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proofs/87825.json
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{"schema": 2, "epoch": 87825, "nonce": "9c594d1e7ee8ac4f", "hotkey": "5FRE1kqMJz8ddModcXUsCquuWhB8QHnjYib2iK2VfiFetkC5", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "8c94d6e58cc59dda7770ab9ad7aa0be90adc8cbedcd419ef4c539a075d371e1f", "model_id": "router", "total_cost_usd": 0.004945137, "n_calls": 6, "call_log_hash": "1161731e1d9fee2f680873cae3fc9d3bc8766b84e7e23fb170a83b9f43659e57", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 212.082, "tokens_in": 1354, "tokens_out": 23866, "results": [{"benchmark": "mmlu", "task_id": "mmlu-11319", "answer": "C", "cost_usd": 0.00018292, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.999831, 0.000162, 1e-06, 1e-06, 3e-06, 1e-06, 1e-06]}, {"benchmark": "mmlu", "task_id": "mmlu-1448", "answer": "To find the speed of observer $O'$ relative to observer $O$, we apply the principles of special relativity, specifically the relativity of simultaneity.\n\n### 1. Analysis of the Setup\n* **Frame $S$ (Observer $O$):** Observer $O$ is at rest at the midpoint ($x = 5$ m) between two sources at $x_1 = 0$ m and $x_2 = 10$ m. Because $O$ is equidistant from the sources and observes the flashes simultaneously, the emission events are simultaneous in this frame.\n * Time interval: $\\Delta t = 0$\n * Spatial separation: $\\Delta x = 10$ m\n* **Frame $S'$ (Observer $O'$):** Observer $O'$ moves at speed $v$ parallel to the x-axis. In this frame, the time difference between the flashes is $\\Delta t' = 13 \\text{ ns} = 13 \\times 10^{-9}$ s.\n\n### 2. Lorentz Transformation\nThe relationship between the time intervals in the two frames is given by the Lorentz transformation:\n$$\\Delta t' = \\gamma \\left( \\Delta t - \\frac{v \\Delta x}{c^2} \\right)$$\nSubstituting $\\Delta t = 0$ and taking the magnitude:\n$$|\\Delta t'| = \\gamma \\frac{v \\Delta x}{c^2}$$\nUsing $\\beta = v/c$ and $\\gamma = \\frac{1}{\\sqrt{1-\\beta^2}}$, we rearrange to solve for $\\beta$:\n$$\\frac{|\\Delta t'| c}{\\Delta x} = \\frac{\\beta}{\\sqrt{1-\\beta^2}}$$\n\n### 3. Calculation\nUsing $c \\approx 3 \\times 10^8$ m/s:\n$$\\frac{(13 \\times 10^{-9})(3 \\times 10^8)}{10} = \\frac{3.9}{10} = 0.39$$\nSetting $\\frac{\\beta}{\\sqrt{1-\\beta^2}} = 0.39$ and squaring both sides:\n$$\\frac{\\beta^2}{1-\\beta^2} = 0.1521 \\implies \\beta^2 = 0.1521(1 - \\beta^2)$$\n$$\\beta^2(1 + 0.1521) = 0.1521 \\implies \\beta^2 = \\frac{0.1521}{1.1521} \\approx 0.132$$\n$$\\beta = \\sqrt{0.132} \\approx 0.363$$\n\nThe speed of $O'$ relative to $O$ is approximately $0.36c$.\n\nC", "cost_usd": 0.00140604, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.995668, 0.004295, 5e-06, 5e-06, 1.6e-05, 5e-06, 5e-06]}, {"benchmark": "math", "task_id": "gsm8k-1180", "answer": "Mr. Marcellus purchased 4000 gallons of milk at $3.50 per gallon, for a total cost of $14,000. Since 2/5 of the milk was sour, he returned 1600 gallons. The refund is 1600 \u00d7 $3.50 = $5,600.\n\n5600", "cost_usd": 7.329600000000004e-05, "chosen_rung": 1, "rungs_used": [1], "distribution": [0.001508, 0.99847, 3e-06, 3e-06, 9e-06, 3e-06, 3e-06]}, {"benchmark": "math", "task_id": "gsm8k-1291", "answer": "27", "cost_usd": 2.043300000000004e-05, "chosen_rung": 1, "rungs_used": [1], "distribution": [0.00054, 0.999447, 2e-06, 2e-06, 5e-06, 2e-06, 2e-06]}, {"benchmark": "code", "task_id": "lcb-abc400_b", "answer": "```python\nimport sys\n\ndef solve() -> None:\n data = sys.stdin.read().strip().split()\n if not data:\n return\n N = int(data[0])\n M = int(data[1])\n\n limit = 10**9\n total = 0\n term = 1\n for _ in range(M + 1):\n total += term\n if total > limit:\n print(\"inf\")\n return\n term *= N\n print(total)\n\nif __name__ == \"__main__\":\n solve()\n```", "cost_usd": 0.00025004, "chosen_rung": 1, "rungs_used": [1], "distribution": [3.4e-05, 0.999963, 0.0, 0.0, 1e-06, 0.0, 0.0]}, {"benchmark": "code", "task_id": "lcb-abc388_d", "answer": "import sys\n\ndef main():\n data = list(map(int, sys.stdin.buffer.read().split()))\n if not data:\n return\n\n N = data[0]\n diff = [0] * (N + 2)\n ans = [0] * N\n\n active = 0\n for t in range(1, N + 1):\n active += diff[t]\n x = data[t] + t + active\n\n b = x - N\n if b < 0:\n b = 0\n ans[t - 1] = b\n\n l = t + 1\n if l <= N:\n r = x if x < N else N\n if l <= r:\n diff[l] += 1\n diff[r + 1] -= 1\n\n sys.stdout.write(\" \".join(map(str, ans)))\n\nif __name__ == \"__main__\":\n main()", "cost_usd": 0.0030124079999999994, "chosen_rung": 1, "rungs_used": [1], "distribution": [0.000249, 0.999742, 1e-06, 1e-06, 4e-06, 1e-06, 1e-06]}], "quote": {"measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "e602aa952aef9846597f81ad61654c2be4e1191024f3f437d1eefe5e01a677a5", "platform_sig": 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