Assignment 1 notebook

.gitignore

@@ -0,0 +1,2 @@
+.ipynb_checkpoints
+*.pdf

.idea/CS2333.iml

@@ -2,9 +2,10 @@
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     <content url="file://$MODULE_DIR$">
-      <excludeFolder url="file://$MODULE_DIR$/.venv" />
+      <excludeFolder url="file://$MODULE_DIR$/.idea" />
+      <excludeFolder url="file://$MODULE_DIR$/.ipynb_checkpoints" />
     </content>
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+    <orderEntry type="jdk" jdkName="Python 3.11 (jupyter)" jdkType="Python SDK" />
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@@ -7,5 +7,5 @@
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     <option name="description" value="" />
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-  <component name="ProjectRootManager" version="2" project-jdk-name="Python 3.11 (CS2333)" project-jdk-type="Python SDK" />
+  <component name="ProjectRootManager" version="2" project-jdk-name="Python 3.11 (jupyter)" project-jdk-type="Python SDK" />
 </project>

.idea/runConfigurations/2024_1_12.xml

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+    <option name="PARAMETERS" value="" />
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.idea/runConfigurations/2024_1_15.xml

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+  <configuration default="false" name="2024-1-15" type="PythonConfigurationType" factoryName="Python" nameIsGenerated="true">
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+    <option name="ADD_SOURCE_ROOTS" value="true" />
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+  <configuration default="false" name="Export A1" type="ShConfigurationType">
+    <option name="SCRIPT_TEXT" value="jupyter-nbconvert.exe &quot;Assignment 1.ipynb&quot; --to webpdf --no-input" />
+    <option name="INDEPENDENT_SCRIPT_PATH" value="true" />
+    <option name="SCRIPT_PATH" value="" />
+    <option name="SCRIPT_OPTIONS" value="" />
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+    <option name="EXECUTE_SCRIPT_FILE" value="false" />
+    <envs />
+    <method v="2" />
+  </configuration>
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2024-1-15.py

@@ -48,10 +48,70 @@ def p3():
 	return strings
 
 
+def valid():
+	alphabet = ['a', 'b']
+	strings = get_all_strings_with_a_given_alphabet_and_length(alphabet, 5)
+
+	def number_of_a_is_equal_to_3(s):
+		return s.count('a') == 3
+
+	def number_of_b_is_equal_to_3(s):
+		return s.count('b') == 3
+
+	# A game stops when a player has 3 a's or 3 b's, so there should be no trailing wins from the losing player
+	def valid_game(s):
+		if number_of_a_is_equal_to_3(s):
+			return not s.endswith('b')
+		elif number_of_b_is_equal_to_3(s):
+			return not s.endswith('a')
+
+	strings = list(filter(valid_game, strings))
+
+	return strings
+
+
+def p5():
+	alphabet = ['a', 'b']
+	valid_strings = valid()
+	strings = get_all_strings_with_a_given_alphabet_and_length(alphabet, 5)
+
+	def number_of_a_or_b_is_equal_to_3(s):
+		return s.count('a') == 3 or s.count('b') == 3
+
+	strings = list(filter(number_of_a_or_b_is_equal_to_3, strings))
+	strings = [i for i in strings if i not in valid_strings]  # strings - valid_strings
+	return strings
+
+
+def p6():
+	alphabet = ['a', 'b']
+	valid_strings = valid()
+	strings = get_all_strings_with_a_given_alphabet_and_length(alphabet, 5)
+
+	def number_of_a_is_equal_to_3(s):
+		return s.count('a') == 3
+
+	def number_of_b_is_equal_to_3(s):
+		return s.count('b') == 3
+
+	def opposite_number_of_winner_is_less_than_3(s):
+		if number_of_a_is_equal_to_3(s):
+			return s.count('b') < 3
+		elif number_of_b_is_equal_to_3(s):
+			return s.count('a') < 3
+
+	strings = list(filter(opposite_number_of_winner_is_less_than_3, strings))
+	strings = [i for i in strings if i not in valid_strings]  # strings - valid_strings
+	return strings
+
+
 def main():
 	print_strings(p1())
 	print_strings(p2())
 	print_strings(p3())
+	print_strings(valid())
+	print_strings(p5())
+	print_strings(p6())
 
 
 if __name__ == '__main__':

Assignment 1.ipynb

@@ -0,0 +1,563 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "source": [
+    "### <center> Assignment 1 \n",
+    "#### <center> January 18, 2024\n",
+    "##### <center> Isaac Shoebottom (3429069)"
+   ],
+   "metadata": {
+    "collapsed": false
+   },
+   "id": "9ec6bd31bcc8447b"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "7709a8186ebbb39a",
+   "metadata": {
+    "collapsed": false,
+    "ExecuteTime": {
+     "end_time": "2024-01-18T20:59:35.482159200Z",
+     "start_time": "2024-01-18T20:59:35.470889900Z"
+    }
+   },
+   "outputs": [],
+   "source": [
+    "# util contains functions that make it easier to form strings \n",
+    "# from languages, and give output that matches with expected \n",
+    "# output for the homework, (replace empty string with epsilon)\n",
+    "from util import *"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "source": [
+    "<div style=\"page-break-after: always;\"></div>\n",
+    "Question 1"
+   ],
+   "metadata": {
+    "collapsed": false
+   },
+   "id": "cf6146860173f94f"
+  },
+  {
+   "cell_type": "markdown",
+   "source": [
+    "(a) $L_1 = \\{0^m 1^n | m,n \\in \\mathbb{Z}^*, n \\leq 2, m \\leq n\\}$"
+   ],
+   "metadata": {
+    "collapsed": false
+   },
+   "id": "b12165c136baa772"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "a3ebc6e89e38cc93",
+   "metadata": {
+    "collapsed": false,
+    "ExecuteTime": {
+     "end_time": "2024-01-18T20:59:35.494611400Z",
+     "start_time": "2024-01-18T20:59:35.477374Z"
+    }
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "['ε', '1', '01', '11', '011', '0011']\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Language rule 1, n is less than or equal to 2\n",
+    "n = 2\n",
+    "# Language rule 2, m is less than or equal to n\n",
+    "m = n\n",
+    "strings = []\n",
+    "# Language rule 3, n and m are both non-negative integers\n",
+    "for i in range(0, n + 1):\n",
+    "\tfor j in range(0, m + 1):\n",
+    "\t\tif j > i:\n",
+    "\t\t\tcontinue\n",
+    "\t\tstrings.append('0' * j + '1' * i)\n",
+    "print_strings(strings)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "b821cc2e0085663a",
+   "metadata": {
+    "collapsed": false
+   },
+   "source": [
+    "(b) $L_2 = \\{a^m b^n | m,n \\in \\mathbb{Z}^*, 2 \\leq n \\leq 6, n \\text{ is a multiple of } m\\}$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "1d32fddeec3be96f",
+   "metadata": {
+    "collapsed": false,
+    "ExecuteTime": {
+     "end_time": "2024-01-18T20:59:35.536014100Z",
+     "start_time": "2024-01-18T20:59:35.486263100Z"
+    }
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "['abb', 'aabb', 'abbb', 'aaabbb', 'abbbb', 'aabbbb', 'aaaabbbb', 'abbbbb', 'aaaaabbbbb', 'abbbbbb', 'aabbbbbb', 'aaabbbbbb', 'aaaaaabbbbbb']\n"
+     ]
+    }
+   ],
+   "source": [
+    "possible_n = [2, 3, 4, 5, 6]\n",
+    "\n",
+    "strings = []\n",
+    "for i in possible_n:\n",
+    "\tinteger_divisors = []\n",
+    "\tfor j in range(1, i + 1):\n",
+    "\t\tif i % j == 0:\n",
+    "\t\t\tinteger_divisors.append(j)\n",
+    "\tfor j in integer_divisors:\n",
+    "\t\tstrings.append('a' * j + 'b' * i)\n",
+    "print_strings(strings)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "2056f304860447ea",
+   "metadata": {
+    "collapsed": false
+   },
+   "source": [
+    "(c) $L_3 = \\{a^n b^{2n-1} c^{3n-2} | n \\in \\mathbb{Z}^+, n \\leq 3\\}$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "a5ba1ad20a2b94e0",
+   "metadata": {
+    "collapsed": false,
+    "ExecuteTime": {
+     "end_time": "2024-01-18T20:59:35.536014100Z",
+     "start_time": "2024-01-18T20:59:35.491094800Z"
+    }
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "['abc', 'aabbbcccc', 'aaabbbbbccccccc']\n"
+     ]
+    }
+   ],
+   "source": [
+    "possible_n = [1, 2, 3]\n",
+    "\n",
+    "strings = []\n",
+    "for i in possible_n:\n",
+    "\tan = i\n",
+    "\tbn = 2 * i - 1\n",
+    "\tcn = 3 * i - 2\n",
+    "\tstrings.append('a' * an + 'b' * bn + 'c' * cn)\n",
+    "print_strings(strings)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "520023ded45b27da",
+   "metadata": {
+    "collapsed": false
+   },
+   "source": [
+    "(d) $L_4 = \\{w \\in \\{a,b\\}^* | |w| \\leq 3 \\text{ and } w \\text{ is a palindrome} \\}$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "id": "452a62fd365b4f58",
+   "metadata": {
+    "collapsed": false,
+    "ExecuteTime": {
+     "end_time": "2024-01-18T20:59:35.539017300Z",
+     "start_time": "2024-01-18T20:59:35.497609900Z"
+    }
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "['ε', 'a', 'b', 'aa', 'bb', 'aaa', 'aba', 'bab', 'bbb']\n"
+     ]
+    }
+   ],
+   "source": [
+    "alphabet = ['a', 'b']\n",
+    "# include empty string\n",
+    "strings = get_all_strings_with_a_given_alphabet_and_length(alphabet, 3)\n",
+    "def is_palindrome(s):\n",
+    "\treturn s == s[::-1]\n",
+    "strings = list(filter(is_palindrome, strings))\n",
+    "print_strings(strings)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "9dff8b1af013d957",
+   "metadata": {
+    "collapsed": false
+   },
+   "source": [
+    "<div style=\"page-break-after: always;\"></div>\n",
+    "Question 2"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ad64b6ae580ba4dd",
+   "metadata": {
+    "collapsed": false
+   },
+   "source": [
+    "(a) $L_5 = \\{w \\in \\{a,b\\}^* | |w| \\leq 3 \\text{ and } n_a(w) \\text{ is even} \\}$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "id": "d5167f62fda41135",
+   "metadata": {
+    "collapsed": false,
+    "ExecuteTime": {
+     "end_time": "2024-01-18T20:59:35.581083800Z",
+     "start_time": "2024-01-18T20:59:35.542564Z"
+    }
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "['ε', 'b', 'aa', 'bb', 'aab', 'aba', 'baa', 'bbb']\n"
+     ]
+    }
+   ],
+   "source": [
+    "alphabet = ['a', 'b']\n",
+    "strings = get_all_strings_with_a_given_alphabet_and_length(alphabet, 3)\n",
+    "\n",
+    "def number_of_a_is_even(s):\n",
+    "\treturn s.count('a') % 2 == 0\n",
+    "\n",
+    "strings = list(filter(number_of_a_is_even, strings))\n",
+    "print_strings(strings)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "9f8b05d94c453058",
+   "metadata": {
+    "collapsed": false
+   },
+   "source": [
+    "(b) $L_6 = \\{a^m b^n c^p | m,n,p \\in \\mathbb{Z}^*, m + n + p \\leq 4, m > n, n = p\\}$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "id": "9b3b29ce2400f4ab",
+   "metadata": {
+    "collapsed": false,
+    "ExecuteTime": {
+     "end_time": "2024-01-18T20:59:35.594528300Z",
+     "start_time": "2024-01-18T20:59:35.585083400Z"
+    }
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "['a', 'aa', 'aabc', 'aaa', 'aaaa']\n"
+     ]
+    }
+   ],
+   "source": [
+    "possible_m = [0, 1, 2, 3, 4]\n",
+    "possible_n = [0, 1, 2, 3, 4]\n",
+    "possible_p = [0, 1, 2, 3, 4]\n",
+    "\n",
+    "strings = []\n",
+    "for m in possible_m:\n",
+    "\tfor n in possible_n:\n",
+    "\t\tfor p in possible_p:\n",
+    "\t\t\t# First rule\n",
+    "\t\t\tif m + n + p <= 4:\n",
+    "\t\t\t\t# Second rule\n",
+    "\t\t\t\tif m > n:\n",
+    "\t\t\t\t\t# Third rule\n",
+    "\t\t\t\t\tif n == p:\n",
+    "\t\t\t\t\t\tstrings.append('a' * m + 'b' * n + 'c' * p)\n",
+    "print_strings(strings)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7626dea00c2aeda4",
+   "metadata": {
+    "collapsed": false
+   },
+   "source": [
+    "(c) $L_7 = \\{w \\in \\{a,b,c\\}^* | |w| \\leq 4, n_a(w) > n_b(w), n_b(w) = n_c(w)\\}$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "id": "8cbf5de6f2655673",
+   "metadata": {
+    "collapsed": false,
+    "ExecuteTime": {
+     "end_time": "2024-01-18T20:59:35.649706300Z",
+     "start_time": "2024-01-18T20:59:35.597530600Z"
+    }
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "['a', 'aa', 'aaa', 'aaaa', 'aabc', 'aacb', 'abac', 'abca', 'acab', 'acba', 'baac', 'baca', 'bcaa', 'caab', 'caba', 'cbaa']\n"
+     ]
+    }
+   ],
+   "source": [
+    "alphabet = ['a', 'b', 'c']\n",
+    "strings = get_all_strings_with_a_given_alphabet_and_length(alphabet, 4)\n",
+    "\n",
+    "def number_of_a_is_greater_than_number_of_b(s):\n",
+    "\treturn s.count('a') > s.count('b')\n",
+    "\n",
+    "def number_of_b_is_equal_to_number_of_c(s):\n",
+    "\treturn s.count('b') == s.count('c')\n",
+    "\n",
+    "strings = list(filter(number_of_a_is_greater_than_number_of_b, strings))\n",
+    "strings = list(filter(number_of_b_is_equal_to_number_of_c, strings))\n",
+    "\n",
+    "print_strings(strings)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "88faa9e0347d6fd8",
+   "metadata": {
+    "collapsed": false
+   },
+   "source": [
+    "<div style=\"page-break-after: always;\"></div>\n",
+    "Question 3"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "efdcdcb4f9a0298",
+   "metadata": {
+    "collapsed": false
+   },
+   "source": [
+    "First it could be helpful to find the set of all valid games to compare against. This is a little tricky as first you need to know the rules of the game and encode them correctly. The rules that I have discerned are as follows:\n",
+    "1. Each valid game has 3 a's or 3 b's, denoting a victory for one side\n",
+    "2. Once it has either victories on either side, the game does not continue, and the string is complete\n",
+    "\n",
+    "$$L = \\{w \\in \\{a,b\\}^* | n_a(w) = 3 \\text{ or } n_b(w) = 3 \\text{ and } w \\text{ does not end with the opposite letter}\\}$$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "id": "660a123373b6ff39",
+   "metadata": {
+    "collapsed": false,
+    "ExecuteTime": {
+     "end_time": "2024-01-18T20:59:35.650706300Z",
+     "start_time": "2024-01-18T20:59:35.603524100Z"
+    }
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "['aaa', 'bbb', 'aaba', 'abaa', 'abbb', 'baaa', 'babb', 'bbab', 'aabba', 'aabbb', 'ababa', 'ababb', 'abbaa', 'abbab', 'baaba', 'baabb', 'babaa', 'babab', 'bbaaa', 'bbaab']\n"
+     ]
+    }
+   ],
+   "source": [
+    "alphabet = ['a', 'b']\n",
+    "strings = get_all_strings_with_a_given_alphabet_and_length(alphabet, 5)\n",
+    "\n",
+    "def number_of_a_is_equal_to_3(s):\n",
+    "\treturn s.count('a') == 3\n",
+    "\n",
+    "def number_of_b_is_equal_to_3(s):\n",
+    "\treturn s.count('b') == 3\n",
+    "\n",
+    "# A game stops when a player has 3 a's or 3 b's, so there should be no trailing wins from the losing player\n",
+    "def valid_game(s):\n",
+    "\tif number_of_a_is_equal_to_3(s):\n",
+    "\t\treturn not s.endswith('b')\n",
+    "\telif number_of_b_is_equal_to_3(s):\n",
+    "\t\treturn not s.endswith('a')\n",
+    "\n",
+    "valid_strings = list(filter(valid_game, strings))\n",
+    "print_strings(valid_strings)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "bcbd29f7048b2af2",
+   "metadata": {
+    "collapsed": false
+   },
+   "source": [
+    "(a) $L_1 = \\{w \\in \\{a,b\\}^* | n_a(w) = 3 \\text{ or } n_b(w) = 3 \\}$\n",
+    "\n",
+    "This one is not representable with code, as the set is produces is infinite. This language places no restrictions on the length of the string, so \"aaabbbbbbbbbbbbbbbbbbbbbbbbbbb\" is a valid string in this language. So here is a few examples of valid strings in this language that are not valid in the original language L but are in this language:\n",
+    "\n",
+    "1. \"aaabbbbbbbbbbbbbbbbbbbbbbbbbbb\"\n",
+    "2. \"bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbaaa\"\n",
+    "3. \"bababaaaaaaaaaaaaaaaaaaaaaaaaaa\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7afad368b9ad3a09",
+   "metadata": {
+    "collapsed": false
+   },
+   "source": [
+    "(b) $L_2 = \\{w \\in \\{a,b\\}^* | |w| \\leq 5, \\text{ and } n_a(w) = 3 \\text{ or } n_b(w) = 3 \\}$\n",
+    "\n",
+    "This language has the problem is not encoding the rule that the game stops once a player has 3 a's or 3 b's. This can be seen in these strings that are valid in this language but not in the original language L:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "id": "53dc46669ffac28c",
+   "metadata": {
+    "collapsed": false,
+    "ExecuteTime": {
+     "end_time": "2024-01-18T20:59:35.652706Z",
+     "start_time": "2024-01-18T20:59:35.612375700Z"
+    }
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "['aaab', 'bbba', 'aaabb', 'aabab', 'abaab', 'abbba', 'baaab', 'babba', 'bbaba', 'bbbaa']\n"
+     ]
+    }
+   ],
+   "source": [
+    "strings = get_all_strings_with_a_given_alphabet_and_length(alphabet, 5)\n",
+    "\n",
+    "def number_of_a_or_b_is_equal_to_3(s):\n",
+    "\treturn s.count('a') == 3 or s.count('b') == 3\n",
+    "\n",
+    "strings = list(filter(number_of_a_or_b_is_equal_to_3, strings))\n",
+    "strings = [i for i in strings if i not in valid_strings] # strings - valid_strings\n",
+    "print_strings(strings)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "990482f3b9b9e5b6",
+   "metadata": {
+    "collapsed": false
+   },
+   "source": [
+    "(c) $L_3 = \\{w \\in \\{a,b\\}^* | (n_a(w) = 3 \\text{ and } n_b(w) < 3) \\text{ or } (n_b(w) = 3 \\text{ and } n_a(w) < 3) \\}$\n",
+    "\n",
+    "This language has a similar problem to the previous one, in that it does not encode the rule that the game stops once a player has 3 a's or 3 b's. The difference in this language is that the limit of string length 5 is not explicitly stated, however it is encoded to be that way by saying that if there are 3 a's then there must be less than 3 b's, and vice versa. This means that the effective maximum length of the string is 5, and thus the language is equivalent to the language from the previous question.\n",
+    "$$\n",
+    "x, y \\in \\mathbb{Z}^* \\\\\n",
+    "x = 3 \\\\\n",
+    "y = < 3 \\\\\n",
+    "< 3 = \\{0, 1, 2\\} \\\\\n",
+    "x + y = \\{3, 4, 5\\}\n",
+    "$$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "id": "397defd138f4789a",
+   "metadata": {
+    "collapsed": false,
+    "ExecuteTime": {
+     "end_time": "2024-01-18T20:59:35.691666Z",
+     "start_time": "2024-01-18T20:59:35.656707400Z"
+    }
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "['aaab', 'bbba', 'aaabb', 'aabab', 'abaab', 'abbba', 'baaab', 'babba', 'bbaba', 'bbbaa']\n"
+     ]
+    }
+   ],
+   "source": [
+    "strings = get_all_strings_with_a_given_alphabet_and_length(alphabet, 5)\n",
+    "\n",
+    "def opposite_number_of_winner_is_less_than_3(s):\n",
+    "\tif number_of_a_is_equal_to_3(s):\n",
+    "\t\treturn s.count('b') < 3\n",
+    "\telif number_of_b_is_equal_to_3(s):\n",
+    "\t\treturn s.count('a') < 3\n",
+    "\n",
+    "strings = list(filter(opposite_number_of_winner_is_less_than_3, strings))\n",
+    "strings = [i for i in strings if i not in valid_strings] # strings - valid_strings\n",
+    "print_strings(strings)"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 3 (ipykernel)",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.11.5"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}

util.py

@@ -9,5 +9,15 @@ def get_all_strings_with_a_given_alphabet_and_length(alphabet, length):
 	return strings
 
 
+def replace_empty_string_with_symbol(strings):
+	symbol = 'ε'
+	for i in range(len(strings)):
+		if strings[i] == '':
+			strings[i] = symbol
+	return strings
+
+
 def print_strings(strings):
-	print(strings, len(strings))
+	strings = replace_empty_string_with_symbol(strings)
+	# print(len(strings))
+	print(strings)