blank symbol. Then continue. + + If a_2 is "R" then we go left one (to the ~*~), write s_2, and go right one. + + If a_2 is "L" then we go back 2 (to the symbol preceding ~*~) and from a new set of states constructed from all symbols in the alphabet go right and copy that symbol. + Then we go back one and print "*". + + Else, we print a_2. + + Finally, go left and goto 4. +4. Move left until we find ~*~ (we're now at the head of the first "tape") again and perform the same algorithm as 3 but with $s_2 \rightarrow s_1 , a_2 \rightarrow a_1$. Then goto 5. +5. Move left until we find the ~+~ symbol (we're at the head of the first tape) and move to the state corresponding to 1 with q_{\text{to}}. + +* Problem Five +We can write the control quadruples for the 3D turing machine in the form: $(q_{\text{from}}, s, a, q_{\text{to}})$ +where $s$ is the expected symbol and $a$ is the movement action $\{ +x, -x, +y, -y, +z, -z \}$ or a symbol to print; +where each movement action is a single step on the specified axis by either positive or negative one. + +At a high level we will apply our strategy in Problem 4 of delimination to construct several rows of tape into a +"matrix" and then scale those into several matrices to produce a 3-dimensional tape; as one could emulate with +~(make-array (list z y x))~. + +Then if we construct a one dimensional tape such that each "row" is a "tape" in the z axis then we "paste" the input +into a one dimensional tape in the form ~- = + * _ <...[(0, 0, 0), (0, 0, 1)...]>~ with the first five unique symbols not in the alphabet; +the ~-~ indicating a new "matrix" in the z-y plane, and each ~+~ symbol a new row in the z direction. But now, we need to keep track of +where we are in the $z$ and $y$ planes for every $x$. To do so we can also create another three new symbols not in the +alphabet; say ~*~ to stick with Problem 4 for the $z$ plane, ~=~ for the $y$ plane, and ~_~ for the actual placement of the +current head of the machine. + +Now, for each control tuple of the 3D turing machine we can construct a new set of states that will, at a high level: + +1. Search until we find ~_~ (we're at the head of the tape), at which we go right one, and if we read $s$ goto 2. +2. If $a$ is $\pm z$ then we move right or left. + + If this symbol is ~+~ we must then traverse to the very beginning of the tape and "increase the size" of every single $z$ since the + tape has the potential to now go in the $x$ or $y$ axes; we do this by searching for the first ~+~, move right, shift the entire tape + right one and print a blank symbol (to add an extra cell at the beginning), then continue searching for ~+~ and for each: shift the entire tape right one, + move left, print a blank symbol, move right two, shift everything to the right one, move left, and print a blank symbol (adding a blank cell before and after + each ~+~). + + Then we go back to the beginning of the tape and continue moving right until we see ~_~. +3. If $a$ is $\pm y$ or $\pm x$ then move left or right until the first instance of either ~=~ or ~-~ respectively and perform a similar kind of algorithm to copy + the rows or matrices with blank $z$ rows of the same size, adding rows / matrices if necessary should we hit the beginning or end of the tape (for $x$) or + a new "matrix" (~-~ for $y$).. diff --git a/Homework/cs5000/hw06/hw06.pdf b/Homework/cs5000/hw06/hw06.pdf Binary files differnew file mode 100644 index 0000000..d898ce4 --- /dev/null +++ b/Homework/cs5000/hw06/hw06.pdf diff --git a/Homework/cs5000/hw06/hw06.tex b/Homework/cs5000/hw06/hw06.tex new file mode 100644 index 0000000..3d0af5f --- /dev/null +++ b/Homework/cs5000/hw06/hw06.tex @@ -0,0 +1,220 @@ +% Created 2023-11-02 Thu 12:14 +% Intended LaTeX compiler: pdflatex +\documentclass[11pt]{article} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +\usepackage{graphicx} +\usepackage{longtable} +\usepackage{wrapfig} +\usepackage{rotating} +\usepackage[normalem]{ulem} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{capt-of} +\usepackage{hyperref} +\notindent \notag \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} +\author{Elizabeth Hunt (A02364151)} +\date{\today} +\title{HW 06} +\hypersetup{ + pdfauthor={Elizabeth Hunt (A02364151)}, + pdftitle={HW 06}, + pdfkeywords={}, + pdfsubject={}, + pdfcreator={Emacs 28.2 (Org mode 9.7-pre)}, + pdflang={English}} +\begin{document} + +\maketitle +\setlength\parindent{20pt} + +\section{Problem One} +\label{sec:orgf4c2128} +\subsection{1 - \{ a\textsuperscript{i} b\textsuperscript{j} c\textsuperscript{k} | i > j > k \}} +\label{sec:orgd221679} +We assume that there exists a Context Free Language, \(L\) such that \(L = { a^i b^j c^k | i > j > k }\). +Then by the Pumping Lemma for Context Free Languages, for each \(z \in L\) of sufficient length \(n\) we can +split \(z\) into the sequence of substrings \(uvwxy\) with \(|vx| \geq 1\) and \(|vwx| \le n\). Any such "pumped" +string \(uv^i w x^i y \forall i \ge 0\) is also in \(L\). + +Consider the string \(a^{n+2} b^{n+1} c^{}^n \in L\) which is always longer than \(n\) (the integer of the Pumping Lemma) +so it is of sufficient length. Splitting this up according to the aforementioned Lemma, we know +\(|vwx| \le n\) and thus the string \(vx\) contains either all of one letter or two different letters. + +If \(vx\) contains only \(a\)'s or \(b\)'s then we can "pump" \(vx\) 0 times to produce \(z^* = u v^0 w x^0 y\) and +since \(|vx| \geq 1\) then we lose \(p \geq 1\) \(a\)'s or \(b\)'s. Thus, in the case of \(a\), \((n+2)-p \leq n+1\) and +\(z^*\) is not in \(L\), and in the case of \(b\), \((n+1)-p \leq n\), with the same conclusion. + +If \(vx\) contains only \(c\)'s then we can "pump" \(vx\) \(n\) times to produce \(z^* = u v^n w x^n y\) at least +\(2n\) the number of \(c's\) (which is always \(\ge n\)), and there would be equal or more \(c\)'s to +\(b\)'s and thus \(z^* \notin L\). + +If \(vx\) contains both \(a\)'s and \(b\)'s then we can "pump" \(vx\) 0 times similar to the case of only \(a\)'s +and \(b\)'s to produce a string with an equal or less number of \(b\)'s than \(c\)'s. + +If \(vx\) contains both \(b\)'s and \(c\)'s then we can "pump" \(vx\) \(3n\) times to come up with a string that +will always have \(b\)'s and \(c\)'s greater or eqaul to the number of \(a\)'s. + +\subsection{2 - \{ a\textsuperscript{i} b\textsuperscript{j} | i = j\textsuperscript{2} \}} +\label{sec:org5f8cc3e} + +We assume that there exists a Context Free Language, \(L\) such that \(L = { a^i b^j | i = j^2 }\). + +Then by the Pumping Lemma for Context Free Languages, for each \(z \in L\) of sufficient length \(n\) we can +split \(z\) into the sequence of substrings \(uvwxy\) with \(|vx| \geq 1\) and \(|vwx| \le n\). Any such "pumped" +string \(uv^i w x^i y \forall i \ge 0\) is also in \(L\). + +Consider the string \(a^{n^2}^{} b^{}^n \in L\) which is always longer than \(n\) (the integer of the Pumping Lemma) +so it is of sufficient length. Splitting this up according to the aforementioned Lemma, we know +\(|vwx| \le n\) and thus the string \(vx\) contains either all of one letter or two different letters. + +\begin{enumerate} +\item If \(vx\) contains only \(a\)'s then we can "pump" \(vx\) 0 times to produce \(z^* = u v^0 w x^0 y\) and +since \(|vx| \geq 1\) then we lose \(p \geq 1\) \(a\)'s and thus \(z^* = a^{n^2 - p} b^n \notin L\) as \(n^2 - p \ne n^2\). + +\item If \(vx\) contains only \(b\)'s then we can "pump" \(vx\) 2 times to produce \(z^* = u v^2 w x^2 y\) and +since \(|vx| \geq 1\) then we add \(p \geq 2\) \(b\)'s and thus \(z^* = a^{n^2} b^{n+p} \notin L\) as \(n^2 \ne (n + p)^2\). + +\item If \(vx\) contains both \(b\)'s and \(a\)'s then we can prove that we can construct a number of times to pump +such that the relationship for \(i = j^2\) cannot be satisfied: +\begin{itemize} +\item \(v\) must contain only \(a\)'s and \(x\) must contain only \(b\)'s; else we could pump and obtain a string +in which the order of \(a\)'s and \(b\)'s is disrupted. +\item Let \(y = |v|\) and \(u = |x|\) with \(1 \leq y + u \le n\) since \(|vwx| \le n\) and \(|vx| \geq 1\). +Then we can pump \(p \geq 1\) times to create the string +\(z^* = a^{(n^2 - y) + py} b^{(n - u) + pu} = a^{n^2 + p(y-1)} b^{n + p(u-1)}\). +\item Then for any such \(n, u, y\) we can find a \(p\) such that +\((n + p(u - 1))(n + p(u - 1)) = n^2 + 2pn(u-1) + p^2(u-1)^2 \neq n^2 + p(y-1)\) +since \(u\) and \(y\) are positive. +\end{itemize} +\end{enumerate} + +\subsection{3 - \{ a\textsuperscript{i} | i is prime \}} +\label{sec:orgc7ac705} + +We assume that there exists a Context Free Language, \(L\) such that \(L = { a^i b^j | i = j^2 }\). + +Then by the Pumping Lemma for Context Free Languages, for each \(z \in L\) of sufficient length \(n\) we can +split \(z\) into the sequence of substrings \(uvwxy\) with \(|vx| \geq 1\) and \(|vwx| \le n\). Any such "pumped" +string \(uv^i w x^i y \forall i \ge 0\) is also in \(L\). + +Consider the string \(z = a^{p(n)} \in L\) where \(p(n)\) is a function returning the first prime number greater +than \(n\); this string is always longer than \(n\) (the integer of the Pumping Lemma) so +it is of sufficient length. Splitting this up according to the aforementioned Lemma, we know +\(|vwx| \le n\) and thus the string \(vx\) is made up of one or more \(a\)'s. + +Given \(t = |v|\) and \(u = |x|\), and \(1 \le t + u \le n\); we can simply pump \(vx\) \(p(n)\) times to obtain +the string \(z^* = uv^{tp(n)} w x^{up(n)}y\) which is \(a^{p(n) + tp(n) + up(n)}\). And \(p(n) + tp(n) + up(n)\) is not prime +since it can be factored into \(p(n)(1 + t + u)\); as either \(t\) or \(u\) is greater than one, then +\(z^* \notin L\). + +\section{Problem Two} +\label{sec:org0e7b7e5} + +\url{https://bit.ly/cs5000-simponic-hw06-p02} + + +\section{Problem Three} +\label{sec:org3b85eb1} + +\url{https://bit.ly/cs5000-simponic-hw06-p03} + +In english(ish): + +\begin{enumerate} +\item If the current symbol is \(B\), move right and goto 2 + +\item If the current symbol is \(a, b, c\), move right and goto 2. If the current symbol is \(B\), move Left and goto 3. +If the current symbol is * we move left and goto 6. + +\item We're at the last \(c\) in the string. If the current symbol is \(c\) then we print \$ and go left until +we see an \(a\). Then, move Left and go to 4. + +\item If the current symbol is \(b\) then we print * and go right until we see \$. At which we move left and +goto 3. If the symbol is * then we move right and goto 5. + +\item If the current symbol is b, c, *, we move right and goto 5. If it's \$ we move left and goto 2 to do the next +iteration. + +\item We move left until we hit \(B\) at which we move right and goto 7. + +\item We're now at the beginning of the \(a\)'s after replacing all \(c\)'s with \$ and \(b\)'s with *. If we read +an \(a\) we print \(+\) and goto 8. If we see a \(-\) (which we replace \(*\)'s with), we accept. + +\item If the current symbol is \(+, a, *\), we move right and goto 8. If we see \$ or \(-\) we move left and +goto 9. + +\item If the current symbol is \(*\) then we print \(-\) and goto 9. If we see \(-\) we move left and goto 10. + +\item If the current symbol is \(*, a\) then we move left and goto 10. If we see \(+\) we move right and goto +7 to do another iteration. +\end{enumerate} + + +\section{Problem Four} +\label{sec:orgdacb3cd} +We can write the control tuples for the two-tape turing machine in the form: \((q_{\text{from}} , s_1, s_2, a_1, a_2, q_{\text{to}})\) +where s\textsubscript{n} is the expected symbol on tape \(n\), \(a_n\) is the action (R / L / new symbol) to perform on tape \(n\), \(q_{\text{from}}\) +is the current state, and \(q_{\text{to}}\) is the next state. + +At a high level we can translate these control tuples to a single tape by separating the single tape into two sections; the +first section simulates the first tape, and the other the second tape. These can be split by a single symbol \texttt{+} not in the alphabet (with the +first cell also being \texttt{+}, representing the beginning of the input on the first tape). Additionally, the position of the tape +head of the turing machine being simulated in its section can be delimited by some symbol not in the input alphabet in each section; call this +symbol \texttt{*}. + +Thus, to build the initial tape we can construct the tape \texttt{+ * <input on tape 1> + * <input on tape 2>} (the head being at the beginning). +Then, for each tuple in the two-tape turing machine \((q_{\text{from}} , s_1, s_2, a_1, a_2, q_{\text{to}})\) we construct a set of new states that will: + +\begin{enumerate} +\item Search until we find \texttt{*} (we're at the head of the first "tape"), at which we go right one, and if we read s\textsubscript{1} goto 2. +\item Search until we find \texttt{*} (we're at the head of the second "tape"), at which we go right one, and if we read s\textsubscript{2} goto 3. +\item We perform a\textsubscript{2}. +\begin{itemize} +\item If \(s_2\) is the \texttt{+} symbol, we perform a new set of states to analyze the next symbol after; if it's non-\texttt{+} we shift all symbols proceeding +to the right (this is a catch for 4 in the case we need to overwrite the \texttt{+} cell delimiting the two sections) and print a blank symbol. Then continue. +\item If a\textsubscript{2} is "R" then we go left one (to the \texttt{*}), write s\textsubscript{2}, and go right one. +\item If a\textsubscript{2} is "L" then we go back 2 (to the symbol preceding \texttt{*}) and from a new set of states constructed from all symbols in the alphabet go right and copy that symbol. +Then we go back one and print "*". +\item Else, we print a\textsubscript{2}. +\item Finally, go left and goto 4. +\end{itemize} +\item Move left until we find \texttt{*} (we're now at the head of the first "tape") again and perform the same algorithm as 3 but with \(s_2 \rightarrow s_1 , a_2 \rightarrow a_1\). Then goto 5. +\item Move left until we find the \texttt{+} symbol (we're at the head of the first tape) and move to the state corresponding to 1 with q\textsubscript{\text{to}}. +\end{enumerate} + +\section{Problem Five} +\label{sec:orgab00621} +We can write the control quadruples for the 3D turing machine in the form: \((q_{\text{from}}, s, a, q_{\text{to}})\) +where \(s\) is the expected symbol and \(a\) is the movement action \(\{ +x, -x, +y, -y, +z, -z \}\) or a symbol to print; +where each movement action is a single step on the specified axis by either positive or negative one. + +At a high level we will apply our strategy in Problem 4 of delimination to construct several rows of tape into a +"matrix" and then scale those into several matrices to produce a 3-dimensional tape; as one could emulate with +\texttt{(make-array (list z y x))}. + +Then if we construct a one dimensional tape such that each "row" is a "tape" in the z axis then we "paste" the input +into a one dimensional tape in the form \texttt{- = + * \_ <...[(0, 0, 0), (0, 0, 1)...]>} with the first five unique symbols not in the alphabet; +the \texttt{-} indicating a new "matrix" in the z-y plane, and each \texttt{+} symbol a new row in the z direction. But now, we need to keep track of +where we are in the \(z\) and \(y\) planes for every \(x\). To do so we can also create another three new symbols not in the +alphabet; say \texttt{*} to stick with Problem 4 for the \(z\) plane, \texttt{=} for the \(y\) plane, and \texttt{\_} for the actual placement of the +current head of the machine. + +Now, for each control tuple of the 3D turing machine we can construct a new set of states that will, at a high level: + +\begin{enumerate} +\item Search until we find \texttt{\_} (we're at the head of the tape), at which we go right one, and if we read \(s\) goto 2. +\item If \(a\) is \(\pm z\) then we move right or left. +\begin{itemize} +\item If this symbol is \texttt{+} we must then traverse to the very beginning of the tape and "increase the size" of every single \(z\) since the +tape has the potential to now go in the \(x\) or \(y\) axes; we do this by searching for the first \texttt{+}, move right, shift the entire tape +right one and print a blank symbol (to add an extra cell at the beginning), then continue searching for \texttt{+} and for each: shift the entire tape right one, +move left, print a blank symbol, move right two, shift everything to the right one, move left, and print a blank symbol (adding a blank cell before and after +each \texttt{+}). +\item Then we go back to the beginning of the tape and continue moving right until we see \texttt{\_}. +\end{itemize} +\item If \(a\) is \(\pm y\) or \(\pm x\) then move left or right until the first instance of either \texttt{=} or \texttt{-} respectively and perform a similar kind of algorithm to copy +the rows or matrices with blank \(z\) rows of the same size, adding rows / matrices if necessary should we hit the beginning or end of the tape (for \(x\)) or +a new "matrix" (\texttt{-} for \(y\)).. +\end{enumerate} +\end{document}
\ No newline at end of file diff --git a/Homework/cs5000/hw07/hw07.org b/Homework/cs5000/hw07/hw07.org new file mode 100644 index 0000000..b88e959 --- /dev/null +++ b/Homework/cs5000/hw07/hw07.org @@ -0,0 +1,58 @@ +#+TITLE: HW 07 +#+AUTHOR: Elizabeth Hunt (A02364151) +#+STARTUP: entitiespretty fold inlineimages +#+LATEX_HEADER: \notindent \notag \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} +#+LATEX: \setlength\parindent{20pt} +#+OPTIONS: toc:nil + +* Problem One + +\begin{verbatim} +1. [A1] Y <- Y - 1 +2. IF Y != 0 GOTO A +3. [B1] IF X1 != 0 GOTO C +4. GOTO E +5. [C1] X1 <- X1 - 1 +6. Y <- Y + 1 +7. Y <- Y + 1 +8. Y <- Y + 1 +9. GOTO B1 +\end{verbatim} + +* Problem Two +1. $(1, \sigma) | \sigma = \{X_1 = 2, Y = 0, Z_1 = 0\}$ +2. $(4, \sigma) | \sigma = \{X_1 = 2, Y = 0, Z_1 = 0\}$ +3. $(5, \sigma) | \sigma = \{X_1 = 1, Y = 0, Z_1 = 0\}$ +4. $(6, \sigma) | \sigma = \{X_1 = 1, Y = 1, Z_1 = 0\}$ +5. $(7, \sigma) | \sigma = \{X_1 = 1, Y = 1, Z_1 = 1\}$ +6. $(1, \sigma) | \sigma = \{X_1 = 1, Y = 1, Z_1 = 1\}$ +7. $(4, \sigma) | \sigma = \{X_1 = 1, Y = 1, Z_1 = 1\}$ +8. $(5, \sigma) | \sigma = \{X_1 = 0, Y = 1, Z_1 = 1\}$ +9. $(6, \sigma) | \sigma = \{X_1 = 0, Y = 2, Z_1 = 1\}$ +10. $(7, \sigma) | \sigma = \{X_1 = 0, Y = 2, Z_1 = 2\}$ +11. $(1, \sigma) | \sigma = \{X_1 = 0, Y = 2, Z_1 = 2\}$ +12. $(2, \sigma) | \sigma = \{X_1 = 0, Y = 2, Z_1 = 2\}$ +13. $(3, \sigma) | \sigma = \{X_1 = 0, Y = 2, Z_1 = 3\}$ +14. $(8, \sigma) | \sigma = \{X_1 = 0, Y = 2, Z_1 = 3\}$ + +* Problem Three +\begin{verbatim} +1. [A1] Y <- Y +2. Y <- Y +3. Y <- Y +4. Y <- Y +5. Y <- Y +6. GOTO E +\end{verbatim} + +* Problem Four +Let $P$ be a program in $L$ that computes $g(x_1, x_2, \cdots, x_n)$; a list of instructions $[I_1, I_2, \cdots, I_k]$, +where $I_1$ is the first instruction and $I_k$ the last. + +Then, define $P^i | i \in N$ to be a new program such that each instruction $I_n$ replaces $I_{n+i}$ (when $n=0$ +we perform no operation), appending +to the end of the instruction list if necessary. We then replace the sublist $[I_1, \cdots, I_i]$ with +$[Y \leftarrow Y]^i$ in the program $P$. As $Y \leftarrow Y$ produces no side effects then $P^i$ still computes $g$. + +Finally, for all $i \in N$ the length of $P^i$ is greater than $k$ and thus there are countably infinitely +many $L$ -programs to compute $g$. diff --git a/Homework/cs5000/hw07/hw07.pdf b/Homework/cs5000/hw07/hw07.pdf Binary files differnew file mode 100644 index 0000000..6d555a1 --- /dev/null +++ b/Homework/cs5000/hw07/hw07.pdf diff --git a/Homework/cs5000/hw07/hw07.tex b/Homework/cs5000/hw07/hw07.tex new file mode 100644 index 0000000..53c92bd --- /dev/null +++ b/Homework/cs5000/hw07/hw07.tex @@ -0,0 +1,88 @@ +% Created 2023-11-04 Sat 18:02 +% Intended LaTeX compiler: pdflatex +\documentclass[11pt]{article} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +\usepackage{graphicx} +\usepackage{longtable} +\usepackage{wrapfig} +\usepackage{rotating} +\usepackage[normalem]{ulem} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{capt-of} +\usepackage{hyperref} +\notindent \notag \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} +\author{Elizabeth Hunt (A02364151)} +\date{\today} +\title{HW 07} +\hypersetup{ + pdfauthor={Elizabeth Hunt (A02364151)}, + pdftitle={HW 07}, + pdfkeywords={}, + pdfsubject={}, + pdfcreator={Emacs 28.2 (Org mode 9.7-pre)}, + pdflang={English}} +\begin{document} + +\maketitle +\setlength\parindent{20pt} + +\section{Problem One} +\label{sec:orgd2348b9} + +\begin{verbatim} +1. [A1] Y <- Y - 1 +2. IF Y != 0 GOTO A +3. [B1] IF X1 != 0 GOTO C +4. GOTO E +5. [C1] X1 <- X1 - 1 +6. Y <- Y + 1 +7. Y <- Y + 1 +8. Y <- Y + 1 +9. GOTO B1 +\end{verbatim} + +\section{Problem Two} +\label{sec:org07c7432} +\begin{enumerate} +\item \((1, \sigma) | \sigma = \{X_1 = 2, Y = 0, Z_1 = 0\}\) +\item \((4, \sigma) | \sigma = \{X_1 = 2, Y = 0, Z_1 = 0\}\) +\item \((5, \sigma) | \sigma = \{X_1 = 1, Y = 0, Z_1 = 0\}\) +\item \((6, \sigma) | \sigma = \{X_1 = 1, Y = 1, Z_1 = 0\}\) +\item \((7, \sigma) | \sigma = \{X_1 = 1, Y = 1, Z_1 = 1\}\) +\item \((1, \sigma) | \sigma = \{X_1 = 1, Y = 1, Z_1 = 1\}\) +\item \((4, \sigma) | \sigma = \{X_1 = 1, Y = 1, Z_1 = 1\}\) +\item \((5, \sigma) | \sigma = \{X_1 = 0, Y = 1, Z_1 = 1\}\) +\item \((6, \sigma) | \sigma = \{X_1 = 0, Y = 2, Z_1 = 1\}\) +\item \((7, \sigma) | \sigma = \{X_1 = 0, Y = 2, Z_1 = 2\}\) +\item \((1, \sigma) | \sigma = \{X_1 = 0, Y = 2, Z_1 = 2\}\) +\item \((2, \sigma) | \sigma = \{X_1 = 0, Y = 2, Z_1 = 2\}\) +\item \((3, \sigma) | \sigma = \{X_1 = 0, Y = 2, Z_1 = 3\}\) +\item \((8, \sigma) | \sigma = \{X_1 = 0, Y = 2, Z_1 = 3\}\) +\end{enumerate} + +\section{Problem Three} +\label{sec:orgf9c18d1} +\begin{verbatim} +1. [A1] Y <- Y +2. Y <- Y +3. Y <- Y +4. Y <- Y +5. Y <- Y +6. GOTO E +\end{verbatim} + +\section{Problem Four} +\label{sec:org49ea029} +Let \(P\) be a program in \(L\) that computes \(g(x_1, x_2, \cdots, x_n)\); a list of instructions \([I_1, I_2, \cdots, I_k]\), +where \(I_1\) is the first instruction and \(I_k\) the last. + +Then, define \(P^i | i \in N\) to be a new program such that each instruction \(I_n\) replaces \(I_{n+i}\) (when \(n=0\) +we perform no operation), appending +to the end of the instruction list if necessary. We then replace the sublist \([I_1, \cdots, I_i]\) with +\([Y \leftarrow Y]^i\) in the program \(P\). As \(Y \leftarrow Y\) produces no side effects then \(P^i\) still computes \(g\). + +Finally, for all \(i \in N\) the length of \(P^i\) is greater than \(k\) and thus there are countably infinitely +many \(L\) -programs to compute \(g\). +\end{document}
\ No newline at end of file diff --git a/Homework/cs5000/hw08/hw08.org b/Homework/cs5000/hw08/hw08.org new file mode 100644 index 0000000..fddc47b --- /dev/null +++ b/Homework/cs5000/hw08/hw08.org @@ -0,0 +1,89 @@ +#+TITLE: HW 08 +#+AUTHOR: Elizabeth Hunt (A02364151) +#+STARTUP: entitiespretty fold inlineimages +#+LATEX_HEADER: \notindent \notag \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} +#+LATEX: \setlength\parindent{0pt} +#+OPTIONS: toc:nil + +* Problem One +Let $m$ be the number of statements that don't produce an effect on $Y$ i.e. +$\{ Y \leftarrow Y, X1 \leftarrow X1 + 1, \cdots \}$, $n$ be the number of statements $Y \leftarrow Y - 1$, and +$o$ be the number of statements $Y \leftarrow Y + 1$. A straightline program of length $k$ +must have $m + n + o = k$ instructions. Then, the terminal snapshot ends with +$Y = o - n + 0m$. Thus, $Y$ is at a maximum if and only if $o = k$, thus $Y = k$, +and at a minimum if and only if $o = n$, thus $Y = 0$ (we cannot have $n > o$ as +by definition of $L$, for a valid program $P$ $Y$ is a non-negative number). + +* Problem Two +Let $P$ be a program in $L$ that computes $f$. Then $f$ is partially computable +by definition, and is partially computable in $L++$ trivially since $L++$ is a +"superset" of $L$. ($L++ \Leftarrow L$). + +Then let $Q$ be a program in $L++$ that computes $f$, $f$ is partially computable +if we can convert $Q$ into $L$. For each instruction $i \in Q$, if $i$ is in the form +$V \leftarrow k | k \in N$ then we replace $i$ with the sequence of instructions and unique labels +(with the notation \cdots representing a reapeating instruction until the instruction is the +following index), else write $i$ ($L++ \Rightarrow L$). + +\begin{verbatim} +1. [A1] IF V != 0 GOTO B1 +2. GOTO C1 +3. [B1] V <- V - 1 +4. GOTO A1 +5. [C1] V <- V + 1 +... V <- V + 1 +(5 + k - 1). V <- V + 1 +\end{verbatim} + +And thus any program in $L++$ is also partially computable. + +* Problem Three +For $n=0$, $i(n, x) = f^0(x) = x$ is primitive recursive since it can be represented as +a projection function $u_1^1(x) = x$. If $f^n$ is primitive recursive, then so is $f^{n+1}$, as $f^{n+1}$ +is a finite composition $f(f^n)$. By induction, for any $n \in N$, $i(n, x)$ is primitive recursive, and thus computable. + +* Problem Four +** One +Let $m \in N$ represent the number of compositions in a function such that the set containing all $n$ -ary +functions with "composition number" $m$ are of the form + +$f(x_1, \cdots, x_n) = g(f_1(x_1, \cdots, x_n), \cdots, f_j(x_1, \cdots, x_n))$ +where each $f_i$ is in the set of functions with a "composition number" $\le m - 1$. Except for the case in which +$m = 0$ - the initial functions; which are all of the form $k$ or $x_i + k$. + +Then, let $F^m | m \in N$ represent the set of all functions of composition number $m$ or less, +and assume that $\forall f \in F^m$ are of either form $k$ or $x_i + k$. Then, $g \in F^{m+1} | m \in N$'s elements +are $g(x_1, \cdots, x_n) = f(f_1(x_1, \cdots, x_n), \cdots, f_j(x_1, \cdots, x_n))$ where $f$ is an initial function, by definition, and each +$f_i | 1 \le i \le j \in F^m$. +1. If $f$ is the Zero function $g$ is in the form $k = 0$ +2. If $f$ is the Successor function $g$ is some $f_i(\cdots) + 1$ which by assumption is thus $(x_i + k) + 1 \Rightarrow x_i + k$ or $k + 1 \Rightarrow k$. +3. If $f$ is the Projection function $g$ is some $f_i(\cdots)$ which by assumption is $(x_i + k)$ or $k$. + +Finally, by induction, $\forall f(x_1, \cdots, x_n) \in$ ~COMP~, $f$ is thus in the form $k$ or $x_i + k | 1 \le i \le n$. + +** Two +By what we showed in One, $f(x_1, \cdots, x_n)$ is dependent on a single $x_i \in {x_1, \cdots, x_n}$ and produces $r = (x_i + k)$ or $r = k$. Then $f(y_1, \cdots, y_n) = (y_i + k) | k$ +for the same $i$. So if for all $i | 1 \le i \le n$, $x_i \le y_i$ then $(y_i + k) \ge (x_i + k)$ or $k = k$. Thus in both cases $f(y_1, \cdots, y_n) \ge f(x_1, \cdots, x_n)$ +and is thus monotonic. + +** Three +Yes. Firstly, every $f \in$ ~COMP~ is primitive recursive, just without the recursion, so it is a subset of the set of all primitive recursive +functions. + +Consider the (proven in class) primitive recursive function $f(x_1, x_2) = x_1 * x_2$, and the equivalent $g \in$ ~COMP~. $g(x_1, x_2)$ is equivalent to +$x_1 + k$, $x_2 + k$, or $k$. Then there all elements of $\{f(x_1, x_2 + 1), f(x_1 + 1, x_2), f(x_1, x_2)\} | x_1, x_2 > 1$ cannot be present in +$\{g(x_1, x_2 + 1), g(x_1 + 1, x_2), g(x_1, x_2)\}$ for the same $x_1, x_2$ due to the dependence relation, which is a contradiction to $f$ and $g$ +being equivalent. + +** Four +Yes. By the fact that all primitive recursive functions are computable, ~COMP~ being a set of primitive recursive functions from Three, is a subset of computable functions. +However, we also showed in Three that there exists a computable (primitive recursive) function which is not in ~COMP~, so ~COMP~ is not equivalent to the set +of all computable functions. + +* Problem Five +1. Let $a(x_1, x_2) = x_1 + x_2$ and $m(x_1, x_2) = x_1 * x_2$ which are primitive recursive by proofs in class. +2. Let $t(x_1)$ be the composition of the successor function $s$ 10 times on the zero function $z$ function: $t(x_1) = s(s(\cdots(n(x_1))))$, and is primitive recursive. +3. Let $v(x_1)$ be $v(x_1) = a(x_1, a(x_1, a(x_1, a(x_1, u_1^1(x_1)))))$, which is primitive recursive. +4. Let $w(x_1)$ be $w(x_1) = a(t(x_1), v(x_1))$, which is primitive recursive. +5. Let $y(x_1)$ be $y(x_1) = m(m(x_1, x_1), s(s(n(x_1))))$, which is primitive recursive. +6. Then $f(x_1)$ is $a(w(x_1), y(x_1))$, which is primitive recursive. diff --git a/Homework/cs5000/hw08/hw08.pdf b/Homework/cs5000/hw08/hw08.pdf Binary files differnew file mode 100644 index 0000000..f297511 --- /dev/null +++ b/Homework/cs5000/hw08/hw08.pdf diff --git a/Homework/cs5000/hw08/hw08.tex b/Homework/cs5000/hw08/hw08.tex new file mode 100644 index 0000000..c20ca1f --- /dev/null +++ b/Homework/cs5000/hw08/hw08.tex @@ -0,0 +1,118 @@ +% Created 2023-11-11 Sat 20:04 +% Intended LaTeX compiler: pdflatex +\documentclass[11pt]{article} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +\usepackage{graphicx} +\usepackage{longtable} +\usepackage{wrapfig} +\usepackage{rotating} +\usepackage[normalem]{ulem} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{capt-of} +\usepackage{hyperref} +\notindent \notag \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} +\author{Elizabeth Hunt (A02364151)} +\date{\today} +\title{HW 08} +\hypersetup{ + pdfauthor={Elizabeth Hunt (A02364151)}, + pdftitle={HW 08}, + pdfkeywords={}, + pdfsubject={}, + pdfcreator={Emacs 29.1 (Org mode 9.7-pre)}, + pdflang={English}} +\begin{document} + +\maketitle +\setlength\parindent{0pt} +\section{Problem One} +\label{sec:org7aa3a78} +Let \(m\) be the number of statements that don't produce an effect on \(Y\) i.e. +\(\{ Y \leftarrow Y, X1 \leftarrow X1 + 1, \cdots \}\), \(n\) be the number of statements \(Y \leftarrow Y - 1\), and +\(o\) be the number of statements \(Y \leftarrow Y + 1\). A straightline program of length \(k\) +must have \(m + n + o = k\) instructions. Then, the terminal snapshot ends with +\(Y = o - n + 0m\). Thus, \(Y\) is at a maximum if and only if \(o = k\), thus \(Y = k\), +and at a minimum if and only if \(o = n\), thus \(Y = 0\) (we cannot have \(n > o\) as +by definition of \(L\), for a valid program \(P\) \(Y\) is a non-negative number). +\section{Problem Two} +\label{sec:orgb9efdfb} +Let \(P\) be a program in \(L\) that computes \(f\). Then \(f\) is partially computable +by definition, and is partially computable in \(L++\) trivially since \(L++\) is a +"superset" of \(L\). (\(L++ \Leftarrow L\)). + +Then let \(Q\) be a program in \(L++\) that computes \(f\), \(f\) is partially computable +if we can convert \(Q\) into \(L\). For each instruction \(i \in Q\), if \(i\) is in the form +\(V \leftarrow k | k \in N\) then we replace \(i\) with the sequence of instructions and unique labels +(with the notation \(\cdots{}\) representing a reapeating instruction until the instruction is the +following index), else write \(i\) (\(L++ \Rightarrow L\)). + +\begin{verbatim} +1. [A1] IF V != 0 GOTO B1 +2. GOTO C1 +3. [B1] V <- V - 1 +4. GOTO A1 +5. [C1] V <- V + 1 +... V <- V + 1 +(5 + k - 1). V <- V + 1 +\end{verbatim} + +And thus any program in \(L++\) is also partially computable. +\section{Problem Three} +\label{sec:org53e881b} +For \(n=0\), \(i(n, x) = f^0(x) = x\) is primitive recursive since it can be represented as +a projection function \(u_1^1(x) = x\). If \(f^n\) is primitive recursive, then so is \(f^{n+1}\), as \(f^{n+1}\) +is a finite composition \(f(f^n)\). By induction, for any \(n \in N\), \(i(n, x)\) is primitive recursive, and thus computable. +\section{Problem Four} +\label{sec:orgbc1477c} +\subsection{One} +\label{sec:org96da465} +Let \(m \in N\) represent the number of compositions in a function such that the set containing all \(n\) -ary +functions with "composition number" \(m\) are of the form + +\(f(x_1, \cdots, x_n) = g(f_1(x_1, \cdots, x_n), \cdots, f_j(x_1, \cdots, x_n))\) +where each \(f_i\) is in the set of functions with a "composition number" \(\le m - 1\). Except for the case in which +\(m = 0\) - the initial functions; which are all of the form \(k\) or \(x_i + k\). + +Then, let \(F^m | m \in N\) represent the set of all functions of composition number \(m\) or less, +and assume that \(\forall f \in F^m\) are of either form \(k\) or \(x_i + k\). Then, \(g \in F^{m+1} | m \in N\)'s elements +are \(g(x_1, \cdots, x_n) = f(f_1(x_1, \cdots, x_n), \cdots, f_j(x_1, \cdots, x_n))\) where \(f\) is an initial function, by definition, and each +\(f_i | 1 \le i \le j \in F^m\). +\begin{enumerate} +\item If \(f\) is the Zero function \(g\) is in the form \(k = 0\) +\item If \(f\) is the Successor function \(g\) is some \(f_i(\cdots) + 1\) which by assumption is thus \((x_i + k) + 1 \Rightarrow x_i + k\) or \(k + 1 \Rightarrow k\). +\item If \(f\) is the Projection function \(g\) is some \(f_i(\cdots)\) which by assumption is \((x_i + k)\) or \(k\). +\end{enumerate} + +Finally, by induction, \(\forall f(x_1, \cdots, x_n) \in\) \texttt{COMP}, \(f\) is thus in the form \(k\) or \(x_i + k | 1 \le i \le n\). +\subsection{Two} +\label{sec:org74c3eb4} +By what we showed in One, \(f(x_1, \cdots, x_n)\) is dependent on a single \(x_i \in {x_1, \cdots, x_n}\) and produces \(r = (x_i + k)\) or \(r = k\). Then \(f(y_1, \cdots, y_n) = (y_i + k) | k\) +for the same \(i\). So if for all \(i | 1 \le i \le n\), \(x_i \le y_i\) then \((y_i + k) \ge (x_i + k)\) or \(k = k\). Thus in both cases \(f(y_1, \cdots, y_n) \ge f(x_1, \cdots, x_n)\) +and is thus monotonic. +\subsection{Three} +\label{sec:org8b3878c} +Yes. Firstly, every \(f \in\) \texttt{COMP} is primitive recursive, just without the recursion, so it is a subset of the set of all primitive recursive +functions. + +Consider the (proven in class) primitive recursive function \(f(x_1, x_2) = x_1 * x_2\), and the equivalent \(g \in\) \texttt{COMP}. \(g(x_1, x_2)\) is equivalent to +\(x_1 + k\), \(x_2 + k\), or \(k\). Then there all elements of \(\{f(x_1, x_2 + 1), f(x_1 + 1, x_2), f(x_1, x_2)\} | x_1, x_2 > 1\) cannot be present in +\(\{g(x_1, x_2 + 1), g(x_1 + 1, x_2), g(x_1, x_2)\}\) for the same \(x_1, x_2\) due to the dependence relation, which is a contradiction to \(f\) and \(g\) +being equivalent. +\subsection{Four} +\label{sec:orgcea60eb} +Yes. By the fact that all primitive recursive functions are computable, \texttt{COMP} being a set of primitive recursive functions from Three, is a subset of computable functions. +However, we also showed in Three that there exists a computable (primitive recursive) function which is not in \texttt{COMP}, so \texttt{COMP} is not equivalent to the set +of all computable functions. +\section{Problem Five} +\label{sec:orgab79ce2} +\begin{enumerate} +\item Let \(a(x_1, x_2) = x_1 + x_2\) and \(m(x_1, x_2) = x_1 * x_2\) which are primitive recursive by proofs in class. +\item Let \(t(x_1)\) be the composition of the successor function \(s\) 10 times on the zero function \(z\) function: \(t(x_1) = s(s(\cdots(n(x_1))))\), and is primitive recursive. +\item Let \(v(x_1)\) be \(v(x_1) = a(x_1, a(x_1, a(x_1, a(x_1, u_1^1(x_1)))))\), which is primitive recursive. +\item Let \(w(x_1)\) be \(w(x_1) = a(t(x_1), v(x_1))\), which is primitive recursive. +\item Let \(y(x_1)\) be \(y(x_1) = m(m(x_1, x_1), s(s(n(x_1))))\), which is primitive recursive. +\item Then \(f(x_1)\) is \(a(w(x_1), y(x_1))\), which is primitive recursive. +\end{enumerate} +\end{document} diff --git a/Homework/cs5000/midterm/cs5000_midterm_01.pdf b/Homework/cs5000/midterm/cs5000_midterm_01.pdf Binary files differnew file mode 100644 index 0000000..fd2ea3c --- /dev/null +++ b/Homework/cs5000/midterm/cs5000_midterm_01.pdf diff --git a/Homework/cs5000/midterm/img/6b.png b/Homework/cs5000/midterm/img/6b.png Binary files differnew file mode 100644 index 0000000..aec6bad --- /dev/null +++ b/Homework/cs5000/midterm/img/6b.png diff --git a/Homework/cs5000/midterm/img/7.png b/Homework/cs5000/midterm/img/7.png Binary files differnew file mode 100644 index 0000000..7abf636 --- /dev/null +++ b/Homework/cs5000/midterm/img/7.png diff --git a/Homework/cs5000/midterm/img/p6.png b/Homework/cs5000/midterm/img/p6.png Binary files differnew file mode 100644 index 0000000..f1a268a --- /dev/null +++ b/Homework/cs5000/midterm/img/p6.png diff --git a/Homework/cs5000/midterm/img/prob_2_parse_tree.png b/Homework/cs5000/midterm/img/prob_2_parse_tree.png Binary files differnew file mode 100644 index 0000000..5600029 --- /dev/null +++ b/Homework/cs5000/midterm/img/prob_2_parse_tree.png diff --git a/Homework/cs5000/midterm/midterm.org b/Homework/cs5000/midterm/midterm.org new file mode 100644 index 0000000..1ad41f7 --- /dev/null +++ b/Homework/cs5000/midterm/midterm.org @@ -0,0 +1,188 @@ +#+TITLE: Theory of Computability Midterm 1 +#+AUTHOR: Elizabeth Hunt (A02364151) +#+STARTUP: entitiespretty fold inlineimages +#+LATEX_HEADER: \notindent \notag \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} +#+LATEX: \setlength\parindent{20pt} +#+OPTIONS: toc:nil + +* Problem 1 +** Stage 1 +We skip Stage 1; there are no productions in the form $A \rightarrow BC$ or $A \rightarrow s$. + +$P' = \{ \}$ + +** Stage 2 +$P' = \{ C_a \rightarrow a , C_b \rightarrow b, C_c \rightarrow c, C_d \rightarrow d \}$ + +And our new productions are $\{ S \rightarrow C_a S C_b C_b , S \rightarrow C_a S C_a , S \rightarrow C_b S C_a C_a , S \rightarrow C_b S C_b , S \rightarrow C_c C_d \}$ + +** Stage 3 + +We replace $S \rightarrow C_a S C_b C_b$ with $\{ S \rightarrow C_a D_1 , D_1 \rightarrow S D_2 , D_2 \rightarrow C_b C_b \}$ + +We replace $S \rightarrow C_a S C_a$ with $\{ S \rightarrow C_a D_3, D_3 \rightarrow S C_a \}$ + +We replace $S \rightarrow C_b S C_a C_a$ with $\{ S \rightarrow C_b D_4 , D_4 \rightarrow S D_5 , D_5 \rightarrow C_a C_a \}$ + +We replace $S \rightarrow C_b S C_b$ with $\{ S \rightarrow C_b D_6 , D_6 \rightarrow S C_b \}$. + +We add $S \rightarrow C_c C_d$ as it is in CNF already. + +Thus, + +\begin{align*} +P' &= \{ C_a \rightarrow a , C_b \rightarrow b, C_c \rightarrow c, C_d \rightarrow d \} \\ + & \cup \{ S \rightarrow C_a D_1 , D_1 \rightarrow S D_2 , D_2 \rightarrow C_b C_b \} \\ + & \cup \{ S \rightarrow C_a D_3, D_3 \rightarrow S C_a \} \\ + & \cup \{ S \rightarrow C_b D_4 , D_4 \rightarrow S D_5 , D_5 \rightarrow C_a C_a \} \\ + & \cup \{ S \rightarrow C_b D_6 , D_6 \rightarrow S C_b \} \\ + & \cup \{ S \rightarrow C_c C_d \} +\end{align*} + +* Problem 2 + +#+attr_latex: :width 150px +[[./img/prob_2_parse_tree.png]] + +Yes, we can recognize the string by this derivation. + +* Problem 3 + +Because strings in $L(M_1)$ and $L(M_2)$ are recognized by Discrete Finite Automata, +they must be regular languages. + +By the Myhill-Nerode theorem, if $L$ is a regular language it can be recognized by a unique DFA +with a minimal number of states. In other words, we know that if two DFA recognize +the same language, they must have the same minimal DFA. + +Let $\text{minimize}(D)$ be the minimization algorithm given in Lecture 04 returning a deterministic +set of states. + +Then, we know $M_1$ is equivalent to $M_2$ when $\text{minimize}(M_1)$ is congruent to +$\text{minimize}(M_2)$. This is only true when all descriptors (\Sigma, q_0, \delta, etc...) are also +equivalent. + +In the below pseduo code we just check the equivalence of the set of states, alphabet, and start +state. Then we perform a search to see if $(\delta_1) = M_1$ is $\subseteq$ of $(\delta_2) = M_2$ and +$\delta_2 \subseteq \delta_1$, and if both are true, then $\delta_1 = \delta_2$. + +If all are equivalent, then the languages recognize the same strings! + +#+BEGIN_SRC python + def minimize(dfa): + minimized = dfa.copy() + # ... mutate minimized according to minimize() + return minimized + + def delta_subseteq(start_state, sigma, delta1, delta2): + visited = set() + for transition in delta2.keys(): + if transition not in delta1 or \ + delta1[transition] != delta2[transition]: + return False + return True + + def equivalent(m1, m2): + minimized_m1 = minimize(m1) + minimized_m2 = minimize(m2) + if minimized_m1.Q != minimized_m2.Q or \ + minimzed_m1.sigma != minimized_m2.sigma or \ + minimized_m1.q0 != minimized_m2.q0 or \ + minimized_m1.F != minimized_m2.F: + return False + + m2_delta_includes_m1_delta = delta_subseteq(minimized_m1.q0, \ + minimized_m1.sigma, \ + minimized_m1.delta, \ + minimized_m2.delta) + + m1_delta_includes_m2_delta = delta_subseteq(minimized_m2.q0, \ + minimized_m2.sigma, \ + minimized_m2.delta, \ + minimized_m1.delta) + + return m2_delta_includes_m1_delta and m1_delta_includes_m2_delta +#+END_SRC + + +* Problem 4 +We can construct a CFG: + +$S \rightarrow aSbbb | abbb$ + +Which we convert to a stack machine: + +| read | pop | push | +| \epsilon | S | aSbbb | +| \epsilon | S | abbb | +| a | a | \epsilon | +| b | b | \epsilon | + +$M = (\{a, b, S\}, \{a, b\}, S, \delta)$ + +where + +1. $\delta(\epsilon, S) = \{aSbbb, abbb\}$ +2. $\delta(a, a) = \{ \epsilon \}$ +3. $\delta(b, b) = \{ \epsilon \}$ + +* Problem 5 + +1. $S \rightarrow 0 | 0T | 1T$ +2. $T \rightarrow 1S | 0S$ + +Is a right linear grammar, and is thus regular. + +* Problem 6 +** One +#+attr_latex: :width 200px +[[./img/p6.png]] + ++ $Q = \{p_0, p_1\}$ ++ $F = \{p_1\}$ ++ $\Sigma = \{1\}$ ++ $S = p_0$ ++ $\delta(p_0, 1) = p_1$ ++ $\delta(p_1, 1) = p_0$ + +** Two + +#+attr_latex: :width 200px +[[./img/6b.png]] + ++ $Q = \{p_0, p_1, p_2, p_3, p_4, p_5\}$ ++ $F = \{p_2, p_4, p_6\}$ ++ $\Sigma = \{a, b\}$ ++ $S = p_0$ ++ $\delta(p_0, a) = p_1$ ++ $\delta(p_0, b) = p_3$ ++ $\delta(p_1, a) = p_6$ ++ $\delta(p_1, b) = p_2$ ++ $\delta(p_2, b) = p_5$ ++ $\delta(p_5, b) = p_2$ ++ $\delta(p_3, b) = p_4$ ++ $\delta(p_4, b) = p_3$ ++ $\delta(p_3, a) = \delta(p_4, a) = \delta(p_2, a) = \delta(p_5, a) = \emptyset$ + +* Problem 7 +#+attr_latex: :width 200px +[[./img/7.png]] + +Because the magnitude of each element in the range of $\delta$ is 1 then, intuitively, if we follow +the subset construction algorithm with queue optimization we will only end up with new states +identical to the ones present in the current NFA (i.e. we will visit no elements of the powerset +whose magnitude is greater than one also). + +Thus the DFA is: + ++ $Q = \{q_0, q_1, q_2, q_3, q_4\}$ ++ $F = \{q_4\}$ ++ $\Sigma = \{a, b\}$ ++ $S = q_0$ ++ $\delta(q_0, a) = q_1$ ++ $\delta(p_1, a) = q_2$ ++ $\delta(q_2, a) = q_2$ ++ $\delta(q_2, b) = q_3$ ++ $\delta(q_3, b) = q_4$ ++ $\delta(p_4, b) = q_4$ ++ $\delta(q_0, b) = \delta(q_1, b) = \delta(q_3, a) = \delta(p_4, a) = \emptyset$ diff --git a/Homework/cs5000/midterm/midterm.pdf b/Homework/cs5000/midterm/midterm.pdf Binary files differnew file mode 100644 index 0000000..29aa5bc --- /dev/null +++ b/Homework/cs5000/midterm/midterm.pdf diff --git a/Homework/cs5000/midterm/midterm.tex b/Homework/cs5000/midterm/midterm.tex new file mode 100644 index 0000000..20335ef --- /dev/null +++ b/Homework/cs5000/midterm/midterm.tex @@ -0,0 +1,242 @@ +% Created 2023-10-06 Fri 20:58 +% Intended LaTeX compiler: pdflatex +\documentclass[11pt]{article} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +\usepackage{graphicx} +\usepackage{longtable} +\usepackage{wrapfig} +\usepackage{rotating} +\usepackage[normalem]{ulem} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{capt-of} +\usepackage{hyperref} +\notindent \notag \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} +\author{Elizabeth Hunt (A02364151)} +\date{\today} +\title{Theory of Computability Midterm 1} +\hypersetup{ + pdfauthor={Elizabeth Hunt (A02364151)}, + pdftitle={Theory of Computability Midterm 1}, + pdfkeywords={}, + pdfsubject={}, + pdfcreator={Emacs 28.2 (Org mode 9.7-pre)}, + pdflang={English}} +\begin{document} + +\maketitle +\setlength\parindent{20pt} + +\section{Problem 1} +\label{sec:orgb0784e8} +\subsection{Stage 1} +\label{sec:org855b93a} +We skip Stage 1; there are no productions in the form \(A \rightarrow BC\) or \(A \rightarrow s\). + +\(P' = \{ \}\) + +\subsection{Stage 2} +\label{sec:org325eccc} +\(P' = \{ C_a \rightarrow a , C_b \rightarrow b, C_c \rightarrow c, C_d \rightarrow d \}\) + +And our new productions are \(\{ S \rightarrow C_a S C_b C_b , S \rightarrow C_a S C_a , S \rightarrow C_b S C_a C_a , S \rightarrow C_b S C_b , S \rightarrow C_c C_d \}\) + +\subsection{Stage 3} +\label{sec:orgeecaa22} + +We replace \(S \rightarrow C_a S C_b C_b\) with \(\{ S \rightarrow C_a D_1 , D_1 \rightarrow S D_2 , D_2 \rightarrow C_b C_b \}\) + +We replace \(S \rightarrow C_a S C_a\) with \(\{ S \rightarrow C_a D_3, D_3 \rightarrow S C_a \}\) + +We replace \(S \rightarrow C_b S C_a C_a\) with \(\{ S \rightarrow C_b D_4 , D_4 \rightarrow S D_5 , D_5 \rightarrow C_a C_a \}\) + +We replace \(S \rightarrow C_b S C_b\) with \(\{ S \rightarrow C_b D_6 , D_6 \rightarrow S C_b \}\). + +We add \(S \rightarrow C_c C_d\) as it is in CNF already. + +Thus, + +\begin{align*} +P' &= \{ C_a \rightarrow a , C_b \rightarrow b, C_c \rightarrow c, C_d \rightarrow d \} \\ + & \cup \{ S \rightarrow C_a D_1 , D_1 \rightarrow S D_2 , D_2 \rightarrow C_b C_b \} \\ + & \cup \{ S \rightarrow C_a D_3, D_3 \rightarrow S C_a \} \\ + & \cup \{ S \rightarrow C_b D_4 , D_4 \rightarrow S D_5 , D_5 \rightarrow C_a C_a \} \\ + & \cup \{ S \rightarrow C_b D_6 , D_6 \rightarrow S C_b \} \\ + & \cup \{ S \rightarrow C_c C_d \} +\end{align*} + +\section{Problem 2} +\label{sec:orgde63c33} + +\begin{center} +\includegraphics[width=150px]{./img/prob_2_parse_tree.png} +\end{center} + +Yes, we can recognize the string by this derivation. + +\section{Problem 3} +\label{sec:org4d6de8f} + +Because strings in \(L(M_1)\) and \(L(M_2)\) are recognized by Discrete Finite Automata, +they must be regular languages. + +By the Myhill-Nerode theorem, if \(L\) is a regular language it can be recognized by a unique DFA +with a minimal number of states. In other words, we know that if two DFA recognize +the same language, they must have the same minimal DFA. + +Let \(\text{minimize}(D)\) be the minimization algorithm given in Lecture 04 returning a deterministic +set of states. + +Then, we know \(M_1\) is equivalent to \(M_2\) when \(\text{minimize}(M_1)\) is congruent to +\(\text{minimize}(M_2)\). This is only true when all descriptors (\(\Sigma\), q\textsubscript{0}, \(\delta\), etc\ldots{}) are also +equivalent. + +In the below pseduo code we just check the equivalence of the set of states, alphabet, and start +state. Then we perform a search to see if \((\delta_1) = M_1\) is \(\subseteq\) of \((\delta_2) = M_2\) and +\(\delta_2 \subseteq \delta_1\), and if both are true, then \(\delta_1 = \delta_2\). + +If all are equivalent, then the languages recognize the same strings! + +\begin{verbatim} +def minimize(dfa): + minimized = dfa.copy() + # ... mutate minimized according to minimize() + return minimized + +def delta_subseteq(start_state, sigma, delta1, delta2): + visited = set() + for transition in delta2.keys(): + if transition not in delta1 or \ + delta1[transition] != delta2[transition]: + return False + return True + +def equivalent(m1, m2): + minimized_m1 = minimize(m1) + minimized_m2 = minimize(m2) + if minimized_m1.Q != minimized_m2.Q or \ + minimzed_m1.sigma != minimized_m2.sigma or \ + minimized_m1.q0 != minimized_m2.q0 or \ + minimized_m1.F != minimized_m2.F: + return False + + m2_delta_includes_m1_delta = delta_subseteq(minimized_m1.q0, \ + minimized_m1.sigma, \ + minimized_m1.delta, \ + minimized_m2.delta) + + m1_delta_includes_m2_delta = delta_subseteq(minimized_m2.q0, \ + minimized_m2.sigma, \ + minimized_m2.delta, \ + minimized_m1.delta) + + return m2_delta_includes_m1_delta and m1_delta_includes_m2_delta +\end{verbatim} + + +\section{Problem 4} +\label{sec:orgf8f3fbd} +We can construct a CFG: + +\(S \rightarrow aSbbb | abbb\) + +Which we convert to a stack machine: + +\begin{center} +\begin{tabular}{lll} +read & pop & push\\[0pt] +\(\epsilon\) & S & aSbbb\\[0pt] +\(\epsilon\) & S & abbb\\[0pt] +a & a & \(\epsilon\)\\[0pt] +b & b & \(\epsilon\)\\[0pt] +\end{tabular} +\end{center} + +\(M = (\{a, b, S\}, \{a, b\}, S, \delta)\) + +where + +\begin{enumerate} +\item \(\delta(\epsilon, S) = \{aSbbb, abbb\}\) +\item \(\delta(a, a) = \{ \epsilon \}\) +\item \(\delta(b, b) = \{ \epsilon \}\) +\end{enumerate} + +\section{Problem 5} +\label{sec:org0f801f2} + +\begin{enumerate} +\item \(S \rightarrow 0 | 0T | 1T\) +\item \(T \rightarrow 1S | 0S\) +\end{enumerate} + +Is a right linear grammar, and is thus regular. + +\section{Problem 6} +\label{sec:org690d7be} +\subsection{One} +\label{sec:orgc0da8de} +\begin{center} +\includegraphics[width=200px]{./img/p6.png} +\end{center} + +\begin{itemize} +\item \(Q = \{p_0, p_1\}\) +\item \(F = \{p_1\}\) +\item \(\Sigma = \{1\}\) +\item \(S = p_0\) +\item \(\delta(p_0, 1) = p_1\) +\item \(\delta(p_1, 1) = p_0\) +\end{itemize} + +\subsection{Two} +\label{sec:org0e05810} + +\begin{center} +\includegraphics[width=200px]{./img/6b.png} +\end{center} + +\begin{itemize} +\item \(Q = \{p_0, p_1, p_2, p_3, p_4, p_5\}\) +\item \(F = \{p_2, p_4, p_6\}\) +\item \(\Sigma = \{a, b\}\) +\item \(S = p_0\) +\item \(\delta(p_0, a) = p_1\) +\item \(\delta(p_0, b) = p_3\) +\item \(\delta(p_1, a) = p_6\) +\item \(\delta(p_1, b) = p_2\) +\item \(\delta(p_2, b) = p_5\) +\item \(\delta(p_5, b) = p_2\) +\item \(\delta(p_3, b) = p_4\) +\item \(\delta(p_4, b) = p_3\) +\item \(\delta(p_3, a) = \delta(p_4, a) = \delta(p_2, a) = \delta(p_5, a) = \emptyset\) +\end{itemize} + +\section{Problem 7} +\label{sec:org4149a63} +\begin{center} +\includegraphics[width=200px]{./img/7.png} +\end{center} + +Because the magnitude of each element in the range of \(\delta\) is 1 then, intuitively, if we follow +the subset construction algorithm with queue optimization we will only end up with new states +identical to the ones present in the current NFA (i.e. we will visit no elements of the powerset +whose magnitude is greater than one also). + +Thus the DFA is: + +\begin{itemize} +\item \(Q = \{q_0, q_1, q_2, q_3, q_4\}\) +\item \(F = \{q_4\}\) +\item \(\Sigma = \{a, b\}\) +\item \(S = q_0\) +\item \(\delta(q_0, a) = q_1\) +\item \(\delta(p_1, a) = q_2\) +\item \(\delta(q_2, a) = q_2\) +\item \(\delta(q_2, b) = q_3\) +\item \(\delta(q_3, b) = q_4\) +\item \(\delta(p_4, b) = q_4\) +\item \(\delta(q_0, b) = \delta(q_1, b) = \delta(q_3, a) = \delta(p_4, a) = \emptyset\) +\end{itemize} +\end{document}
\ No newline at end of file diff --git a/Homework/cs5000/midterm02/compile_l_program.js b/Homework/cs5000/midterm02/compile_l_program.js new file mode 100644 index 0000000..e69de29 --- /dev/null +++ b/Homework/cs5000/midterm02/compile_l_program.js diff --git a/Homework/cs5000/midterm02/midterm.org b/Homework/cs5000/midterm02/midterm.org new file mode 100644 index 0000000..7d1312f --- /dev/null +++ b/Homework/cs5000/midterm02/midterm.org @@ -0,0 +1,218 @@ +#+TITLE: HW 08 +#+AUTHOR: Elizabeth Hunt (A02364151) +#+STARTUP: entitiespretty fold inlineimages +#+LATEX_HEADER: \notindent \notag \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} +#+LATEX: \setlength\parindent{0pt} + +* Problem One +#+attr_latex: :width 7cm +[[./p1.png]] + +* Problem Two +#+attr_latex: :width 7cm +[[./p2.png]] + +* Problem Three +Using the following code proceeding the appendix we receive + +$l(111) = 4$ + +$r(111) = 3$ + +$lt(111) = 12$ + +#+BEGIN_SRC js + const p3 = () => { + const x = 111; + const { l, r } = lr(x); + const lt = length(x); + + [`l(${x}) = ${l}`, `r(${x}) = ${r}`, `lt(${x}) = ${lt}`].forEach((s) => + console.log(s) + ); + }; + p3(); +#+END_SRC + +* Problem Four +Using the following code proceeding the appendix we receive + +$(17)_0 = 0$ + +$(17)_1 = 0$ + +$(17)_2 = 0$ + +$(17)_3 = 0$ + +$(17)_4 = 0$ + +$(17)_5 = 0$ + +$(17)_6 = 0$ + +$(17)_7 = 1$ + +$(17)_8 = 0$ + +#+BEGIN_SRC js + const p4 = () => { + const x = 17; + + for (let i = 0; i <= 8; i++) + console.log(`(${x})_${i} = ${access(x, i)}`) + }; + p4(); +#+END_SRC + + +And for all $i > 8$, $p_i > 17$ and thus $p_i^{(t+1)} \nmid x$ for all $t \ge 0$, and thus the valid set of $t$'s, +$T$, has $\text{min}(T) = 0$, so $(17)_i = 0$. + +* Problem Five +We compute the new code: + +$[\#(I_1), \#(I_2)]$ + +For $\#(I_1)$: + +$\#(I_1) = \langle a, \langle b, c \rangle \rangle$ where $a = \#(B1) = 2$, $b = 1$, $c = \#(X1) - 1 = 1$, so +$\#(I_1) = \langle 2, \langle 1, 1 \rangle \rangle = 2^2(2\langle1, 1\rangle + 1) - 1 = 4(11) - 1 = 43$. + +For $\#(I_2)$: + +$\#(I_2) = \langle a, \langle b, c \rangle \rangle$ where $a = 0$ as there is no label for the instruction, $b = \#(B1) + 2 = 4$, +$c = \#(X1) - 1 = 1$, so $\#(I_2) = \langle 0, \langle 4, 1 \rangle \rangle = 2^0(2\langle 4, 1 \rangle + 1) - 1 = 94$. + +Thus: + +$[43, 94] = (2^{43})(3^{94}) - 1 = 6218530334586699211614548872374762259672175972138197450751$. + +* Problem Six +** \phi_5^1(x) +\phi_5^1(x) has source $5 + 1$ = $6$ which corresponds to the godel sequence $2^1 * 3^1 = [1, 1]$. 1 = +$\langle 1, \langle 0, 0 \rangle \rangle$ which corresponds to an instruction with $\#(L) = 1$, $\#(V) = 0$, and an operation +of $0$: + +\begin{verbatim} +[ A1 ] Y <- Y +[ A1 ] Y <- Y +\end{verbatim} + +** \phi_7^1(x) +\phi_7^1(x) has source $7 + 1$ = $8$ which corresponds to the godel sequence $2^3 = [3]$. 3 = +$\langle 2, \langle 0, 0 \rangle \rangle$ which corresponds to an instruction with $\#(L) = 2$, $\#(V) = 0$, and an +operation of $0$: + +\begin{verbatim} +[ B1 ] Y <- Y +\end{verbatim} + +** \phi_11^1(x) +\phi_11^1(x) has source $11 + 1$ = $12$ which corresponds to the godel sequence $2^2 * 3^1 = [2, 1]$. 2 = +$\langle 0, \langle 1, 0 \rangle \rangle$ which corresponds to an instruction with $\#(L) = 0$, $\#(V) = 0$, and an +operation of $1$. + +And, we already found $1$ in \phi_5^1(x): + +\begin{verbatim} +Y <- Y + 1 +[ A1 ] Y <- Y +\end{verbatim} + +** \phi_13^1(x) +\phi_13^1(x) has source $13 + 1$ = $14$ which corresponds to the godel sequence $2^1 * 7^1 = [1, 0, 0, 1]$. +And, we already found $1$ in $\phi_5^1(x)$, $0$ is trivially ~Y <- Y~ (unlabeled, $\#(V) = 0$, op = 0). + +\begin{verbatim} +[ A1 ] Y <- Y +Y <- Y +Y <- Y +[ A1 ] Y <- Y +\end{verbatim} + +** \phi_17^1(x) +\phi_17(x) has source $17 + 1$ = $18$ which corresponds to the godel sequence $2^1 * 3^2 = [1, 2]$. +And, we already found $1$ in $\phi_5^1(x)$, and $2$ in \phi_11^1(x) + +\begin{verbatim} +[ A1 ] Y <- Y +Y <- Y + 1 +\end{verbatim} + +* Problem Seven +1. Let $m(x_1, x_2) = x_1 * x_2$ which is primitive recursive by proofs in class. +2. Let $four(x_1) = s(s(s(s(n(x_1)))))$; the successor function composed 4 times on the null function. +3. Then $f(x_1)$ is $m(four(x_1), x_1)$ which is an application of composition of primitive recursive functions. + +Thus, $f$ is primitive recursive, and thus computable. + +* Problem Eight +We can use the handy identity that $lcm(x_1, x_2) = \frac{(a * b)}{gcd(a, b)} = \lfloor \frac{(a * b)}{gcd(a, b)} \rfloor$ +since $gcd(a, b)$ by definition divides $a * b$. + +1. Define $gcd(x_1, 0) = x_1$ +2. Let $R(x_1, x_2)$ be the remainder function when $x_1$ divides $x_2$ which is primitive recursive by + the proof found on page 56 of the book. +3. We construct the informal recursion $gcd(x_1, x_2) = gcd(x_2, R(x_1, x_2))$ by Euclid's Algorithm. +4. Let $floordiv(x_1, x_2)$ be the floor of the result of division $\frac{x_1}{x_2}$ which is primitive + recursive by the proof found on page 56 of the book. +5. Let $m(x_1, x_2) = x_1 * x_2$ which is primitive recursive by proofs in class. +6. Then $lcm(x_1, x_2) = floordiv(m(x_1, x_2), gcd(x_1, x_2))$ which is an application of composotion of primitive recursive functions. + +Then $lcm$ is primitive recursive. + + +* Appendix +#+BEGIN_SRC js + const isPrime = (n) => + !Array(Math.ceil(Math.sqrt(n))) + .fill(0) + .map((_, i) => i + 2) // first prime is 2 + .some((i) => n !== i && n % i === 0); + + const primesCache = [2]; + const p = (i) => { + if (primesCache.length <= i) { + let x = primesCache.at(-1); + while (primesCache.length <= i) { + if (isPrime(++x)) primesCache.push(x); + } + } + return primesCache.at(i - 1); + }; + + const lr = (z, maxSearch = 100) => { + let x = 0; + for (let i = 0; i < maxSearch; ++i) + if ((z + 1) % Math.pow(2, i) === 0) x = Math.max(i, x); + + const y = ((z + 1) / Math.pow(2, x) - 1) / 2; + return { l: x, r: y }; + }; + + const access = (x, i) => { + if (i === 0 || x === 0) return 0; + + const p_i = p(i); + let minT = x; + for (let t = x; t >= 0; t--) + if (x % Math.pow(p_i, t + 1) !== 0) minT = Math.min(t, minT); + return minT; + }; + + const length = (x) => { + let minI = x; + for (let i = x; i >= 0; i--) + if ( + access(x, i) !== 0 && + Array(x) + .fill(0) + .map((_, j) => j + 1) + .every((j) => j <= i || access(x, j) == 0) + ) + minI = Math.min(i, minI); + + return minI; + }; +#+END_SRC diff --git a/Homework/cs5000/midterm02/midterm.pdf b/Homework/cs5000/midterm02/midterm.pdf Binary files differnew file mode 100644 index 0000000..f79a624 --- /dev/null +++ b/Homework/cs5000/midterm02/midterm.pdf diff --git a/Homework/cs5000/midterm02/midterm.tex b/Homework/cs5000/midterm02/midterm.tex new file mode 100644 index 0000000..6848ed7 --- /dev/null +++ b/Homework/cs5000/midterm02/midterm.tex @@ -0,0 +1,265 @@ +% Created 2023-11-17 Fri 13:57 +% Intended LaTeX compiler: pdflatex +\documentclass[11pt]{article} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +\usepackage{graphicx} +\usepackage{longtable} +\usepackage{wrapfig} +\usepackage{rotating} +\usepackage[normalem]{ulem} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{capt-of} +\usepackage{hyperref} +\notindent \notag \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} +\author{Elizabeth Hunt (A02364151)} +\date{\today} +\title{HW 08} +\hypersetup{ + pdfauthor={Elizabeth Hunt (A02364151)}, + pdftitle={HW 08}, + pdfkeywords={}, + pdfsubject={}, + pdfcreator={Emacs 28.2 (Org mode 9.7-pre)}, + pdflang={English}} +\begin{document} + +\maketitle +\tableofcontents + +\setlength\parindent{0pt} + +\section{Problem One} +\label{sec:orgbeb25aa} +\begin{center} +\includegraphics[width=7cm]{./p1.png} +\end{center} + +\section{Problem Two} +\label{sec:orgc078de2} +\begin{center} +\includegraphics[width=7cm]{./p2.png} +\end{center} + +\section{Problem Three} +\label{sec:orga508990} +Using the following code proceeding the appendix we receive + +\(l(111) = 4\) + +\(r(111) = 3\) + +\(lt(111) = 12\) + +\begin{verbatim} +const p3 = () => { + const x = 111; + const { l, r } = lr(x); + const lt = length(x); + + [`l(${x}) = ${l}`, `r(${x}) = ${r}`, `lt(${x}) = ${lt}`].forEach((s) => + console.log(s) + ); +}; +p3(); +\end{verbatim} + +\section{Problem Four} +\label{sec:org7ca236e} +Using the following code proceeding the appendix we receive + +\((17)_0 = 0\) + +\((17)_1 = 0\) + +\((17)_2 = 0\) + +\((17)_3 = 0\) + +\((17)_4 = 0\) + +\((17)_5 = 0\) + +\((17)_6 = 0\) + +\((17)_7 = 1\) + +\((17)_8 = 0\) + +\begin{verbatim} +const p4 = () => { + const x = 17; + + for (let i = 0; i <= 8; i++) + console.log(`(${x})_${i} = ${access(x, i)}`) +}; +p4(); +\end{verbatim} + + +And for all \(i > 8\), \(p_i > 17\) and thus \(p_i^{(t+1)} \nmid x\) for all \(t \ge 0\), and thus the valid set of \(t\)'s, +\(T\), has \(\text{min}(T) = 0\), so \((17)_i = 0\). + +\section{Problem Five} +\label{sec:org25e3a57} +We compute the new code: + +\([\#(I_1), \#(I_2)]\) + +For \(\#(I_1)\): + +\(\#(I_1) = \langle a, \langle b, c \rangle \rangle\) where \(a = \#(B1) = 2\), \(b = 1\), \(c = \#(X1) - 1 = 1\), so +\(\#(I_1) = \langle 2, \langle 1, 1 \rangle \rangle = 2^2(2\langle1, 1\rangle + 1) - 1 = 4(11) - 1 = 43\). + +For \(\#(I_2)\): + +\(\#(I_2) = \langle a, \langle b, c \rangle \rangle\) where \(a = 0\) as there is no label for the instruction, \(b = \#(B1) + 2 = 4\), +\(c = \#(X1) - 1 = 1\), so \(\#(I_2) = \langle 0, \langle 4, 1 \rangle \rangle = 2^0(2\langle 4, 1 \rangle + 1) - 1 = 94\). + +Thus: + +\([43, 94] = (2^{43})(3^{94}) - 1 = 6218530334586699211614548872374762259672175972138197450751\). + +\section{Problem Six} +\label{sec:orgc5e0177} +\subsection{\(\phi\)\textsubscript{5}\textsuperscript{1}(x)} +\label{sec:orgf652342} +\(\phi\)\textsubscript{5}\textsuperscript{1}(x) has source \(5 + 1\) = \(6\) which corresponds to the godel sequence \(2^1 * 3^1 = [1, 1]\). 1 = +\(\langle 1, \langle 0, 0 \rangle \rangle\) which corresponds to an instruction with \(\#(L) = 1\), \(\#(V) = 0\), and an operation +of \(0\): + +\begin{verbatim} +[ A1 ] Y <- Y +[ A1 ] Y <- Y +\end{verbatim} + +\subsection{\(\phi\)\textsubscript{7}\textsuperscript{1}(x)} +\label{sec:orgd9de496} +\(\phi\)\textsubscript{7}\textsuperscript{1}(x) has source \(7 + 1\) = \(8\) which corresponds to the godel sequence \(2^3 = [3]\). 3 = +\(\langle 2, \langle 0, 0 \rangle \rangle\) which corresponds to an instruction with \(\#(L) = 2\), \(\#(V) = 0\), and an +operation of \(0\): + +\begin{verbatim} +[ B1 ] Y <- Y +\end{verbatim} + +\subsection{\(\phi\)\textsubscript{11}\textsuperscript{1}(x)} +\label{sec:orge5e7392} +\(\phi\)\textsubscript{11}\textsuperscript{1}(x) has source \(11 + 1\) = \(12\) which corresponds to the godel sequence \(2^2 * 3^1 = [2, 1]\). 2 = +\(\langle 0, \langle 1, 0 \rangle \rangle\) which corresponds to an instruction with \(\#(L) = 0\), \(\#(V) = 0\), and an +operation of \(1\). + +And, we already found \(1\) in \(\phi\)\textsubscript{5}\textsuperscript{1}(x): + +\begin{verbatim} +Y <- Y + 1 +[ A1 ] Y <- Y +\end{verbatim} + +\subsection{\(\phi\)\textsubscript{13}\textsuperscript{1}(x)} +\label{sec:org65f2245} +\(\phi\)\textsubscript{13}\textsuperscript{1}(x) has source \(13 + 1\) = \(14\) which corresponds to the godel sequence \(2^1 * 7^1 = [1, 0, 0, 1]\). +And, we already found \(1\) in \(\phi_5^1(x)\), \(0\) is trivially \texttt{Y <- Y} (unlabeled, \(\#(V) = 0\), op = 0). + +\begin{verbatim} +[ A1 ] Y <- Y +Y <- Y +Y <- Y +[ A1 ] Y <- Y +\end{verbatim} + +\subsection{\(\phi\)\textsubscript{17}\textsuperscript{1}(x)} +\label{sec:orgacbaf8a} +\(\phi\)\textsubscript{17}(x) has source \(17 + 1\) = \(18\) which corresponds to the godel sequence \(2^1 * 3^2 = [1, 2]\). +And, we already found \(1\) in \(\phi_5^1(x)\), and \(2\) in \(\phi\)\textsubscript{11}\textsuperscript{1}(x) + +\begin{verbatim} +[ A1 ] Y <- Y +Y <- Y + 1 +\end{verbatim} + +\section{Problem Seven} +\label{sec:org15f3033} +\begin{enumerate} +\item Let \(m(x_1, x_2) = x_1 * x_2\) which is primitive recursive by proofs in class. +\item Let \(four(x_1) = s(s(s(s(n(x_1)))))\); the successor function composed 4 times on the null function. +\item Then \(f(x_1)\) is \(m(four(x_1), x_1)\) which is an application of composition of primitive recursive functions. +\end{enumerate} + +Thus, \(f\) is primitive recursive, and thus computable. + +\section{Problem Eight} +\label{sec:org1d07ae0} +We can use the handy identity that \(lcm(x_1, x_2) = \frac{(a * b)}{gcd(a, b)} = \lfloor \frac{(a * b)}{gcd(a, b)} \rfloor\) +since \(gcd(a, b)\) by definition divides \(a * b\). + +\begin{enumerate} +\item Define \(gcd(x_1, 0) = x_1\) +\item Let \(R(x_1, x_2)\) be the remainder function when \(x_1\) divides \(x_2\) which is primitive recursive by +the proof found on page 56 of the book. +\item We construct the informal recursion \(gcd(x_1, x_2) = gcd(x_2, R(x_1, x_2))\) by Euclid's Algorithm. +\item Let \(floordiv(x_1, x_2)\) be the floor of the result of division \(\frac{x_1}{x_2}\) which is primitive +recursive by the proof found on page 56 of the book. +\item Let \(m(x_1, x_2) = x_1 * x_2\) which is primitive recursive by proofs in class. +\item Then \(lcm(x_1, x_2) = floordiv(m(x_1, x_2), gcd(x_1, x_2))\) which is an application of composotion of primitive recursive functions. +\end{enumerate} + +Then \(lcm\) is primitive recursive. + + +\section{Appendix} +\label{sec:org7f26251} +\begin{verbatim} +const isPrime = (n) => + !Array(Math.ceil(Math.sqrt(n))) + .fill(0) + .map((_, i) => i + 2) // first prime is 2 + .some((i) => n !== i && n % i === 0); + +const primesCache = [2]; +const p = (i) => { + if (primesCache.length <= i) { + let x = primesCache.at(-1); + while (primesCache.length <= i) { + if (isPrime(++x)) primesCache.push(x); + } + } + return primesCache.at(i - 1); +}; + +const lr = (z, maxSearch = 100) => { + let x = 0; + for (let i = 0; i < maxSearch; ++i) + if ((z + 1) % Math.pow(2, i) === 0) x = Math.max(i, x); + + const y = ((z + 1) / Math.pow(2, x) - 1) / 2; + return { l: x, r: y }; +}; + +const access = (x, i) => { + if (i === 0 || x === 0) return 0; + + const p_i = p(i); + let minT = x; + for (let t = x; t >= 0; t--) + if (x % Math.pow(p_i, t + 1) !== 0) minT = Math.min(t, minT); + return minT; +}; + +const length = (x) => { + let minI = x; + for (let i = x; i >= 0; i--) + if ( + access(x, i) !== 0 && + Array(x) + .fill(0) + .map((_, j) => j + 1) + .every((j) => j <= i || access(x, j) == 0) + ) + minI = Math.min(i, minI); + + return minI; +}; +\end{verbatim} +\end{document}
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