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Solved 16, 17, and 67 in ipython.

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Felix Martin
2018-02-05 21:56:25 +01:00
parent f884b5e8ed
commit 0e8aaa4646
10 changed files with 1936 additions and 1 deletions
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2
ipython/EulerProblem015.ipynb
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}
],
"metadata": {
"completion_date": "137846528820",
"completion_date": "Sun, 31 Aug 2014, 20:05",
"kernelspec": {
"display_name": "Python 3",
"language": "python",

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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Euler Problem 16\n",
"\n",
"$2^{15}$ = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.\n",
"\n",
"What is the sum of the digits of the number $2^{1000}$?"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Seriously? Okay, probably more difficult in C."
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"1366\n"
]
}
],
"source": [
"s = sum(map(int, str(2**1000)))\n",
"assert(s == 1366)\n",
"print(s)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Okay, let's do it without big ints."
]
},
{
"cell_type": "code",
"execution_count": 18,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"def double_number_string(number_string):\n",
" number_string = number_string[::-1] # reverse\n",
" result = []\n",
" carriage = 0\n",
" for c in number_string:\n",
" s = carriage + int(c) + int(c)\n",
" carriage = s // 10\n",
" next_digit = s % 10\n",
" result.append(str(next_digit))\n",
" if carriage > 0:\n",
" result.append(str(carriage))\n",
" result = \"\".join(result)[::-1]\n",
" return result\n",
"\n",
"assert(double_number_string(\"132\") == \"264\")\n",
"assert(double_number_string(\"965\") == \"1930\")"
]
},
{
"cell_type": "code",
"execution_count": 19,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"1366\n"
]
}
],
"source": [
"def powers_of_two(n):\n",
" \"\"\" Retuns nth power of two as a string. \"\"\"\n",
" assert(n >= 0)\n",
" if n == 0:\n",
" return \"1\"\n",
" s = \"2\"\n",
" for _ in range(n - 1):\n",
" s = double_number_string(s)\n",
" return s\n",
"\n",
"assert(powers_of_two(3) == \"8\")\n",
"assert(powers_of_two(10) == \"1024\")\n",
"\n",
"number_string = powers_of_two(1000)\n",
"print(sum(map(int, number_string)))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Here we go. The conversion to integer and the sum would be easy in C."
]
}
],
"metadata": {
"completion_date": "Sun, 31 Aug 2014, 20:49",
"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": [
"c",
"manual"
]
},
"nbformat": 4,
"nbformat_minor": 0
}

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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Euler Problem 17\n",
"\n",
"If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.\n",
"\n",
"If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?\n",
"\n",
"\n",
"NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of \"and\" when writing out numbers is in compliance with British usage."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"I can see us doing a semi-automated approach here or we write a nice function. We probably write a nice function because we are geeks."
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"def single_digit_integer_to_spoken_language(n):\n",
" if n == 0:\n",
" return \"\"\n",
" assert(n > 0 and n < 10)\n",
" return {1: 'one', 2: 'two', 3: 'three', 4: 'four', 5: 'five',\n",
" 6: 'six', 7: 'seven', 8: 'eight', 9: 'nine'}[n]\n",
"\n",
"def double_digit_integer_to_spoken_language(n):\n",
" assert(n > 9 and n < 100)\n",
" try:\n",
" return {\n",
" 10: 'ten', 11: 'eleven', 12: 'twelve', 13: 'thirteen', \n",
" 14: 'fourteen', 15: 'fifteen', 16: 'sixteen',\n",
" 17: 'seventeen', 18: 'eighteen', 19: 'nineteen'}[n]\n",
" except KeyError:\n",
" pass\n",
" a, b = str(n)\n",
" a = {2: 'twenty', 3: 'thirty', 4: 'forty', 5: 'fifty',\n",
" 6: 'sixty', 7: 'seventy', 8: 'eighty', 9: 'ninety'}[int(a)]\n",
" b = integer_to_spoken_language(int(b))\n",
" return a + '-' + b\n",
"\n",
"def triple_digit_integer_to_spoken_language(n):\n",
" a, b = str(n)[0], str(n)[1:]\n",
" a = single_digit_integer_to_spoken_language(int(a))\n",
" b = integer_to_spoken_language(int(b))\n",
" if not b:\n",
" return a + \" hundred\"\n",
" return a + \" hundred and \" + b\n",
"\n",
"def four_digit_integer_to_spoken_language(n):\n",
" a, b = str(n)[0], str(n)[1:]\n",
" a = single_digit_integer_to_spoken_language(int(a))\n",
" b = integer_to_spoken_language(int(b)) \n",
" return a + \" thousand \" + b\n",
"\n",
"def integer_to_spoken_language(n):\n",
" l = len(str(n))\n",
" if l == 1:\n",
" return single_digit_integer_to_spoken_language(n)\n",
" elif l == 2:\n",
" return double_digit_integer_to_spoken_language(n)\n",
" elif l == 3:\n",
" return triple_digit_integer_to_spoken_language(n)\n",
" elif l == 4:\n",
" return four_digit_integer_to_spoken_language(n)\n",
" else:\n",
" raise Exception(\"Length not supported.\")\n",
"\n",
"assert(integer_to_spoken_language(5) == 'five')\n",
"assert(integer_to_spoken_language(19) == 'nineteen')\n",
"assert(integer_to_spoken_language(21) == 'twenty-one')\n",
"assert(integer_to_spoken_language(210) == 'two hundred and ten')\n",
"assert(integer_to_spoken_language(3000) == 'three thousand ')\n",
"assert(integer_to_spoken_language(8333) == 'eight thousand three hundred and thirty-three')"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Okay, I won't win a code golf contest but at least we can get the solution now."
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"21124\n"
]
}
],
"source": [
"l = len(\"\".join([integer_to_spoken_language(i) for i in range(1, 1001)]).replace(\" \", \"\").replace(\"-\", \"\"))\n",
"assert(l == 21124)\n",
"print(l)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Made the classical fourty/forty error I have already done four years ago. Some things don't change. Also I still do not really like this problem. The Haskell solution is actually way easier to read this time."
]
}
],
"metadata": {
"completion_date": "Sun, 31 Aug 2014, 21:20",
"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": [
"count",
"numbers",
"tricky"
]
},
"nbformat": 4,
"nbformat_minor": 0
}

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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Euler Problem 18\n",
"\n",
"By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.\n",
"\n",
"~~~\n",
" 3\n",
" 7 4\n",
" 2 4 6\n",
" 8 5 9 3\n",
"~~~\n",
"\n",
"That is, 3 + 7 + 4 + 9 = 23.\n",
"\n",
"Find the maximum total from top to bottom of the triangle below:\n",
"\n",
"~~~\n",
"75\n",
"95 64\n",
"17 47 82\n",
"18 35 87 10\n",
"20 04 82 47 65\n",
"19 01 23 75 03 34\n",
"88 02 77 73 07 63 67\n",
"99 65 04 28 06 16 70 92\n",
"41 41 26 56 83 40 80 70 33\n",
"41 48 72 33 47 32 37 16 94 29\n",
"53 71 44 65 25 43 91 52 97 51 14\n",
"70 11 33 28 77 73 17 78 39 68 17 57\n",
"91 71 52 38 17 14 91 43 58 50 27 29 48\n",
"63 66 04 68 89 53 67 30 73 16 69 87 40 31\n",
"04 62 98 27 23 09 70 98 73 93 38 53 60 04 23\n",
"~~~\n",
"\n",
"NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, [Problem 67](https://projecteuler.net/problem=67), is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"This is incredibly simple we simply from bottom to top choosing the higher value for each mini-tree.\n",
"\n",
"For example,\n",
"\n",
"~~~\n",
" 95 64\n",
"17 47 82\n",
"~~~\n",
"\n",
"will become:\n",
"\n",
"~~~\n",
" 142 146\n",
"~~~"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"t = \"\"\"\n",
"75\n",
"95 64\n",
"17 47 82\n",
"18 35 87 10\n",
"20 04 82 47 65\n",
"19 01 23 75 03 34\n",
"88 02 77 73 07 63 67\n",
"99 65 04 28 06 16 70 92\n",
"41 41 26 56 83 40 80 70 33\n",
"41 48 72 33 47 32 37 16 94 29\n",
"53 71 44 65 25 43 91 52 97 51 14\n",
"70 11 33 28 77 73 17 78 39 68 17 57\n",
"91 71 52 38 17 14 91 43 58 50 27 29 48\n",
"63 66 04 68 89 53 67 30 73 16 69 87 40 31\n",
"04 62 98 27 23 09 70 98 73 93 38 53 60 04 23\n",
"\"\"\""
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"def reduce_rows(xs, ys):\n",
" \"\"\" xs is lower row and ys is upper row \"\"\"\n",
" assert(len(xs) == len(ys) + 1)\n",
" return [max([xs[i] + ys[i], xs[i + 1] + ys[i]]) for i in range(len(ys))]\n",
" \n",
"assert(reduce_rows([17, 47, 82], [95, 64]) == [142, 146])"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Okay, now all we have to do is the parsing and then a simple fold."
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"xss = [list(map(int, xs.split())) for xs in t.split(\"\\n\") if xs]\n",
"xss.reverse()\n",
"from functools import reduce\n",
"s = reduce(reduce_rows, xss[1:], xss[0])[0]\n",
"assert(s == 1074)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Okay, let's put this into a nice function an then solve [problem 67](EulerProblem067) right away."
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"1074\n"
]
}
],
"source": [
"def find_greatest_path_sum_in_triangle_string(ts):\n",
" from functools import reduce\n",
" xss = [list(map(int, xs.split())) for xs in ts.split(\"\\n\") if xs]\n",
" xss.reverse()\n",
" r = lambda xs, ys: [max([xs[i] + ys[i], xs[i + 1] + ys[i]]) for i in range(len(ys))]\n",
" return reduce(r, xss[1:], xss[0])[0]\n",
"\n",
"print(find_greatest_path_sum_in_triangle_string(t))"
]
}
],
"metadata": {
"completion_date": "Thu, 4 Sep 2014, 20:38",
"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": [
"fold",
"reduce",
"search"
]
},
"nbformat": 4,
"nbformat_minor": 0
}

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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Euler Problem 67\n",
"\n",
"By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.\n",
"\n",
"~~~\n",
" 3\n",
" 7 4\n",
" 2 4 6\n",
"8 5 9 3\n",
"~~~\n",
"\n",
"That is, 3 + 7 + 4 + 9 = 23.\n",
"\n",
"Find the maximum total from top to bottom in triangle.txt (right click and 'Save Link/Target As...'), a 15K text file containing a triangle with one-hundred rows.\n",
"\n",
"NOTE: This is a much more difficult version of Problem 18. It is not possible to try every route to solve this problem, as there are 299 altogether! If you could check one trillion (1012) routes every second it would take over twenty billion years to check them all. There is an efficient algorithm to solve it. ;o)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"So let's hope our solution is good. I have saved the triangle in a file EulerProblem067.txt in the same directory as this notebook."
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"7273\n"
]
}
],
"source": [
"def find_greatest_path_sum_in_triangle_string(ts):\n",
" from functools import reduce\n",
" xss = [list(map(int, xs.split())) for xs in ts.split(\"\\n\") if xs]\n",
" xss.reverse()\n",
" r = lambda xs, ys: [max([xs[i] + ys[i], xs[i + 1] + ys[i]]) for i in range(len(ys))]\n",
" return reduce(r, xss[1:], xss[0])[0]\n",
"\n",
"with open('EulerProblem067.txt', 'r') as f:\n",
" triangle_string = f.read()\n",
" \n",
"print(find_greatest_path_sum_in_triangle_string(triangle_string))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Well, that was fast."
]
}
],
"metadata": {
"completion_date": "Fri, 5 Sep 2014, 07:36",
"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": [
"reduce",
"triangle"
]
},
"nbformat": 4,
"nbformat_minor": 0
}

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46 19 77 66 24 18 05 32 02 84 31 99 92 58 96 72 91 36 62 99 55 29 53 42 12 37 26 58 89 50 66 19 82 75 12 48 24 87 91 85 02 07 03 76 86
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