Finished problem 39 and added some new ones.

ipython/EulerProblem039.ipynb

@@ -13,21 +13,116 @@
     "For which value of p ≤ 1000, is the number of solutions maximised?"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "We are looking for right angle triangles so Pythagoras' theorem $a^2 + b^2 = c^2$ can be used. Also it must be true that $a <= b <= c$ is true for every solution. Let's start with a function that tests the given examples."
+   ]
+  },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 13,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def is_right_angle_triangle(a, b, c, p):\n",
+    "    if a + b + c != p:\n",
+    "        return False\n",
+    "    if a**2 + b**2 != c**2:\n",
+    "        return False\n",
+    "    return True\n",
+    "\n",
+    "given = [(20, 48, 52, 120), (24, 45, 51, 120), (30, 40, 50, 120)]\n",
+    "for g in given:\n",
+    "    assert(is_right_angle_triangle(*g))\n",
+    "assert(is_right_angle_triangle(29, 41, 50, 120) == False)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "This seems bruteforceable. Let's try to find the solutions for p = 120."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[(30, 40, 50), (24, 45, 51), (20, 48, 52)]\n"
+     ]
+    }
+   ],
+   "source": [
+    "def brute_force(p):\n",
+    "    solutions = [(a,b, p - a - b)\n",
+    "        for b in range(1, p // 2 + 1)\n",
+    "        for a in range(1, b)\n",
+    "        if a*a + b*b == (p - a - b) * (p - a - b)\n",
+    "    ]\n",
+    "    return solutions\n",
+    "\n",
+    "print(brute_force(120))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Looks good. Let's try for all $p <= 1000$."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
    "metadata": {
     "collapsed": true
    },
    "outputs": [],
-   "source": []
+   "source": [
+    "solutions = [(len(brute_force(p)), p) for p in range(1, 1001)]"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Well, not too fast, but also not too slow. Let's get the max and we have our solution."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 24,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "840\n"
+     ]
+    }
+   ],
+   "source": [
+    "s, p = max(solutions)\n",
+    "print(p)\n",
+    "assert(p == 840)"
+   ]
   }
  ],
  "metadata": {
-  "completion_date": "",
+  "completion_date": "Sun, 20 May 2018, 21:16",
   "kernelspec": {
    "display_name": "Python 3",
-   "language": "python",
+   "language": "python3.6",
    "name": "python3"
   },
   "language_info": {
@@ -40,7 +135,7 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.5.5"
+   "version": "3.6.5"
   },
   "tags": [
    "triangle",
@@ -49,5 +144,5 @@
   ]
  },
  "nbformat": 4,
- "nbformat_minor": 0
+ "nbformat_minor": 1
 }