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    "# [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",
    "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 $\\frac{p}{\\phi(p)}$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "244e05bd",
   "metadata": {},
   "outputs": [],
   "source": [
    "import heapq\n",
    "\n",
    "limit = 10^7\n",
    "\n",
    "primes = prime_range(limit)\n",
    "queue = [(p / (p - 1), (p, p - 1)) for p in primes]\n",
    "heapq.heapify(queue)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "9a057da0",
   "metadata": {},
   "source": [
    "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",
    "\n",
    "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."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "99c4267a",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "8319823"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
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   "source": [
    "answer = None\n",
    "visited = set()\n",
    "while queue != []:\n",
    "    _, (n, totient) = heapq.heappop(queue)\n",
    "        \n",
    "    if n in visited:\n",
    "        continue\n",
    "    visited.add(n)\n",
    "        \n",
    "    if is_permutation_pair(n, totient):\n",
    "        answer = n\n",
    "        break\n",
    "        \n",
    "    for p in primes:\n",
    "        q = n * p\n",
    "        if q >= limit:\n",
    "            break\n",
    "            \n",
    "        if n % p != 0:\n",
    "            new_totient = totient * (p - 1)\n",
    "        else:\n",
    "            new_totient = totient * p\n",
    "        \n",
    "        ratio = q / new_totient\n",
    "        heapq.heappush(queue, (ratio, (q, new_totient)))\n",
    "        \n",
    "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. This solution avoids making that assumption."
   ]
  }
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