283 lines
9.3 KiB
Plaintext
283 lines
9.3 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "markdown",
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"source": [
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"# Question 6\n",
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"Let $S = \\{0,1\\}^*$ be the set of all strings of zero and ones, which includes the empty string $\\epsilon$.\n",
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"Let $h : S \\rightarrow \\mathbb{Z}^*$ be the function defined by $h(x)$ equal the number of zeros in $x$ multiplied by the number of ones in $x$\n",
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"For example, $h(00100011) = 5 \\times 3 = 15, and h(111) = 0 \\times 3 = 0$\n",
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"\n",
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"(a) Is $h$ one-to-one? No, because both strings $001$ and $110$ map to the same value, $2$\n",
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"$2 \\times 1 = 2$ and $1 \\times 2 = 2$\n",
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"\n",
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"(b) Is $h$ onto? Yes, because you can find every non-negative integer by multiplying any number of ones by one zero $(1 \\times 1), (1 \\times 2), ...$\n",
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"Let the number of zeros be exactly 1, and n be the number of ones, and m be any non-negative integer\n",
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"$h(n) = m, 1n = m, n = m$ "
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],
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"metadata": {
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"collapsed": false
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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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],
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"metadata": {
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"collapsed": false
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},
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"id": "ec5269e5b4749d08"
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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 8\n",
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"(a)\n",
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"This function is not one-to-one (111110, 1111100 are both 5)\n",
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"This function is onto (10, 110, 1110, 11...0 is all non-negative integers)\n",
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"\n",
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"(b)\n",
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"This function is not one-to-one (11111, 01111 are both 1)\n",
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"This function is onto (All strings are mapped to, as you can simply pad any character with any 4 characters first)\n",
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"\n",
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"(c)\n",
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"This function is not one-to-one (Quebec, Yukon both are 0)\n",
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"This function is not onto (None of the provinces contain 3 or 5 a's"
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],
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"metadata": {
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"collapsed": false
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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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"metadata": {
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"collapsed": false
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},
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"id": "f27520443479cce9"
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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 9\n",
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"Consider the relation $R$ defined on the set $\\mathbb{Z}$ as follows:\n",
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"$$R = \\{(m,n) | m,n \\in \\mathbb{Z}, mn < 0\\}$$\n",
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"\n",
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"(a) Is the relation reflexive? No\n",
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"For every integer $x$, $x \\times x = x^2$, and by the definition of squares, can never be less than zero\n",
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"\n",
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"(b) Is the relation symmetric? Yes\n",
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"Let $x, y$ be a pair of integers in the relation and $z$ be the product of $x and y$, so $x \\times y = -z$\n",
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"Since multiplication is commutative, the position of $x$ and $y$ do not matter, so $(x,y)$ and $(y,x)$ are in the relation\n",
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"(Both $x \\times y$ and $y \\times x$ equal $-z$)\n",
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"\n",
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"(c) Is the relation transitive? No\n",
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"Let $a = 1, b = -2, c = 3$\n",
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"$1 \\times -2 = -2$ and $-2 \\times 3 = -6$, however $1 \\times 3 = 3$, which is $> 0$\n",
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"$\\therefore$ this relation is not transitive\n",
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"\n",
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"(d) Is this an equivalence relation? No, because these three conditions are not met"
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],
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"metadata": {
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"collapsed": false
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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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},
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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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"\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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"\n",
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"(b) Is this relation symmetric? Yes\n",
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"Let $x, y$ be a pair of integers in the relation and $z$ be the sum of these integers, so $x + y = 2z$\n",
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"Since addition is commutative, the position of $x$ and $y$ do not matter, so $(x,y) and $(y,x) are in the relation\n",
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"(Both $x + y$ and $y + x$ equal $2z$)\n",
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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 $b + c = 2p$, 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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"metadata": {
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"collapsed": false
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},
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"id": "5fe823d05dd08b12"
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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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},
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"id": "889a2a7359618ee7"
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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 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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],
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"metadata": {
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"collapsed": false
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},
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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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"metadata": {
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"collapsed": false,
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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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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 2
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython2",
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"version": "2.7.6"
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"nbformat": 4,
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"nbformat_minor": 5
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