mirror of
https://github.com/donnemartin/interactive-coding-challenges.git
synced 2024-03-22 13:11:13 +08:00
309 lines
7.6 KiB
Python
309 lines
7.6 KiB
Python
{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"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)."
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Solution Notebook"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Problem: Given a list of 2x2 matrices, minimize the cost of matrix multiplication.\n",
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"\n",
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"* [Constraints](#Constraints)\n",
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"* [Test Cases](#Test-Cases)\n",
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"* [Algorithm](#Algorithm)\n",
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"* [Code](#Code)\n",
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"* [Unit Test](#Unit-Test)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Constraints\n",
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"\n",
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"* Do we just want to calculate the cost and not list the actual order of operations?\n",
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" * Yes\n",
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"* Can we assume the inputs are valid?\n",
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" * No\n",
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"* Can we assume this fits memory?\n",
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" * Yes"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Test Cases\n",
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"\n",
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"* None -> Exception\n",
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"* [] -> 0\n",
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"* [Matrix(2, 3), Matrix(3, 6), Matrix(6, 4), Matrix(4, 5)] -> 124"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Algorithm\n",
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"\n",
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"We'll use bottom up dynamic programming to build a table.\n",
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"\n",
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"<pre>\n",
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"\n",
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" 0 1 2 3\n",
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"[2,3][3,6][6,4][4,5]\n",
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"\n",
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"Case: 0 * 1\n",
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"2 * 3 * 6 = 36\n",
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"\n",
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"Case: 1 * 2\n",
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"3 * 6 * 4 = 72\n",
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"\n",
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"Case: 2 * 3\n",
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"6 * 4 * 5 = 120\n",
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"\n",
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"Case: 0 * 1 * 2\n",
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"0 * (1 * 2) = 2 * 3 * 4 + 72 = 96\n",
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"(0 * 1) * 2 = 36 + 2 * 6 * 4 = 84\n",
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"min: 84\n",
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"\n",
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"Case: 1 * 2 * 3\n",
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"1 * (2 * 3) = 3 * 6 * 5 + 120 = 210\n",
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"(1 * 2) * 3 = 72 + 3 * 4 * 5 = 132\n",
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"min: 132\n",
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"\n",
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"Case: 0 * 1 * 2 * 3\n",
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"0 * (1 * 2 * 3) = 2 * 3 * 5 + 132 = 162\n",
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"(0 * 1) * (2 * 3) = 36 + 120 + 2 * 6 * 5 = 216\n",
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"(0 * 1 * 2) * 3 = 84 + 2 * 4 * 5 = 124\n",
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"min: 124\n",
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"\n",
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" ---------------------\n",
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" | 0 | 1 | 2 | 3 |\n",
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" ---------------------\n",
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"0 | 0 | 36 | 84 | 124 |\n",
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"1 | x | 0 | 72 | 132 |\n",
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"2 | x | x | 0 | 120 |\n",
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"3 | x | x | x | 0 |\n",
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" ---------------------\n",
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"\n",
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"min cost = T[0][cols-1] = 124\n",
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"\n",
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"for k in range(i, j):\n",
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" T[i][j] = minimum of (T[i][k] + T[k+1][j] +\n",
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" m[i].first * m[k].second * m[j].second) for all k\n",
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"</pre>"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"### Explanation of k"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"<pre>\n",
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" 0 1 2 3\n",
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"[2,3][3,6][6,4][4,5]\n",
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"\n",
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"Fill in the missing cell, where i = 0, j = 3\n",
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"\n",
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" ---------------------\n",
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" | 0 | 1 | 2 | 3 |\n",
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" ---------------------\n",
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"0 | 0 | 36 | 84 | ??? |\n",
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"1 | x | 0 | 72 | 132 |\n",
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"2 | x | x | 0 | 120 |\n",
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"3 | x | x | x | 0 |\n",
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" ---------------------\n",
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"\n",
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"Case: 0 * (1 * 2 * 3), k = 0\n",
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"i = 0, j = 3\n",
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"\n",
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"0 * (1 * 2 * 3) = 2 * 3 * 5 + 132 = 162\n",
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"T[i][k] + T[k+1][j] + m[i].first * m[k].second * m[j].second\n",
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"T[0][0] + T[1][3] + 2 * 3 * 5\n",
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"0 + 132 + 30 = 162\n",
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"\n",
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"Case: (0 * 1) * (2 * 3), k = 1\n",
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"i = 0, j = 3\n",
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"\n",
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"(0 * 1) * (2 * 3) = 36 + 120 + 2 * 6 * 5 = 216\n",
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"T[i][k] + T[k+1][j] + m[i].first * m[k].second * m[j].second\n",
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"T[0][1] + T[2][3] + 2 * 6 * 5\n",
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"36 + 120 + 60 = 216\n",
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"\n",
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"Case: (0 * 1 * 2) * 3, k = 2\n",
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"i = 0, j = 3\n",
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"\n",
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"(0 * 1 * 2) * 3 = 84 + 2 * 4 * 5 = 124\n",
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"T[i][k] + T[k+1][j] + m[i].first * m[k].second * m[j].second\n",
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"T[0][2] + T[3][3] + 2 * 4 * 5\n",
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"84 + 0 + 40 = 124\n",
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"\n",
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"</pre>\n",
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"\n",
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"Complexity:\n",
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"* Time: O(n^3)\n",
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"* Space: O(n^2)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Code"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"metadata": {},
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"outputs": [],
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"source": [
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"class Matrix(object):\n",
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"\n",
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" def __init__(self, first, second):\n",
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" self.first = first\n",
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" self.second = second"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"metadata": {},
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"outputs": [],
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"source": [
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"import sys\n",
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"\n",
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"\n",
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"class MatrixMultiplicationCost(object):\n",
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"\n",
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" def find_min_cost(self, matrices):\n",
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" if matrices is None:\n",
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" raise TypeError('matrices cannot be None')\n",
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" if not matrices:\n",
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" return 0\n",
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" size = len(matrices)\n",
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" T = [[0] * size for _ in range(size)]\n",
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" for offset in range(1, size):\n",
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" for i in range(size-offset):\n",
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" j = i + offset\n",
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" min_cost = sys.maxsize\n",
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" for k in range(i, j):\n",
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" cost = (T[i][k] + T[k+1][j] +\n",
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" matrices[i].first *\n",
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" matrices[k].second *\n",
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" matrices[j].second)\n",
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" if cost < min_cost:\n",
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" min_cost = cost\n",
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" T[i][j] = min_cost\n",
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" return T[0][size-1]"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Unit Test"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"Overwriting test_find_min_cost.py\n"
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]
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}
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],
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"source": [
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"%%writefile test_find_min_cost.py\n",
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"import unittest\n",
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"\n",
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"\n",
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"class TestMatrixMultiplicationCost(unittest.TestCase):\n",
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"\n",
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" def test_find_min_cost(self):\n",
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" matrix_mult_cost = MatrixMultiplicationCost()\n",
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" self.assertRaises(TypeError, matrix_mult_cost.find_min_cost, None)\n",
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" self.assertEqual(matrix_mult_cost.find_min_cost([]), 0)\n",
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" matrices = [Matrix(2, 3),\n",
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" Matrix(3, 6),\n",
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" Matrix(6, 4),\n",
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" Matrix(4, 5)]\n",
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" expected_cost = 124\n",
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" self.assertEqual(matrix_mult_cost.find_min_cost(matrices), expected_cost)\n",
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" print('Success: test_find_min_cost')\n",
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"\n",
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"\n",
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"def main():\n",
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" test = TestMatrixMultiplicationCost()\n",
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" test.test_find_min_cost()\n",
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"\n",
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"\n",
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"if __name__ == '__main__':\n",
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" main()"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 4,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"Success: test_find_min_cost\n"
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]
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}
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],
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"source": [
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"%run -i test_find_min_cost.py"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.7.2"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 1
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}
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