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+ {"schema": 2, "epoch": 87820, "nonce": "01d8b9ce7399a817", "hotkey": "5FRE1kqMJz8ddModcXUsCquuWhB8QHnjYib2iK2VfiFetkC5", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "8c94d6e58cc59dda7770ab9ad7aa0be90adc8cbedcd419ef4c539a075d371e1f", "model_id": "router", "total_cost_usd": 0.0009510368, "n_calls": 5, "call_log_hash": "e0485521bee9e9abf2dc2567424eaf2251dc62a669f16fea6301c8e55dd4b379", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 96.47, "tokens_in": 994, "tokens_out": 5290, "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 $f(x) \\cdot g(x)$. Using the original forms or the reduced forms yields the same result. Let's use the expanded form directly from the original expressions to avoid confusion, then reduce at the end.\n\n$$ (4x - 5)(2x^2 - 4x + 2) $$\n\nDistribute each term of $f(x)$ into $g(x)$:\n$$ = 4x(2x^2 - 4x + 2) - 5(2x^2 - 4x + 2) $$\n\nCalculate the products:\n1. $4x \\cdot 2x^2 = 8x^3$\n2. $4x \\cdot (-4x) = -16x^2$\n3. $4x \\cdot 2 = 8x$\n4. $-5 \\cdot 2x^2 = -10x^2$\n5. $-5 \\cdot (-4x) = 20x$\n6. $-5 \\cdot 2 = -10$\n\nCombine these terms:\n$$ 8x^3 - 16x^2 + 8x - 10x^2 + 20x - 10 $$\n\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.**\nNow, we take each coefficient modulo 8:\n* Coefficient of $x^3$: $8 \\equiv 0 \\pmod 8$\n* Coefficient of $x^2$: $-26$. Since $-26 = -3 \\times 8 + 2$ is incorrect ($ -24 - 2$), let's do $-26 \\div 8$. $-26 = -4 \\times 8 + 6$. So, $-26 \\equiv 6 \\pmod 8$.\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": 0.00045005, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.99942, 0.000567, 2e-06, 2e-06, 6e-06, 2e-06, 2e-06]}, {"benchmark": "mmlu", "task_id": "mmlu-2586", "answer": "C", "cost_usd": 5.9829999999999974e-05, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.99998, 1.8e-05, 0.0, 0.0, 1e-06, 0.0, 0.0]}, {"benchmark": "math", "task_id": "gsm8k-260", "answer": "11", "cost_usd": 3.326399999999997e-05, "chosen_rung": 1, "rungs_used": [1], "distribution": [0.000483, 0.999505, 2e-06, 2e-06, 5e-06, 2e-06, 2e-06]}, {"benchmark": "math", "task_id": "gsm8k-244", "answer": "Griffin started with 24 fries. After Kyle took 5, he had 19. Billy took twice as many as Kyle (10), leaving 9. Ginger gave him x fries, making it 9 + x. Colby took 3 less than Kyle's 5, which is 2, so Griffin ended with 7 + x. This equals 27, so x = 20.\n\n\\boxed{20}", "cost_usd": 7.749280000000002e-05, "chosen_rung": 1, "rungs_used": [1], "distribution": [0.000211, 0.99978, 1e-06, 1e-06, 3e-06, 1e-06, 1e-06]}, {"benchmark": "code", "task_id": "lcb-abc399_b", "answer": "import sys\n\ndef main():\n data = sys.stdin.read().split()\n if not data:\n return\n N = int(data[0])\n scores = list(map(int, data[1:1+N]))\n pairs = [(scores[i], i) for i in range(N)]\n pairs.sort(key=lambda x: -x[0])\n ranks = [0] * N\n cur_rank = 1\n i = 0\n while i < N:\n score = pairs[i][0]\n j = i\n while j < N and pairs[j][0] == score:\n j += 1\n for k in range(i, j):\n idx = pairs[k][1]\n ranks[idx] = cur_rank\n cur_rank += (j - i)\n i = j\n sys.stdout.write('\\n'.join(str(r) for r in ranks))\n\nif __name__ == '__main__':\n main()", "cost_usd": 0.0003304, "chosen_rung": 1, "rungs_used": [1], "distribution": [0.000281, 0.99971, 1e-06, 1e-06, 4e-06, 1e-06, 1e-06]}, {"benchmark": "code", "task_id": "lcb-abc390_d", "answer": "", "cost_usd": 0.0, "chosen_rung": 1, "rungs_used": [], "distribution": [0.000368, 0.999622, 2e-06, 2e-06, 4e-06, 2e-06, 2e-06]}], "quote": {"measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "09a8911f21a2fef7432971ca5155afc1bc7319fba1fedf82dca62b41e26eeb75", "platform_sig": 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