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   "source": [
    "# [Path Sum: Two Ways](https://projecteuler.net/problem=81)\n",
    "\n",
    "First things first, we'll read in the matrix."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "id": "ef846cb1",
   "metadata": {},
   "outputs": [],
   "source": [
    "with open(\"txt/0081_matrix.txt\") as f:\n",
    "    mat = matrix((int(n) for n in line.split(',')) for line in f)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "af2036f4",
   "metadata": {},
   "source": [
    "This problem is fundamentally a [shortest path problem](https://en.wikipedia.org/wiki/Shortest_path_problem), a well-studied problem with lots of algorithms to choose from. We'll employ a variant of [Dijkstra's algorithm](https://en.wikipedia.org/wiki/Dijkstra%27s_algorithm) - we would need a different algorithm if the matrix had negative entries.\n",
    "\n",
    "In short, this algorithm starts with the top-left entry of the matrix and adds its neighbors below and right of it to a search queue. The queue emits entries in the order of smallest path sum from the top-left entry, so the first time we visit an entry, we know the path taken to it is its minimal path. This means when we visit the bottom-right entry, we'll have computed its minimal path and can exit. We also keep track of nodes we've already visited so we don't waste time re-visiting them.\n",
    "\n",
    "Note that we could improve this even further by implementing a [Fibonacci heap](https://en.wikipedia.org/wiki/Fibonacci_heap) for our priority queue, but using a binary heap - [built-in to Python!](https://docs.python.org/3/library/heapq.html) - is plenty fast for this problem."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "100b1cf6",
   "metadata": {},
   "outputs": [],
   "source": [
    "import heapq\n",
    "\n",
    "def minimal_path_sum(mat):\n",
    "    m, n = mat.dimensions()\n",
    "    \n",
    "    visited = set()\n",
    "    queue = [(0, (0, 0))]\n",
    "    while queue != []:\n",
    "        cost, (i, j) = heapq.heappop(queue)\n",
    "        \n",
    "        if (i, j) in visited:\n",
    "            continue\n",
    "        visited.add((i, j))\n",
    "        \n",
    "        cost += mat[i, j]\n",
    "        \n",
    "        if (i, j) == (m - 1, n - 1):\n",
    "            break\n",
    "        \n",
    "        if i + 1 < m:\n",
    "            heapq.heappush(queue, (cost, (i + 1, j)))\n",
    "            \n",
    "        if j + 1 < n:\n",
    "            heapq.heappush(queue, (cost, (i, j + 1)))\n",
    "            \n",
    "    return cost"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "9fa7d67d",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "427337"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "minimal_path_sum(mat)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "aff1f323",
   "metadata": {},
   "source": [
    "#### Copyright (C) 2025 filifa\n",
    "\n",
    "This work is licensed under the [Creative Commons Attribution-ShareAlike 4.0 International license](https://creativecommons.org/licenses/by-sa/4.0/) and the [BSD Zero Clause license](https://spdx.org/licenses/0BSD.html)."
   ]
  }
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