Solved 42 and 43.

ipython/EulerProblem043.ipynb

@@ -13,6 +13,8 @@
     "collapsed": true
    },
    "source": [
+    "[https://projecteuler.net/problem=43](https://projecteuler.net/problem=43)\n",
+    "\n",
     "The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property.\n",
     "\n",
     "Let $d_1$ be the 1st digit, $d_2$ be the 2nd digit, and so on. In this way, we note the following:\n",
@@ -34,6 +36,103 @@
     "Find the sum of all 0 to 9 pandigital numbers with this property."
    ]
   },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "We write a function that tests for the sub-string divisibility property. Then we simply brute force."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def is_sub_string_divisible(s):\n",
+    "    s = \"\".join(s)\n",
+    "    if int(s[1:4]) % 2 != 0:\n",
+    "        return False\n",
+    "    if int(s[2:5]) % 3 != 0:\n",
+    "        return False\n",
+    "    if int(s[3:6]) % 5 != 0:\n",
+    "        return False\n",
+    "    if int(s[4:7]) % 7 != 0:\n",
+    "        return False\n",
+    "    if int(s[5:8]) % 11 != 0:\n",
+    "        return False\n",
+    "    if int(s[6:9]) % 13 != 0:\n",
+    "        return False\n",
+    "    if int(s[7:10]) % 17 != 0:\n",
+    "        return False\n",
+    "    return True\n",
+    "\n",
+    "assert(is_sub_string_divisible(\"1406357289\"))\n",
+    "assert(is_sub_string_divisible(\"1406357298\") ==  False)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "import itertools\n",
+    "\n",
+    "s = sum([int(\"\".join(p)) for p in itertools.permutations(\"0123456789\") if is_sub_string_divisible(p)])"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "16695334890\n"
+     ]
+    }
+   ],
+   "source": [
+    "print(s)\n",
+    "assert(s == 16695334890)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "I just want to write my own permutations functions to see that I am able to do it quickly. We pick each element once and combine it with the permutations for the remaining items:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 32,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def permutations(items):\n",
+    "    items = list(items)\n",
+    "    if not items:\n",
+    "        return [[]]\n",
+    "    perms = []\n",
+    "    for item in items:\n",
+    "        items_new = list(items)\n",
+    "        items_new.remove(item)\n",
+    "        permutations_new = permutations(items_new)\n",
+    "        for perm in permutations_new:\n",
+    "            perms.append([item] + perm)\n",
+    "    return perms\n",
+    "\n",
+    "\n",
+    "assert(permutations([]) == [[]])\n",
+    "assert(permutations([1]) == [[1]])\n",
+    "assert(permutations([1,2]) == [[1,2], [2,1]])\n",
+    "assert(permutations([1,2,3]) == [[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1]])\n"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": null,
@@ -45,7 +144,7 @@
   }
  ],
  "metadata": {
-  "completion_date": "",
+  "completion_date": "Fri, 21 Dec 2018, 16:28",
   "kernelspec": {
    "display_name": "Python 3",
    "language": "python3.6",
@@ -65,7 +164,8 @@
   },
   "tags": [
    "pandigital",
-   "divisibility"
+   "divisibility",
+   "permutations"
   ]
  },
  "nbformat": 4,