Finish a2
This commit is contained in:
parent
0b83fe9bfb
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"output_type": "execute_result"
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}
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],
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"source": [
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"from automata.fa.dfa import DFA\n",
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"\n",
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"# DFA which matches all binary strings ending in an odd number of '1's\n",
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"my_dfa = DFA(\n",
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"\tstates={'q0', 'q1', 'q2'},\n",
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"\tinput_symbols={'0', '1'},\n",
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"\ttransitions={\n",
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"\t\t'q0': {'0': 'q0', '1': 'q1'},\n",
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"\t\t'q1': {'0': 'q0', '1': 'q2'},\n",
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"\t\t'q2': {'0': 'q2', '1': 'q1'}\n",
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"\t},\n",
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"\tinitial_state='q0',\n",
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"\tfinal_states={'q1'}\n",
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")\n",
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"\n",
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"my_dfa.show_diagram()\n"
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],
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"end_time": "2024-01-26T19:01:38.983537200Z",
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"start_time": "2024-01-26T19:01:38.652214500Z"
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@ -61,6 +20,16 @@
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},
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"id": "b67364155fcb1072"
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},
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{
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"cell_type": "markdown",
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"source": [
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"<div style=\"page-break-after: always;\"></div>"
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@ -82,6 +51,16 @@
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},
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"id": "ead9052998c5edf6"
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},
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{
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"cell_type": "markdown",
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"source": [
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"<div style=\"page-break-after: always;\"></div>"
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],
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"collapsed": false
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@ -109,12 +88,22 @@
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},
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"id": "874c8dbcc345edad"
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},
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{
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"cell_type": "markdown",
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"source": [
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"<div style=\"page-break-after: always;\"></div>"
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],
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"metadata": {
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"collapsed": false
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"id": "ca477f12a96f397a"
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},
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{
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"cell_type": "markdown",
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"source": [
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"# Question 11\n",
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"Consider the relation $R$ defined on the set $\\mathbb{Z}$ as follows:\n",
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"$$\\forall m,n \\in \\mathbb{Z}, (m,n \\in R \\text{ if and only if } m + n = 2k \\text{ for some integer } k$$\n",
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"$$\\forall m,n \\in \\mathbb{Z}, (m,n) \\in R \\text{ if and only if } m + n = 2k \\text{ for some integer } k$$\n",
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"\n",
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"(a) Is this relation reflexive? Yes\n",
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"For every integer $x$, $x + x = 2x$, therefore it is reflexive\n",
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@ -126,12 +115,147 @@
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"\n",
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"(c) Is this relation transitive? Yes\n",
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"Let $(a,b)$ and $(b,c)$ be valid pairs of integers of the relation $R$\n",
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"$a + b = 2n$, and $b + c = 2p$"
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"$a + b = 2n$, and $b + c = 2p$, then $b = 2n - a$ and $b = 2p - c$\n",
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"$-2n + a = 2p - c$\n",
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"$a + c = 2p + 2n$\n",
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"Since the sum of a and c is the sum of two even numbers (numbers multiplied by 2 must be even), then the result must be even, so the result is divisible by 2\n",
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"$\\therefore$ The relationship is transitive"
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],
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"cell_type": "markdown",
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"<div style=\"page-break-after: always;\"></div>"
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],
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"metadata": {
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"collapsed": false
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},
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"id": "889a2a7359618ee7"
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{
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"cell_type": "markdown",
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"source": [
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"# Question 14\n",
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"There are 7 equivalence classes as the results are grouped by the integer returned by $n_0(x) - n_1(x)$/$n_0(y) - n_1(y)$ for any relation pair, the results are that being set of differences $\\{-3, -2, -1, 0, 1, 2, 3\\}$"
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"id": "fd69c73a15e8e6a0"
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},
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{
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"cell_type": "code",
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"{-3: [('111', '111')],\n",
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" -2: [('11', '11')],\n",
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" -1: [('1', '1'),\n",
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" ('1', '011'),\n",
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" ('1', '101'),\n",
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" ('1', '110'),\n",
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" ('011', '1'),\n",
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" ('011', '011'),\n",
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" ('011', '101'),\n",
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" ('011', '110'),\n",
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" ('101', '1'),\n",
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" ('101', '011'),\n",
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" ('101', '101'),\n",
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" ('101', '110'),\n",
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" ('110', '1'),\n",
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" ('110', '011'),\n",
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" ('110', '101'),\n",
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" ('110', '110')],\n",
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" 0: [('', ''),\n",
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" ('', '01'),\n",
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" ('', '10'),\n",
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" ('01', ''),\n",
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" ('01', '01'),\n",
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" ('01', '10'),\n",
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" ('10', ''),\n",
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" ('10', '01'),\n",
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" ('10', '10')],\n",
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" 1: [('0', '0'),\n",
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" ('0', '001'),\n",
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" ('0', '010'),\n",
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" ('0', '100'),\n",
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" ('001', '0'),\n",
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" ('001', '001'),\n",
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" ('001', '010'),\n",
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" ('001', '100'),\n",
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" ('010', '0'),\n",
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" ('010', '001'),\n",
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" ('010', '010'),\n",
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" ('010', '100'),\n",
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" ('100', '0'),\n",
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" ('100', '001'),\n",
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" ('100', '010'),\n",
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" ('100', '100')],\n",
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" 2: [('00', '00')],\n",
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" 3: [('000', '000')]}\n"
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]
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}
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],
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"source": [
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"# Set of all strings containing 0 and 1 up to length 3\n",
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"def get_all_strings_with_a_given_alphabet_and_length(_alphabet, _length):\n",
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"\t_strings = ['']\n",
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"\tfor i in range(_length):\n",
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"\t\ts = [s + c for s in _strings for c in _alphabet]\n",
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"\t\t_strings += s\n",
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"\n",
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"\t# Remove duplicates (set() isn't as nice as it doesn't preserve order, at least for verification purposes)\n",
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"\t_strings = list(dict.fromkeys(_strings))\n",
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"\treturn _strings\n",
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"\n",
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"alphabet = ['0', '1']\n",
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"strings = get_all_strings_with_a_given_alphabet_and_length(alphabet, 3)\n",
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"\n",
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"# The relation is valid if the number of zeros from string x minus the number of ones from string x is equal to the number of zeros from string y minus the number of ones from string y\n",
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"def is_valid_relation(_x, _y):\n",
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"\t# Count the number of zeros and ones in each string\n",
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"\tx_zeros = _x.count('0')\n",
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"\tx_ones = _x.count('1')\n",
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"\ty_zeros = _y.count('0')\n",
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"\ty_ones = _y.count('1')\n",
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"\n",
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"\t# Return true if the difference between the number of zeros and ones is equal\n",
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"\treturn (x_zeros - x_ones) == (y_zeros - y_ones)\n",
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"\n",
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"from collections import defaultdict\n",
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"\n",
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"# Print all valid relations\n",
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"equivalence_classes = defaultdict(list)\n",
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"for x in strings:\n",
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"\tfor y in strings:\n",
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"\t\tif is_valid_relation(x, y):\n",
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"\t\t\t# create a tuple of the two strings\n",
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"\t\t\tt = (x, y)\n",
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"\t\t\t# calculate the difference between the number of zeros and ones\n",
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"\t\t\td = x.count('0') - x.count('1')\n",
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"\t\t\t# add the tuple a list under the key of the difference\n",
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"\t\t\tequivalence_classes[d].append(t)\n",
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"\t\t\t\t\n",
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"# Print the dict\n",
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"equivalence_classes = dict(equivalence_classes)\n",
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"from pprint import pprint\n",
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"pprint(equivalence_classes)"
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],
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"ExecuteTime": {
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"end_time": "2024-01-26T22:03:01.411977200Z",
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"start_time": "2024-01-26T22:03:01.364977700Z"
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}
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},
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"id": "94712ae197e8639c",
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"execution_count": 5
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}
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],
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