| {"schema": 2, "epoch": 87687, "nonce": "b6ec415e1ae3c4f4", "hotkey": "5GuVFWnG62s4r1AGAfrhhHBMnsMPMXKNxrmBpeJdFGyCQEaX", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "095ffbac36e314358948a86370be2e0be5384b2278d5ee9bc924d7f155c7d474", "model_id": "router", "total_cost_usd": 0.044834099999999995, "n_calls": 6, "call_log_hash": "21f25899d286d511b95f34b9e7d922696a07d97dcc5f63e36e728772dd30854e", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 39.144, "tokens_in": 2003, "tokens_out": 6082, "results": [{"benchmark": "mmlu", "task_id": "mmlu-8094", "answer": "A", "cost_usd": 8.8e-06, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.003934, 0.000939, 0.000849, 0.000839, 0.979019, 0.009712, 0.004708]}, {"benchmark": "mmlu", "task_id": "mmlu-3982", "answer": "D", "cost_usd": 8.8e-06, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.007755, 0.002021, 0.001918, 0.001914, 0.935221, 0.036216, 0.014955]}, {"benchmark": "math", "task_id": "gsm8k-266", "answer": "Selling price: $400,000\n\nTransfer fee: \n\\(3\\% \\times 400,000 = 12,000\\)\n\nBrokerage fee: \n\\(5\\% \\times 400,000 = 20,000\\)\n\nTotal deductions: \n\\(12,000 + 20,000 + 250,000 = 282,000\\)\n\nNet proceeds: \n\\(400,000 - 282,000 = 118,000\\)\n\n118000", "cost_usd": 0.0001091, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.012769, 0.003655, 0.003425, 0.003499, 0.86893, 0.065166, 0.042556]}, {"benchmark": "math", "task_id": "gsm8k-820", "answer": "First day: 8 hours \nSecond day: \\(8 \\div 2 = 4\\) hours \nTotal: \\(8 + 4 = 12\\) hours \nIn minutes: \\(12 \\times 60 = 720\\)\n\n720", "cost_usd": 7.23e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.011407, 0.003088, 0.002915, 0.002985, 0.894377, 0.059658, 0.02557]}, {"benchmark": "code", "task_id": "lcb-abc388_b", "answer": "import sys\n\ndef main():\n data = list(map(int, sys.stdin.buffer.read().split()))\n n, d = data[0], data[1]\n snakes = [(data[i], data[i + 1]) for i in range(2, 2 + 2 * n, 2)]\n\n ans = []\n for k in range(1, d + 1):\n ans.append(str(max(t * (l + k) for t, l in snakes)))\n\n sys.stdout.write(\"\\n\".join(ans))\n\nif __name__ == \"__main__\":\n main()\n", "cost_usd": 0.0001151, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.006808, 0.001895, 0.001757, 0.001753, 0.762751, 0.019101, 0.205935]}, {"benchmark": "code", "task_id": "lcb-abc400_d", "answer": "Answering the problem using 0-1 BFS (or Dijkstra's algorithm).\n\n### Method Explanation\n\nWe can model the problem as finding the shortest path on a grid where:\n1. Moving between adjacent **road** cells costs `0` front kicks.\n2. Performing a front kick from a reachable cell allows Takahashi to turn cells up to $2$ steps away in any of the 4 cardinal directions into roads and step onto them. This costs `1` front kick.\n\nSpecifically, from any cell $(r, c)$ that Takahashi can reach:\n- He can walk to an adjacent cell $(r', c')$ for cost `0` if $(r', c')$ is already a road.\n- He can perform a kick to break walls and step directly onto any cell $(r', c')$ at relative displacement $(\\pm 1, 0), (\\pm 2, 0), (0, \\pm 1), (0, \\pm 2)$ for cost `1`. Once he steps onto such a cell, it becomes a road, and he can freely walk to any adjacent existing roads from there at cost `0`.\n\nWe can use a **0-1 BFS** using a double-ended queue (`collections.deque`):\n- 0-cost moves are pushed to the **front** of the deque.\n- 1-cost moves (kicks) are pushed to the **back** of the deque.\n\n### Python 3 Source Code\n\n```python\nimport sys\nfrom collections import deque\n\ndef solve():\n input = sys.stdin.read\n data = input().split()\n if not data:\n return\n \n H = int(data[0])\n W = int(data[1])\n \n grid = data[2:2+H]\n \n A = int(data[2+H]) - 1\n B = int(data[2+H+1]) - 1\n C = int(data[2+H+2]) - 1\n D = int(data[2+H+3]) - 1\n \n dist = [[float('inf')] * W for _ in range(H)]\n dist[A][B] = 0\n \n dq = deque([(A, B)])\n \n # Kicks can reach these 8 relative positions\n kick_offsets = [\n (-1, 0), (-2, 0),\n (1, 0), (2, 0),\n (0, -1), (0, -2),\n (0, 1), (0, 2)\n ]\n \n # Normal moves (cost 0)\n walk_offsets = [(-1, 0), (1, 0), (0, -1), (0, 1)]\n \n while dq:\n r, c = dq.popleft()\n d = dist[r][c]\n \n if r == C and c == D:\n print(d)\n return\n \n # 1. Try 0-cost moves to adjacent existing roads\n for dr, dc in walk_offsets:\n nr, nc = r + dr, c + dc\n if 0 <= nr < H and 0 <= nc < W:\n if grid[nr][nc] == '.':\n if dist[nr][nc] > d:\n dist[nr][nc] = d\n dq.appendleft((nr, nc))\n \n # 2. Try 1-cost moves via front kick\n for dr, dc in kick_offsets:\n nr, nc = r + dr, c + dc\n if 0 <= nr < H and 0 <= nc < W:\n if dist[nr][nc] > d + 1:\n dist[nr][nc] = d + 1\n dq.append((nr, nc))\n\nif __name__ == '__main__':\n solve()\n```", "cost_usd": 0.04452, "chosen_rung": 5, "rungs_used": [5], "distribution": [0.003914, 0.001721, 0.002135, 0.002377, 0.003852, 0.981493, 0.004508]}], "quote": {"measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "80724b4a0b6f2d448a9dbb4e4ac691ce1faff356c1fa59d9f38df66fc20bb817", "platform_sig": 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