{
  "arm": "Qwen3.7 Max",
  "provider": "alibaba",
  "benchmark": "postcutoff_pilot_v1",
  "freeze_tag": "postcutoff-pilot-v1-freeze-20260705",
  "note": "Frontier comparator. Single public model; full per-task output included.",
  "tasks": [
    {
      "task_id": "PILOT-N2",
      "task_type": "math",
      "answer": "30907",
      "gold": "30907",
      "correct": true,
      "tokens_in": 147,
      "tokens_out": 48353,
      "cost_usd": 0.363015,
      "latency_s": 873.84,
      "timed_out": false,
      "solution": "Looking at this problem, I need to count tilings of a 3\u00d710 rectangle using exactly 2 monominoes (1\u00d71 tiles) and 14 dominoes (1\u00d72 tiles).\n\n## Setting Up the Transfer Matrix\n\nI process the board column by column. At each column boundary, the **state** is a 3-bit mask indicating which cells in the next column are already covered by horizontal dominoes extending from the current column. There are 8 possible profiles: 000, 001, 010, 011, 100, 101, 110, 111.\n\nI track the number of monominoes used via a variable $x$, building a generating function where the coefficient of $x^k$ gives tilings with $k$ monominoes.\n\n**Exploiting symmetry:** Swapping top and bottom rows maps profiles 001\u2194100 and 011\u2194110. This reduces the system to 6 states: $A=000$, $B=\\{001,100\\}$, $C=010$, $D=\\{011,110\\}$, $E=101$, $F=111$.\n\n## Transition Matrix (mod $x^3$)\n\nAfter carefully enumerating all ways to fill each column's empty cells with vertical dominoes, horizontal dominoes, and monominoes:\n\n$$M \\equiv \\begin{pmatrix} 2x & 2x^2+2 & x^2 & 2x & x & 1 \\\\ x^2+1 & x & x & 1 & 0 & 0 \\\\ x^2 & 2x & 0 & 0 & 1 & 0 \\\\ x & 1 & 0 & 0 & 0 & 0 \\\\ x & 0 & 1 & 0 & 0 & 0 \\\\ 1 & 0 & 0 & 0 & 0 & 0 \\end{pmatrix} \\pmod{x^3}$$\n\n## Computing $M^{10} \\cdot e_1$ mod $x^3$\n\nStarting with $v^{(0)} = (1, 0, 0, 0, 0, 0)$, I iterate $v^{(n+1)} = M \\cdot v^{(n)}$:\n\n| $n$ | $v_A^{(n)}$ (mod $x^3$) | Constant term check |\n|---|---|---|\n| 1 | $2x$ | $T(1)=0$ \u2713 |\n| 2 | $11x^2 + 3$ | $T(2)=3$ \u2713 |\n| 3 | $18x$ | $T(3)=0$ \u2713 |\n| 4 | $123x^2 + 11$ | $T(4)=11$ \u2713 |\n| 5 | $106x$ | |\n| 6 | $908x^2 + 41$ | $T(6)=41$ \u2713 |\n| 7 | $540x$ | |\n| 8 | $5580x^2 + 153$ | $T(8)=153$ \u2713 |\n| 9 | $2554x$ | |\n| 10 | $30907x^2 + 571$ | $T(10)=571$ \u2713 |\n\n## Final Step Detail\n\nFor $v^{(10)}_A$:\n$$v^{(10)}_A = 2x(2554x) + (2x^2+2)(8769x^2+209) + x^2(2859x^2) + 2x(951x) + x(361x) + (5580x^2+153)$$\n\nComputing mod $x^3$:\n- $2x \\cdot 2554x = 5108x^2$\n- $(2x^2+2)(8769x^2+209) \\equiv 17956x^2 + 418$\n- $x^2 \\cdot 2859x^2 \\equiv 0$\n- $2x \\cdot 951x = 1902x^2$\n- $x \\cdot 361x = 361x^2$\n- $5580x^2 + 153$\n\nSum of $x^2$ coefficients: $5108 + 17956 + 1902 + 361 + 5580 = 30907$\n\nThe constant term $571 = T(10)$ confirms correctness, as does the $x^2$ coefficient of $v^{(2)}$ matching a manual count of 11 tilings for the 3\u00d72 case.\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,
      "latency_s": 0,
      "timed_out": true,
      "solution": ""
    },
    {
      "task_id": "PILOT-Q1",
      "task_type": "simple_qa",
      "answer": "10152",
      "gold": "10152",
      "correct": true,
      "tokens_in": 163,
      "tokens_out": 1135,
      "cost_usd": 0.00892,
      "latency_s": 22.35,
      "timed_out": false,
      "solution": "# Solving the Depot Delivery Problem\n\n## Step 1: Crates Shipped Each Day\n\nThe depot ships 48 crates on Monday, increasing by 6 each day:\n\n| Day | Crates |\n|-----|--------|\n| Monday | 48 |\n| Tuesday | 48 + 6 = 54 |\n| Wednesday | 54 + 6 = 60 |\n| Thursday | 60 + 6 = 66 |\n| Friday | 66 + 6 = 72 |\n\n## Step 2: Units Delivered Each Day\n\n**Monday:** 48 \u00d7 36 = 1,728 units\n\n**Tuesday:** 54 \u00d7 36 = 1,944 units\n\n**Wednesday:** One third of 60 crates = 20 crates are half-full (18 units). The remaining 40 crates are full (36 units).\n- 40 \u00d7 36 = 1,440\n- 20 \u00d7 18 = 360\n- Total: 1,440 + 360 = 1,800 units\n\n**Thursday:** 66 \u00d7 36 = 2,376 units\n\n**Friday:** 72 crates shipped, but 8 are returned unopened, so only 64 count.\n- 64 \u00d7 36 = 2,304 units\n\n## Step 3: Weekly Total\n\n1,728 + 1,944 + 1,800 + 2,376 + 2,304 = 10,152 units\n\nANSWER: 10152"
    },
    {
      "task_id": "PILOT-R1",
      "task_type": "reasoning",
      "answer": "12",
      "gold": "12",
      "correct": true,
      "tokens_in": 181,
      "tokens_out": 4596,
      "cost_usd": 0.034923,
      "latency_s": 84.23,
      "timed_out": false,
      "solution": "Looking at this problem, I need to find the finishing positions of 5 runners where exactly 2 are liars (in consecutive positions) and 3 are truth-tellers.\n\n**Setting up the statements:**\n- **Ava:** \"I finished immediately ahead of Ben\" \u2192 A = B - 1\n- **Ben:** \"Cleo finished last\" \u2192 C = 5\n- **Cleo:** \"I finished ahead of Ava\" \u2192 C < A\n- **Dan:** \"Eve did not finish first\" \u2192 E \u2260 1\n- **Eve:** \"Dan finished third\" \u2192 D = 3\n\n**Key constraints:** A < D (Ava ahead of Dan), 2 liars in consecutive positions.\n\n**Testing liar pairs systematically:**\n\nI'll check the pair **Ben and Eve as liars** (with Ava, Cleo, Dan as truth-tellers):\n\nFrom truth-tellers:\n- Ava (T): A = B - 1\n- Cleo (T): C < A\n- Dan (T): E \u2260 1\n\nFrom liars:\n- Ben (L): C \u2260 5\n- Eve (L): D \u2260 3\n\nSince C < A < D, I need at least 3 positions. Let me try **A = 2**:\n- C < 2 \u2192 C = 1\n- B = A + 1 = 3\n- Remaining positions: {4, 5} for D and E\n- D > A = 2, so D \u2208 {4, 5}\n- Eve (L) says D = 3, which is false for both \u2014 satisfied either way\n- Try **D = 5, E = 4**: Dan (T) says E \u2260 1 \u2192 4 \u2260 1 \u2713\n\n**Resulting order:** C=1, A=2, B=3, E=4, D=5\n\n**Verification:**\n| Runner | Position | Role | Statement | Truth Value |\n|--------|----------|------|-----------|-------------|\n| Ava | 2 | Truth | \"Immediately ahead of Ben\" (2 = 3-1) | \u2713 True |\n| Ben | 3 | **Liar** | \"Cleo finished last\" (C=1, not 5) | \u2713 False |\n| Cleo | 1 | Truth | \"Ahead of Ava\" (1 < 2) | \u2713 True |\n| Dan | 5 | Truth | \"Eve not first\" (E=4 \u2260 1) | \u2713 True |\n| Eve | 4 | **Liar** | \"Dan finished third\" (D=5 \u2260 3) | \u2713 False |\n\n- Exactly 2 liars (Ben, Eve) \u2713\n- Liars in positions 3 and 4 \u2014 **consecutive** \u2713\n- Ava (2) ahead of Dan (5) \u2713\n\n**Product of liar positions:** 3 \u00d7 4 = 12\n\nANSWER: 12"
    },
    {
      "task_id": "PILOT-C1",
      "task_type": "code",
      "answer": null,
      "gold": "300062",
      "correct": false,
      "tokens_in": 0,
      "tokens_out": 0,
      "cost_usd": 0,
      "latency_s": 0,
      "timed_out": true,
      "solution": ""
    }
  ]
}