{
  "arm": "GPT-5.6 Sol",
  "provider": "openai",
  "benchmark": "postcutoff_pilot_v1",
  "freeze_tag": "postcutoff-pilot-v1-freeze-20260705",
  "note": "Frontier comparator (released 2026-07-09). Single-agent gpt-5.6-sol at max reasoning, run via OpenRouter because the direct OpenAI API had not finished its gradual rollout at run time. \"Ultra\" is a separate four-agent product mode, not measured here. Cost is a lower bound: one task returned an empty provider response billed at $0.",
  "tasks": [
    {
      "task_id": "PILOT-N2",
      "task_type": "math",
      "answer": "30907",
      "gold": "30907",
      "correct": true,
      "tokens_in": 139,
      "tokens_out": 16847,
      "cost_usd": 0.506105,
      "latency_s": 242.55,
      "timed_out": false,
      "solution": "Let \\(x\\) mark each \\(1\\times1\\) square tile. Process the rectangle one column at a time, recording which cells of the next column are already occupied by horizontal dominoes.\n\nBy reflection symmetry, use the profile states\n\\[\nA=000,\\quad B=001\\text{ or }100,\\quad C=010,\\quad\nD=011\\text{ or }110,\\quad E=101,\\quad F=111,\n\\]\nwhere \\(B_n,D_n\\) denote the common value for either reflected profile.\n\nSince only the coefficient of \\(x^2\\) is needed, computations may be done modulo \\(x^3\\). Filling one column gives\n\\[\n\\begin{aligned}\nA_{n+1}&=2xA_n+2(1+x^2)B_n+x^2C_n+2xD_n+xE_n+F_n,\\\\\nB_{n+1}&=(1+x^2)A_n+xB_n+xC_n+D_n,\\\\\nC_{n+1}&=x^2A_n+2xB_n+E_n,\\\\\nD_{n+1}&=xA_n+B_n,\\\\\nE_{n+1}&=xA_n+C_n,\\\\\nF_{n+1}&=A_n.\n\\end{aligned}\n\\]\n(The omitted \\(x^3A_n\\) term in the first equation corresponds to filling a column with three squares.)\n\nStarting with \\(A_0=1\\) and all other states zero, iteration gives, modulo \\(x^3\\),\n\\[\n\\begin{array}{c|cccccc}\nn&A_n&B_n&C_n&D_n&E_n&F_n\\\\ \\hline\n1&2x&1+x^2&x^2&x&x&1\\\\\n3&18x&4+24x^2&14x^2&7x&6x&3+11x^2\\\\\n5&106x&15+219x^2&101x^2&40x&25x&11+123x^2\\\\\n7&540x&56+1490x^2&570x^2&202x&96x&41+908x^2\\\\\n9&2554x&209+8769x^2&2859x^2&951x&361x&153+5580x^2\n\\end{array}\n\\]\n\nThus\n\\[\n\\begin{aligned}\n[x^2]A_{10}\n&=5108+2(8769+209)+1902+361+5580\\\\\n&=30907.\n\\end{aligned}\n\\]\nWith two squares, the remaining \\(28\\) cells necessarily use fourteen dominoes.\n\nANSWER: 30907"
    },
    {
      "task_id": "PILOT-K1",
      "task_type": "code",
      "answer": null,
      "gold": "121090",
      "correct": false,
      "tokens_in": 0,
      "tokens_out": 0,
      "cost_usd": 0.0,
      "latency_s": 310.42,
      "timed_out": false,
      "solution": "",
      "note": "Empty provider response after 310.4s with no tokens billed. Counted as a miss under the single-attempt policy; the arm cost is therefore a lower bound."
    },
    {
      "task_id": "PILOT-Q1",
      "task_type": "simple_qa",
      "answer": "10152",
      "gold": "10152",
      "correct": true,
      "tokens_in": 150,
      "tokens_out": 359,
      "cost_usd": 0.01152,
      "latency_s": 6.5,
      "timed_out": false,
      "solution": "Daily crates: \\(48, 54, 60, 66, 72\\).\n\n- Monday: \\(48\\cdot36=1728\\)\n- Tuesday: \\(54\\cdot36=1944\\)\n- Wednesday: \\(40\\cdot36+20\\cdot18=1800\\)\n- Thursday: \\(66\\cdot36=2376\\)\n- Friday: \\((72-8)\\cdot36=2304\\)\n\nTotal:\n\\[\n1728+1944+1800+2376+2304=10152.\n\\]\n\nANSWER: 10152"
    },
    {
      "task_id": "PILOT-R1",
      "task_type": "reasoning",
      "answer": "12",
      "gold": "12",
      "correct": true,
      "tokens_in": 174,
      "tokens_out": 5398,
      "cost_usd": 0.16281,
      "latency_s": 85.46,
      "timed_out": false,
      "solution": "Let \\(a,b,c,d,e\\) be the finishing positions of Ava, Ben, Cleo, Dan, and Eve.\n\nThe statements are true exactly when:\n\n\\[\nb=a+1,\\qquad c=5,\\qquad c<a,\\qquad e\\ne1,\\qquad d=3.\n\\]\n\nFirst, \\(e\\neq1\\). Otherwise Dan lies.\n\n- If \\(d=3\\), Eve tells the truth, so two of Ava, Ben, and Cleo must tell the truth. Since \\(e=1\\), \\(d=3\\), and \\(a<d\\), we have \\(a=2\\), making Ava's statement false. Ben and Cleo would then both be truthful, requiring \\(c=5\\) and \\(c<2\\), impossible.\n- If \\(d\\neq3\\), both Dan and Eve lie, so the first three statements must all be true. But \\(c=5\\) and \\(c<a\\) are incompatible.\n\nThus Dan tells the truth.\n\nSuppose \\(d=3\\). Then Eve also tells the truth, so exactly one of Ava, Ben, and Cleo tells the truth. Since \\(a<3\\), \\(a=1\\) or \\(2\\). Checking these possibilities:\n\n- If \\(a=1\\), Cleo lies. The possible liar-position pairs are \\((2,4)\\) or \\((1,5)\\).\n- If \\(a=2\\), Ava lies. The second liar finishes in position \\(4\\) or \\(5\\).\n\nNone are consecutive, so \\(d\\neq3\\). Therefore Eve lies, and exactly one of Ava, Ben, and Cleo lies.\n\n- Ava cannot be the liar, since Ben and Cleo would then require \\(c=5<c<a\\), impossible.\n- If Cleo is the liar, then \\(c=5\\), so consecutive liar positions force \\(e=4\\). The remaining positions \\(1,2,3\\) cannot satisfy both \\(b=a+1\\), \\(a<d\\), and \\(d\\neq3\\).\n- Hence Ben is the other liar. Ava and Cleo are truthful, so\n  \\[\n  b=a+1,\\qquad c<a.\n  \\]\n  The liar positions \\(b\\) and \\(e\\) must be consecutive.\n\nIf \\(a=3\\), then \\(b=4,d=5,e=2\\), which does not make \\(b,e\\) consecutive. Thus \\(a=2\\), giving\n\n\\[\n(c,a,b,e,d)=(1,2,3,4,5).\n\\]\n\nThe liars are Ben and Eve, finishing third and fourth. Their position product is \\(3\\cdot4=12\\).\n\nANSWER: 12"
    },
    {
      "task_id": "PILOT-C1",
      "task_type": "code",
      "answer": null,
      "gold": "300062",
      "correct": false,
      "tokens_in": 24942,
      "tokens_out": 7943,
      "cost_usd": 0.363,
      "latency_s": 486.37,
      "timed_out": false,
      "solution": "",
      "note": "Reasoning budget (32,768 tokens) exhausted with no final answer emitted (finish_reason=length). 24,942 input tokens reflect server-side reasoning re-billed as input."
    }
  ]
}