path: root/src/Lecture7/slides/.ipynb_checkpoints/X1-ComputationalComplexity-notebook-checkpoint.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Nested loops\n",
    "\n",
    "The following two functions compute sum and product of matrices, respectively.\n",
    "\n",
    "By counting the nested loops it is easy to see that `add()` is $O(n^2)$ while `prod()` is $O(n^3)$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Time for add:  0.005766554000000035\n",
      "Time for prod: 1.3871021639999999\n"
     ]
    }
   ],
   "source": [
    "from random import randint\n",
    "import time\n",
    "\n",
    "def add(A, B):\n",
    "    S = [[0] * len(A) for i in range(len(A))]\n",
    "    for i in range(len(A)):\n",
    "        for j in range(len(A)):\n",
    "            S[i][j] = A[i][j] + B[i][j]\n",
    "    return S\n",
    "\n",
    "def prod(A, B):\n",
    "    S = [[0] * len(A) for i in range(len(A))]\n",
    "    for i in range(len(A)):\n",
    "        for j in range(len(A)):\n",
    "            for k in range(len(A)):\n",
    "                S[i][j] = S[i][j] + A[i][k] * B[k][j]\n",
    "    return S\n",
    "\n",
    "N = 200\n",
    "A = [ [randint(0,100) for i in range(N)] for j in range(N) ]\n",
    "B = [ [randint(0,100) for i in range(N)] for j in range(N) ]\n",
    "\n",
    "t0 = time.process_time()\n",
    "add(A,B)\n",
    "t1 = time.process_time()\n",
    "prod(A,B)\n",
    "t2 = time.process_time()\n",
    "\n",
    "print(\"Time for add: \", t1-t0)\n",
    "print(\"Time for prod:\", t2-t1)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Sorting a list, slow version\n",
    "\n",
    "The following code implements a slow version of the so-called *insertion sort* alogithm\n",
    "\n",
    "Complexity: $O(n^2)$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Running time: 1.1191449070000001\n"
     ]
    }
   ],
   "source": [
    "from random import randint\n",
    "import time\n",
    "\n",
    "def correct_position(e, S):\n",
    "    for i in range(len(S)):\n",
    "        if S[i] > e:\n",
    "            return i\n",
    "    return len(S)\n",
    "\n",
    "def sort_list(L):\n",
    "    S = []\n",
    "    for e in L:\n",
    "        cp = correct_position(e, S)\n",
    "        S.insert(cp, e)\n",
    "    return S\n",
    "\n",
    "N = 10000\n",
    "L = [randint(0,10**9) for i in range(N)]\n",
    "\n",
    "t0 = time.process_time()\n",
    "sort_list(L)\n",
    "t1 = time.process_time()\n",
    "\n",
    "print(\"Running time:\", t1-t0)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Binary search\n",
    "\n",
    "The following code implements a binary search.\n",
    "\n",
    "Complexity: $O(\\log_2(n))$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The correct position of e = 216197744 in L is:\n",
      "... 216196218 216197540 e 216198673 216198962 ...\n",
      "\n",
      "Time for sorting:   0.26413054400000036\n",
      "Time for searching: 9.616099999965044e-05\n"
     ]
    }
   ],
   "source": [
    "from random import randint\n",
    "import time\n",
    "\n",
    "def binary_search(e, S, start, end):\n",
    "    if start == end:\n",
    "        return start\n",
    "    midpoint = (start+end) // 2\n",
    "    if e < S[midpoint]:\n",
    "        return binary_search(e, S, start, midpoint)\n",
    "    else:\n",
    "        return binary_search(e, S, midpoint+1, end)\n",
    "    \n",
    "N = 1000000\n",
    "L = [randint(0,10**9) for i in range(N)]\n",
    "e = randint(0,10**9)\n",
    "\n",
    "t0 = time.process_time()\n",
    "L.sort() # Using Python's sort()\n",
    "t1 = time.process_time()\n",
    "i = binary_search(e, L, 0, len(L))\n",
    "t2 = time.process_time()\n",
    "print(\"The correct position of e =\", e, \"in L is:\")\n",
    "print(\"...\", L[i-2], L[i-1], \"e\", L[i], L[i+1], \"...\")\n",
    "print(\"\")\n",
    "print(\"Time for sorting:  \", t1-t0)\n",
    "print(\"Time for searching:\", t2-t1)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Sorting a list, fast version (with binary_search)\n",
    "\n",
    "The following code uses the function `binary_search()` above instead of `correct_position()` in our insertion sort algorithm.\n",
    "\n",
    "Complexity: $O(n\\log_2(n))$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 69,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Running time: 0.03710268399998995\n"
     ]
    }
   ],
   "source": [
    "from random import randint\n",
    "import time\n",
    "\n",
    "def binary_search(e, S, start, end):\n",
    "    if start == end:\n",
    "        return start\n",
    "    midpoint = (start+end) // 2\n",
    "    if e < S[midpoint]:\n",
    "        return binary_search(e, S, start, midpoint)\n",
    "    else:\n",
    "        return binary_search(e, S, midpoint+1, end)\n",
    "    \n",
    "def sort_list(L):\n",
    "    S = []\n",
    "    for e in L:\n",
    "        cp = binary_search(e, S, 0, len(S)) # Changed here\n",
    "        S.insert(cp, e)\n",
    "    return S\n",
    "    \n",
    "N = 10000\n",
    "L = [randint(0,10**9) for i in range(N)]\n",
    "\n",
    "t0 = time.process_time()\n",
    "sort_list(L)\n",
    "t1 = time.process_time()\n",
    "\n",
    "print(\"Running time:\", t1-t0)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Fast exponentiation\n",
    "\n",
    "The following cell contains two functions for computing $a^n$ ($n$ non-negative integer): a slow one that runs in $O(n)$ and a fast one that runs in $O(\\log_2(n))$. We compare these two also with Python's built-in operator `**`.\n",
    "\n",
    "Complexity: $O(n)$ for the slow algorithm, $O(\\log_2(n))$ for the other two."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 30,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "2.71828179834636\n",
      "2.7182817863957984\n",
      "2.7182817983473577\n",
      "Time for slow_power(): 3.234879998000004\n",
      "Time for fast_power(): 9.059099999575437e-05\n",
      "Time for Python's **:  0.00010159500000384014\n"
     ]
    }
   ],
   "source": [
    "import time\n",
    "\n",
    "def slow_power(a, n):\n",
    "    r = 1\n",
    "    for i in range(n):\n",
    "        r = r * a\n",
    "    return r\n",
    "\n",
    "def fast_power(a, n):\n",
    "    if n == 0:\n",
    "        return 1\n",
    "    if n%2 == 0:\n",
    "        return fast_power(a*a, n//2)\n",
    "    else:\n",
    "        return a * fast_power(a, n-1)\n",
    "\n",
    "a = 1.00000001\n",
    "n = 100000000\n",
    "\n",
    "t0 = time.process_time()\n",
    "print(slow_power(a, n))\n",
    "t1 = time.process_time()\n",
    "print(fast_power(a, n))\n",
    "t2 = time.process_time()\n",
    "print(a**n)\n",
    "t3 = time.process_time()\n",
    "\n",
    "print(\"Time for slow_power():\", t1-t0)\n",
    "print(\"Time for fast_power():\", t2-t1)\n",
    "print(\"Time for Python's **: \", t3-t2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Fast gcd\n",
    "\n",
    "Complexity: $O(\\log_2(n))$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 31,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "126\n",
      "Running time: 0.00017707599999994272\n"
     ]
    }
   ],
   "source": [
    "import time\n",
    "\n",
    "def gcd(a, b):\n",
    "    if b == 0:\n",
    "        return a\n",
    "    else:\n",
    "        return gcd(b, a%b)\n",
    "\n",
    "t0 = time.process_time()\n",
    "print(gcd(155275387236018, 572335397352432))\n",
    "t1 = time.process_time()\n",
    "\n",
    "print(\"Running time:\", t1-t0)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Fibonacci numbers\n",
    "\n",
    "In the following cell there are two functions that compute the $n$-th Fibonacci number. They are almost the same, but the second one memorizes the results in a list to avoid computing them multiple times, and it is much much faster.\n",
    "\n",
    "Complexity: $O\\left(\\left(\\frac{1+\\sqrt 5}{2}\\right)^n\\right)\\sim O(1.6^n)$ for the slow version, $O(n)$ for the fast version."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 37,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "9227465\n",
      "9227465\n",
      "Time for F_slow: 2.3301570169999906\n",
      "Time for F_fast: 8.848800000293977e-05\n"
     ]
    }
   ],
   "source": [
    "import time\n",
    "\n",
    "F_memorized = [-1] * (10**6)\n",
    "\n",
    "def F_slow(n):\n",
    "    if n <= 1:\n",
    "        return n\n",
    "    else:\n",
    "        return F_slow(n-1) + F_slow(n-2)\n",
    "    \n",
    "def F_fast(n):\n",
    "    if F_memorized[n] == -1:\n",
    "        if n <= 1:\n",
    "            F_memorized[n] = n\n",
    "        else:\n",
    "            F_memorized[n] = F_fast(n-1) + F_fast(n-2)\n",
    "    \n",
    "    return F_memorized[n]\n",
    "\n",
    "n = 35\n",
    "\n",
    "t0 = time.process_time()\n",
    "print(F_slow(n))\n",
    "t1 = time.process_time()\n",
    "print(F_fast(n))\n",
    "t2 = time.process_time()\n",
    "\n",
    "print(\"Time for F_slow:\", t1-t0)\n",
    "print(\"Time for F_fast:\", t2-t1)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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