| {"schema": 2, "epoch": 87859, "nonce": "2a32c17250154688", "hotkey": "5FRE1kqMJz8ddModcXUsCquuWhB8QHnjYib2iK2VfiFetkC5", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "2e4d6a88063177a0a7a564f24e10b347ef48e87c9595a209162369a4fa4d4477", "model_id": "router", "total_cost_usd": 0.0034765777, "n_calls": 6, "call_log_hash": "8af99d0f0b876b622f266cb725c8c51060729d75bccf55b15a1d48d796a333c6", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 444.245, "tokens_in": 1040, "tokens_out": 19916, "results": [{"benchmark": "mmlu", "task_id": "mmlu-4", "answer": "To find the product of the polynomials $f(x) = 4x - 5$ and $g(x) = 2x^2 - 4x + 2$ in the polynomial ring $\\mathbb{Z}_8[x]$, we perform polynomial multiplication and reduce the coefficients modulo 8.\n\n**Step 1: Write down the polynomials.**\n$f(x) = 4x - 5$\n$g(x) = 2x^2 - 4x + 2$\n\nNote that in $\\mathbb{Z}_8$, coefficients are integers modulo 8.\n$-5 \\equiv 3 \\pmod 8$\n$-4 \\equiv 4 \\pmod 8$\nSo, we can also write:\n$f(x) = 4x + 3$\n$g(x) = 2x^2 + 4x + 2$\n\n**Step 2: Multiply the polynomials.**\nWe compute the product $(4x - 5)(2x^2 - 4x + 2)$.\nUsing the distributive property:\n$$ (4x - 5)(2x^2 - 4x + 2) = 4x(2x^2 - 4x + 2) - 5(2x^2 - 4x + 2) $$\n\nExpand the first part:\n$$ 4x(2x^2) + 4x(-4x) + 4x(2) = 8x^3 - 16x^2 + 8x $$\n\nExpand the second part:\n$$ -5(2x^2) - 5(-4x) - 5(2) = -10x^2 + 20x - 10 $$\n\nCombine the results:\n$$ 8x^3 - 16x^2 + 8x - 10x^2 + 20x - 10 $$\nGroup like terms:\n$$ 8x^3 + (-16 - 10)x^2 + (8 + 20)x - 10 $$\n$$ 8x^3 - 26x^2 + 28x - 10 $$\n\n**Step 3: Reduce coefficients modulo 8.**\n* Coefficient of $x^3$: $8 \\equiv 0 \\pmod 8$\n* Coefficient of $x^2$: $-26$. Since $-26 = -3 \\times 8 + 6$, $-26 \\equiv 6 \\pmod 8$.\n* Coefficient of $x$: $28$. Since $28 = 3 \\times 8 + 4$, $28 \\equiv 4 \\pmod 8$.\n* Constant term: $-10$. Since $-10 = -2 \\times 8 + 6$, $-10 \\equiv 6 \\pmod 8$.\n\nSubstituting these back into the polynomial:\n$$ 0x^3 + 6x^2 + 4x + 6 $$\n$$ 6x^2 + 4x + 6 $$\n\n**Step 4: Match with the given options.**\nA) $2x^2 + 5$\nB) $6x^2 + 4x + 6$\nC) $0$\nD) $x^2 + 1$\n\nThe calculated product matches option B.\n\nB", "cost_usd": 0.00069835, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.999234, 0.00036, 0.0, 0.0, 0.000405, 2e-06, 0.0]}, {"benchmark": "mmlu", "task_id": "mmlu-10423", "answer": "B", "cost_usd": 5.8219999999999995e-05, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.997983, 0.001389, 0.0, 0.0, 0.000626, 3e-06, 0.0]}, {"benchmark": "math", "task_id": "gsm8k-453", "answer": "Jenna starts with 8 sapphires. After trading 3 sapphires for 2 rubies, she has 5 sapphires and 2 rubies. The value of the sapphires is 5 \u00d7 $800 = $4,000, and the value of the rubies is 2 \u00d7 $1,200 = $2,400. The total value is $4,000 + $2,400 = $6,400.\n\n\\boxed{6400}", "cost_usd": 7.955639999999997e-05, "chosen_rung": 1, "rungs_used": [1], "distribution": [0.003225, 0.996518, 0.0, 0.0, 0.000168, 8.9e-05, 0.0]}, {"benchmark": "math", "task_id": "gsm8k-119", "answer": "Adrien's salary four years ago was $40,000, and it was 30% higher than Lylah's, so Lylah's salary four years ago was $40,000 / 1.3 = $40,000 \u00d7 10/13 = $400,000/13. Four years later, Adrien's salary increased by 40%, becoming $40,000 \u00d7 1.4 = $56,000. Since both salaries increased simultaneously, it is assumed they increased by the same percentage, so Lylah's salary also increased by 40%, becoming ($400,000/13) \u00d7 1.4 = $560,000/13. The total salary four years later is $56,000 + $560,000/13 = ($56,000 \u00d7 13 + $560,000)/13 = ($728,000 + $560,000)/13 = $1,288,000/13 \u2248 $99,076.92.\n\n\\boxed{99076.92}", "cost_usd": 0.0023648513, "chosen_rung": 1, "rungs_used": [1], "distribution": [0.000225, 0.999661, 0.0, 0.0, 9.6e-05, 1.8e-05, 0.0]}, {"benchmark": "code", "task_id": "lcb-abc400_b", "answer": "N, M = map(int, input().split())\n\nlimit = 10**9\ntotal = 0\npower = 1\n\nfor _ in range(M + 1):\n total += power\n if total > limit:\n print(\"inf\")\n break\n power *= N\nelse:\n print(total)\n", "cost_usd": 7.140000000000011e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [8e-06, 7.9e-05, 0.0, 0.0, 0.999896, 1.7e-05, 0.0]}, {"benchmark": "code", "task_id": "lcb-abc399_c", "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, m = data[0], data[1]\n parent = list(range(n + 1))\n size = [1] * (n + 1)\n\n def find(x):\n while parent[x] != x:\n parent[x] = parent[parent[x]]\n x = parent[x]\n return x\n\n kept = 0\n idx = 2\n for _ in range(m):\n u, v = data[idx], data[idx + 1]\n idx += 2\n ru, rv = find(u), find(v)\n if ru != rv:\n if size[ru] < size[rv]:\n ru, rv = rv, ru\n parent[rv] = ru\n size[ru] += size[rv]\n kept += 1\n\n print(m - kept)\n\nif __name__ == \"__main__\":\n main()\n", "cost_usd": 0.00020419999999999987, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.000534, 0.000281, 0.0, 0.0, 0.997795, 0.00139, 0.0]}], "quote": {"measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "ffa001ef9738920f40b7a49c031185776bd626d9941cee4a055156454aabb70c", "platform_sig": 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