83 lines
2.9 KiB
Plaintext
83 lines
2.9 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "59fa6654",
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"metadata": {},
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"source": [
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"# [Combinatoric Selections](https://projecteuler.net/problem=53)\n",
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"\n",
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"Python/SageMath's [unlimited precision integers](https://docs.python.org/3/library/stdtypes.html#numeric-types-int-float-complex) make this trivial."
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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": "abf93cda",
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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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"4075"
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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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"binoms = (binomial(n, r) for n in range(1, 101) for r in range(0, n + 1))\n",
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"len([b for b in binoms if b > 1000000])"
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]
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},
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{
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"cell_type": "markdown",
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"id": "6d7f4409",
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"metadata": {},
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"source": [
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"If it weren't for unlimited precision, we would have a problem, since a lot of these [binomial coefficients](https://en.wikipedia.org/wiki/Binomial_coefficient) overflow a 64 bit integer - the largest one we encounter here is $\\binom{100}{50} = 100891344545564193334812497256$.\n",
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"\n",
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"However, even in that case there's a pretty simple workaround: finding $n$ and $r$ such that\n",
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"$$\\binom{n}{r} > 1000000$$\n",
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"is the same as finding $n$ and $r$ such that\n",
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"$$\\log{\\binom{n}{r}} > \\log{1000000}$$\n",
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"By applying the definition of the binomial coefficients and [logarithmic identities](https://en.wikipedia.org/wiki/List_of_logarithmic_identities), this turns to\n",
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"$$\\log{n!} - \\log{r!} - \\log{(n-r)!} > \\log{1000000}$$\n",
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"$\\log{100!} \\approx 363.739$, nowhere close to overflowing a float.\n",
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"\n",
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"$\\log{n!}$ can be computed in a number of ways. You can use the [log-gamma function](https://en.wikipedia.org/wiki/Gamma_function), or you can implement a function yourself, either by applying logarithmic identities again:\n",
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"$$\\log{n!} = \\log{1} + \\log{2} + \\log{3} \\cdots + \\log{n}$$\n",
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"or by using [Stirling's approximation](https://en.wikipedia.org/wiki/Stirling%27s_approximation):\n",
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"$$\\log{n!} \\approx \\left(n + \\frac{1}{2}\\right)\\log{n} - n + \\frac{1}{2}\\log{2\\pi}$$\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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"nbformat": 4,
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"nbformat_minor": 5
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