Lab-sheet-01-Variables-Conditional-Statements-and-Loops.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Variables, conditional statements and loops"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Q1: Building variables"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 46,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "20\n"
     ]
    }
   ],
   "source": [
    "age = 20 \n",
    "print(age)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 47,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "21\n"
     ]
    }
   ],
   "source": [
    "age = age + 1\n",
    "print(age)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 48,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "22\n"
     ]
    }
   ],
   "source": [
    "age += 1\n",
    "print(age)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 49,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "int"
      ]
     },
     "execution_count": 49,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "type(age)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 50,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "float"
      ]
     },
     "execution_count": 50,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "type(5.4)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The questions asks students to experiment with this further. Here are some further examples:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 51,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "28\n"
     ]
    }
   ],
   "source": [
    "age = 32\n",
    "years_in_education = age - 4\n",
    "print(years_in_education)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 52,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "784"
      ]
     },
     "execution_count": 52,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "years_in_education **= 2  # Squaring\n",
    "years_in_education"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 53,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "0"
      ]
     },
     "execution_count": 53,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "years_in_education % 4  # Remainder when dividing by 4"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 54,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "196.0"
      ]
     },
     "execution_count": 54,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "years_in_education / 4  # Check that it is a multiple of 4"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Q2: Creating character variables"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "It is possible to create character variables:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 55,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Vince\n"
     ]
    }
   ],
   "source": [
    "firstname = \"Vince\"\n",
    "print(firstname)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "These are called **strings** in Python (`str` for short):"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 56,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "<class 'str'>\n"
     ]
    }
   ],
   "source": [
    "print(type(firstname))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "We can combine strings:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 57,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "'Vince Knight'"
      ]
     },
     "execution_count": 57,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "lastname = \"Knight\"\n",
    "fullname = firstname + \" \" + lastname\n",
    "fullname  # Note that when using Jupyter we don't need to use 'print'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Q3: Boolean variables and if statements\n",
    "\n",
    "This is a type of variable that is either `True` or `False`"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 58,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "True"
      ]
     },
     "execution_count": 58,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a = 4\n",
    "b = 8 / 2\n",
    "a == b "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 59,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "False"
      ]
     },
     "execution_count": 59,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a != b"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 60,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "True"
      ]
     },
     "execution_count": 60,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a >= b"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 61,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "True"
      ]
     },
     "execution_count": 61,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a < b + 1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "We can (just like with any other variable) assign a boolean value to a variable:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 62,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "True"
      ]
     },
     "execution_count": 62,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "statement = a == b\n",
    "statement"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 63,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "bool"
      ]
     },
     "execution_count": 63,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "type(statement)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "We can use this in conjunction with `if`, `elif` and `else` statements to create code that is equivalent to the following mathematical function:\n",
    "\n",
    "$$f(n)=\\begin{cases}\n",
    "1&\\text{ if } n\\leq 5\\\\\n",
    "2&\\text{ if } n> 5\\text{ and } n \\text{ even}\\\\\n",
    "3&\\text{ otherwise }\\\\\n",
    "\\end{cases}$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 64,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "3"
      ]
     },
     "execution_count": 64,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "n = 11  # Experiment by changing the value of n\n",
    "if n <= 5:\n",
    "    value = 1\n",
    "elif n % 2 == 0:  # Otherwise if (else if)\n",
    "    value = 2\n",
    "else:  # Otherwise\n",
    "    value = 3\n",
    "value"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Q4: While loops\n",
    "\n",
    "It is possible to repeat code until a particular boolean variable becomes `False`:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 65,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "55"
      ]
     },
     "execution_count": 65,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "count = 0  # A variable to count\n",
    "total = 0  # We will sum the first ten numbers\n",
    "while count < 10:  # Keep repeating until count if >= 10\n",
    "    count += 1  # Adding 1 to count\n",
    "    total += count  # Adding the count to the total\n",
    "total"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Q5: Worked example\n",
    "\n",
    "This example gets a lot of numerical evidence for the following equality:\n",
    "\n",
    "$$\n",
    "\\sum_{i=0}^n i ^ 2 = \\frac{n(n+1)(2n+1)}{6}\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Here is the right hand side (for $n=20$):"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 66,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "2870.0"
      ]
     },
     "execution_count": 66,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "n = 20\n",
    "rhs = n * (n +  1) * (2 * n + 1) / 6\n",
    "rhs"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Here is the left hand side, note that we make use of a `while` loop to keep 'summing':"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 67,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "2870"
      ]
     },
     "execution_count": 67,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "lhs = 0\n",
    "i = 0\n",
    "while i < n:\n",
    "    i += 1\n",
    "    lhs += i ** 2\n",
    "lhs"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Finally here is a boolean to check if the two values are equal (so that we don't have to check the numbers ourselves):"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 68,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "True"
      ]
     },
     "execution_count": 68,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "lhs == rhs"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Now we can automate the checking of the above for various values of $n$ using another `while` loop:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 28,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "max_n = 2000\n",
    "n = 0\n",
    "while n < max_n:  # checking the equality for n up until 2000\n",
    "    n += 1\n",
    "    rhs = n * (n +  1) * (2 * n + 1) / 6\n",
    "    lhs = 0\n",
    "    i = 0\n",
    "    while i < n:\n",
    "        i += 1\n",
    "        lhs += i ** 2\n",
    "    if lhs != rhs:\n",
    "        print(False)  # Will print a False if it ever finds an error"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Q6 Debugging exercise\n",
    "\n",
    "Here is the correct version of the code:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "max_n = 2000\n",
    "n = 0\n",
    "#while n > max_n:  This inequality is the wrong way around:\n",
    "while n < max_n:\n",
    "    #n += 2  We should be incrementing by 1\n",
    "    n += 1\n",
    "    #rhs = ((n ** 2 + 2 * n) ** 2) / 4 This is the incorrect formula\n",
    "    rhs = ((n ** 2 + n) ** 2) / 4\n",
    "    lhs = 0\n",
    "    i = 0\n",
    "    while i < n:\n",
    "        i += 1\n",
    "        #lhs += i ** 2 Incrementing the incorrect amount\n",
    "        lhs += i ** 3\n",
    "    if lhs != rhs:\n",
    "        print(False)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Q7\n",
    "\n",
    "This is the same as the first part of the previous question but with a simpler equality:\n",
    "\n",
    "$$\n",
    "\\sum_{i=0}^{n}i=\\frac{n(n+1)}{2}\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Here is the right hand side:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 30,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "210.0"
      ]
     },
     "execution_count": 30,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "n = 20\n",
    "rhs = n * (n +  1) / 2\n",
    "rhs"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Here is the left hand side:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 33,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "210"
      ]
     },
     "execution_count": 33,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "lhs = 0\n",
    "i = 0\n",
    "while i < n:\n",
    "    i += 1\n",
    "    lhs += i  # This is the only line that is different to before\n",
    "lhs"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 34,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "True"
      ]
     },
     "execution_count": 34,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "lhs == rhs"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Q8: TICKABLE"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "We do the same as in Q5: wrap the code in another `while` loop:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 35,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "max_n = 2000\n",
    "n = 0\n",
    "while n < max_n:  # checking the equality for n up until 2000\n",
    "    n += 1\n",
    "    rhs = n * (n +  1) / 2\n",
    "    lhs = 0\n",
    "    i = 0\n",
    "    while i < n:\n",
    "        i += 1\n",
    "        lhs += i\n",
    "    if lhs != rhs:\n",
    "        print(False)  # Will print a False if it ever finds an error"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Nothing is output when we run the above cell so the equality has been check for the first 2000 values of n."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Q9\n",
    "\n",
    "Here we use a `while` loop to get the sum of the first numbers less than 1000 that are divisible by 3:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "333667"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "upper_limit = 1000\n",
    "\n",
    "total = 0\n",
    "number = 0\n",
    "\n",
    "while number < upper_limit:\n",
    "    number += 1  # Increment the number\n",
    "    if number % 3 != 0:  # Check if it is divisible by 3\n",
    "        total += number\n",
    "total"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Q10\n",
    "\n",
    "Here we use a `while` loop to get the sum of the first 1000 numbers that are divisible by 3:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "750000"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "upper_count = 1000\n",
    "\n",
    "total = 0\n",
    "number = 0\n",
    "count = 0\n",
    "\n",
    "while count < upper_count:\n",
    "    number += 1  # Increment the number\n",
    "    if number % 3 != 0:  # Check if it is divisible by 3\n",
    "        total += number\n",
    "        # Keep track of how many numbers we have that are divisible by 3\n",
    "        count += 1  \n",
    "total "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Q11\n",
    "\n",
    "Implement the following algorithm for calculating the square root of a number:\n",
    "\n",
    "$$\n",
    "x_{n+1}=\\frac{x_n + K / x_{n}}{2}\n",
    "$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "13.5\n",
      "10.157407407407408\n",
      "9.607418380093858\n",
      "9.591675965316126\n"
     ]
    }
   ],
   "source": [
    "epsilon = 0.001\n",
    "K = 92  #This sets an initial value to take the square root of\n",
    "X = K / 4.0  # This picks an initial value for the sequence\n",
    "while abs(X**2 - K) > epsilon:  # abs is the absolute value\n",
    "    X = (X + K / X) / 2\n",
    "    print(X)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Let us check:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "92.00024782362304"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "X ** 2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "That is correct up to a difference of 0.001 (which was the precision we required)."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": []
  }
 ],
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