{ "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", "
\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",
""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Explanation of k"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"\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",
"\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": {},
"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": {},
"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": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Overwriting test_find_min_cost.py\n"
]
}
],
"source": [
"%%writefile test_find_min_cost.py\n",
"import unittest\n",
"\n",
"\n",
"class TestMatrixMultiplicationCost(unittest.TestCase):\n",
"\n",
" def test_find_min_cost(self):\n",
" matrix_mult_cost = MatrixMultiplicationCost()\n",
" self.assertRaises(TypeError, matrix_mult_cost.find_min_cost, None)\n",
" self.assertEqual(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",
" self.assertEqual(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": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Success: test_find_min_cost\n"
]
}
],
"source": [
"%run -i test_find_min_cost.py"
]
}
],
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