130 lines
5.2 KiB
Plaintext
130 lines
5.2 KiB
Plaintext
{
|
|
"cells": [
|
|
{
|
|
"cell_type": "markdown",
|
|
"id": "b7b7784e",
|
|
"metadata": {},
|
|
"source": [
|
|
"# [Reciprocal Cycles](https://projecteuler.net/problem=26)\n",
|
|
"\n",
|
|
"Another easy problem with SageMath."
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": 1,
|
|
"id": "20a3987a",
|
|
"metadata": {},
|
|
"outputs": [
|
|
{
|
|
"data": {
|
|
"text/plain": [
|
|
"983"
|
|
]
|
|
},
|
|
"execution_count": 1,
|
|
"metadata": {},
|
|
"output_type": "execute_result"
|
|
}
|
|
],
|
|
"source": [
|
|
"max(range(2, 1000), key=lambda d: (1/d).period())"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"id": "6dc68724",
|
|
"metadata": {},
|
|
"source": [
|
|
"But let's talk about how we would write our own algorithm for calculating the period.\n",
|
|
"\n",
|
|
"If $d = 2^a 5^b n$ where $n$ is coprime to 2 and 5, then the [period](https://mathworld.wolfram.com/DecimalPeriod.html) of $\\frac{1}{d}$ is the [multiplicative order](https://en.wikipedia.org/wiki/Multiplicative_order) of 10 modulo $n$. This is the same as finding the smallest positive $k$ such that\n",
|
|
"$$10^k \\equiv 1 \\pmod{n}$$\n",
|
|
"(Wondering why this is called multiplicative order? It has to do with the mathematical concept of [groups](https://en.wikipedia.org/wiki/Group_(mathematics)), but you don't need to be familiar with them to apply the formula.)\n",
|
|
"\n",
|
|
"Based on the definition, we can easily write a function that computes multiplicative order that is efficient enough for this problem, but it's not very efficient in general. The computation is a special case of the [discrete logarithm](https://en.wikipedia.org/wiki/Discrete_logarithm), which has no known efficient algorithm in general."
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": 2,
|
|
"id": "d967be38",
|
|
"metadata": {},
|
|
"outputs": [],
|
|
"source": [
|
|
"def multiplicative_order(a, n):\n",
|
|
" if n == 1:\n",
|
|
" return 1\n",
|
|
" \n",
|
|
" g = 1\n",
|
|
" for k in range(1, n):\n",
|
|
" g *= a\n",
|
|
" g %= n\n",
|
|
" if g == 1:\n",
|
|
" return k"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": 3,
|
|
"id": "dc51eb0e",
|
|
"metadata": {},
|
|
"outputs": [],
|
|
"source": [
|
|
"def period(d):\n",
|
|
" while d % 2 == 0:\n",
|
|
" d //= 2\n",
|
|
" while d % 5 == 0:\n",
|
|
" d //= 5\n",
|
|
" \n",
|
|
" return multiplicative_order(10, d)"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "markdown",
|
|
"id": "14f53b42",
|
|
"metadata": {},
|
|
"source": [
|
|
"Personally, I don't feel it's immediately obvious *why* multiplicative order can be used to calculate these periods. Here's a deeper dive into why this works, if you're curious.\n",
|
|
"\n",
|
|
"## Explanation\n",
|
|
"As above, let $d = 2^a 5^b n$, where $n$ is coprime to 2 and 5, and consider the unit fraction $u=\\frac{1}{d}$. The decimal representation of this fraction (as with any fraction) may have a [fractional part](https://en.wikipedia.org/wiki/Fractional_part) with $q$ leading digits, followed by a repetend of $r$ digits.\n",
|
|
"\n",
|
|
"Let's make note of a trick you probably already know: multiplying a number by 10 gives the same result as if you just moved the original number's decimal point to the right one digit (e.g. $1.25 \\times 10 = 12.5$). This means that $10^q u$ has a decimal representation with only the repetend in its fractional part. If you were to then multiply it by $10^r$, the integer part would increase, but the fractional part would stay the same since it repeats every $r$ digits.\n",
|
|
"\n",
|
|
"The above is important for understanding the following: $10^q 10^r u - 10^q u$ *is an integer*. How do we know? Because by the logic above, the fractional part of both $10^q 10^r u$ and $10^q u$ only have the repetend, so they cancel out when we subtract, leaving us with an integer. It immediately follows that $d(10^q 10^r u - 10^q u) = 10^q 10^r - 10^q$ is also an integer that has $d$ as a factor. We can express this as\n",
|
|
"$$10^{q+r} \\equiv 10^q \\pmod{d}$$\n",
|
|
"Through the properties of [modular arithmetic](https://en.wikipedia.org/wiki/Modular_arithmetic), we can cancel common factors of 2 and 5 in $10^{q+r}$, $10^q$, and $d$ until we end up with\n",
|
|
"$$10^r \\equiv 1 \\pmod{n}$$\n",
|
|
"By definition, $r$ is the multiplicative order of 10 modulo $n$.\n",
|
|
"\n",
|
|
"As a concrete example of the above, consider $d = 2^4 \\times 5 \\times 63 = 5040$. The decimal representation of $u = \\frac{1}{d}$ is $0.0001(984126)$, where 984126 is the repetend. Therefore $q=4$ and $r=6$. Sure enough, $10^q 10^r u - 10^q u = 1984125$ is an integer; therefore, $10^{10} \\equiv 10^4 \\pmod{5040}$, and $10^6 \\equiv 1 \\pmod{63}$.\n",
|
|
"\n",
|
|
"## Relevant sequences\n",
|
|
"* Periods of reciprocals: [A007732](https://oeis.org/A007732)"
|
|
]
|
|
}
|
|
],
|
|
"metadata": {
|
|
"kernelspec": {
|
|
"display_name": "SageMath 9.5",
|
|
"language": "sage",
|
|
"name": "sagemath"
|
|
},
|
|
"language_info": {
|
|
"codemirror_mode": {
|
|
"name": "ipython",
|
|
"version": 3
|
|
},
|
|
"file_extension": ".py",
|
|
"mimetype": "text/x-python",
|
|
"name": "python",
|
|
"nbconvert_exporter": "python",
|
|
"pygments_lexer": "ipython3",
|
|
"version": "3.11.2"
|
|
}
|
|
},
|
|
"nbformat": 4,
|
|
"nbformat_minor": 5
|
|
}
|