277 lines
7.7 KiB
Plaintext
277 lines
7.7 KiB
Plaintext
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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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"# Diophantine equation (Euler Problem 66)"
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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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"Consider quadratic Diophantine equations of the form:\n",
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"\n",
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"$x^2 – Dy^2 = 1$\n",
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"\n",
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"For example, when D=13, the minimal solution in x is 6492 – 13×1802 = 1.\n",
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"\n",
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"It can be assumed that there are no solutions in positive integers when D is square.\n",
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"\n",
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"By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the following:\n",
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"\n",
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"$3^2 – 2×2^2 = 1$\n",
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"\n",
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"$2^2 – 3×1^2 = 1$\n",
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"\n",
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"$9^2 – 5×4^2 = 1$\n",
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"\n",
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"$5^2 – 6×2^2 = 1$\n",
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"\n",
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"$8^2 – 7×3^2 = 1$\n",
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"\n",
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"Hence, by considering minimal solutions in x for D ≤ 7, the largest x is obtained when D=5.\n",
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"\n",
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"Find the value of D ≤ 1000 in minimal solutions of x for which the largest value of x is obtained.\n"
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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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"collapsed": true
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},
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"outputs": [],
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"source": [
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"import math\n",
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"\n",
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"def get_floor_sqrt(n):\n",
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" return math.floor(math.sqrt(n))\n",
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"\n",
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"assert(get_floor_sqrt(5) == 2)\n",
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"assert(get_floor_sqrt(23) == 4)\n",
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"\n",
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"def next_expansion_1(current_a, current_nominator, current_denominator, original_number):\n",
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" cn = current_nominator\n",
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" cd = current_denominator\n",
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" cn, cd = cd, (original_number - cd * cd) // cn\n",
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" next_a = math.floor((math.sqrt(original_number) + cn) // cd)\n",
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" cn = cn - next_a * cd\n",
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" cn *= -1\n",
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" \n",
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" return next_a, cn, cd\n",
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"\n",
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"def get_continued_fraction_sequence(n):\n",
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" \n",
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" # If number is a square number we return it.\n",
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" floor_sqrt = get_floor_sqrt(n)\n",
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" if n == floor_sqrt * floor_sqrt:\n",
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" return ((floor_sqrt), [])\n",
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" \n",
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" # Otherwise, we calculate the next expansion till we\n",
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" # encounter a step a second time. \n",
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" a = floor_sqrt\n",
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" cn = a\n",
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" cd = 1\n",
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" sequence = []\n",
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" previous_steps = []\n",
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" \n",
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" while not (a, cn, cd) in previous_steps :\n",
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" #print(\"a: {} cn: {} cd: {}\".format(a, cn, cd))\n",
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" previous_steps.append((a, cn, cd))\n",
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" a, cn, cd = next_expansion_1(a, cd, cn, n)\n",
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" sequence.append(a)\n",
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" sequence.pop()\n",
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" return ((floor_sqrt), sequence)\n",
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"\n",
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"def get_continued_fraction_sequence(n):\n",
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" \n",
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" # If number is a square number we return it.\n",
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" floor_sqrt = get_floor_sqrt(n)\n",
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" if n == floor_sqrt * floor_sqrt:\n",
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" return ((floor_sqrt), [])\n",
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" \n",
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" # Otherwise, we calculate the next expansion till we\n",
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" # encounter a step a second time. \n",
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" a = floor_sqrt\n",
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" cn = a\n",
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" cd = 1\n",
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" sequence = []\n",
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" previous_steps = []\n",
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" \n",
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" while not (a, cn, cd) in previous_steps :\n",
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" #print(\"a: {} cn: {} cd: {}\".format(a, cn, cd))\n",
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" previous_steps.append((a, cn, cd))\n",
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" a, cn, cd = next_expansion_1(a, cd, cn, n)\n",
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" sequence.append(a)\n",
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" sequence.pop()\n",
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" return ((floor_sqrt), sequence)\n",
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"\n",
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"assert(get_continued_fraction_sequence(1) == ((1), []))\n",
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"assert(get_continued_fraction_sequence(4) == ((2), []))\n",
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"assert(get_continued_fraction_sequence(25) == ((5), []))\n",
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"assert(get_continued_fraction_sequence(2) == ((1), [2]))\n",
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"assert(get_continued_fraction_sequence(3) == ((1), [1,2]))\n",
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"assert(get_continued_fraction_sequence(5) == ((2), [4]))\n",
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"assert(get_continued_fraction_sequence(13) == ((3), [1,1,1,1,6]))"
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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": 43,
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"metadata": {},
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"outputs": [],
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"source": [
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"def gcd(a, b):\n",
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" if b > a:\n",
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" a, b = b, a\n",
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" while a % b != 0:\n",
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" a, b = b, a % b\n",
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" return b\n",
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"\n",
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"def add_fractions(n1, d1, n2, d2):\n",
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" d = d1 * d2\n",
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" n1 = n1 * (d // d1)\n",
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" n2 = n2 * (d // d2)\n",
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" n = n1 + n2\n",
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" p = gcd(n, d)\n",
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" return (n // p, d // p)\n",
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"\n",
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"def next_expansion_2(previous_numerator, previous_denumerator, value):\n",
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" if previous_numerator == 0:\n",
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" return (value, 1)\n",
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" return add_fractions(previous_denumerator, previous_numerator, value, 1)\n",
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" \n",
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"def get_fractions(n, x):\n",
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" #print(\"get_fractions(n={}, x={})\".format(n, x))\n",
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" # Get sequence of a_x\n",
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" first_value, sequence = get_continued_fraction_sequence(n)\n",
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" sequence = [first_value] + math.ceil(x / len(sequence)) * sequence\n",
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" sequence = sequence[:x + 1]\n",
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" sequence = sequence[::-1]\n",
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" \n",
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" n, d = 0, 1\n",
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" for s in sequence:\n",
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" n, d = next_expansion_2(n, d, s)\n",
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" return (n, d)\n",
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"\n",
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"assert(get_fractions(7, 0) == (2, 1))\n",
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"assert(get_fractions(7, 1) == (3, 1))\n",
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"assert(get_fractions(7, 2) == (5, 2))\n",
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"assert(get_fractions(7, 3) == (8, 3))"
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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": 48,
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"metadata": {},
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"outputs": [],
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"source": [
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"def get_minimal_solution(d):\n",
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" for i in range(0, 100):\n",
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" x, y = get_fractions(d, i)\n",
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" if x * x - d * y * y == 1:\n",
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" return((x, y))\n",
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"\n",
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"assert(get_minimal_solution(2) == (3, 2))\n",
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"assert(get_minimal_solution(3) == (2, 1))\n",
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"assert(get_minimal_solution(85) == (285769, 30996))\n",
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"assert(get_minimal_solution(109) == (158070671986249, 15140424455100))"
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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": 52,
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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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"True\n"
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]
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}
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],
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"source": [
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"from math import sqrt\n",
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"\n",
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"def is_square(n):\n",
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" return sqrt(n).is_integer()\n",
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"\n",
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"print(is_square(4))"
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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": 55,
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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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"d: 661 x: 16421658242965910275055840472270471049\n"
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]
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}
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],
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"source": [
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"x_max = 0\n",
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"d_max = 0\n",
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"\n",
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"for d in range(2, 1001):\n",
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" if is_square(d):\n",
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" continue\n",
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" \n",
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" x, y = get_minimal_solution(d)\n",
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" if x > x_max:\n",
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" x_max = x\n",
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" d_max = d\n",
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"\n",
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"print(\"d: {} x: {}\".format(d_max, x_max))\n",
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"s = d_max\n",
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"assert(s == 661)"
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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": "Tue, 22 Jan 2019, 05:10",
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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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"fractions",
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"square roots",
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"sequence",
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"periodic"
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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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