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filifa 2025-07-23 22:30:40 -04:00
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notebooks/problem0072.ipynb
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@ -50,7 +50,7 @@
"id": "8e268820",
"metadata": {},
"source": [
"If you try to implement the totient function yourself, you might find it difficult to make this approach fast enough. An alternative is to sieve the values of totient - see [problem 70](https://projecteuler.net/problem=70). However, there's a few more *very* interesting methods to compute totient sums.\n",
"If you try to implement the totient function yourself, you might find it difficult to make this approach fast enough. An alternative is to \"sieve\" the values of totient - see [problem 70](https://projecteuler.net/problem=70). However, there's a few more *very* interesting methods to compute totient sums.\n",
"\n",
"## Recursive totient sum\n",
"\n",
@ -159,7 +159,7 @@
"\n",
"$$M(n) = 1 + \\lfloor b\\rfloor M(\\lfloor a\\rfloor) - \\sum_{x=1}^a \\mu(x)\\left\\lfloor \\frac{n}{x}\\right\\rfloor - \\sum_{y=2}^b M\\left(\\left\\lfloor \\frac{n}{y}\\right\\rfloor\\right)$$\n",
"\n",
"Now we have a recursive implementation of $M(n)$. All that's left is to calculate the values of $\\mu(n)$ that we need. We can do this with a sieve, since $\\mu$ is a [multiplicative function](https://en.wikipedia.org/wiki/Multiplicative_function)."
"Now we have a recursive implementation of $M(n)$. All that's left is to calculate the values of $\\mu(n)$ that we need. By taking advantage of $\\mu$ being [multiplicative](https://en.wikipedia.org/wiki/Multiplicative_function), we can compute values using a similar strategy to the sieve of Eratosthenes - see [problem 10](https://projecteuler.net/problem=10)."
]
},
{
@ -169,9 +169,7 @@
"metadata": {},
"outputs": [],
"source": [
"from math import isqrt\n",
"\n",
"def mobius_sieve(limit):\n",
"def mobius_range(limit):\n",
" ms = [n for n in range(0, limit)]\n",
"\n",
" for n in range(0, limit):\n",
@ -193,7 +191,7 @@
"id": "f12a8b30",
"metadata": {},
"source": [
"We'll use $a = \\sqrt{1000000} = 1000$ as our upper bound on the sieve."
"We'll use $a = \\sqrt{1000000} = 1000$ as our upper bound."
]
},
{
@ -203,7 +201,8 @@
"metadata": {},
"outputs": [],
"source": [
"mu = list(mobius_sieve(isqrt(1000000) + 1))"
"from math import isqrt\n",
"mu = list(mobius_range(isqrt(1000000) + 1))"
]
},
{