lab1.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# DSCI 512 Lab 1"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [],
   "source": [
    "import numpy as np\n",
    "import pandas as pd\n",
    "import matplotlib.pyplot as plt\n",
    "import urllib.request\n",
    "from collections import defaultdict, Counter\n",
    "%config InlineBackend.figure_formats = ['svg']"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Submission instructions\n",
    "rubric={mechanics:10}\n",
    "\n",
    "- Follow the [general lab instructions](https://ubc-mds.github.io/resources_pages/general_lab_instructions/).\n",
    "- Make at least three commits in your lab's GitHub repository.\n",
    "- Push the final `.ipynb` file and the rendered `pdf` file to your GitHub repository for this lab.\n",
    "- Before submitting your lab, run all cells in your notebook to make sure there are no errors by doing `Kernel -> Restart Kernel and Clear All Outputs` and then `Run -> Run All Cells`. Notebooks without the output displayed may not be graded at all (because we need to see the output in order to grade your work).\n",
    "- Make sure that your plots/output are rendered properly in Gradescope.\n",
    "- Upload a ZIP file to Gradescope containing **only** the following two files:\n",
    "    1. Your completed and executed `.ipynb` file\n",
    "    2. A PDF-exported version of the `.ipynb` file\n",
    "- Include a clickable link to your GitHub repo for the lab just below this cell."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "_YOUR REPO LINK GOES HERE_"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "## Exercise 1: time complexity\n",
    "\n",
    "For each of the following functions, determine the time complexity as a function of the input $n$ using big-O notation and briefly justify your answer. Assume $n$ is a positive integer. If you get stuck, it's fair game to test things empirically and then try to understand what you observe. **Please state your assumptions if you don’t know how long some operation in Python takes.** \n",
    "\n",
    "The first question is done for you, as an example."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [],
   "source": [
    "def example(n):\n",
    "    for i in range(n):\n",
    "        print(i)\n",
    "        print(i**2)\n",
    "        x = 9\n",
    "        y = 10"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "example(5)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Sample answer**: The time complexity of `example` is  $O(n)$ because the function loops over $n$ elements and only performs constant-time operations inside the loop. "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### 1(a)\n",
    "rubric={reasoning:2}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q1_a\n",
    "points: 2\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [],
   "source": [
    "def loopy(n):\n",
    "    for i in range(n):\n",
    "        for j in range(n):\n",
    "            print('i =', i, '  j =', j)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "i = 0   j = 0\n",
      "i = 0   j = 1\n",
      "i = 0   j = 2\n",
      "i = 0   j = 3\n",
      "i = 1   j = 0\n",
      "i = 1   j = 1\n",
      "i = 1   j = 2\n",
      "i = 1   j = 3\n",
      "i = 2   j = 0\n",
      "i = 2   j = 1\n",
      "i = 2   j = 2\n",
      "i = 2   j = 3\n",
      "i = 3   j = 0\n",
      "i = 3   j = 1\n",
      "i = 3   j = 2\n",
      "i = 3   j = 3\n"
     ]
    }
   ],
   "source": [
    "loopy(4)"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "The time complexity of `loopy` is  $O(n^2)$ because of it is a loop within a loop, each loop with n iterations."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "#### 1(b)\n",
    "rubric={reasoning:2}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q1_b\n",
    "points: 2\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "outputs": [],
   "source": [
    "def triangle(n):\n",
    "    for i in range(n):\n",
    "        for j in range(i):\n",
    "            print(\"+\", end='')\n",
    "        print(\"\")"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      "+\n",
      "++\n",
      "+++\n",
      "++++\n",
      "+++++\n",
      "++++++\n"
     ]
    }
   ],
   "source": [
    "triangle(7)"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "Outer loop has n iterations. Inner loop has 1,2,...(n-1) iterations. Critical section (code within inner loop) is constant time.\n",
    "We can count the total number of critical section iterations ('+'s printed) with $\\sum_{i=1}^{n-1}i=\\frac{n(n-1)}{2} \\in O(n^2)$.\n",
    "\n",
    "Therefore the time complexity of `broken` is $O(n^2)$."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### 1(c)\n",
    "rubric={reasoning:2}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q1_c\n",
    "points: 2\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [],
   "source": [
    "def foo(n):\n",
    "    x = np.zeros(n)\n",
    "    x = x + 1000\n",
    "    return x"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "size of x:  100000\n"
     ]
    }
   ],
   "source": [
    "print('size of x: ', len(foo(100000)))"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The time complexity of `foo` is  $O(n)$ because 1000 is added to each element of x, which is of size $n$. "
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### 1(d)\n",
    "rubric={reasoning:3}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q1_d\n",
    "points: 3\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [],
   "source": [
    "def bar(n):\n",
    "    x = np.zeros(1000)\n",
    "    x = x + n\n",
    "    return x"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "size of x:  1000\n"
     ]
    }
   ],
   "source": [
    "print('size of x: ', len(bar(100000)))"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The time complexity of `bar` is  $O(1)$ because the size of `x` is constant. We are adding $n$ to each element of `x`, but since `x` has a constant size (1000), the running time is independent of $n$, and therefore $O(1)$.\n",
    "\n",
    "In other words, the number of additions is constant (1000) and whether you're adding a small number or a large number doesn't affect the running time."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### 1(e)\n",
    "rubric={reasoning:3}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q1_e\n",
    "points: 3\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {},
   "outputs": [],
   "source": [
    "def broken(n):\n",
    "    for i in range(n**2):\n",
    "        if i == n:\n",
    "            break  # \"break\" exits the innermost loop\n",
    "        print(i)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "0\n",
      "1\n",
      "2\n",
      "3\n"
     ]
    }
   ],
   "source": [
    "broken(4)"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The time complexity of `broken` is $O(n)$ because the loop iterates $n$ times; it is broken when $i$ equals $n$. "
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### 1(f)\n",
    "rubric={reasoning:3}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q1_f\n",
    "points: 3\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [],
   "source": [
    "def cabin(n):\n",
    "    i = n\n",
    "    while i > 1:\n",
    "        print('i = ', i)\n",
    "        i = i // 2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Note: the `//` operator performs integer division, meaning the result is rounded *down* to the nearest integer."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "i =  2048\n",
      "i =  1024\n",
      "i =  512\n",
      "i =  256\n",
      "i =  128\n",
      "i =  64\n",
      "i =  32\n",
      "i =  16\n",
      "i =  8\n",
      "i =  4\n",
      "i =  2\n"
     ]
    }
   ],
   "source": [
    "cabin(2048)"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The time complexity of `cabin` is $O(\\log n)$. At each iteration `i` is divided by 2, so the number is steps is the number of times you need to divide a number by 2 to get to 1, which is roughly $\\log_2 n$. The base of the log doesn't matter in big O, so we write $O(\\log n)$."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "tags": []
   },
   "source": [
    "#### 1(g)\n",
    "rubric={reasoning:3}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q1_g\n",
    "points: 3\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {},
   "outputs": [],
   "source": [
    "def cabin10(n):\n",
    "    i = n\n",
    "    while i > 1:\n",
    "        print('i = ', i)\n",
    "        i = i // 10"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "i =  2048\n",
      "i =  204\n",
      "i =  20\n",
      "i =  2\n"
     ]
    }
   ],
   "source": [
    "cabin10(2048)"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The time complexity of `cabin10` is also $O(\\log n)$.\n",
    "\n",
    "Note that $O(\\log n)$ is the same thing as $O(\\log_2 n)$ or $O(\\log_{10} n)$ because they only differ by a constant factor."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "tags": []
   },
   "source": [
    "#### 1(h)\n",
    "rubric={reasoning:3}\n",
    "\n",
    "For this question, answer in terms of both $m$ and $n$."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q1_h\n",
    "points: 3\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {},
   "outputs": [],
   "source": [
    "def blahblah(n, m):\n",
    "    x = 0\n",
    "\n",
    "    for i in range(n):\n",
    "        for j in range(m):\n",
    "            x = x + 1\n",
    "\n",
    "    for i in range(n):\n",
    "        x = x + 1\n",
    "\n",
    "    for i in range(m):\n",
    "        x = x + 1\n",
    "        \n",
    "    return x"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "11"
      ]
     },
     "execution_count": 20,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "blahblah(2,3)"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The time complexity is $O(nm)$ because we have an $nm$ loop, an $n$ loop and an $m$ loop, and the latter two are smaller + can be neglegected."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "tags": []
   },
   "source": [
    "#### 1(i)\n",
    "rubric={reasoning:3}\n",
    "\n",
    "For this question, answer in terms of both $m$ and $n$."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q1_i\n",
    "points: 3\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "metadata": {},
   "outputs": [],
   "source": [
    "def bllllergh(n, m):\n",
    "    x = 0\n",
    "    for i in range(n):\n",
    "        for j in range(m):\n",
    "            for k in range(m):\n",
    "                x = x + 1\n",
    "    for i in range(n):\n",
    "        for j in range(n):\n",
    "            for k in range(m):\n",
    "                x = x + 1\n",
    "    return x"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 22,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "30"
      ]
     },
     "execution_count": 22,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "bllllergh(2,3)"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$O(nm^2 + n^2m)$. Neither term can be removed because we can't know which one will be bigger without knowing $n$ and $m$."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### 1(j)\n",
    "rubric={reasoning:3}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q1_j\n",
    "points: 1\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {},
   "outputs": [],
   "source": [
    "def log_cabin(n):\n",
    "    for i in range(n):\n",
    "        print('i = ', i)\n",
    "        for j in range(n//3):\n",
    "            print('j = ', j)\n",
    "            cabin(n)\n",
    "        print('-----------')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "i =  0\n",
      "j =  0\n",
      "i =  4\n",
      "i =  2\n",
      "-----------\n",
      "i =  1\n",
      "j =  0\n",
      "i =  4\n",
      "i =  2\n",
      "-----------\n",
      "i =  2\n",
      "j =  0\n",
      "i =  4\n",
      "i =  2\n",
      "-----------\n",
      "i =  3\n",
      "j =  0\n",
      "i =  4\n",
      "i =  2\n",
      "-----------\n"
     ]
    }
   ],
   "source": [
    "log_cabin(4)"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The time complexity of `log_cabin` is  $O(n^2 \\log n)$. The outer loop runs $n$ times, the inner loops runs $n/3$ times, and the body of the inner loop is $O(\\log n)$ time. Multiplying these together and discarding the constant factor of $1/3$ gives us the answer."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exercise 2: time complexity of recursive functions\n",
    "\n",
    "For each of the following recursive functions, determine the time complexity as a function of the input $n$ and briefly justify your answer. Assume $n$ is a positive integer."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### 2(a)\n",
    "rubric={reasoning:3}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q2_a\n",
    "points: 3\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 25,
   "metadata": {},
   "outputs": [],
   "source": [
    "def titled(n):\n",
    "    if n >= 0:\n",
    "        print('n: ', n)\n",
    "        return titled(n-1)\n",
    "    else:\n",
    "        return \"sandwich\""
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 26,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "n:  15\n",
      "n:  14\n",
      "n:  13\n",
      "n:  12\n",
      "n:  11\n",
      "n:  10\n",
      "n:  9\n",
      "n:  8\n",
      "n:  7\n",
      "n:  6\n",
      "n:  5\n",
      "n:  4\n",
      "n:  3\n",
      "n:  2\n",
      "n:  1\n",
      "n:  0\n"
     ]
    },
    {
     "data": {
      "text/plain": [
       "'sandwich'"
      ]
     },
     "execution_count": 26,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "titled(15)"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "The time complexity of `titled` is  $O(n)$ because there are $n-1$ recursive calls to the function."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### 2(b)\n",
    "rubric={reasoning:3}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q2_b\n",
    "points: 3\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 27,
   "metadata": {},
   "outputs": [],
   "source": [
    "def untitled(n):\n",
    "    if n < 0:\n",
    "        return \"sandwich\"\n",
    "    else:\n",
    "        print('n: ', n)\n",
    "        return untitled(n-2)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 28,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "n:  8\n",
      "n:  6\n",
      "n:  4\n",
      "n:  2\n",
      "n:  0\n"
     ]
    },
    {
     "data": {
      "text/plain": [
       "'sandwich'"
      ]
     },
     "execution_count": 28,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "untitled(8)"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "toc-hr-collapsed": true
   },
   "source": [
    "The time complexity of `untitled` is  $O(n)$ because there are $n/2$ recursive calls to the function and $O(n/2)$ comes down to $O(n)$, as constants are ignored in big-O. "
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 2(c)\n",
    "rubric={reasoning:3}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q2_c\n",
    "points: 3\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 29,
   "metadata": {},
   "outputs": [],
   "source": [
    "def does_nothing(n):\n",
    "    print('n:', n)\n",
    "    if n == 0:\n",
    "        return\n",
    "    does_nothing(n-1)\n",
    "    does_nothing(n-1)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 30,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "n: 3\n",
      "n: 2\n",
      "n: 1\n",
      "n: 0\n",
      "n: 0\n",
      "n: 1\n",
      "n: 0\n",
      "n: 0\n",
      "n: 2\n",
      "n: 1\n",
      "n: 0\n",
      "n: 0\n",
      "n: 1\n",
      "n: 0\n",
      "n: 0\n"
     ]
    }
   ],
   "source": [
    "does_nothing(3)"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "toc-hr-collapsed": true
   },
   "source": [
    "The time complexity of `does_nothing` is $O(2^n)$ because each call of the function has two recursive calls. If you draw the calls in a tree fashion, the depth of the tree (the number of levels) is $n$, and at each level we see a branching factor of two.\n",
    "\n",
    "So there will be total $2^0 + 2^1 + 2^2 + \\ldots + 2^{n-1} + 2^n = 2^{n+1} - 1$ function calls. In big O, this is equivalent to $O(2^n)$."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### 2(d)\n",
    "rubric={reasoning:3}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q2_d\n",
    "points: 3\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 31,
   "metadata": {},
   "outputs": [],
   "source": [
    "def does_nothing_more_slowly(n):\n",
    "    print(n)\n",
    "    if n == 0:\n",
    "        return\n",
    "    does_nothing_more_slowly(n-1)\n",
    "    does_nothing_more_slowly(n-1)\n",
    "    does_nothing_more_slowly(n-1)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 32,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "3\n",
      "2\n",
      "1\n",
      "0\n",
      "0\n",
      "0\n",
      "1\n",
      "0\n",
      "0\n",
      "0\n",
      "1\n",
      "0\n",
      "0\n",
      "0\n",
      "2\n",
      "1\n",
      "0\n",
      "0\n",
      "0\n",
      "1\n",
      "0\n",
      "0\n",
      "0\n",
      "1\n",
      "0\n",
      "0\n",
      "0\n",
      "2\n",
      "1\n",
      "0\n",
      "0\n",
      "0\n",
      "1\n",
      "0\n",
      "0\n",
      "0\n",
      "1\n",
      "0\n",
      "0\n",
      "0\n"
     ]
    }
   ],
   "source": [
    "does_nothing_more_slowly(3)"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "toc-hr-collapsed": true
   },
   "source": [
    "The time complexity of `does_nothing_more_slowly` is $O(3^n)$ because each call of the function has three recursive calls. If you draw the calls in a tree fashion, the depth of the tree (the number of levels) is $n$, and at each level we see a branching factor of three.\n",
    "\n",
    "So there will be total $3^0 + 3^1 + 3^2 + \\ldots + 3^{n}$ function calls. In big O, this is equivalent to $O(3^n)$."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "#### 2(e)\n",
    "rubric={reasoning:3}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q2_e\n",
    "points: 3\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 33,
   "metadata": {
    "collapsed": false,
    "is_executing": true,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "outputs": [],
   "source": [
    "def recursive_cabin(n):\n",
    "    if n == 0:\n",
    "        return\n",
    "    print(\"n = %d\" % n)\n",
    "    recursive_cabin(n//2)\n",
    "    "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 34,
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "n = 100\n",
      "n = 50\n",
      "n = 25\n",
      "n = 12\n",
      "n = 6\n",
      "n = 3\n",
      "n = 1\n"
     ]
    }
   ],
   "source": [
    "recursive_cabin(100)"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "At each call, n is halved, resulting in a total of $O(log(n))$ calls to `recursive_cabin`. Within each call, we perform only constant-time operations. Therefore, the total time complexity is $O(log(n))$."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "#### 2(f)\n",
    "rubric={reasoning:3}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q2_f\n",
    "points: 3\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 35,
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "outputs": [],
   "source": [
    "def does_nothing_more_quickly(n):\n",
    "    print(n)\n",
    "    if n == 0:\n",
    "        return\n",
    "    for i in range(n):\n",
    "        does_nothing_more_quickly(0)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 36,
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "10\n",
      "0\n",
      "0\n",
      "0\n",
      "0\n",
      "0\n",
      "0\n",
      "0\n",
      "0\n",
      "0\n",
      "0\n"
     ]
    }
   ],
   "source": [
    "does_nothing_more_quickly(10)"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "The loop has $n$ iterations. In each iteration of the loop, we make a recursive call to `does_nothing_more_quickly`, however, since we only recursively call with `n=0`, we always hit the base case and immediately return. In other words, all recursive calls are constant-time. Therefore we have time complexity of $O(n)$."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### (challenging) 2(g)\n",
    "rubric={reasoning:1}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q2_g\n",
    "points: 1\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 37,
   "metadata": {},
   "outputs": [],
   "source": [
    "def looprec(n):\n",
    "    print(\"Hello!\")\n",
    "    print('N: ', n)\n",
    "    for i in range(n):\n",
    "        looprec(n-1)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 38,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Hello!\n",
      "N:  3\n",
      "Hello!\n",
      "N:  2\n",
      "Hello!\n",
      "N:  1\n",
      "Hello!\n",
      "N:  0\n",
      "Hello!\n",
      "N:  1\n",
      "Hello!\n",
      "N:  0\n",
      "Hello!\n",
      "N:  2\n",
      "Hello!\n",
      "N:  1\n",
      "Hello!\n",
      "N:  0\n",
      "Hello!\n",
      "N:  1\n",
      "Hello!\n",
      "N:  0\n",
      "Hello!\n",
      "N:  2\n",
      "Hello!\n",
      "N:  1\n",
      "Hello!\n",
      "N:  0\n",
      "Hello!\n",
      "N:  1\n",
      "Hello!\n",
      "N:  0\n"
     ]
    }
   ],
   "source": [
    "looprec(3)"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "toc-hr-collapsed": true
   },
   "source": [
    "The tree generated by this recursive function has $n!$ leaves, because the root branches $n$ times, then $n-1$ times, then $n-2$ times, etc. However, the running time is technically the total number of nodes, not the number of leaves in the tree. Usually these are the same in terms of big-O, but here it's not quite. The total number of nodes is $\\sum_{i=1}^n \\frac{n!}{i!}$. Since big-O is just an upper bound, we can upper-bound each term in the sum by $n!$ and thus the whole sum by $n\\times n!$ or $(n+1)!$. This would be a fine answer, as would be $O(n^{n+1})$, which is an even looser bound. In reality, the running time is so huge that anything this slow would be impractical, so finding a very tight bound isn't that critical here."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exercise 3: space complexity\n",
    "\n",
    "For each of the following functions, determine the space complexity as a function of the input $n$ using big-O notation and briefly justify your answer. "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### 3(a)\n",
    "rubric={reasoning:3}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q3_a\n",
    "points: 3\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 39,
   "metadata": {},
   "outputs": [],
   "source": [
    "def foo(n):\n",
    "    x = np.random.rand(n)\n",
    "    y = np.random.rand(n)\n",
    "    total = 0\n",
    "    for x_i in x:\n",
    "        for y_i in y:\n",
    "            total += x_i*y_i\n",
    "    return total"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The space complexity of `foo` is $O(n)$ because we are storing $x$ and $y$, each of which has $n$ elements."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### 3(b)\n",
    "rubric={reasoning:3}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q3_b\n",
    "points: 3\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 40,
   "metadata": {},
   "outputs": [],
   "source": [
    "def bar(n):\n",
    "    x = np.zeros(1000)\n",
    "    x = x + n\n",
    "    return x"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The space complexity of `bar` is $O(1)$ because $x$ has only 1000 elements, which is a constant. "
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### 3(c)\n",
    "rubric={reasoning:3}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q3_c\n",
    "points: 3\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 41,
   "metadata": {
    "is_executing": true
   },
   "outputs": [],
   "source": [
    "def FUNCTION(n):\n",
    "    x = set()\n",
    "    for i in range(n):\n",
    "        for j in range(n):\n",
    "            x.add(j)\n",
    "    return x"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The space complexity of `FUNCTION` is $O(n)$ because `x` is a set, which cannot contain duplicates. While the time complexity is $O(n^2)$, the set only grows to contain the integers from $0$ to $n-1$."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "#### 3(d)\n",
    "rubric={reasoning:3}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q3_d\n",
    "points: 3\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 42,
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "outputs": [],
   "source": [
    "def hello(n):\n",
    "    x = list()\n",
    "    for i in range(n):\n",
    "        y = [j for j in range(1000)]\n",
    "        x.append(y)\n",
    "    return x"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "The space complexity of `hello` is $O(n)$ because we store `1000n` elements, each constant-size, to `x` (1000 elements for each loop). We can discard the constant factor, yielding $O(n)$."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### (challenging) 3(f)\n",
    "rubric={reasoning:1}\n",
    "\n",
    "In this exercise, determine the **space** complexity of `hello` in terms of $n$."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q3_f\n",
    "points: 1\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 43,
   "metadata": {},
   "outputs": [],
   "source": [
    "def hello(n):\n",
    "    if n == 0:\n",
    "        return 1\n",
    "    return hello(n-1) + hello(n-1)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 44,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "16"
      ]
     },
     "execution_count": 44,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "hello(4)"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$O(n)$. \n",
    "\n",
    "While the time complexity of `hello` is $O(2^n)$, the space complexity is only $O(n)$. The function call stack stores functions within functions within functions. E.g. `hello(10)` calls `hello(9)` calls `hello(8)` calls `hello(7)`, etc. The depth of the recursion is $n$, so we never have a stack of function calls more than $n$ deep. What this means practically  is that the function at the top of the stack finishes before the next function is called, so the stack isn't actually growing once it reaches $n$ function calls, because functions are completing every time they are being called. "
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exercise 4: Timing for hashing"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### 4(a)\n",
    "rubric={reasoning:5}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q4_a\n",
    "points: 5\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "We observed in class that Python's `dict` data structure grows in size when it becomes too full. In this question, we will investigate exactly when this growth occurs.\n",
    "\n",
    "- First, create a plot showing the memory size of a dictionary using [`sys.getsizeof`](https://docs.python.org/3/library/sys.html#sys.getsizeof) as a function of the number of elements it contains. Consider dictionary sizes ranging from 1 to 1,000,000 elements.\n",
    "- Using the data you generate, identify the points where the dictionary size increases. Discuss your result."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 64,
   "metadata": {},
   "outputs": [],
   "source": [
    "## IGNORE ##\n",
    "import sys\n",
    "\n",
    "d = dict()\n",
    "l = list()\n",
    "nmax = 1_000_000\n",
    "d_sizes = np.zeros(nmax)\n",
    "l_sizes = np.zeros(nmax)\n",
    "for i in range(nmax):\n",
    "    key = np.random.randint(0, 1e15)\n",
    "    d[key] = \"blah\"\n",
    "    d_sizes[i] = sys.getsizeof(d)\n",
    "    l.append(\"blah\")\n",
    "    l_sizes[i] = sys.getsizeof(l)\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 46,
   "metadata": {},
   "outputs": [
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   "source": [
    "## IGNORE ##\n",
    "plt.plot(d_sizes, label='dictionary')\n",
    "plt.plot(l_sizes, label='list')\n",
    "plt.xlabel('number of elements')\n",
    "plt.ylabel('size (bytes)')\n",
    "plt.title('dictionary growth')\n",
    "plt.legend();"
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   "metadata": {},
   "source": [
    "And a log-log version:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 47,
   "metadata": {},
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   ],
   "source": [
    "## IGNORE ##\n",
    "plt.loglog(d_sizes, label='dictionary')\n",
    "plt.loglog(l_sizes, label='list')\n",
    "plt.xlabel('number of elements')\n",
    "plt.ylabel('size (bytes)')\n",
    "plt.title('dictionary growth')\n",
    "plt.legend()"
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  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Here are the sizes where the hash table grew:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 48,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "(array([     4,      9,     20,     41,     84,    169,    340,    681,\n",
       "          1364,   2729,   5460,  10921,  21844,  43689,  87380, 174761,\n",
       "        349524, 699049]),)"
      ]
     },
     "execution_count": 48,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "## IGNORE ##\n",
    "np.where(np.diff(d_sizes) != 0)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "These are roughly powers of 2:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 49,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[ 2.        ,  3.169925  ,  4.32192809,  5.357552  ,  6.39231742,\n",
       "         7.40087944,  8.40939094,  9.41151099, 10.41362793, 11.41415668,\n",
       "        12.41468524, 13.41481734, 14.41494944, 15.41498246, 16.41501549,\n",
       "        17.41502374, 18.415032  , 19.41503406]])"
      ]
     },
     "execution_count": 49,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "## IGNORE ##\n",
    "np.log2(np.where(np.diff(d_sizes) != 0))"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
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   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### (challenging) 4(b)\n",
    "rubric={reasoning:1}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
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   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q4_b\n",
    "points: 1\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
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   },
   "source": [
    "Repeat the experiment, but this time with a Python `list`, instead of a `dict`. Plot the memory size of a list (using `sys.getsizeof`) versus the number of elements, ranging from 1 to 1,000,000. Then compare the growth behavior of the list to that of the dict and discuss your findings.\n"
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   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
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   },
   "source": [
    "# BEGIN SOLUTION"
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   "execution_count": 50,
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   "source": [
    "## IGNORE ##\n",
    "plt.plot(l_sizes)\n",
    "plt.xlabel('number of elements');\n",
    "plt.ylabel('size (bytes)');\n",
    "plt.title('list growth');"
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The dictionary appears to double from time to time, the list grows more frequently and grows by a smaller amount; it seems to grow by a small percentage of the size:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 51,
   "metadata": {},
   "outputs": [
    {
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       "       0.11111802, 0.11111546, 0.1111171 , 0.11111454, 0.11111542])"
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     },
     "execution_count": 51,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "## IGNORE ##\n",
    "z = np.diff(l_sizes) / l_sizes[1:]\n",
    "z[z > 0]"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "So it grows by roughly 11%."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
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   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
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   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exercise 5: Set implementation with BSTs\n",
    "\n",
    "In this exercise, you will implement a set data structure based on a binary search tree. You will write the tree as a Python class. We are providing some starter code for you below. "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### 5(a)\n",
    "rubric={accuracy:6,quality:6}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
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   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q5_a\n",
    "points: 12\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
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   "source": [
    "\n",
    "Implement a recursive method `insert` that takes a new element and inserts it into the tree. Your function should work by recursively calling `insert` on the left or right subtree depending on whether the new value is less than or greater than the tree's value, respectively. If the element is already in the tree, then the call to `insert` should do nothing. As usual, a docstring is required.\n",
    "\n",
    "Hint: When inserting something into the tree, you should be creating a new `TreeSet` object with `TreeSet()`, then inserting the value into this newly created `TreeSet`, and then making sure this new `TreeSet` is stored in your current `TreeSet` as either `self.left` or `self.right`."
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   "metadata": {
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   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 52,
   "metadata": {},
   "outputs": [],
   "source": [
    "class TreeSet:\n",
    "    \"\"\"\n",
    "    A set implementation based on a binary tree.\n",
    "    \"\"\"\n",
    "\n",
    "    def __init__(self):\n",
    "        self.value = None\n",
    "        self.left = None\n",
    "        self.right = None\n",
    "\n",
    "    def insert(self, value):\n",
    "        \"\"\"\n",
    "        Insert the given value into the TreeSet\n",
    "\n",
    "        Parameters\n",
    "        ----------\n",
    "        value : object\n",
    "            the value to insert\n",
    "        \"\"\"\n",
    "        # YOUR CODE HERE\n",
    "        # BEGIN SOLUTION\n",
    "        if self.value is None:\n",
    "            self.value = value\n",
    "            return\n",
    "\n",
    "        # Traverse the left subtree\n",
    "        if value < self.value:\n",
    "            if self.left is None:\n",
    "                self.left = TreeSet()\n",
    "            self.left.insert(value)\n",
    "        # Traverse the right subtree\n",
    "        elif value > self.value:\n",
    "            if self.right is None:\n",
    "                self.right = TreeSet()\n",
    "            self.right.insert(value)\n",
    "        # else if they are equal, do nothing and exit\n",
    "        # END SOLUTION\n",
    "\n",
    "    def insert(self, item):\n",
    "        # pick a random side to insert on\n",
    "        if item == left:\n",
    "            return\n",
    "        if item < left:\n",
    "            if self.left is None:\n",
    "                self.left = BinaryTree(item)\n",
    "            else:\n",
    "                self.left.insert(item)\n",
    "        else:\n",
    "            if self.right is None:\n",
    "                self.right = BinaryTree(item)\n",
    "            else:\n",
    "                self.right.insert(item)\n",
    "                \n",
    "    def contains(self, value):\n",
    "        \"\"\"\n",
    "        Check to see if the binary tree has a certain value \n",
    "\n",
    "        Parameters\n",
    "        ----------\n",
    "        value : object\n",
    "            the value to search for within the tree\n",
    "\n",
    "        Returns\n",
    "        -------\n",
    "        bool \n",
    "            if contained in the tree returns True, otherwise False  \n",
    "\n",
    "        Example\n",
    "        --------\n",
    "        >>> my_set = TreeSet() \n",
    "        >>> my_set.insert(\"Attempt\") \n",
    "        >>> my_set.contains(\"Failure\")\n",
    "        False\n",
    "        \"\"\"\n",
    "        if self.value is None:\n",
    "            return False\n",
    "            \n",
    "        if value == self.value:\n",
    "            return True\n",
    "\n",
    "        if value < self.value:\n",
    "            if self.left is None:\n",
    "                return False\n",
    "            else:\n",
    "                return self.left.contains(value)\n",
    "        else:\n",
    "            if self.right is None:\n",
    "                return False\n",
    "            else:\n",
    "                return self.right.contains(value)\n",
    "\n",
    "    def __str__(self):\n",
    "        \"\"\"\n",
    "        A crude way to print the tree. A better way would be to print the tree by depth. \n",
    "\n",
    "        Note: __str__ is a special method, like __init__, that returns a string representation of an object.\n",
    "\n",
    "        Parameters\n",
    "        ----------\n",
    "        s : str\n",
    "           the starting string value. Default is empty string\n",
    "\n",
    "        Returns\n",
    "        -------\n",
    "        str \n",
    "            aggregated items in the set\n",
    "\n",
    "        Example\n",
    "        --------\n",
    "        >>> my_set = TreeSet() \n",
    "        >>> my_set.insert(\"try\")\n",
    "        >>> my_set.insert(\"your\")\n",
    "        >>> my_set.insert(\"best\")\n",
    "        >>> print(my_set)\n",
    "        best, try, your,\n",
    "        \"\"\"\n",
    "\n",
    "        if self.value is None:\n",
    "            return \"(An empty tree)\"\n",
    "\n",
    "        s = \"\"\n",
    "        \n",
    "        if self.left is not None:\n",
    "            s += self.left.__str__()\n",
    "\n",
    "        s += str(self.value) + \", \"\n",
    "\n",
    "        if self.right is not None:\n",
    "            s += self.right.__str__()\n",
    "\n",
    "        return s\n",
    "    "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "If your implementation is correct, it should pass the following tests:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 53,
   "metadata": {},
   "outputs": [],
   "source": [
    "# First test\n",
    "my_set = TreeSet()\n",
    "my_set.insert(\"today\")\n",
    "my_set.insert(\"hello\")\n",
    "my_set.insert(\"data science\")\n",
    "my_set.insert(\"jerry\")\n",
    "my_set.insert(\"apple\")\n",
    "my_set.insert(\"17\")\n",
    "my_set.insert(\"hello\")"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 54,
   "metadata": {},
   "outputs": [],
   "source": [
    "assert my_set.contains(\"data science\")\n",
    "assert my_set.contains(\"apple\")\n",
    "assert not my_set.contains(\"18\")\n",
    "assert not my_set.contains(\"blah\")"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 55,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Second test\n",
    "my_set = TreeSet()\n",
    "my_set.insert(3)\n",
    "my_set.insert(5)\n",
    "my_set.insert(10)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 56,
   "metadata": {},
   "outputs": [],
   "source": [
    "assert my_set.contains(3)\n",
    "assert not my_set.contains(18)"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### 5(b)\n",
    "rubric={accuracy:6}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q5_b\n",
    "points: 6\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "The following code measures the execution time of Python's `in` operation when used with a built-in `set`:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 57,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "List size:  100\n",
      "List size:  1000\n",
      "List size:  10000\n",
      "List size:  100000\n",
      "List size:  1000000\n"
     ]
    }
   ],
   "source": [
    "list_sizes = [100, 1000, 10_000, 100_000, 1_000_000]\n",
    "\n",
    "results = defaultdict(list)\n",
    "results[\"size\"] = list_sizes\n",
    "\n",
    "key = -1\n",
    "x_all = np.random.randint(1e8, size=max(list_sizes))\n",
    "\n",
    "for list_size in list_sizes:\n",
    "    print('List size: ', list_size)\n",
    "    x = x_all[:list_size]\n",
    "    \n",
    "    x_set = set(x)\n",
    "    time = %timeit -q -o -r 1 (key in x_set)\n",
    "    results[\"Python set in\"].append(time.average)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 58,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "<div>\n",
       "<style scoped>\n",
       "    .dataframe tbody tr th:only-of-type {\n",
       "        vertical-align: middle;\n",
       "    }\n",
       "\n",
       "    .dataframe tbody tr th {\n",
       "        vertical-align: top;\n",
       "    }\n",
       "\n",
       "    .dataframe thead th {\n",
       "        text-align: right;\n",
       "    }\n",
       "</style>\n",
       "<table border=\"1\" class=\"dataframe\">\n",
       "  <thead>\n",
       "    <tr style=\"text-align: right;\">\n",
       "      <th></th>\n",
       "      <th>size</th>\n",
       "      <th>Python set in</th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <th>0</th>\n",
       "      <td>100</td>\n",
       "      <td>1.016000e-08</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>1</th>\n",
       "      <td>1000</td>\n",
       "      <td>1.016985e-08</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>2</th>\n",
       "      <td>10000</td>\n",
       "      <td>1.018987e-08</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>3</th>\n",
       "      <td>100000</td>\n",
       "      <td>1.010127e-08</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>4</th>\n",
       "      <td>1000000</td>\n",
       "      <td>1.012881e-08</td>\n",
       "    </tr>\n",
       "  </tbody>\n",
       "</table>\n",
       "</div>"
      ],
      "text/plain": [
       "      size  Python set in\n",
       "0      100   1.016000e-08\n",
       "1     1000   1.016985e-08\n",
       "2    10000   1.018987e-08\n",
       "3   100000   1.010127e-08\n",
       "4  1000000   1.012881e-08"
      ]
     },
     "execution_count": 58,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "df = pd.DataFrame(results)\n",
    "df"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Now, measure the speed of the `contains` operation in your `TreeSet` implementation. Add the results to the DataFrame and print it out.\n",
    "\n",
    "(Note: for efficiency, we only go up to $n=10^6$ here. Populating the `TreeSet` objects with $10^7$ items would take a long time.)"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 59,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "List size:  100\n",
      "  Creating TreeSet object\n",
      "  Creating set object\n",
      "  Starting TreeSet timing\n",
      "  Starting set timing\n",
      "List size:  1000\n",
      "  Creating TreeSet object\n",
      "  Creating set object\n",
      "  Starting TreeSet timing\n",
      "  Starting set timing\n",
      "List size:  10000\n",
      "  Creating TreeSet object\n",
      "  Creating set object\n",
      "  Starting TreeSet timing\n",
      "  Starting set timing\n",
      "List size:  100000\n",
      "  Creating TreeSet object\n",
      "  Creating set object\n",
      "  Starting TreeSet timing\n",
      "  Starting set timing\n",
      "List size:  1000000\n",
      "  Creating TreeSet object\n",
      "  Creating set object\n",
      "  Starting TreeSet timing\n",
      "  Starting set timing\n"
     ]
    }
   ],
   "source": [
    "## IGNORE ##\n",
    "list_sizes = [100, 1000, 10_000, 100_000, 1_000_000]\n",
    "\n",
    "results = defaultdict(list)\n",
    "results[\"size\"] = list_sizes\n",
    "\n",
    "key = -1\n",
    "\n",
    "# x_all = np.random.randint(1e8, size=max(list_sizes))\n",
    "x_all = np.arange(max(list_sizes))\n",
    "np.random.shuffle(x_all)\n",
    "# above: this avoids duplicates, whereas the above is with replacement\n",
    "# and results in *slightly* too few elements in the set\n",
    "# since the set doesn't allow duplicates\n",
    "\n",
    "for list_size in list_sizes:\n",
    "    \n",
    "    print('List size: ', list_size) \n",
    "    x = x_all[:list_size]\n",
    "    \n",
    "    print(\"  Creating TreeSet object\")\n",
    "    my_set = TreeSet()\n",
    "    for elem in x:\n",
    "        my_set.insert(elem)\n",
    "        \n",
    "    print(\"  Creating set object\")\n",
    "    python_set = set()\n",
    "    for elem in x:\n",
    "        python_set.add(elem)\n",
    "    \n",
    "    print(\"  Starting TreeSet timing\")\n",
    "    time = %timeit -q -o -r 3 my_set.contains(key)\n",
    "    results[\"TreeSet contains\"].append(time.average)\n",
    "\n",
    "    print(\"  Starting set timing\")\n",
    "    time = %timeit -q -o -r 3 key in python_set\n",
    "    results[\"Python set in\"].append(time.average)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 60,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "<div>\n",
       "<style scoped>\n",
       "    .dataframe tbody tr th:only-of-type {\n",
       "        vertical-align: middle;\n",
       "    }\n",
       "\n",
       "    .dataframe tbody tr th {\n",
       "        vertical-align: top;\n",
       "    }\n",
       "\n",
       "    .dataframe thead th {\n",
       "        text-align: right;\n",
       "    }\n",
       "</style>\n",
       "<table border=\"1\" class=\"dataframe\">\n",
       "  <thead>\n",
       "    <tr style=\"text-align: right;\">\n",
       "      <th></th>\n",
       "      <th>size</th>\n",
       "      <th>TreeSet contains</th>\n",
       "      <th>Python set in</th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <th>0</th>\n",
       "      <td>100</td>\n",
       "      <td>2.376131e-07</td>\n",
       "      <td>1.011378e-08</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>1</th>\n",
       "      <td>1000</td>\n",
       "      <td>2.436729e-07</td>\n",
       "      <td>1.014531e-08</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>2</th>\n",
       "      <td>10000</td>\n",
       "      <td>3.802051e-07</td>\n",
       "      <td>1.014467e-08</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>3</th>\n",
       "      <td>100000</td>\n",
       "      <td>4.641805e-07</td>\n",
       "      <td>1.013072e-08</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>4</th>\n",
       "      <td>1000000</td>\n",
       "      <td>6.545841e-07</td>\n",
       "      <td>1.015326e-08</td>\n",
       "    </tr>\n",
       "  </tbody>\n",
       "</table>\n",
       "</div>"
      ],
      "text/plain": [
       "      size  TreeSet contains  Python set in\n",
       "0      100      2.376131e-07   1.011378e-08\n",
       "1     1000      2.436729e-07   1.014531e-08\n",
       "2    10000      3.802051e-07   1.014467e-08\n",
       "3   100000      4.641805e-07   1.013072e-08\n",
       "4  1000000      6.545841e-07   1.015326e-08"
      ]
     },
     "execution_count": 60,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "## IGNORE ##\n",
    "df = pd.DataFrame(results)\n",
    "df"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Convert to milliseconds:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 61,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "<div>\n",
       "<style scoped>\n",
       "    .dataframe tbody tr th:only-of-type {\n",
       "        vertical-align: middle;\n",
       "    }\n",
       "\n",
       "    .dataframe tbody tr th {\n",
       "        vertical-align: top;\n",
       "    }\n",
       "\n",
       "    .dataframe thead th {\n",
       "        text-align: right;\n",
       "    }\n",
       "</style>\n",
       "<table border=\"1\" class=\"dataframe\">\n",
       "  <thead>\n",
       "    <tr style=\"text-align: right;\">\n",
       "      <th></th>\n",
       "      <th>TreeSet contains</th>\n",
       "      <th>Python set in</th>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>size</th>\n",
       "      <th></th>\n",
       "      <th></th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <th>100</th>\n",
       "      <td>0.000238</td>\n",
       "      <td>0.00001</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>1000</th>\n",
       "      <td>0.000244</td>\n",
       "      <td>0.00001</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>10000</th>\n",
       "      <td>0.000380</td>\n",
       "      <td>0.00001</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>100000</th>\n",
       "      <td>0.000464</td>\n",
       "      <td>0.00001</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>1000000</th>\n",
       "      <td>0.000655</td>\n",
       "      <td>0.00001</td>\n",
       "    </tr>\n",
       "  </tbody>\n",
       "</table>\n",
       "</div>"
      ],
      "text/plain": [
       "         TreeSet contains  Python set in\n",
       "size                                    \n",
       "100              0.000238        0.00001\n",
       "1000             0.000244        0.00001\n",
       "10000            0.000380        0.00001\n",
       "100000           0.000464        0.00001\n",
       "1000000          0.000655        0.00001"
      ]
     },
     "execution_count": 61,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "## IGNORE ##\n",
    "df.set_index(\"size\")*1000"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### 5(c)\n",
    "rubric={reasoning:6}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q5_c\n",
    "points: 5\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "\n",
    "\n",
    "Discuss your results from the previous part. How do Python's `set` and your `TreeSet` compare? Specifically:\n",
    "\n",
    "- Which method is faster?\n",
    "- What is the theoretical time complexity of `in` with a `set`, and `contains` with a `TreeSet`?\n",
    "- Are the empirical results consistent with the theoretical time complexities?\n",
    "- Are the results what you expected, overall?"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The first observation is that the set implementations are fast, like binary search. The timings of `TreeSet` are pretty unreliable, because it depends on the order of the random numbers; in some cases, if the root happens to be a small number, then it will be very fast to search for $-1$. Overall, using `in` with Python's `set` is by far the fastest. Crucially, the same `in` operator is way faster with a `set` than a `list`. So we cannot simply say \"`in` is fast\" or \"`in` is slow\" - it depends on what data structure is holding the data.\n",
    "\n",
    "Searching in a hash table should take $O(1)$ time, which we can see empirically in the last column. The `TreeSet` situation is messier because the tree implementation isn't balanced. If balanced we'd expect $O(\\log n)$ and worst cast would be $O(n)$. "
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### (challenging) 5(d)\n",
    "rubric={reasoning:1}"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN QUESTION\n",
    "name: q5_d\n",
    "points: 1\n",
    "manual: true"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "\n",
    "Now, also time the `insert` function from `TreeSet` and compare it to the speed of `add` from Python's `set`. This time, $n$ is **not** the number of elements we are inserting. Rather, you are measuring the speed of inserting _one_ value into the set, and $n$ is the current size of the set before insertion. \n",
    "\n",
    "Note: you'll have to be a bit careful setting this up. If you repeatedly insert the same value into the set, is the experiment valid?\n",
    "\n",
    "Discuss the results."
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# BEGIN SOLUTION"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 62,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "List size:  100\n",
      "Creating TreeSet object\n",
      "Creating set object\n",
      "Timing TreeSet insert\n",
      "362 ns ± 1.76 ns per loop (mean ± std. dev. of 3 runs, 1,000,000 loops each)\n",
      "Timing set add\n",
      "57.5 ns ± 0.399 ns per loop (mean ± std. dev. of 3 runs, 10,000,000 loops each)\n",
      "List size:  1000\n",
      "Creating TreeSet object\n",
      "Creating set object\n",
      "Timing TreeSet insert\n",
      "642 ns ± 5.03 ns per loop (mean ± std. dev. of 3 runs, 1,000,000 loops each)\n",
      "Timing set add\n",
      "58.9 ns ± 1.24 ns per loop (mean ± std. dev. of 3 runs, 10,000,000 loops each)\n",
      "List size:  10000\n",
      "Creating TreeSet object\n",
      "Creating set object\n",
      "Timing TreeSet insert\n",
      "902 ns ± 6.86 ns per loop (mean ± std. dev. of 3 runs, 1,000,000 loops each)\n",
      "Timing set add\n",
      "58.8 ns ± 1.3 ns per loop (mean ± std. dev. of 3 runs, 10,000,000 loops each)\n",
      "List size:  100000\n",
      "Creating TreeSet object\n",
      "Creating set object\n",
      "Timing TreeSet insert\n",
      "1.02 μs ± 5.87 ns per loop (mean ± std. dev. of 3 runs, 1,000,000 loops each)\n",
      "Timing set add\n",
      "58.3 ns ± 1.54 ns per loop (mean ± std. dev. of 3 runs, 10,000,000 loops each)\n",
      "List size:  1000000\n",
      "Creating TreeSet object\n",
      "Creating set object\n",
      "Timing TreeSet insert\n",
      "1.23 μs ± 1.66 ns per loop (mean ± std. dev. of 3 runs, 1,000,000 loops each)\n",
      "Timing set add\n",
      "58.2 ns ± 1 ns per loop (mean ± std. dev. of 3 runs, 10,000,000 loops each)\n"
     ]
    }
   ],
   "source": [
    "## IGNORE ##\n",
    "insertion_list_sizes = [100, 1_000, 10_000, 100_000, 1_000_000]\n",
    "\n",
    "insertion_results = defaultdict(list)\n",
    "insertion_results[\"size\"] = insertion_list_sizes\n",
    "\n",
    "key = -1\n",
    "\n",
    "x_all = np.random.randint(1e8, size=max(insertion_list_sizes)+1)\n",
    "\n",
    "\n",
    "for list_size in insertion_list_sizes:\n",
    "    print('List size: ', list_size)\n",
    "    x = x_all[:list_size]\n",
    "\n",
    "    print(\"Creating TreeSet object\")\n",
    "    my_set = TreeSet()\n",
    "    for elem in x:\n",
    "        my_set.insert(elem)\n",
    "\n",
    "    print(\"Creating set object\")\n",
    "    python_set = set()\n",
    "    for elem in x:\n",
    "        python_set.add(elem)\n",
    "\n",
    "    print(\"Timing TreeSet insert\")\n",
    "    # assume randint is fast\n",
    "    time = %timeit -o -r 3 my_set.insert(x_all[list_size])\n",
    "    insertion_results['TreeSet insert'].append(time.average)\n",
    "\n",
    "    print(\"Timing set add\")\n",
    "    # assume randint is fast\n",
    "    time = %timeit -o -r 3 python_set.add(x_all[list_size])\n",
    "    insertion_results['Python set add'].append(time.average)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 63,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Converted to milliseconds:\n"
     ]
    },
    {
     "data": {
      "text/html": [
       "<div>\n",
       "<style scoped>\n",
       "    .dataframe tbody tr th:only-of-type {\n",
       "        vertical-align: middle;\n",
       "    }\n",
       "\n",
       "    .dataframe tbody tr th {\n",
       "        vertical-align: top;\n",
       "    }\n",
       "\n",
       "    .dataframe thead th {\n",
       "        text-align: right;\n",
       "    }\n",
       "</style>\n",
       "<table border=\"1\" class=\"dataframe\">\n",
       "  <thead>\n",
       "    <tr style=\"text-align: right;\">\n",
       "      <th></th>\n",
       "      <th>TreeSet insert</th>\n",
       "      <th>Python set add</th>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>size</th>\n",
       "      <th></th>\n",
       "      <th></th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <th>100</th>\n",
       "      <td>0.000362</td>\n",
       "      <td>0.000058</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>1000</th>\n",
       "      <td>0.000642</td>\n",
       "      <td>0.000059</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>10000</th>\n",
       "      <td>0.000902</td>\n",
       "      <td>0.000059</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>100000</th>\n",
       "      <td>0.001022</td>\n",
       "      <td>0.000058</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>1000000</th>\n",
       "      <td>0.001234</td>\n",
       "      <td>0.000058</td>\n",
       "    </tr>\n",
       "  </tbody>\n",
       "</table>\n",
       "</div>"
      ],
      "text/plain": [
       "         TreeSet insert  Python set add\n",
       "size                                   \n",
       "100            0.000362        0.000058\n",
       "1000           0.000642        0.000059\n",
       "10000          0.000902        0.000059\n",
       "100000         0.001022        0.000058\n",
       "1000000        0.001234        0.000058"
      ]
     },
     "execution_count": 63,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "## IGNORE ##\n",
    "df = pd.DataFrame(insertion_results).set_index(\"size\")\n",
    "print(\"Converted to milliseconds:\")\n",
    "df*1000"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "In the above, we can see that `set`, which is based on a hash table, is basically $O(1)$ time. The `TreeSet` insertion looks reasonably logarithmic, in the sense that the time goes up by a roughly fixed amount each time. "
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END SOLUTION"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {
    "collapsed": false,
    "jupyter": {
     "outputs_hidden": false
    }
   },
   "source": [
    "# END QUESTION"
   ]
  }
 ],
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