sieve instead
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filifa
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@ -32,39 +32,57 @@
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"As in [problem 69](https://projecteuler.net/problem=69),\n",
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"$$\\phi(n) = n\\prod_{p | n} \\left(1 - \\frac{1}{p}\\right)$$\n",
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"\n",
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"Rather than calculate the totients of every single number up to $10^7$, we'll start with just the primes - for a prime $p$, $\\phi(p) = p-1$. We'll store these primes in a [min-heap](https://en.wikipedia.org/wiki/Heap_(data_structure)) ordered by $\\frac{p}{\\phi(p)}$."
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"We can calculate the totients of the numbers up to $10^7$ using a very similar approach to the [sieve of Eratosthenes](https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes) for generating prime numbers (see [problem 10](https://projecteuler.net/problem=10)).\n",
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"\n",
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"We'll initialize a list of numbers from 0 to $10^7$."
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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": "244e05bd",
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"id": "524fca40",
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"metadata": {},
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"outputs": [],
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"source": [
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"import heapq\n",
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"\n",
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"limit = 10^7\n",
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"\n",
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"primes = prime_range(limit)\n",
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"queue = [(p / (p - 1), (p, p - 1)) for p in primes]\n",
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"heapq.heapify(queue)"
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"totients = list(range(0, limit))"
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]
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},
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{
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"cell_type": "markdown",
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"id": "9a057da0",
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"id": "88ef58dd",
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"metadata": {},
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"source": [
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"Then we'll search to find the value $n$ with the smallest ratio that is also a digit permutation of its totient. We'll check composite values by pushing $np$ to the queue for each prime $p$.\n",
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"\n",
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"When we find a value that is a digit permutation of its totient, we'll know that it also has the smallest ratio and can stop."
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"Iterating $n$ from 2 to $10^7$, if `totients[n] == n`, then $n$ is prime, and we'll update its totient and all its multiples using the above formula. If `totients[n] != n`, then we'll check if $n/\\phi(n)$ is small and if $\\phi(n)$ is a permutation of $n$, keeping track of the best answer so far."
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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": "99c4267a",
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"id": "aa071cc4",
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"metadata": {},
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"outputs": [],
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"source": [
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"answer = None\n",
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"ratio = float('inf')\n",
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"\n",
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"for n in range(2, limit):\n",
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" if totients[n] != n:\n",
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" r = n / totients[n]\n",
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" if r < ratio and is_permutation_pair(n, totients[n]):\n",
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" ratio = r\n",
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" answer = n\n",
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"\n",
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" continue\n",
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"\n",
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" for p in range(n, limit, n):\n",
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" totients[p] -= totients[p] // n"
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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": "d4819aa5",
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"metadata": {},
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"outputs": [
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{
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@ -73,38 +91,12 @@
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"8319823"
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]
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},
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"execution_count": 3,
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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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"answer = None\n",
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"visited = set()\n",
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"while queue != []:\n",
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" _, (n, totient) = heapq.heappop(queue)\n",
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" \n",
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" if n in visited:\n",
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" continue\n",
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" visited.add(n)\n",
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" \n",
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" if is_permutation_pair(n, totient):\n",
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" answer = n\n",
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" break\n",
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" \n",
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" for p in primes:\n",
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" q = n * p\n",
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" if q >= limit:\n",
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" break\n",
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" \n",
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" if n % p != 0:\n",
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" new_totient = totient * (p - 1)\n",
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" else:\n",
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" new_totient = totient * p\n",
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" \n",
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" ratio = q / new_totient\n",
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" heapq.heappush(queue, (ratio, (q, new_totient)))\n",
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" \n",
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"answer"
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]
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},
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@ -113,7 +105,11 @@
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"id": "40cf1e01",
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"metadata": {},
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"source": [
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"Note: lots of people in the problem thread make the assumption that the answer must be a [semiprime](https://en.wikipedia.org/wiki/Semiprime). However, Steendor points out that for certain upper bounds, this assumption does not hold. This solution avoids making that assumption."
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"Note: lots of people in the problem thread make the assumption that the answer must be a [semiprime](https://en.wikipedia.org/wiki/Semiprime). However, Steendor points out that for certain upper bounds, this assumption does not hold. For instance, $2817 = 3^2 \\times 313$ and $\\phi(2817) = 1872$, and 2817/1872 is the lowest ratio until 2991. 2817 may be an exception rather than the norm (all the other numbers in A102018 up to $10^8$ are semiprimes); nevertheless, this solution avoids making the assumption.\n",
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"\n",
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"## Relevant sequences\n",
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"* All numbers $n$ such that $\\phi(n)$ is a digit permutation: [A115921](https://oeis.org/A115921)\n",
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"* Subsequence of A115921 such that $n/\\phi(n)$ is a record low: [A102018](https://oeis.org/A102018)"
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]
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}
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],
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