143 lines
4.4 KiB
Plaintext
143 lines
4.4 KiB
Plaintext
{
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"cells": [
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"# Euler Problem 27\n",
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"\n",
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"Euler discovered the remarkable quadratic formula:\n",
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"\n",
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"$n^2 + n + 41$\n",
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"\n",
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"It turns out that the formula will produce 40 primes for the consecutive integer values 0≤n≤39. However, when $n=40$, $40^2 + 40 + 41 = 40(40+1)+41$ is divisible by $41$, and certainly when $n=41,41^2+41+41$ is clearly divisible by 41.\n",
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"\n",
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"The incredible formula $n^2−79n+1601$ was discovered, which produces 80 primes for the consecutive values 0≤n≤79. The product of the coefficients, $−79$ and $1601$, is $−126479$.\n",
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"\n",
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"Considering quadratics of the form:\n",
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"\n",
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"$n^2 + an +b$, where |a|<1000 and |b|≤1000\n",
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"\n",
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"where |n| is the modulus/absolute value of n e.g. |11|=11 and |−4|=4.\n",
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"\n",
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"Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n=0."
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]
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},
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"cell_type": "markdown",
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"Okay, bruteforcing this complete thing is definitely hard. The interesting thing is that euler provided two examples. If we calculate the primes for both terms we see that there is a certain overlap. This indicates that there is a relation between the two. Sure enough if we put $n-40$ into the first term we get the following.\n",
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"\n",
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"$(n - 40)^2 + (n - 40) + 41 = n^2 - 80n + 1600 + n - 40 + 41 = n^2 - 79n +1601$\n",
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"\n",
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"Let's assume that all incredible formulas can be derived by inserting $(n - p)$ into the formula. Of course, what ever value we choose for p the resulting terms must not exceed the boundaries for a or b. We calculate the boundaries.\n",
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"\n",
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"$(n-p)^2 + (n-p) + 41 = n^2 + n(-2p + 1) + p^2 - p + 41$\n",
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"\n",
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"Where $(-2p + 1) = a$ and $p^2 -p + 41 = b$. We can now calulate the bounds for a:\n",
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"\n",
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"$-2p + 1 > -1000 \\rightarrow p < 500.5$\n",
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"\n",
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"$-2 p + 1 < 1000 \\rightarrow p > -499.5$\n",
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"\n",
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"And b:\n",
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"\n",
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"$p^2 - p + 41 <= 1000$\n",
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"\n",
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"$-30.472 < p < 31.472$\n",
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"\n",
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"$p^2 - p + 41 >= 1000$ True for $\\forall p \\in \\mathbb{N}$\n",
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"\n",
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"So now we only have to check for the values p in range(-30, 32). Alternatively, for the example $p = 40$ was used, maybe the next smaller value $p = 31$ yields the correct solution:\n",
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"\n",
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"$s = a\\times b = (-2 * 31 + 1) * (31^2 - 31 + 41) = -61 \\times 971 = -59231$"
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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": false
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},
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"outputs": [
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"data": {
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"text/plain": [
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"-59231"
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]
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},
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"execution_count": 1,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"from functools import lru_cache\n",
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"\n",
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"@lru_cache(maxsize=1000)\n",
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"def is_prime(n):\n",
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" if n < 2:\n",
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" return False\n",
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" from math import sqrt\n",
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" for s in range(2, int(sqrt(n) + 1)):\n",
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" if n % s == 0:\n",
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" return False\n",
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" if s * s > n:\n",
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" return True\n",
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" return True\n",
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" \n",
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"assert(is_prime(41) == True)\n",
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"\n",
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"def get_quadratic(n, p):\n",
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" return n*n + n * (-2* p + 1) + p*p - p + 41\n",
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"\n",
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"n_max = 0\n",
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"p_max = 0\n",
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"\n",
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"for p in range(-30, 32):\n",
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" for n in range(0, 10000):\n",
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" if not is_prime(get_quadratic(n, p)) and n > n_max:\n",
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" n_max = n\n",
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" p_max = p\n",
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" break\n",
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"\n",
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"p = p_max\n",
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"s = (-2 * p + 1) * (p*p - p + 41)\n",
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"assert(s == -59231)\n",
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"s"
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]
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}
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],
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"metadata": {
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"completion_date": "Mon, 21 Aug 2017, 21:11",
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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": "3.5.4"
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},
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"tags": [
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"quadratic",
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"primes",
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"formula",
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"brain"
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