Added some further ipython solutions.
This commit is contained in:
parent
2e825d4f78
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@ -10,31 +10,41 @@
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"\n",
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"\n",
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"$2^2=4, 2^3=8, 2^4=16, 2^5=32$\n",
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"$2^2=4, 2^3=8, 2^4=16, 2^5=32$\n",
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"\n",
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"\n",
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"32=9, 33=27, 34=81, 35=243\n",
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"$3^2=9, 3^3=27, 3^4=81, 3^5=243$\n",
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"\n",
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"\n",
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"42=16, 43=64, 44=256, 45=1024\n",
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"$4^2=16, 4^3=64, 4^4=256, 4^5=1024$\n",
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"\n",
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"\n",
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"52=25, 53=125, 54=625, 55=3125\n",
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"$5^2=25, 5^3=125, 5^4=625, 5^5=3125$\n",
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"\n",
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"\n",
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"If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:\n",
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"If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:\n",
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"\n",
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"\n",
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"4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125\n",
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"$4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125$\n",
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"\n",
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"\n",
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"How many distinct terms are in the sequence generated by ab for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?"
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"How many distinct terms are in the sequence generated by $a^b$ for $2 ≤ a ≤ 100$ and $2 ≤ b ≤ 100$?"
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]
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]
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},
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},
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{
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{
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"cell_type": "code",
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"cell_type": "code",
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"execution_count": null,
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"execution_count": 1,
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"metadata": {
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"metadata": {},
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"collapsed": true
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"outputs": [
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},
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{
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"outputs": [],
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"name": "stdout",
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"source": []
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"output_type": "stream",
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"text": [
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"9183\n"
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]
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}
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],
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"source": [
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"s = len(set([a**b for a in range(2, 101) for b in range(2, 101)]))\n",
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"assert(s == 9183)\n",
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"print(s)"
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]
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}
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}
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],
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],
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"metadata": {
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"metadata": {
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"comopletion_date": "",
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"comopletion_date": "Fri, 25 Aug 2017, 10:03",
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"kernelspec": {
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"kernelspec": {
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"display_name": "Python 3",
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"display_name": "Python 3",
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"language": "python",
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"language": "python",
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@ -52,7 +62,10 @@
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"pygments_lexer": "ipython3",
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"pygments_lexer": "ipython3",
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"version": "3.6.3"
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"version": "3.6.3"
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},
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},
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"tags": []
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"tags": [
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"distinct",
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"powers"
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]
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},
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},
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"nbformat": 4,
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"nbformat": 4,
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"nbformat_minor": 2
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"nbformat_minor": 2
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83
ipython/EulerProblem030.ipynb
Executable file
83
ipython/EulerProblem030.ipynb
Executable file
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{
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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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"# Euler Problem 30\n",
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"\n",
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"Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:\n",
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"\n",
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"$1634 = 1^4 + 6^4 + 3^4 + 4^4$\n",
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"\n",
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"$8208 = 8^4 + 2^4 + 0^4 + 8^4$\n",
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"\n",
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"$9474 = 9^4 + 4^4 + 7^4 + 4^4$\n",
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"\n",
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"As $1 = 1^4$ is not a sum it is not included.\n",
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"\n",
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"The sum of these numbers is $1634 + 8208 + 9474 = 19316$.\n",
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"\n",
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"Find the sum of all the numbers that can be written as the sum of fifth powers of their digits."
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]
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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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"This is an eays brute force. We look up the fifth powers."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 7,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"443839\n"
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]
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}
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],
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"source": [
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"fifth_power_lookup = {str(i): i**5 for i in range(0,10)}\n",
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"\n",
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"def is_number_sum_of_fiths_powers_of_digits(n):\n",
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" return n == sum([fifth_power_lookup[d] for d in str(n)])\n",
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"\n",
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"s = sum([i for i in range(1, 1000000) if is_number_sum_of_fiths_powers_of_digits(i) and not i is 1])\n",
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"print(s)\n",
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"assert(s == 443839)"
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]
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}
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],
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"metadata": {
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"completion_date": "Fri, 25 Aug 2017, 12:13",
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.6.3"
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},
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"tags": [
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"digit",
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"powers",
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"brute force"
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]
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}
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65
ipython/EulerProblem031.ipynb
Executable file
65
ipython/EulerProblem031.ipynb
Executable file
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{
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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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"# Euler Problem 31\n",
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"\n",
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"In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:\n",
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"\n",
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"1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).\n",
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"\n",
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"It is possible to make £2 in the following way:\n",
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"\n",
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"1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p\n",
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"\n",
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"How many different ways can £2 be made using any number of coins?"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"collapsed": true
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},
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"outputs": [],
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"source": [
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"I really have mental issues to this in a non recursive way."
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"collapsed": true
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},
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"outputs": [],
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"source": [
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"\n",
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" "
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.6.3"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}
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