add problem 50
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filifa
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{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "1c5feb53",
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"metadata": {},
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"source": [
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"# [Consecutive Prime Sum](https://projecteuler.net/problem=50)\n",
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"\n",
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"Another way to phrase this problem is that we're looking for the longest subsequence of primes $p_i, p_{i+1}, p_{i+2}, \\ldots, p_{j-1}$ that sum to a prime below 1000000.\n",
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"\n",
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"If we let $p_0 = 2$, then $p_{41537} = 499979$, $p_{41538} = 500009$, and $p_{41539} = 500029$. Since $p_{41538} + p_{41539} > 1000000$, there's no point in checking any sum that's composed with a prime greater than 500010."
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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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"id": "8a9f568e",
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"metadata": {},
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"outputs": [],
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"source": [
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"primes = prime_range(500010)"
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]
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},
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{
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"cell_type": "markdown",
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"id": "c98ea076",
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"metadata": {},
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"source": [
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"Now here's an algorithm for finding the longest consecutive prime sum. The clever bit about this algorithm is that it avoids repeatedly recalculating large sums - instead, we slide through the list of primes, growing and shrinking a running total. (As much as I'd like to take credit for this algorithm, the steps were outlined by the user tzaman on the problem thread.)"
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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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"id": "c7014e59",
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"metadata": {},
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"outputs": [],
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"source": [
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"def longest_consecutive_prime_sum(primes):\n",
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" total = 0\n",
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" end = 0\n",
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" best = (0, 0)\n",
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" for begin in range(0, len(primes)):\n",
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" # grow the sum until it's larger than 1000000\n",
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" for maximum in range(end, len(primes)):\n",
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" total += primes[maximum]\n",
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" if total >= 1000000:\n",
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" break\n",
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"\n",
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" # shrink the sum until it's prime (or shorter than our best sum so far)\n",
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" for end in reversed(range(begin, maximum + 1)):\n",
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" total -= primes[end]\n",
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" \n",
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" if end - begin <= best[1] - best[0]:\n",
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" break\n",
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"\n",
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" if is_prime(total):\n",
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" best = (begin, end)\n",
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" break\n",
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"\n",
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" # try again, starting one prime after our current start point\n",
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" total -= primes[begin]\n",
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" \n",
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" return best"
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]
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},
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{
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"cell_type": "markdown",
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"id": "f274e457",
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"metadata": {},
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"source": [
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"The algorithm returns the indices of the slice that's the longest sum."
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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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"id": "472dfc59",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"(3, 546)"
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]
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},
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"execution_count": 3,
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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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"i, j = longest_consecutive_prime_sum(primes)\n",
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"i, j"
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]
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},
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{
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"cell_type": "markdown",
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"id": "c495528f",
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"metadata": {},
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"source": [
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"The longest prime sum starts at $p_3$ and ends with $p_{545}$."
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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": 4,
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"id": "325f7bb3",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"997651"
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]
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},
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"execution_count": 4,
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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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"sum(primes[i:j])"
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]
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},
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{
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"cell_type": "markdown",
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"id": "16a496a9",
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"metadata": {},
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"source": [
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"## Relevant sequences\n",
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"* Primes expressible as the sum of (at least two) consecutive primes in at least 1 way: [A067377](https://oeis.org/A067377)"
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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": "SageMath 9.5",
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"language": "sage",
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"name": "sagemath"
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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": "3.11.2"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 5
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}
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