114 lines
3.5 KiB
Plaintext
114 lines
3.5 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "3aeab8bf",
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"metadata": {},
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"source": [
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"# [Square Root Digital Expansion](https://projecteuler.net/problem=80)\n",
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"\n",
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"Unfortunately, just calling Python's `sqrt` won't cut it - floats don't have nearly enough digits of precision.\n",
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"\n",
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"The easiest approach here is to use SageMath's arbitrary precision routines, or Python's [decimal](https://docs.python.org/3/library/decimal.html) package. We need to watch for rounding issues, so we'll calculate slightly more than 100 digits and truncate."
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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": "b1d58be8",
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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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"40886"
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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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"def digital_sum(n):\n",
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" s = sqrt(n).n(digits=110)\n",
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" digits = (int(d) for d in str(s)[:101] if d != '.')\n",
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" return sum(digits)\n",
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" \n",
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"\n",
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"sum(digital_sum(n) for n in range(1, 101) if not is_square(n))"
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]
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},
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{
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"cell_type": "markdown",
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"id": "ad2b9aee",
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"metadata": {},
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"source": [
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"If you'd rather not use those tools for some reason, another simple approach is to take advantage of Python's arbitrary precision integers and calculate the [integer square root](https://en.wikipedia.org/wiki/Integer_square_root) of $n \\times 10^k$ for a sufficiently large $k$ ($k \\geq 200$ for this problem)."
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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": "6bd0cb0d",
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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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"40886"
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]
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},
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"execution_count": 2,
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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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"def digital_sum(n):\n",
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" s = isqrt(n * 10^200)\n",
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" digits = (int(d) for d in str(s)[:100])\n",
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" return sum(digits)\n",
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" \n",
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"\n",
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"sum(digital_sum(n) for n in range(1, 101) if not is_square(n))"
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]
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},
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{
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"cell_type": "markdown",
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"id": "c193ba51",
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"metadata": {},
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"source": [
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"If you want to go the extra mile, there are several [algorithms for computing square roots](https://en.wikipedia.org/wiki/Square_root_algorithms) that you can implement yourself, such as Heron's method, which is a special case of [Newton's method](https://en.wikipedia.org/wiki/Newton%27s_method) for solving $x^2 - n = 0$. The method works by starting with an initial estimate $x_0$ (such as $\\frac{n}{2}$), then repeatedly calculating\n",
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"$$x_{k+1} = \\frac{1}{2}\\left(x_k + \\frac{n}{x_k}\\right)$$\n",
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"until $|x_{k+1} - x_k|$ is sufficiently small. For computing the integer square root, this can be when $|x_{k+1} - x_k| < 1$.\n",
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"\n",
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"#### Copyright (C) 2025 filifa\n",
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"\n",
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"This work is licensed under the [Creative Commons Attribution-ShareAlike 4.0 International license](https://creativecommons.org/licenses/by-sa/4.0/) and the [BSD Zero Clause license](https://spdx.org/licenses/0BSD.html)."
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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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