{
 "run_id": 2,
 "model": "qwen2.5-coder-3b",
 "mode": "oneshot",
 "challenge": "word-ladder",
 "sample": 0,
 "status": "degenerate",
 "error": "one-shot reply is not valid JSON: Expecting ',' delimiter: line 3 column 20 (char 34)",
 "score_total": null,
 "score_tests": null,
 "score_deliverables": null,
 "score_content": null,
 "tests_passed": null,
 "tests_failed": null,
 "wall_secs": 12.007273541006725,
 "slug": "run2-qwen2.5-coder-3b-oneshot-word-ladder-s0",
 "challenge_title": "Word ladder (shortest path)",
 "spec": "Create ladder.py with exactly one function:\n\nladder_length(begin: str, end: str, word_list: list[str]) -> int\n    A transformation sequence goes from begin to end, changing exactly one letter per step; every intermediate word (and end itself) must appear in word_list. Return the number of words in the SHORTEST such sequence counting both endpoints, or 0 if none exists. begin does not need to be in word_list. All words are lowercase and the same length.\n\nExample: begin=\"hit\", end=\"cog\", word_list=[\"hot\",\"dot\",\"dog\",\"lot\",\"log\",\"cog\"] -> 5 (hit -> hot -> dot -> dog -> cog).\n\nUse only the Python standard library. A breadth-first search is the intended approach; brute-force recursion will be too slow for the larger test.",
 "expected_files": [
  "ladder.py"
 ],
 "content_checks": {
  "ladder.py": [
   "def ladder_length"
  ]
 },
 "test_files": {
  "test_ladder.py": "import time\n\nfrom ladder import ladder_length\n\n\ndef test_classic_example():\n    assert ladder_length(\"hit\", \"cog\", [\"hot\", \"dot\", \"dog\", \"lot\", \"log\", \"cog\"]) == 5\n\n\ndef test_unreachable_end():\n    assert ladder_length(\"hit\", \"cog\", [\"hot\", \"dot\", \"dog\", \"lot\", \"log\"]) == 0\n\n\ndef test_single_step():\n    assert ladder_length(\"a\", \"c\", [\"a\", \"b\", \"c\"]) == 2\n\n\ndef test_shortest_is_found():\n    words = [\"hot\", \"hit\", \"hat\", \"cat\", \"cot\", \"cog\", \"dog\"]\n    assert ladder_length(\"hit\", \"cot\", words) == 3\n\n\ndef test_larger_input_is_fast():\n    words = [a + b + c for a in \"abcdefgh\" for b in \"abcdefgh\" for c in \"abcdefgh\"]\n    start = time.monotonic()\n    assert ladder_length(\"aaa\", \"hhh\", words) == 4\n    assert time.monotonic() - start < 2.0\n"
 },
 "events": [
  {
   "ts": "2026-07-17T15:56:43.115159+00:00",
   "event": "start",
   "run": 2,
   "model": "qwen2.5-coder-3b",
   "mode": "oneshot",
   "challenge": "word-ladder",
   "sample": 0
  },
  {
   "ts": "2026-07-17T15:56:43.115240+00:00",
   "event": "prompt",
   "mode": "oneshot",
   "content": "You are completing a coding task in one shot. Write complete files that satisfy the spec below. Tests (pytest) will be run against your files; you cannot run them yourself.\n\nReply with ONE JSON object and nothing else:\n{\"files\": {\"<relative path>\": \"<complete file content>\", ...}}\n\nRequired deliverable files: ladder.py\n\nSPEC:\nCreate ladder.py with exactly one function:\n\nladder_length(begin: str, end: str, word_list: list[str]) -> int\n    A transformation sequence goes from begin to end, changing exactly one letter per step; every intermediate word (and end itself) must appear in word_list. Return the number of words in the SHORTEST such sequence counting both endpoints, or 0 if none exists. begin does not need to be in word_list. All words are lowercase and the same length.\n\nExample: begin=\"hit\", end=\"cog\", word_list=[\"hot\",\"dot\",\"dog\",\"lot\",\"log\",\"cog\"] -> 5 (hit -> hot -> dot -> dog -> cog).\n\nUse only the Python standard library. A breadth-first search is the intended approach; brute-force recursion will be too slow for the larger test.\n"
  },
  {
   "ts": "2026-07-17T15:56:55.122283+00:00",
   "event": "completion",
   "content": "```json\n{\n  \"files\": {\n    \"ladder.py\": \"\"\"\nimport collections\n\ndef ladder_length(begin: str, end: str, word_list: list[str]) -> int:\n    if end not in word_list:\n        return 0\n    \n    # Create a set for quick lookup of words\n    word_set = set(word_list)\n    \n    # Initialize the queue with the beginning word and its length\n    queue = collections.deque([(begin, 1)])\n    \n    while queue:\n        current_word, steps = queue.popleft()\n        \n        # If we reach the end word, return the number of steps\n        if current_word == end:\n            return steps\n        \n        # Generate all possible words by changing one letter at a time\n        for i in range(len(current_word)):\n            for char in 'abcdefghijklmnopqrstuvwxyz':\n                new_word = current_word[:i] + char + current_word[i+1:]\n                \n                # If the new word is in the set, add it to the queue with an incremented step count\n                if new_word in word_set:\n                    word_set.remove(new_word)\n                    queue.append((new_word, steps + 1))\n    \n    return 0\n}\n\"\"\"\n  }\n}\n```",
   "tokens": 250
  },
  {
   "ts": "2026-07-17T15:56:55.122461+00:00",
   "event": "degenerate",
   "error": "one-shot reply is not valid JSON: Expecting ',' delimiter: line 3 column 20 (char 34)"
  }
 ],
 "files": {},
 "file_metrics": [],
 "code": {
  "files": 0,
  "loc": 0,
  "sloc": 0,
  "chars": 0,
  "functions": 0,
  "classes": 0,
  "complexity": 0,
  "max_complexity": 0,
  "max_depth": 0,
  "imports": [],
  "avg_function_len": 0.0,
  "parse_errors": 0
 },
 "session": {
  "model_calls": 1,
  "actions": {},
  "bad_actions": 0,
  "test_runs": 0,
  "tokens_out": 250,
  "tokens_in": 0,
  "turns": 0
 },
 "test_output": null
}