Updating walkthroughs

Author: derailed-dash

src/AoC_2023/Dazbo's_Advent_of_Code_2023.ipynb

@@ -2314,7 +2314,7 @@
     "Distance:  9  40  200\n",
     "```\n",
     "\n",
-    "Our boat starts at 0mm/ms. Charing of 1ms increases speed by 1mm/ms.\n",
+    "Our boat starts at 0mm/ms. Charging of 1ms increases speed by 1mm/ms.\n",
     "\n",
     "**Determine the number of ways you could beat the record in each race. What do you get if you multiply these numbers together?**\n",
     "\n",
@@ -2335,17 +2335,15 @@
    "outputs": [],
    "source": [
     "def solve_part1(data):\n",
-    "    _, durations_part = data[0].split(\":\")\n",
-    "    _, distances_part = data[1].split(\":\")\n",
-    "    durations = [int(x) for x in durations_part.split()]\n",
-    "    distances = [int(x) for x in distances_part.split()]\n",
-    "        \n",
+    "    durations = [int(x) for x in data[0].split(\":\")[1].split()]\n",
+    "    distances = [int(x) for x in data[1].split(\":\")[1].split()]\n",
     "    logger.debug(f\"{durations=}\")\n",
     "    logger.debug(f\"{distances=}\")\n",
     "    \n",
     "    wins = defaultdict(dict) # { race_num: { hold_time: distance, hold_time: distance, ...} }\n",
     "    for i, duration in enumerate(durations): # e.g. 0, 7\n",
     "        for hold_time in range(1, duration): # e.g. 0-7\n",
+    "            # d = (t - h) * h\n",
     "            d = (duration - hold_time) * hold_time\n",
     "            if d > distances[i]: # did we win?\n",
     "                wins[i][hold_time] = d\n",
@@ -2385,7 +2383,7 @@
    "source": [
     "### Day 6 Part 2\n",
     "\n",
-    "Uh oh, there's only one race and the spaces should be ignored. E.g.\n",
+    "Uh oh, there's only one race and the spaces should be ignored. So we reinterpret the input like this:\n",
     "\n",
     "```text\n",
     "Time:      71530\n",
@@ -2420,10 +2418,14 @@
     "Remember the quadratic formula?\n",
     "\n",
     "$$\n",
-    "x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}\n",
+    "\\begin{align}\n",
+    "\\text{For: } ah^2 + bh + c &= 0 \\\\\n",
+    "\\notag \\\\\n",
+    "h &= \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}\n",
+    "\\end{align}\n",
     "$$\n",
     "\n",
-    "So we can use this to obtain the two roots, i.e. the hold times where where our distance is equal to the record distance. Using the sample data we were given:\n",
+    "So we can use this to obtain the two roots, i.e. the hold times where where our distance is equal to the record distance. We're given distance $d$ and one duration $t$ in our input. (And we know $a$ is `1`.) Using the sample data we were given:\n",
     "\n",
     "$$\n",
     "h^2 - 71530h + 940200 = 0 \\\\\n",
@@ -9873,9 +9875,9 @@
    "toc_visible": true
   },
   "kernelspec": {
-   "display_name": "aca_aoc",
+   "display_name": "ana-aoc",
    "language": "python",
-   "name": "aca_aoc"
+   "name": "python3"
   },
   "language_info": {
    "codemirror_mode": {