Solved 44 and 45.
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dd8208ce12
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@ -24,6 +24,79 @@
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"Find the pair of pentagonal numbers, $P_j$ and $P_k$, for which their sum and difference are pentagonal and $D = |P_k − P_j|$ is minimised; what is the value of D?"
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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": 3,
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"metadata": {},
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"outputs": [],
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"source": [
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"def pentagonal(n):\n",
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" return n * (3 * n - 1) // 2\n",
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"\n",
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"assert(pentagonal(1) == 1)\n",
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"assert(pentagonal(4) == 22)"
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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": 31,
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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 = 10000\n",
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"p = [pentagonal(n) for n in range(1, n)]\n",
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"p_set = set(p)"
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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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"Okay, I was honestly just going for plain brute force here and the first solution that came up was the solution to the problem... Not really happy with that."
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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": 33,
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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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"i = 1020, j = 2167, d = 5482660.\n"
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]
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}
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],
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"source": [
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"for i in range(1, n):\n",
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" for j in range(i + 1, n):\n",
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" if p[i - 1] + p[j - 1] in p_set and p[j - 1] - p[i - 1] in p_set:\n",
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" d = pentagonal(j) - pentagonal(i)\n",
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" s = d\n",
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" print(\"i = {}, j = {}, d = {}.\".format(i, j, d))"
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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": 35,
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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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"5482660\n"
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]
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}
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],
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"source": [
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"print(s)\n",
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"assert(s == 5482660)"
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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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@ -35,7 +108,7 @@
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}
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],
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"metadata": {
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"completion_date": "",
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"completion_date": "Sat, 22 Dec 2018, 22:57",
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python3.6",
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@ -53,7 +126,11 @@
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"pygments_lexer": "ipython3",
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"version": "3.6.5"
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},
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"tags": []
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"tags": [
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"pentagonal",
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"brute force",
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"improve"
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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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@ -0,0 +1,128 @@
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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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"# Triangular, pentagonal, and hexagonal (Euler Problem 45)"
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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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"collapsed": true
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},
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"source": [
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"[https://projecteuler.net/problem=45](https://projecteuler.net/problem=45)\n",
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"\n",
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"Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:\n",
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"\n",
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"Triangle\t \t$T_n=n(n+1)/2 = 1, 3, 6, 10, 15, ...$\n",
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"\n",
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"Pentagonal\t \t$P_n=n(3n−1)/2 = 1, 5, 12, 22, 35, ...$\n",
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"\n",
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"Hexagonal\t \t$H_n=n(2n−1) = 1, 6, 15, 28, 45, ...$\n",
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"\n",
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"It can be verified that T285 = P165 = H143 = 40755.\n",
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"\n",
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"Find the next triangle number that is also pentagonal and hexagonal."
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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": 1,
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"metadata": {},
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"outputs": [],
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"source": [
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"def T(n):\n",
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" return n * (n + 1) // 2\n",
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"\n",
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"assert(T(3) == 6)\n",
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"\n",
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"def P(n):\n",
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" return n * (3 * n - 1) // 2\n",
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"\n",
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"assert(P(3) == 12)\n",
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"\n",
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"def H(n):\n",
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" return n * (2 * n - 1)\n",
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"\n",
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"assert(H(3) == 15)"
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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": 2,
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"metadata": {},
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"outputs": [],
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"source": [
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"n_max = 100000\n",
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"s = 0\n",
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"\n",
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"T_list = [T(n) for n in range(1, n_max + 1)]\n",
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"P_set = {P(n) for n in range(1, n_max + 1)}\n",
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"H_set = {H(n) for n in range(1, n_max + 1)}\n",
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"\n",
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"for n in range(1, n_max + 1):\n",
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" if T_list[n - 1] in P_set and T_list[n - 1] in H_set:\n",
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" t = T_list[n - 1]\n",
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" if t > 40755:\n",
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" s = t"
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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": 3,
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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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"1533776805\n"
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]
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}
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],
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"source": [
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"print(s)\n",
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"assert(s == 1533776805)"
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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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}
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],
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"metadata": {
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"completion_date": "Sat, 22 Dec 2018, 23:17",
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python3.6",
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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": "3.6.5"
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},
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"tags": [
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"triangular",
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"pentagona",
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"hexagonal"
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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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@ -11778,6 +11778,7 @@ div#notebook {
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</div>
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<div class="inner_cell">
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<div class="text_cell_render border-box-sizing rendered_html">
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<p><a href="https://projecteuler.net/problem=43">https://projecteuler.net/problem=43</a></p>
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<p>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.</p>
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<p>Let $d_1$ be the 1st digit, $d_2$ be the 2nd digit, and so on. In this way, we note the following:</p>
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<p>$d_2d_3d_4$=406 is divisible by 2</p>
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@ -11791,6 +11792,116 @@ div#notebook {
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</div>
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</div>
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</div>
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<div class="cell border-box-sizing text_cell rendered"><div class="prompt input_prompt">
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</div>
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<div class="inner_cell">
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<div class="text_cell_render border-box-sizing rendered_html">
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<p>We write a function that tests for the sub-string divisibility property. Then we simply brute force.</p>
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</div>
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</div>
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</div>
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<div class="cell border-box-sizing code_cell rendered">
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<div class="input">
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<div class="prompt input_prompt">In [14]:</div>
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<div class="inner_cell">
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<div class="input_area">
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<div class=" highlight hl-ipython3"><pre><span></span><span class="k">def</span> <span class="nf">is_sub_string_divisible</span><span class="p">(</span><span class="n">s</span><span class="p">):</span>
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<span class="n">s</span> <span class="o">=</span> <span class="s2">""</span><span class="o">.</span><span class="n">join</span><span class="p">(</span><span class="n">s</span><span class="p">)</span>
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<span class="k">if</span> <span class="nb">int</span><span class="p">(</span><span class="n">s</span><span class="p">[</span><span class="mi">1</span><span class="p">:</span><span class="mi">4</span><span class="p">])</span> <span class="o">%</span> <span class="mi">2</span> <span class="o">!=</span> <span class="mi">0</span><span class="p">:</span>
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<span class="k">return</span> <span class="kc">False</span>
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<span class="k">if</span> <span class="nb">int</span><span class="p">(</span><span class="n">s</span><span class="p">[</span><span class="mi">2</span><span class="p">:</span><span class="mi">5</span><span class="p">])</span> <span class="o">%</span> <span class="mi">3</span> <span class="o">!=</span> <span class="mi">0</span><span class="p">:</span>
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<span class="k">return</span> <span class="kc">False</span>
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<span class="k">if</span> <span class="nb">int</span><span class="p">(</span><span class="n">s</span><span class="p">[</span><span class="mi">3</span><span class="p">:</span><span class="mi">6</span><span class="p">])</span> <span class="o">%</span> <span class="mi">5</span> <span class="o">!=</span> <span class="mi">0</span><span class="p">:</span>
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<span class="k">return</span> <span class="kc">False</span>
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<span class="k">if</span> <span class="nb">int</span><span class="p">(</span><span class="n">s</span><span class="p">[</span><span class="mi">4</span><span class="p">:</span><span class="mi">7</span><span class="p">])</span> <span class="o">%</span> <span class="mi">7</span> <span class="o">!=</span> <span class="mi">0</span><span class="p">:</span>
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<span class="k">return</span> <span class="kc">False</span>
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<span class="k">if</span> <span class="nb">int</span><span class="p">(</span><span class="n">s</span><span class="p">[</span><span class="mi">5</span><span class="p">:</span><span class="mi">8</span><span class="p">])</span> <span class="o">%</span> <span class="mi">11</span> <span class="o">!=</span> <span class="mi">0</span><span class="p">:</span>
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<span class="k">return</span> <span class="kc">False</span>
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<span class="k">if</span> <span class="nb">int</span><span class="p">(</span><span class="n">s</span><span class="p">[</span><span class="mi">6</span><span class="p">:</span><span class="mi">9</span><span class="p">])</span> <span class="o">%</span> <span class="mi">13</span> <span class="o">!=</span> <span class="mi">0</span><span class="p">:</span>
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<span class="k">return</span> <span class="kc">False</span>
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<span class="k">if</span> <span class="nb">int</span><span class="p">(</span><span class="n">s</span><span class="p">[</span><span class="mi">7</span><span class="p">:</span><span class="mi">10</span><span class="p">])</span> <span class="o">%</span> <span class="mi">17</span> <span class="o">!=</span> <span class="mi">0</span><span class="p">:</span>
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<span class="k">return</span> <span class="kc">False</span>
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<span class="k">return</span> <span class="kc">True</span>
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<span class="k">assert</span><span class="p">(</span><span class="n">is_sub_string_divisible</span><span class="p">(</span><span class="s2">"1406357289"</span><span class="p">))</span>
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<span class="k">assert</span><span class="p">(</span><span class="n">is_sub_string_divisible</span><span class="p">(</span><span class="s2">"1406357298"</span><span class="p">)</span> <span class="o">==</span> <span class="kc">False</span><span class="p">)</span>
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</pre></div>
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</div>
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</div>
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</div>
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<div class="cell border-box-sizing code_cell rendered">
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<div class="input">
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<div class="prompt input_prompt">In [15]:</div>
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<div class="inner_cell">
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<div class="input_area">
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<div class=" highlight hl-ipython3"><pre><span></span><span class="kn">import</span> <span class="nn">itertools</span>
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<span class="n">s</span> <span class="o">=</span> <span class="nb">sum</span><span class="p">([</span><span class="nb">int</span><span class="p">(</span><span class="s2">""</span><span class="o">.</span><span class="n">join</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="n">itertools</span><span class="o">.</span><span class="n">permutations</span><span class="p">(</span><span class="s2">"0123456789"</span><span class="p">)</span> <span class="k">if</span> <span class="n">is_sub_string_divisible</span><span class="p">(</span><span class="n">p</span><span class="p">)])</span>
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</pre></div>
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</div>
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</div>
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</div>
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<div class="cell border-box-sizing code_cell rendered">
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<div class="prompt input_prompt">In [17]:</div>
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<div class="input_area">
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<div class=" highlight hl-ipython3"><pre><span></span><span class="nb">print</span><span class="p">(</span><span class="n">s</span><span class="p">)</span>
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<span class="k">assert</span><span class="p">(</span><span class="n">s</span> <span class="o">==</span> <span class="mi">16695334890</span><span class="p">)</span>
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</pre></div>
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</div>
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</div>
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<div class="prompt"></div>
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<div class="output_subarea output_stream output_stdout output_text">
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<pre>16695334890
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</pre>
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<div class="text_cell_render border-box-sizing rendered_html">
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<p>I just want to write my own permutations functions to see that I am able to do it quickly. We pick each element once and combine it with the permutations for the remaining items:</p>
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</div>
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<div class="input">
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<div class="prompt input_prompt">In [32]:</div>
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<div class="input_area">
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<div class=" highlight hl-ipython3"><pre><span></span><span class="k">def</span> <span class="nf">permutations</span><span class="p">(</span><span class="n">items</span><span class="p">):</span>
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<span class="n">items</span> <span class="o">=</span> <span class="nb">list</span><span class="p">(</span><span class="n">items</span><span class="p">)</span>
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<span class="k">if</span> <span class="ow">not</span> <span class="n">items</span><span class="p">:</span>
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<span class="k">return</span> <span class="p">[[]]</span>
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<span class="n">perms</span> <span class="o">=</span> <span class="p">[]</span>
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<span class="k">for</span> <span class="n">item</span> <span class="ow">in</span> <span class="n">items</span><span class="p">:</span>
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<span class="n">items_new</span> <span class="o">=</span> <span class="nb">list</span><span class="p">(</span><span class="n">items</span><span class="p">)</span>
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<span class="n">items_new</span><span class="o">.</span><span class="n">remove</span><span class="p">(</span><span class="n">item</span><span class="p">)</span>
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<span class="n">permutations_new</span> <span class="o">=</span> <span class="n">permutations</span><span class="p">(</span><span class="n">items_new</span><span class="p">)</span>
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<span class="k">for</span> <span class="n">perm</span> <span class="ow">in</span> <span class="n">permutations_new</span><span class="p">:</span>
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<span class="n">perms</span><span class="o">.</span><span class="n">append</span><span class="p">([</span><span class="n">item</span><span class="p">]</span> <span class="o">+</span> <span class="n">perm</span><span class="p">)</span>
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<span class="k">return</span> <span class="n">perms</span>
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<span class="k">assert</span><span class="p">(</span><span class="n">permutations</span><span class="p">([])</span> <span class="o">==</span> <span class="p">[[]])</span>
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<span class="k">assert</span><span class="p">(</span><span class="n">permutations</span><span class="p">([</span><span class="mi">1</span><span class="p">])</span> <span class="o">==</span> <span class="p">[[</span><span class="mi">1</span><span class="p">]])</span>
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<span class="k">assert</span><span class="p">(</span><span class="n">permutations</span><span class="p">([</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">])</span> <span class="o">==</span> <span class="p">[[</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">],</span> <span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">]])</span>
|
||||
<span class="k">assert</span><span class="p">(</span><span class="n">permutations</span><span class="p">([</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">])</span> <span class="o">==</span> <span class="p">[[</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">],</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">3</span><span class="p">,</span><span class="mi">2</span><span class="p">],</span> <span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">3</span><span class="p">],</span> <span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span> <span class="p">[</span><span class="mi">3</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">],</span> <span class="p">[</span><span class="mi">3</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">]])</span>
|
||||
</pre></div>
|
||||
</div>
|
||||
</div>
|
||||
</div>
|
||||
</div>
|
||||
<div class="cell border-box-sizing code_cell rendered">
|
||||
<div class="input">
|
||||
<div class="prompt input_prompt">In [ ]:</div>
|
||||
|
|
|
|||
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Load Diff
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Load Diff
|
|
@ -677,16 +677,50 @@
|
|||
</tr>
|
||||
|
||||
|
||||
<tr class="table-warning">
|
||||
<tr>
|
||||
|
||||
<td><a href="EulerProblem043.html">Problem 043</a></td>
|
||||
<td></td>
|
||||
<td>Fri, 21 Dec 2018, 16:28</td>
|
||||
<td>
|
||||
|
||||
<kbd>pandigital</kbd>
|
||||
|
||||
<kbd>divisibility</kbd>
|
||||
|
||||
<kbd>permutations</kbd>
|
||||
|
||||
</td>
|
||||
</tr>
|
||||
|
||||
|
||||
<tr>
|
||||
|
||||
<td><a href="EulerProblem044.html">Problem 044</a></td>
|
||||
<td>Sat, 22 Dec 2018, 22:57</td>
|
||||
<td>
|
||||
|
||||
<kbd>pentagonal</kbd>
|
||||
|
||||
<kbd>brute force</kbd>
|
||||
|
||||
<kbd>improve</kbd>
|
||||
|
||||
</td>
|
||||
</tr>
|
||||
|
||||
|
||||
<tr>
|
||||
|
||||
<td><a href="EulerProblem045.html">Problem 045</a></td>
|
||||
<td>Sat, 22 Dec 2018, 23:17</td>
|
||||
<td>
|
||||
|
||||
<kbd>triangular</kbd>
|
||||
|
||||
<kbd>pentagona</kbd>
|
||||
|
||||
<kbd>hexagonal</kbd>
|
||||
|
||||
</td>
|
||||
</tr>
|
||||
|
||||
|
|
|
|||
Loading…
Reference in New Issue