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   "source": [
    "# [Totient Permutation](https://projecteuler.net/problem=70)\n",
    "\n",
    "SageMath's implementation of $\\phi(n)$ is fast enough that you could brute force this if you wanted, but if we're clever, we can solve more quickly.\n",
    "\n",
    "We'll write a simple function for determining if two numbers are digit permutations of each other."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "id": "e6757165",
   "metadata": {},
   "outputs": [],
   "source": [
    "def is_permutation_pair(a, b):\n",
    "    s, t = str(a), str(b)\n",
    "    return sorted(s) == sorted(t)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "bdeb8c77",
   "metadata": {},
   "source": [
    "As in [problem 69](https://projecteuler.net/problem=69),\n",
    "$$\\phi(n) = n\\prod_{p | n} \\left(1 - \\frac{1}{p}\\right)$$\n",
    "\n",
    "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",
    "\n",
    "We'll initialize a list of numbers from 0 to $10^7$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "524fca40",
   "metadata": {},
   "outputs": [],
   "source": [
    "limit = 10^7\n",
    "totients = list(range(0, limit))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "88ef58dd",
   "metadata": {},
   "source": [
    "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."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "aa071cc4",
   "metadata": {},
   "outputs": [],
   "source": [
    "answer = None\n",
    "ratio = float('inf')\n",
    "\n",
    "for n in range(2, limit):\n",
    "    if totients[n] != n:\n",
    "        r = n / totients[n]\n",
    "        if r < ratio and is_permutation_pair(n, totients[n]):\n",
    "            ratio = r\n",
    "            answer = n\n",
    "\n",
    "        continue\n",
    "\n",
    "    for p in range(n, limit, n):\n",
    "        totients[p] -= totients[p] // n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "id": "d4819aa5",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "8319823"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "answer"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "40cf1e01",
   "metadata": {},
   "source": [
    "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",
    "\n",
    "## Relevant sequences\n",
    "* All numbers $n$ such that $\\phi(n)$ is a digit permutation: [A115921](https://oeis.org/A115921)\n",
    "* Subsequence of A115921 such that $n/\\phi(n)$ is a record low: [A102018](https://oeis.org/A102018)"
   ]
  }
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