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Implemented solutions 5-7 in ipython.

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Felix Martin
2018-02-03 20:57:46 +01:00
parent 3f5ee98ee8
commit e5685b8f07
6 changed files with 1145 additions and 0 deletions
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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Euler Problem\n",
"\n",
"2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.\n",
"\n",
"What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"My easiest guess is to multiply all prime numbers till the number."
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"def get_primes_smaller(number):\n",
" primes = []\n",
" prospects = [n for n in range(2, number)]\n",
" while prospects:\n",
" p = prospects[0]\n",
" prospects = [x for x in prospects if x % p != 0]\n",
" primes.append(p)\n",
" return primes"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"210\n"
]
}
],
"source": [
"from operator import mul\n",
"from functools import reduce\n",
"\n",
"def get_number_which_is_divisible_by_all_numbers_from_one_to(n):\n",
" ps= get_primes_smaller(n + 1)\n",
" return reduce(mul, ps, 1)\n",
"\n",
"print(get_number_which_is_divisible_by_all_numbers_from_one_to(10))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"That obviously didn't work. The reason is that the same prime can occur multiple times in the factorization of a divisor. For example $2^{3} = 8$. We can always brute force of course. We do a smart brute force and only check multiples from the product of primes because this factor must be part of the solution."
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"232792560\n"
]
}
],
"source": [
"def is_divisible_by_numbers_smaller_or_equal(number, maximum_number):\n",
" for n in range(2, maximum_number + 1):\n",
" if number % n != 0:\n",
" return False\n",
" return True\n",
"\n",
"def get_number_which_is_divisible_by_all_numbers_from_one_to(n):\n",
" ps = get_primes_smaller(n + 1)\n",
" factor = reduce(mul, ps, 1)\n",
" multiples_of_factor = factor\n",
" while True:\n",
" if is_divisible_by_numbers_smaller_or_equal(multiples_of_factor, n):\n",
" return multiples_of_factor\n",
" multiples_of_factor += factor\n",
"\n",
"assert(get_number_which_is_divisible_by_all_numbers_from_one_to(10) == 2520)\n",
"print(get_number_which_is_divisible_by_all_numbers_from_one_to(20))"
]
}
],
"metadata": {
"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"
}
},
"nbformat": 4,
"nbformat_minor": 0
}

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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Euler Problem 6\n",
"\n",
"The sum of the squares of the first ten natural numbers is,\n",
"\n",
"$1^2 + 2^2 + ... + 10^2 = 385$\n",
"\n",
"The square of the sum of the first ten natural numbers is,\n",
"\n",
"$(1 + 2 + ... + 10)^2 = 55^2 = 3025$\n",
"\n",
"Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is $3025 − 385 = 2640$.\n",
"\n",
"Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Okay, this is as straightforward as it can get."
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"25164150\n"
]
}
],
"source": [
"s = sum([x for x in range(1, 101)])**2 - sum([x**2 for x in range(1, 101)])\n",
"assert(s == 25164150)\n",
"print(s)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"General solution."
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"def diff_between_sum_of_squares_and_square_sum(n):\n",
" return sum([x for x in range(1, n + 1)])**2 - sum([x**2 for x in range(1, n + 1)])\n",
"\n",
"assert(diff_between_sum_of_squares_and_square_sum(100) == 25164150)\n",
"assert(diff_between_sum_of_squares_and_square_sum(10) == 2640)"
]
}
],
"metadata": {
"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"
}
},
"nbformat": 4,
"nbformat_minor": 0
}

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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Euler Problem 7\n",
"\n",
"By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.\n",
"\n",
"What is the 10 001st prime number?"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"We reuse our prime functions and use a trick from [stackoverflow](https://stackoverflow.com/questions/2300756/get-the-nth-item-of-a-generator-in-python) to get the nth element."
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"104743\n"
]
}
],
"source": [
"import itertools\n",
"\n",
"def is_prime(n, smaller_primes):\n",
" for s in smaller_primes:\n",
" if n % s == 0:\n",
" return False\n",
" if s * s > n:\n",
" return True\n",
" return True\n",
"\n",
"def prime_generator_function():\n",
" primes = [2, 3, 5, 7]\n",
" for p in primes:\n",
" yield p\n",
" while True:\n",
" p += 2\n",
" if is_prime(p, primes):\n",
" primes.append(p)\n",
" yield p\n",
"\n",
"def get_nth_prime(n):\n",
" ps = prime_generator_function()\n",
" return next(itertools.islice(ps, n - 1, n))\n",
"\n",
"assert(get_nth_prime(6) == 13)\n",
"print(get_nth_prime(10001))"
]
}
],
"metadata": {
"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"
}
},
"nbformat": 4,
"nbformat_minor": 0
}