use a generator

notebooks/problem0070.ipynb

@@ -32,51 +32,60 @@
     "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 $10^7$."
+    "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)). Iterating over values of $n$, if `totients[n] == n - 1`, then $n$ is prime, and we'll update all its multiples using the above formula."
    ]
   },
   {
    "cell_type": "code",
    "execution_count": 2,
-   "id": "524fca40",
+   "id": "54065139",
    "metadata": {},
    "outputs": [],
    "source": [
-    "limit = 10^7\n",
-    "totients = list(range(0, limit))"
+    "def totients(limit):\n",
+    "    totients = [n - 1 for n in range(0, limit)]\n",
+    "    totients[0] = 0\n",
+    "    totients[1] = 1\n",
+    "    \n",
+    "    for n in range(0, limit // 2 + 1):\n",
+    "        yield totients[n]\n",
+    "        if n == 0 or n == 1 or totients[n] != n - 1:\n",
+    "            continue\n",
+    "\n",
+    "        for k in range(2 * n, limit, n):\n",
+    "            totients[k] -= totients[k] // n\n",
+    "            \n",
+    "    yield from totients[limit // 2 + 1:]"
    ]
   },
   {
    "cell_type": "markdown",
-   "id": "88ef58dd",
+   "id": "45a80c4b",
    "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."
+    "$n-1$ can't be a permutation of $n$, so our solution won't be prime. If $n$ is composite, 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",
+   "id": "60741588",
    "metadata": {},
    "outputs": [],
    "source": [
+    "limit = 10^7\n",
+    "\n",
     "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",
+    "for (n, totient) in enumerate(totients(limit)):\n",
+    "    if n == 0 or n == 1 or totient == n - 1:\n",
     "        continue\n",
-    "\n",
-    "    for p in range(n, limit, n):\n",
-    "        totients[p] -= totients[p] // n"
+    "        \n",
+    "    r = n / totient\n",
+    "    if r < ratio and is_permutation_pair(n, totient):\n",
+    "        ratio = r\n",
+    "        answer = n"
    ]
   },
   {