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Finished problem 39 and added some new ones.

main
Felix Martin 2018-05-20 16:32:19 -04:00
parent aa14d544d2
commit 894160d1a0
11 changed files with 35915 additions and 9 deletions
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107
ipython/EulerProblem039.ipynb
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@ -13,21 +13,116 @@
"For which value of p ≤ 1000, is the number of solutions maximised?"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"We are looking for right angle triangles so Pythagoras' theorem $a^2 + b^2 = c^2$ can be used. Also it must be true that $a <= b <= c$ is true for every solution. Let's start with a function that tests the given examples."
]
},
{
"cell_type": "code",
"execution_count": null,
"execution_count": 13,
"metadata": {},
"outputs": [],
"source": [
"def is_right_angle_triangle(a, b, c, p):\n",
" if a + b + c != p:\n",
" return False\n",
" if a**2 + b**2 != c**2:\n",
" return False\n",
" return True\n",
"\n",
"given = [(20, 48, 52, 120), (24, 45, 51, 120), (30, 40, 50, 120)]\n",
"for g in given:\n",
" assert(is_right_angle_triangle(*g))\n",
"assert(is_right_angle_triangle(29, 41, 50, 120) == False)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"This seems bruteforceable. Let's try to find the solutions for p = 120."
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"[(30, 40, 50), (24, 45, 51), (20, 48, 52)]\n"
]
}
],
"source": [
"def brute_force(p):\n",
" solutions = [(a,b, p - a - b)\n",
" for b in range(1, p // 2 + 1)\n",
" for a in range(1, b)\n",
" if a*a + b*b == (p - a - b) * (p - a - b)\n",
" ]\n",
" return solutions\n",
"\n",
"print(brute_force(120))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Looks good. Let's try for all $p <= 1000$."
]
},
{
"cell_type": "code",
"execution_count": 18,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": []
"source": [
"solutions = [(len(brute_force(p)), p) for p in range(1, 1001)]"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Well, not too fast, but also not too slow. Let's get the max and we have our solution."
]
},
{
"cell_type": "code",
"execution_count": 24,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"840\n"
]
}
],
"source": [
"s, p = max(solutions)\n",
"print(p)\n",
"assert(p == 840)"
]
}
],
"metadata": {
"completion_date": "",
"completion_date": "Sun, 20 May 2018, 21:16",
"kernelspec": {
"display_name": "Python 3",
"language": "python",
"language": "python3.6",
"name": "python3"
},
"language_info": {
@ -40,7 +135,7 @@
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.5.5"
"version": "3.6.5"
},
"tags": [
"triangle",
@ -49,5 +144,5 @@
]
},
"nbformat": 4,
"nbformat_minor": 0
"nbformat_minor": 1
}

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ipython/EulerProblem041.ipynb Normal file
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@ -0,0 +1,57 @@
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Pandigital prime (Euler Problem 41)"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.\n",
"\n",
"What is the largest n-digit pandigital prime that exists?"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": []
}
],
"metadata": {
"completion_date": "",
"kernelspec": {
"display_name": "Python 3",
"language": "python3.6",
"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": "3.6.5"
},
"tags": [
"pandigital",
"prime"
]
},
"nbformat": 4,
"nbformat_minor": 2
}

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ipython/EulerProblem042.ipynb Normal file
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@ -0,0 +1,59 @@
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Coded triangle numbers (Euler Problem 42)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"The nth term of the sequence of triangle numbers is given by, $t_n = \\frac{1}{2}n(n+1)$; so the first ten triangle numbers are:\n",
"\n",
" 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...\n",
"\n",
"By converting each letter in a word to a number corresponding to its alphabetical position and adding these values we form a word value. For example, the word value for SKY is 19 + 11 + 25 = 55 = $t_{10}$. If the word value is a triangle number then we shall call the word a triangle word.\n",
"\n",
"Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words, how many are triangle words? (The file is saved in the same directory as this notebook file as EulerProblem042.txt)"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": []
}
],
"metadata": {
"completion_date": "",
"kernelspec": {
"display_name": "Python 3",
"language": "python3.6",
"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": "3.6.5"
},
"tags": [
"triangle",
"words"
]
},
"nbformat": 4,
"nbformat_minor": 2
}

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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Sub-string divisibility (Euler Problem 43)"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property.\n",
"\n",
"Let $d_1$ be the 1st digit, $d_2$ be the 2nd digit, and so on. In this way, we note the following:\n",
"\n",
"$d_2d_3d_4$=406 is divisible by 2\n",
"\n",
"$d_3d_4d_5$=063 is divisible by 3\n",
"\n",
"$d_4d_5d_6$=635 is divisible by 5\n",
"\n",
"$d_5d_6d_7=357$ is divisible by 7\n",
"\n",
"$d_6d_7d_8=572$ is divisible by 11\n",
"\n",
"$d_7d_8d_9=728$ is divisible by 13\n",
"\n",
"$d_8d_9d_{10} =289$ is divisible by 17\n",
"\n",
"Find the sum of all 0 to 9 pandigital numbers with this property."
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": []
}
],
"metadata": {
"completion_date": "",
"kernelspec": {
"display_name": "Python 3",
"language": "python3.6",
"name": "python3"
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"name": "ipython",
"version": 3
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.6.5"
},
"tags": [
"pandigital",
"divisibility"
]
},
"nbformat": 4,
"nbformat_minor": 2
}

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@ -11779,21 +11779,152 @@ div#notebook {
<p>$\{20,48,52\}, \{24,45,51\}, \{30,40,50\}$</p>
<p>For which value of p ≤ 1000, is the number of solutions maximised?</p>
</div>
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<div class="cell border-box-sizing text_cell rendered"><div class="prompt input_prompt">
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<div class="text_cell_render border-box-sizing rendered_html">
<p>We are looking for right angle triangles so Pythagoras' theorem $a^2 + b^2 = c^2$ can be used. Also it must be true that $a &lt;= b &lt;= c$ is true for every solution. Let's start with a function that tests the given examples.</p>
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<div class="input">
<div class="prompt input_prompt">In&nbsp;[&nbsp;]:</div>
<div class="prompt input_prompt">In&nbsp;[13]:</div>
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<div class=" highlight hl-ipython3"><pre><span></span>
<div class=" highlight hl-ipython3"><pre><span></span><span class="k">def</span> <span class="nf">is_right_angle_triangle</span><span class="p">(</span><span class="n">a</span><span class="p">,</span> <span class="n">b</span><span class="p">,</span> <span class="n">c</span><span class="p">,</span> <span class="n">p</span><span class="p">):</span>
<span class="k">if</span> <span class="n">a</span> <span class="o">+</span> <span class="n">b</span> <span class="o">+</span> <span class="n">c</span> <span class="o">!=</span> <span class="n">p</span><span class="p">:</span>
<span class="k">return</span> <span class="kc">False</span>
<span class="k">if</span> <span class="n">a</span><span class="o">**</span><span class="mi">2</span> <span class="o">+</span> <span class="n">b</span><span class="o">**</span><span class="mi">2</span> <span class="o">!=</span> <span class="n">c</span><span class="o">**</span><span class="mi">2</span><span class="p">:</span>
<span class="k">return</span> <span class="kc">False</span>
<span class="k">return</span> <span class="kc">True</span>
<span class="n">given</span> <span class="o">=</span> <span class="p">[(</span><span class="mi">20</span><span class="p">,</span> <span class="mi">48</span><span class="p">,</span> <span class="mi">52</span><span class="p">,</span> <span class="mi">120</span><span class="p">),</span> <span class="p">(</span><span class="mi">24</span><span class="p">,</span> <span class="mi">45</span><span class="p">,</span> <span class="mi">51</span><span class="p">,</span> <span class="mi">120</span><span class="p">),</span> <span class="p">(</span><span class="mi">30</span><span class="p">,</span> <span class="mi">40</span><span class="p">,</span> <span class="mi">50</span><span class="p">,</span> <span class="mi">120</span><span class="p">)]</span>
<span class="k">for</span> <span class="n">g</span> <span class="ow">in</span> <span class="n">given</span><span class="p">:</span>
<span class="k">assert</span><span class="p">(</span><span class="n">is_right_angle_triangle</span><span class="p">(</span><span class="o">*</span><span class="n">g</span><span class="p">))</span>
<span class="k">assert</span><span class="p">(</span><span class="n">is_right_angle_triangle</span><span class="p">(</span><span class="mi">29</span><span class="p">,</span> <span class="mi">41</span><span class="p">,</span> <span class="mi">50</span><span class="p">,</span> <span class="mi">120</span><span class="p">)</span> <span class="o">==</span> <span class="kc">False</span><span class="p">)</span>
</pre></div>
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<div class="cell border-box-sizing text_cell rendered"><div class="prompt input_prompt">
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<div class="text_cell_render border-box-sizing rendered_html">
<p>This seems bruteforceable. Let's try to find the solutions for p = 120.</p>
</div>
</div>
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<div class="cell border-box-sizing code_cell rendered">
<div class="input">
<div class="prompt input_prompt">In&nbsp;[16]:</div>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="k">def</span> <span class="nf">brute_force</span><span class="p">(</span><span class="n">p</span><span class="p">):</span>
<span class="n">solutions</span> <span class="o">=</span> <span class="p">[(</span><span class="n">a</span><span class="p">,</span><span class="n">b</span><span class="p">,</span> <span class="n">p</span> <span class="o">-</span> <span class="n">a</span> <span class="o">-</span> <span class="n">b</span><span class="p">)</span>
<span class="k">for</span> <span class="n">b</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">p</span> <span class="o">//</span> <span class="mi">2</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)</span>
<span class="k">for</span> <span class="n">a</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">b</span><span class="p">)</span>
<span class="k">if</span> <span class="n">a</span><span class="o">*</span><span class="n">a</span> <span class="o">+</span> <span class="n">b</span><span class="o">*</span><span class="n">b</span> <span class="o">==</span> <span class="p">(</span><span class="n">p</span> <span class="o">-</span> <span class="n">a</span> <span class="o">-</span> <span class="n">b</span><span class="p">)</span> <span class="o">*</span> <span class="p">(</span><span class="n">p</span> <span class="o">-</span> <span class="n">a</span> <span class="o">-</span> <span class="n">b</span><span class="p">)</span>
<span class="p">]</span>
<span class="k">return</span> <span class="n">solutions</span>
<span class="nb">print</span><span class="p">(</span><span class="n">brute_force</span><span class="p">(</span><span class="mi">120</span><span class="p">))</span>
</pre></div>
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<pre>[(30, 40, 50), (24, 45, 51), (20, 48, 52)]
</pre>
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<p>Looks good. Let's try for all $p &lt;= 1000$.</p>
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<div class="cell border-box-sizing code_cell rendered">
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<div class="prompt input_prompt">In&nbsp;[18]:</div>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="n">solutions</span> <span class="o">=</span> <span class="p">[(</span><span class="nb">len</span><span class="p">(</span><span class="n">brute_force</span><span class="p">(</span><span class="n">p</span><span class="p">)),</span> <span class="n">p</span><span class="p">)</span> <span class="k">for</span> <span class="n">p</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="mi">1001</span><span class="p">)]</span>
</pre></div>
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<p>Well, not too fast, but also not too slow. Let's get the max and we have our solution.</p>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="n">s</span><span class="p">,</span> <span class="n">p</span> <span class="o">=</span> <span class="nb">max</span><span class="p">(</span><span class="n">solutions</span><span class="p">)</span>
<span class="nb">print</span><span class="p">(</span><span class="n">p</span><span class="p">)</span>
<span class="k">assert</span><span class="p">(</span><span class="n">p</span> <span class="o">==</span> <span class="mi">840</span><span class="p">)</span>
</pre></div>
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<pre>840
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@ -576,7 +576,7 @@
<tr>
<td>Problem 039</td>
<td></td>
<td>Sun, 20 May 2018, 21:16</td>
<td><a href="EulerProblem039.html">Problem 039</a></td>
<td>
@ -604,6 +604,45 @@
</td>
</tr>
<tr>
<td>Problem 041</td>
<td></td>
<td><a href="EulerProblem041.html">Problem 041</a></td>
<td>
<kbd>pandigital</kbd>
<kbd>prime</kbd>
</td>
</tr>
<tr>
<td>Problem 042</td>
<td></td>
<td><a href="EulerProblem042.html">Problem 042</a></td>
<td>
<kbd>triangle</kbd>
<kbd>words</kbd>
</td>
</tr>
<tr>
<td>Problem 043</td>
<td></td>
<td><a href="EulerProblem043.html">Problem 043</a></td>
<td>
<kbd>pandigital</kbd>
<kbd>divisibility</kbd>
</td>
</tr>
<tr>
<td>Problem 067</td>
<td>Fri, 5 Sep 2014, 07:36</td>

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#!/usr/bin/env python
import jinja2
import os
import subprocess