interactive-coding-challenges/recursion_dynamic/matrix_mult/find_min_cost_solution.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "This notebook was prepared by [Donne Martin](https://github.com/donnemartin). Source and license info is on [GitHub](https://github.com/donnemartin/interactive-coding-challenges)."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Solution Notebook"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Problem: Given a list of 2x2 matrices, minimize the cost of matrix multiplication.\n",
    "\n",
    "* [Constraints](#Constraints)\n",
    "* [Test Cases](#Test-Cases)\n",
    "* [Algorithm](#Algorithm)\n",
    "* [Code](#Code)\n",
    "* [Unit Test](#Unit-Test)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Constraints\n",
    "\n",
    "* Do we just want to calculate the cost and not list the actual order of operations?\n",
    "    * Yes\n",
    "* Can we assume the inputs are valid?\n",
    "    * No\n",
    "* Can we assume this fits memory?\n",
    "    * Yes"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Test Cases\n",
    "\n",
    "* None -> Exception\n",
    "* [] -> 0\n",
    "* [Matrix(2, 3), Matrix(3, 6), Matrix(6, 4), Matrix(4, 5)] -> 124"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Algorithm\n",
    "\n",
    "We'll use bottom up dynamic programming to build a table.\n",
    "\n",
    "<pre>\n",
    "\n",
    "  0    1    2    3\n",
    "[2,3][3,6][6,4][4,5]\n",
    "\n",
    "Case: 0 * 1\n",
    "2 * 3 * 6 = 36\n",
    "\n",
    "Case: 1 * 2\n",
    "3 * 6 * 4 = 72\n",
    "\n",
    "Case: 2 * 3\n",
    "6 * 4 * 5 = 120\n",
    "\n",
    "Case: 0 * 1 * 2\n",
    "0 * (1 * 2) = 2 * 3 * 4 + 72 = 96\n",
    "(0 * 1) * 2 = 36 + 2 * 6 * 4 = 84\n",
    "min: 84\n",
    "\n",
    "Case: 1 * 2 * 3\n",
    "1 * (2 * 3) = 3 * 6 * 5 + 120 = 210\n",
    "(1 * 2) * 3 = 72 + 3 * 4 * 5 = 132\n",
    "min: 132\n",
    "\n",
    "Case: 0 * 1 * 2 * 3\n",
    "0 * (1 * 2 * 3) = 2 * 3 * 5 + 132 = 162\n",
    "(0 * 1) * (2 * 3) = 36 + 120 + 2 * 6 * 5 = 216\n",
    "(0 * 1 * 2) * 3 = 84 + 2 * 4 * 5 = 124\n",
    "min: 124\n",
    "\n",
    "  ---------------------\n",
    "  | 0 |  1 |  2 |   3 |\n",
    "  ---------------------\n",
    "0 | 0 | 36 | 84 | 124 |\n",
    "1 | x |  0 | 72 | 132 |\n",
    "2 | x |  x |  0 | 120 |\n",
    "3 | x |  x |  x |   0 |\n",
    "  ---------------------\n",
    "\n",
    "min cost = T[0][cols-1] = 124\n",
    "\n",
    "for k in range(i, j):\n",
    "    T[i][j] = minimum of (T[i][k] + T[k+1][j] +\n",
    "                          m[i].first * m[k].second * m[j].second) for all k\n",
    "</pre>"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Explanation of k"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<pre>\n",
    "  0    1    2    3\n",
    "[2,3][3,6][6,4][4,5]\n",
    "\n",
    "Fill in the missing cell, where i = 0, j = 3\n",
    "\n",
    "  ---------------------\n",
    "  | 0 |  1 |  2 |   3 |\n",
    "  ---------------------\n",
    "0 | 0 | 36 | 84 | ??? |\n",
    "1 | x |  0 | 72 | 132 |\n",
    "2 | x |  x |  0 | 120 |\n",
    "3 | x |  x |  x |   0 |\n",
    "  ---------------------\n",
    "\n",
    "Case: 0 * (1 * 2 * 3), k = 0\n",
    "i = 0, j = 3\n",
    "\n",
    "0 * (1 * 2 * 3) = 2 * 3 * 5 + 132 = 162\n",
    "T[i][k] + T[k+1][j] + m[i].first * m[k].second * m[j].second\n",
    "T[0][0] + T[1][3] + 2 * 3 * 5\n",
    "0 + 132 + 30 = 162\n",
    "\n",
    "Case: (0 * 1) * (2 * 3), k = 1\n",
    "i = 0, j = 3\n",
    "\n",
    "(0 * 1) * (2 * 3) = 36 + 120 + 2 * 6 * 5 = 216\n",
    "T[i][k] + T[k+1][j] + m[i].first * m[k].second * m[j].second\n",
    "T[0][1] + T[2][3] + 2 * 6 * 5\n",
    "36 + 120 + 60 = 216\n",
    "\n",
    "Case: (0 * 1 * 2) * 3, k = 2\n",
    "i = 0, j = 3\n",
    "\n",
    "(0 * 1 * 2) * 3 = 84 + 2 * 4 * 5 = 124\n",
    "T[i][k] + T[k+1][j] + m[i].first * m[k].second * m[j].second\n",
    "T[0][2] + T[3][3] + 2 * 4 * 5\n",
    "84 + 0 + 40 = 124\n",
    "\n",
    "</pre>\n",
    "\n",
    "Complexity:\n",
    "* Time: O(n^3)\n",
    "* Space: O(n^2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Code"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "class Matrix(object):\n",
    "\n",
    "    def __init__(self, first, second):\n",
    "        self.first = first\n",
    "        self.second = second"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "import sys\n",
    "\n",
    "\n",
    "class MatrixMultiplicationCost(object):\n",
    "\n",
    "    def find_min_cost(self, matrices):\n",
    "        if matrices is None:\n",
    "            raise TypeError('matrices cannot be None')\n",
    "        if not matrices:\n",
    "            return 0\n",
    "        size = len(matrices)\n",
    "        T = [[0] * size for _ in range(size)]\n",
    "        for offset in range(1, size):\n",
    "            for i in range(size-offset):\n",
    "                j = i + offset\n",
    "                min_cost = sys.maxsize\n",
    "                for k in range(i, j):\n",
    "                    cost = (T[i][k] + T[k+1][j] +\n",
    "                            matrices[i].first *\n",
    "                            matrices[k].second *\n",
    "                            matrices[j].second)\n",
    "                    if cost < min_cost:\n",
    "                        min_cost = cost\n",
    "                T[i][j] = min_cost\n",
    "        return T[0][size-1]"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Unit Test"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Overwriting test_find_min_cost.py\n"
     ]
    }
   ],
   "source": [
    "%%writefile test_find_min_cost.py\n",
    "from nose.tools import assert_equal, assert_raises\n",
    "\n",
    "\n",
    "class TestMatrixMultiplicationCost(object):\n",
    "\n",
    "    def test_find_min_cost(self):\n",
    "        matrix_mult_cost = MatrixMultiplicationCost()\n",
    "        assert_raises(TypeError, matrix_mult_cost.find_min_cost, None)\n",
    "        assert_equal(matrix_mult_cost.find_min_cost([]), 0)\n",
    "        matrices = [Matrix(2, 3),\n",
    "                    Matrix(3, 6),\n",
    "                    Matrix(6, 4),\n",
    "                    Matrix(4, 5)]\n",
    "        expected_cost = 124\n",
    "        assert_equal(matrix_mult_cost.find_min_cost(matrices), expected_cost)\n",
    "        print('Success: test_find_min_cost')\n",
    "\n",
    "\n",
    "def main():\n",
    "    test = TestMatrixMultiplicationCost()\n",
    "    test.test_find_min_cost()\n",
    "\n",
    "\n",
    "if __name__ == '__main__':\n",
    "    main()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Success: test_find_min_cost\n"
     ]
    }
   ],
   "source": [
    "%run -i test_find_min_cost.py"
   ]
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.5.0"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 0
}