A few comments

src/AoC_2023/Dazbo's_Advent_of_Code_2023.ipynb

@@ -4474,28 +4474,32 @@
     "**My solution**:\n",
     "\n",
     "Grow the input according to the duplication rule:\n",
+    "\n",
     "- Turn the 2D grid into a NumPy array, as this is faster for manipulating grids of data. I've converted all `#` to a `1` and all `.` to a `0`. This makes subsequent operations a bit easier. \n",
     "- For rows: iterate through each row check if a row contains `1`. If not, insert an extra row. I've parameterised this so that we can add an arbitrary number of rows. But for Part 1, we only ever need to add one additional row wherever a row contains no galaxies.\n",
     "- For cols: do the same, but transpose the array first, using the `ndarray transpose()` method. After transposing, each column now exists as a row.  So I can use the same function as I used for rows. After inserting the rows, we must transpose back.\n",
     "\n",
-    "Determine all the locations of points. This code is pretty useful:\n",
+    "Now we work out where the galaxies are:\n",
+    "\n",
+    "- Determine all the locations in the array where the value is 1.\n",
     "\n",
     "```python\n",
     "    # Determine all locations with a galaxy and convert to Points\n",
     "    y, x = np.where(array == 1) # remember that 1 = #\n",
     "    hash_points = [Point(x, y) for (y, x) in zip(y, x)]\n",
     "```\n",
     "\n",
-    "- Determine all the locations in the array where the value is 1.\n",
     "- For each of these locations return the list of `y` values and the list of `x` values.\n",
     "- Then I use [zip](https://aoc.just2good.co.uk/python/zip) to combine the two lists, to get a single list of `(y,x)` tuples. Then I use a [list comprehension](https://aoc.just2good.co.uk/python/comprehensions) to create a `Point` from each of these tuples.\n",
     "\n",
     "Next, get all the pairs of points using [`itertools.combinations()`](https://aoc.just2good.co.uk/python/perms_combos).\n",
+    "\n",
     "- The `combinations()` function returns all unique combinations of points, but doesn't care about sequence. \n",
     "- I.e. if we have A -> B, it will not also return B -> A. \n",
     "- The instructions are pretty clear here: _\"Only count each pair once; order within the pair doesn't matter.\"_\n",
     "\n",
-    "Now I find the  _Manhattan distance_ for each pair. \n",
+    "Now I find the  _Manhattan distance_ for each pair:\n",
+    "\n",
     "- The Manhattan distance is the sum of horizontal and vertical distance, which is exactly what we want. \n",
     "- My `Point` class already knows how to do this.\n",
     "- Sum up the distances.\n",
@@ -4512,7 +4516,7 @@
     "def insert_rows(array: np.ndarray, insert_number) -> np.ndarray:\n",
     "    new_rows = []\n",
     "    for row in array:\n",
-    "        new_rows.append(row)\n",
+    "        new_rows.append(row) # add back the original row\n",
     "        if 1 not in row:\n",
     "            new_rows.extend(row for _ in range(insert_number))\n",
     "\n",