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74 files changed, 5098 insertions, 0 deletions

diff --git a/Homework/cs5000/.DS_Store b/Homework/cs5000/.DS_Store
new 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
new 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
new 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
new 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
new 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
new 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
new 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
new 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
new 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
new 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
new 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
new 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
new 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
new 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
new 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
new 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
new 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
new 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
new 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}
\ No newline at end of file
diff --git a/Homework/cs5000/hw03/__pycache__/cs5000_f23_hw03.cpython-311.pyc b/Homework/cs5000/hw03/__pycache__/cs5000_f23_hw03.cpython-311.pyc
new file mode 100644
index 0000000..f3444a4
--- /dev/null
+++ b/Homework/cs5000/hw03/__pycache__/cs5000_f23_hw03.cpython-311.pyc
diff --git a/Homework/cs5000/hw03/hw03.zip b/Homework/cs5000/hw03/hw03.zip
new file mode 100644
index 0000000..1092d16
--- /dev/null
+++ b/Homework/cs5000/hw03/hw03.zip
diff --git a/Homework/cs5000/hw03/hw03/CS5000_F23_hw03.pdf b/Homework/cs5000/hw03/hw03/CS5000_F23_hw03.pdf
new file mode 100644
index 0000000..cb3a987
--- /dev/null
+++ b/Homework/cs5000/hw03/hw03/CS5000_F23_hw03.pdf
diff --git a/Homework/cs5000/hw03/hw03/__pycache__/cs5000_f23_hw03.cpython-311.pyc b/Homework/cs5000/hw03/hw03/__pycache__/cs5000_f23_hw03.cpython-311.pyc
new file mode 100644
index 0000000..d005ebc
--- /dev/null
+++ 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
new 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
new file mode 100644
index 0000000..0252337
--- /dev/null
+++ b/Homework/cs5000/hw04.zip
diff --git a/Homework/cs5000/hw04/BrooksWeber_Ch10_RegularGrammars.pdf b/Homework/cs5000/hw04/BrooksWeber_Ch10_RegularGrammars.pdf
new file mode 100644
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diff --git a/Homework/cs5000/hw04/CS5000_F23_hw04.pdf b/Homework/cs5000/hw04/CS5000_F23_hw04.pdf
new file mode 100644
index 0000000..a4a7963
--- /dev/null
+++ b/Homework/cs5000/hw04/CS5000_F23_hw04.pdf
diff --git a/Homework/cs5000/hw04/hw04.org b/Homework/cs5000/hw04/hw04.org
new file mode 100644
index 0000000..fced560
--- /dev/null
+++ b/Homework/cs5000/hw04/hw04.org
@@ -0,0 +1,106 @@
+#+TITLE: HW 04
+#+AUTHOR: Elizabeth Hunt
+#+STARTUP: entitiespretty fold inlineimages
+#+LATEX_HEADER: \notindent \notag  \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry}
+#+LATEX: \setlength\parindent{20pt}
+#+OPTIONS: toc:nil
+
+* Question One
+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 _regular_ language since there is a FA to describe it:
+
++ $F = \{q_0\}$
++ $\Sigma = \{0,1,2\}$
++ $S = q_0$
++ $\delta(q_0, 0) = q_0$, $\delta(q_0, 1) = q_0$, $\delta(q_0, 2) = q_0$
++ $Q = {q_0}$
+
+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 $L$.
+
+* 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
new 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}
\ No newline at end of file
diff --git a/Homework/cs5000/hw04/img/problem_2_2_dfa.png b/Homework/cs5000/hw04/img/problem_2_2_dfa.png
new file mode 100644
index 0000000..78d8e54
--- /dev/null
+++ b/Homework/cs5000/hw04/img/problem_2_2_dfa.png
diff --git a/Homework/cs5000/hw04/img/problem_3_dfa.png b/Homework/cs5000/hw04/img/problem_3_dfa.png
new file mode 100644
index 0000000..259067e
--- /dev/null
+++ b/Homework/cs5000/hw04/img/problem_3_dfa.png
diff --git a/Homework/cs5000/hw05/.DS_Store b/Homework/cs5000/hw05/.DS_Store
new file mode 100644
index 0000000..9fc8419
--- /dev/null
+++ b/Homework/cs5000/hw05/.DS_Store
diff --git a/Homework/cs5000/hw05/CS5000_F23_HW05.pdf b/Homework/cs5000/hw05/CS5000_F23_HW05.pdf
new file mode 100644
index 0000000..aea985e
--- /dev/null
+++ b/Homework/cs5000/hw05/CS5000_F23_HW05.pdf
diff --git a/Homework/cs5000/hw05/PyCYK/__pycache__/cnfg.cpython-310.pyc b/Homework/cs5000/hw05/PyCYK/__pycache__/cnfg.cpython-310.pyc
new file mode 100644
index 0000000..6866f6c
--- /dev/null
+++ b/Homework/cs5000/hw05/PyCYK/__pycache__/cnfg.cpython-310.pyc
diff --git a/Homework/cs5000/hw05/PyCYK/__pycache__/cyk.cpython-310.pyc b/Homework/cs5000/hw05/PyCYK/__pycache__/cyk.cpython-310.pyc
new file mode 100644
index 0000000..d5fd992
--- /dev/null
+++ b/Homework/cs5000/hw05/PyCYK/__pycache__/cyk.cpython-310.pyc
diff --git a/Homework/cs5000/hw05/PyCYK/cnfg.py b/Homework/cs5000/hw05/PyCYK/cnfg.py
new file mode 100644
index 0000000..bb1ed17
--- /dev/null
+++ b/Homework/cs5000/hw05/PyCYK/cnfg.py
@@ -0,0 +1,82 @@
+###############################################

+# 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
new 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
new 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
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diff --git a/Homework/cs5000/hw08/hw08.tex b/Homework/cs5000/hw08/hw08.tex
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+% 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
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diff --git a/Homework/cs5000/midterm/img/7.png b/Homework/cs5000/midterm/img/7.png
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diff --git a/Homework/cs5000/midterm/img/p6.png b/Homework/cs5000/midterm/img/p6.png
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diff --git a/Homework/cs5000/midterm/img/prob_2_parse_tree.png b/Homework/cs5000/midterm/img/prob_2_parse_tree.png
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diff --git a/Homework/cs5000/midterm/midterm.org b/Homework/cs5000/midterm/midterm.org
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+#+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
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+++ 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
new 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}
\ No newline at end of file
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