Html pages no include tags and completion date.

This commit is contained in:
Felix Martin 2018-02-04 16:54:28 +01:00
parent 210b3ccbc6
commit 5f87a31248
16 changed files with 1266 additions and 22 deletions

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@ -33,7 +33,9 @@
{ {
"cell_type": "code", "cell_type": "code",
"execution_count": 2, "execution_count": 2,
"metadata": {}, "metadata": {
"collapsed": true
},
"outputs": [], "outputs": [],
"source": [ "source": [
"assert(get_sum_of_natural_dividable_by_3_and_5_below(10) == 23)" "assert(get_sum_of_natural_dividable_by_3_and_5_below(10) == 23)"
@ -42,7 +44,9 @@
{ {
"cell_type": "code", "cell_type": "code",
"execution_count": 3, "execution_count": 3,
"metadata": {}, "metadata": {
"collapsed": false
},
"outputs": [ "outputs": [
{ {
"name": "stdout", "name": "stdout",
@ -58,6 +62,7 @@
} }
], ],
"metadata": { "metadata": {
"completion_date": "Tue, 19 Aug 2014, 20:11",
"kernelspec": { "kernelspec": {
"display_name": "Python 3", "display_name": "Python 3",
"language": "python", "language": "python",
@ -73,8 +78,11 @@
"name": "python", "name": "python",
"nbconvert_exporter": "python", "nbconvert_exporter": "python",
"pygments_lexer": "ipython3", "pygments_lexer": "ipython3",
"version": "3.6.3" "version": "3.5.4"
} },
"tags": [
"brute_force"
]
}, },
"nbformat": 4, "nbformat": 4,
"nbformat_minor": 2 "nbformat_minor": 2

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@ -138,6 +138,7 @@
} }
], ],
"metadata": { "metadata": {
"completion_date": "Tue, 19 Aug 2014, 20:36",
"kernelspec": { "kernelspec": {
"display_name": "Python 3", "display_name": "Python 3",
"language": "python", "language": "python",

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@ -176,6 +176,7 @@
} }
], ],
"metadata": { "metadata": {
"completion_date": "Tue, 19 Aug 2014, 23:35",
"kernelspec": { "kernelspec": {
"display_name": "Python 3", "display_name": "Python 3",
"language": "python", "language": "python",

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@ -150,6 +150,7 @@
} }
], ],
"metadata": { "metadata": {
"completion_date": "Wed, 20 Aug 2014, 13:27",
"kernelspec": { "kernelspec": {
"display_name": "Python 3", "display_name": "Python 3",
"language": "python", "language": "python",
@ -169,7 +170,9 @@
}, },
"tags": [ "tags": [
"palindrome", "palindrome",
"combinations" "combinations",
"brute force",
"timeit"
] ]
}, },
"nbformat": 4, "nbformat": 4,

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@ -107,6 +107,7 @@
} }
], ],
"metadata": { "metadata": {
"completion_date": "Wed, 20 Aug 2014, 14:32",
"kernelspec": { "kernelspec": {
"display_name": "Python 3", "display_name": "Python 3",
"language": "python", "language": "python",

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@ -71,6 +71,7 @@
} }
], ],
"metadata": { "metadata": {
"completion_date": "Wed, 20 Aug 2014, 15:13",
"kernelspec": { "kernelspec": {
"display_name": "Python 3", "display_name": "Python 3",
"language": "python", "language": "python",
@ -87,7 +88,8 @@
"nbconvert_exporter": "python", "nbconvert_exporter": "python",
"pygments_lexer": "ipython3", "pygments_lexer": "ipython3",
"version": "3.5.4" "version": "3.5.4"
} },
"tags": []
}, },
"nbformat": 4, "nbformat": 4,
"nbformat_minor": 0 "nbformat_minor": 0

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@ -64,6 +64,7 @@
} }
], ],
"metadata": { "metadata": {
"completion_date": "Wed, 20 Aug 2014, 15:40",
"kernelspec": { "kernelspec": {
"display_name": "Python 3", "display_name": "Python 3",
"language": "python", "language": "python",

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@ -0,0 +1,144 @@
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Euler Problem 8\n",
"\n",
"The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.\n",
"\n",
"~~~\n",
"73167176531330624919225119674426574742355349194934\n",
"96983520312774506326239578318016984801869478851843\n",
"85861560789112949495459501737958331952853208805511\n",
"12540698747158523863050715693290963295227443043557\n",
"66896648950445244523161731856403098711121722383113\n",
"62229893423380308135336276614282806444486645238749\n",
"30358907296290491560440772390713810515859307960866\n",
"70172427121883998797908792274921901699720888093776\n",
"65727333001053367881220235421809751254540594752243\n",
"52584907711670556013604839586446706324415722155397\n",
"53697817977846174064955149290862569321978468622482\n",
"83972241375657056057490261407972968652414535100474\n",
"82166370484403199890008895243450658541227588666881\n",
"16427171479924442928230863465674813919123162824586\n",
"17866458359124566529476545682848912883142607690042\n",
"24219022671055626321111109370544217506941658960408\n",
"07198403850962455444362981230987879927244284909188\n",
"84580156166097919133875499200524063689912560717606\n",
"05886116467109405077541002256983155200055935729725\n",
"71636269561882670428252483600823257530420752963450\n",
"~~~\n",
"\n",
"Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"We need the number as a list of integers first."
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"def remove_newlines(s):\n",
" import re\n",
" return re.sub(r'\\n', '', s)\n",
"\n",
"digits_string = \"\"\"73167176531330624919225119674426574742355349194934\n",
"96983520312774506326239578318016984801869478851843\n",
"85861560789112949495459501737958331952853208805511\n",
"12540698747158523863050715693290963295227443043557\n",
"66896648950445244523161731856403098711121722383113\n",
"62229893423380308135336276614282806444486645238749\n",
"30358907296290491560440772390713810515859307960866\n",
"70172427121883998797908792274921901699720888093776\n",
"65727333001053367881220235421809751254540594752243\n",
"52584907711670556013604839586446706324415722155397\n",
"53697817977846174064955149290862569321978468622482\n",
"83972241375657056057490261407972968652414535100474\n",
"82166370484403199890008895243450658541227588666881\n",
"16427171479924442928230863465674813919123162824586\n",
"17866458359124566529476545682848912883142607690042\n",
"24219022671055626321111109370544217506941658960408\n",
"07198403850962455444362981230987879927244284909188\n",
"84580156166097919133875499200524063689912560717606\n",
"05886116467109405077541002256983155200055935729725\n",
"71636269561882670428252483600823257530420752963450\"\"\"\n",
"\n",
"digits = [int(d) for d in remove_newlines(digits_string)]"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
" Then we can use slicing to do a brute force."
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"23514624000\n"
]
}
],
"source": [
"def product(xs):\n",
" from operator import mul\n",
" from functools import reduce\n",
" return reduce(mul, xs, 1)\n",
"\n",
"def get_largest_product_of_n_digits(digits, n):\n",
" return max((product(digits[i:i + n]) for i in range(len(digits))))\n",
"\n",
"assert(get_largest_product_of_n_digits(digits, 4) == 5832)\n",
"assert(get_largest_product_of_n_digits(digits, 13) == 23514624000)\n",
"print(get_largest_product_of_n_digits(digits, 13))"
]
}
],
"metadata": {
"completion_date": "Wed, 20 Aug 2014, 16:03",
"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": "3.5.4"
},
"tags": [
"product",
"reduce",
"slice"
]
},
"nbformat": 4,
"nbformat_minor": 0
}

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@ -0,0 +1,165 @@
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Euler Problem 9\n",
"\n",
"A Pythagorean triplet is a set of three natural numbers, $a < b < c$, for which,\n",
"\n",
"$a^2 + b^2 = c^2$\n",
"\n",
"For example, $3^2 + 4^2 = 9 + 16 = 25 = 5^2$.\n",
"\n",
"There exists exactly one Pythagorean triplet for which $a + b + c = 1000$.\n",
"Find the product $abc$."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"We start bruteforcing even though it feels like there may be a smart solution."
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"66.96907186399949\n",
"31875000\n"
]
}
],
"source": [
"import timeit\n",
"\n",
"def brute_force():\n",
" for a in range(1, 251):\n",
" for b in range(a + 1, 501):\n",
" for c in range(b + 1, 1001):\n",
" if a + b + c == 1000 and a * a + b * b == c * c:\n",
" return a * b * c\n",
"\n",
"print(timeit.timeit(brute_force, number=10))\n",
"print(brute_force())"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Let's do some optimization by cancelling the loops when we exceed the boundaries. Actually, I have also realized that choosing 251 and 501 is a little bit arbitrary. For example if a, b, c where something like $332, 333, 335$ that could be a solution and the first range was too low. Then again, we would have realized that as soon as we get back None, so it is okay."
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"64.4656584559998\n",
"31875000\n"
]
}
],
"source": [
"def brute_force():\n",
" for a in range(1, 251):\n",
" for b in range(a + 1, 501):\n",
" if a + b > 600:\n",
" break\n",
" for c in range(b + 1, 1001):\n",
" if a + b + c == 1000 and a * a + b * b == c * c:\n",
" return a * b * c\n",
" \n",
"print(timeit.timeit(brute_force, number=10))\n",
"print(brute_force())"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Big time save. Kappa. Okay, I am stupid. If I have a and b I can calculate c and check if it is a solution."
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"0.22822808900036762\n",
"31875000\n"
]
}
],
"source": [
"def smart_brute_force():\n",
" for a in range(1, 251):\n",
" for b in range(a + 1, 501):\n",
" c = 1000 - a - b\n",
" if a * a + b * b == c * c:\n",
" return a * b * c\n",
" \n",
"print(timeit.timeit(smart_brute_force, number=10))\n",
"print(smart_brute_force()) "
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": []
}
],
"metadata": {
"completion_date": "Wed, 20 Aug 2014, 17:06",
"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": "3.5.4"
},
"tags": [
"brute force",
"pythagorean",
"triplet"
]
},
"nbformat": 4,
"nbformat_minor": 0
}

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@ -0,0 +1,37 @@
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Euler Problem 10\n",
"\n"
]
}
],
"metadata": {
"completion_date": "Wed, 20 Aug 2014, 20:28",
"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": "3.5.4"
},
"tags": [
"primes"
]
},
"nbformat": 4,
"nbformat_minor": 0
}

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@ -1,6 +0,0 @@
{
"cells": [],
"metadata": {},
"nbformat": 4,
"nbformat_minor": 0
}

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@ -4,15 +4,26 @@ import subprocess
from operator import itemgetter from operator import itemgetter
from collections import namedtuple from collections import namedtuple
from os.path import getmtime from os.path import getmtime
import json
def extract_metadata(ipynb_file):
Metadata = namedtuple('Metadata', ['tags', 'completion_date'])
with open(ipynb_file, 'r') as f:
j = json.load(f)
tags = j['metadata']['tags']
completion_date = j['metadata']['completion_date']
return Metadata(tags, completion_date)
def file_name_to_solution(name): def file_name_to_solution(name):
Solution = namedtuple('Solution', ["number", "ipynb", "html", "name"]) Solution = namedtuple('Solution', ["number", "ipynb", "html", "name", "metadata"])
number = int(name.replace("EulerProblem", "").replace(".ipynb", "")) number = int(name.replace("EulerProblem", "").replace(".ipynb", ""))
ipynb = name ipynb = name
metadata = extract_metadata(ipynb)
html = name.replace(".ipynb", ".html") html = name.replace(".ipynb", ".html")
name = name.replace("EulerProblem", "Problem ").replace(".ipynb", "") name = name.replace("EulerProblem", "Problem ").replace(".ipynb", "")
return Solution(number, ipynb, html, name) return Solution(number, ipynb, html, name, metadata)
def get_solution_list(directory="./"): def get_solution_list(directory="./"):
@ -56,4 +67,4 @@ if __name__ == "__main__":
solutions = get_solution_list() solutions = get_solution_list()
convert_solutions_to_html(solutions) convert_solutions_to_html(solutions)
render_solutions(solutions) render_solutions(solutions)
ship_to_failx() #ship_to_failx()

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@ -27,11 +27,30 @@
<div class="row" style="padding-top: 40px;"> <div class="row" style="padding-top: 40px;">
<div class="col"> <div class="col">
<ul> <table class="table">
{% for s in solutions %} <thead>
<li><a href="{{s.html}}">{{s.name}}</a></li> <tr>
{% endfor %} <th scope="col">Name</th>
</ul> <th scope="col">Complete on</th>
<th scope="col">Link</th>
<th scope="col">Tags</th>
</tr>
</thead>
<tbody>
{% for s in solutions %}
<tr>
<td>{{s.name}}</td>
<td>{{s.metadata.completion_date}}</td>
<td><a href="{{s.html}}">{{s.name}}</a></td>
<td>
{% for t in s.metadata.tags %}
<kbd>{{t}}</kbd>
{% endfor %}
</td>
</tr>
{% endfor %}
</tbody>
</table>
</div> </div>
</div> </div>