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Renaming b -> c Fixes #14 (#19)
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‎apxb/apxb-exercise-solutions.ipynb renamed to ‎apxc/apxc-exercise-solutions.ipynb

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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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"# [Learn Quantum Computing with Python and Q#](https://www.manning.com/books/learn-quantum-computing-with-python-and-q-sharp?a_aid=learn-qc-granade&a_bid=ee23f338)<br>Appendix B Exercise Solutions\n",
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"# [Learn Quantum Computing with Python and Q#](https://www.manning.com/books/learn-quantum-computing-with-python-and-q-sharp?a_aid=learn-qc-granade&a_bid=ee23f338)<br>Appendix C Exercise Solutions\n",
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"----\n",
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"> Copyright (c) Sarah Kaiser and Chris Granade.\n",
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"> Code sample from the book \"Learn Quantum Computing with Python and Q#\" by\n",
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"> Sarah Kaiser and Chris Granade, published by Manning Publications Co.\n",
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"> Book ISBN 9781617296130.\n",
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"> Code licensed under the MIT License."
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]
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],
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"metadata": {}
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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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"\n",
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"### Preamble"
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]
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],
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"metadata": {}
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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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"import numpy as np"
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]
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],
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"outputs": [],
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"metadata": {}
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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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"\n",
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"### Exercise B.1 "
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]
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],
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"metadata": {}
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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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"**What would 25 meters West and 110 meters North be in feet?**"
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],
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"metadata": {}
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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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"source": [
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"directions_in_meters = np.array([[-25], [110]])\n",
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"directions_in_feet = 3.28 * directions_in_meters\n",
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"directions_in_feet"
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],
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"outputs": [
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{
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"output_type": "execute_result",
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"data": {
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"text/plain": [
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"array([[-82. ],\n",
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" [360.8]])"
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]
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},
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"execution_count": 2,
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"metadata": {},
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"output_type": "execute_result"
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"execution_count": 2
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}
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],
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"source": [
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"directions_in_meters = np.array([[-25], [110]])\n",
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"directions_in_feet = 3.28 * directions_in_meters\n",
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"directions_in_feet"
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]
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"metadata": {}
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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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"----\n",
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"### Exercise B.2"
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]
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],
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"metadata": {}
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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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"**Which of the following functions are linear?**\n",
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"\n",
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"- $f(x) = 2x$\n",
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"- $f(x) = x^2$\n",
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"- $f(x) = 2^x$"
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]
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],
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"metadata": {}
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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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"Recall that a function $f$ is linear if and only if for all scalars $a$ and $b$, $f(ax + by) = af(x) + bf(y)$.\n",
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"Let's check this condition for each of the three functions above:\n",
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"\n",
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"- $f(ax + by) = 2(ax + by) = 2ax + 2by = a \\times 2x + b \\times 2y = af(x) + bf(y)$, thus this function is linear.\n",
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"- $f(ax + by) = (ax + by)^2 = a^2 x^2 + 2abxy + b^2 y^2$, but $af(x) + bf(y) = ax^2 + by^2$. Since $a \\ne a^2$ for all $a$, and since the second expression is missing the $2abxy$ term, you can conclude that this function is **not** linear.\n",
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"- $f(ax + by) = 2^{ax + by} = 2^{ax} \\times 2^{by}$, but $af(x) + bf(y) = a2^x + b2^y$. These two expressions aren't the same, so you can conclude that this function is **not** linear."
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]
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],
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"metadata": {}
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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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"----\n",
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"### Exercise B.3"
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],
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"metadata": {}
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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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"**Suppose that you have a linear function $g$ such that $g([[1], [0]]) = [[2.3], [-3.1]]$ and $g([[0], [1]]) = [[-5.2], [0.7]]$.\n",
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"Compute $g([[2], [-2]])$.**"
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]
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],
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"metadata": {}
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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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"source": [
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"g_horizontal = np.array([[2.3], [-3.1]])\n",
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"g_vertical = np.array([[-5.2], [0.7]])\n",
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"\n",
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"2 * g_horizontal + (-2) * g_vertical"
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],
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"outputs": [
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{
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"output_type": "execute_result",
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"data": {
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"text/plain": [
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"array([[15. ],\n",
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" [-7.6]])"
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]
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},
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"execution_count": 3,
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"metadata": {},
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"output_type": "execute_result"
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"execution_count": 3
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}
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],
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"source": [
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"g_horizontal = np.array([[2.3], [-3.1]])\n",
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"g_vertical = np.array([[-5.2], [0.7]])\n",
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"\n",
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"2 * g_horizontal + (-2) * g_vertical"
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]
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"metadata": {}
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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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"----\n",
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"### Exercise B.4"
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],
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"metadata": {}
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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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"**Let $X$ be the matrix $[[0, 1], [1, 0]]$, and let $\\vec{y}$ be the vector $[[2], [3]]$.\n",
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"Using NumPy, compute $X\\vec{y}$ and $XX$.**"
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]
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],
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"metadata": {}
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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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"source": [
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"X = np.array([\n",
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" [0, 1],\n",
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" [2],\n",
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" [3]\n",
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"])"
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],
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"outputs": [],
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"metadata": {}
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},
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{
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"cell_type": "code",
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"execution_count": 5,
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"metadata": {},
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"source": [
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"X @ y"
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],
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"outputs": [
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{
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"output_type": "execute_result",
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"data": {
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"text/plain": [
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"array([[3],\n",
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" [2]])"
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]
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},
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"execution_count": 5,
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"metadata": {},
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"output_type": "execute_result"
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"execution_count": 5
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}
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],
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"X @ y"
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"metadata": {}
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},
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"cell_type": "code",
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"execution_count": 6,
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"metadata": {},
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"source": [
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"X @ X"
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],
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"outputs": [
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{
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"output_type": "execute_result",
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"data": {
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"text/plain": [
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"array([[1, 0],\n",
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" [0, 1]])"
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]
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},
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"execution_count": 6,
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"metadata": {},
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"output_type": "execute_result"
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"execution_count": 6
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}
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],
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"X @ X"
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"metadata": {}
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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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"----\n",
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"### Exercise B.5"
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],
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"metadata": {}
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},
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{
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"cell_type": "markdown",
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"source": [
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"**Given a vector $[[2], [3]]$, find a vector that points in the same direction but with length 1.**\n",
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"\n",
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"*HINT*: You can either do this by using an inner product, or the `np.linalg.norm` function."
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],
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"metadata": {}
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"cell_type": "code",
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"execution_count": 7,
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"outputs": [],
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"source": [
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"v = np.array([\n",
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" [2],\n",
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" [3]\n",
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"])"
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]
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],
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"outputs": [],
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"metadata": {}
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},
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{
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"cell_type": "markdown",
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"source": [
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"We can try both ways suggested by the hint above to confirm that you get the same answer from each."
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],
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"metadata": {}
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},
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"cell_type": "code",
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"execution_count": 8,
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"source": [
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"v / np.sqrt(v.transpose() @ v)"
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],
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"outputs": [
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{
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"output_type": "execute_result",
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"data": {
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"text/plain": [
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"array([[0.5547002 ],\n",
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" [0.83205029]])"
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},
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"output_type": "execute_result"
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"execution_count": 8
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}
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],
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"v / np.sqrt(v.transpose() @ v)"
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"metadata": {}
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"cell_type": "code",
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"execution_count": 9,
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"source": [
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"v / np.linalg.norm(v)"
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],
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"outputs": [
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{
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"output_type": "execute_result",
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"data": {
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"text/plain": [
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"array([[0.5547002 ],\n",
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" [0.83205029]])"
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},
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"output_type": "execute_result"
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"execution_count": 9
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],
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"v / np.linalg.norm(v)"
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"metadata": {}
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"metadata": {
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},
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"nbformat": 4,
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"nbformat_minor": 4
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}
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}

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