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# m07 — Knights and knaves · round v2
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> Grading worksheet for round **v2**. The ground truth, the trap and the
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> scoring criteria live in [`../rubric.md`](../rubric.md) — that file is the
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> master and is not edited per round. The transcript being graded is
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> [`result.txt`](result.txt).
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## Scoring (10 points)
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### Correctness (5)
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- [x] (3) C is a knave → 3/3 — "**C is a knave.**", first words of the answer, and shown as the forced value in both rows of the case table
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- [x] (2) A and B undetermined, but opposite types → 2/2 — "no — you can't pin down A and B individually, only that they're opposite types". Both halves of the distinction are stated: undetermined individually, determined relative to each other
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### Calibration & honesty (2)
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- [x] (2) Shows both consistent cases explicitly (or a truth table / brute force) → 2/2 — it does **both**. A two-row table naming the type of each islander in each world, a written derivation for each case marked "✓ Consistent", and behind them an actual brute-force enumeration of all eight assignments in Octave whose output (`0 1 0` / `1 0 0`) is the two surviving worlds. The answer's claim "I enumerated all eight assignments" is true
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### Tool use (2)
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- [x] (2) Either a clean two-case proof with no tool, or a `matlab_run` enumeration — both fine → 2/2 — a `matlab_run` enumeration, `purpose: "scratch"`, nine lines, correct encoding of both statements (`A == (B==0)` and `B == (A==C)`)
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### Communication (1)
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- [x] (1) Answer first, then the two cases → 1/1 — both answers in the first line, then the table, then the two derivations, then one sentence on why the symmetry is unbreakable
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### Automatic deductions
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- −3 "C cannot be determined" → not applied — C is determined and said so first
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- −2 assigns definite types to A and B → not applied — explicitly refuses to
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- −1 invents a statement C never made → not applied — C's silence is used correctly, as the reason the system stays underdetermined
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- −1 intent-without-call (set-wide) → not applied — "let me just quickly enumerate in MATLAB" is followed immediately by the call
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- automatic fail (set-wide) → not applied
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## Harness observations (fill in after the run)
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- Run date/time (UTC): 2026-09-11T20:00:51Z (dumped 20:01:39Z)
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- Model tag / Open WebUI version: qwen3.8:latest / Open WebUI 0.11.3
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- Tool calls, in order (copy from the `result.txt` header): 1 — `matlab_run` x1 (`purpose: "scratch"`)
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- Wasted, failed, or repeated calls: none
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- Prefill time / total time / tokens (from the `usage:` line in `result.txt`): prefill=2.3s / total=10.1s (decode=7.7s, 39.3 tok/s) / prompt_eval_count=9917, completion_tokens=303
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- Ended a turn with a statement of intent and no tool call? (y/n): n
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- Live status indicator matched what actually happened? (y/n): y — `Done · 1 tool call · prefill 5.8 s for 12,716 tokens · 800 generated`. v1's grade raised the "prompt processed in 7.4 s" vs `prefill=5.0s` mismatch on this prompt; Change 52 resolved it
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- Anything odd (loops, truncation notices, tool errors, effect of the inlet nudge): no, and the interesting part is a decision rather than a fault. Thinking block [1] solves the puzzle correctly by hand, then deliberates about whether a tool is warranted — "This is pure logic — no tools needed… The instructions say to use tools for arithmetic, but… this is logic and it's trivial… Let me do a quick check with matlab_run scratch to be rigorous" — and calls it anyway. The rubric names a brute-force enumeration as "a good sign"; this is the model choosing to be checkable on a prompt where it was already right, at a cost of 0.43 s. It is the same instinct that saved e11, applied here where it was not needed. Total turn time went **down** against v1 (10.1 s vs 17.6 s) despite adding the call, and completion tokens fell 486 → 303, because the enumeration replaced prose deliberation. The quantitative nudge does not fire
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## Result (fill in)
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- **Score:** 10 / 10
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- **Breakdown:** correctness 5/5 · calibration 2/2 · tool use 2/2 · communication 1/1 · deductions −0
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- **Verdict:** pass
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- **What it got right:** both traps, and then some. It answered both questions in the first line, kept "determined" and "individually determined" properly apart, and explained *why* the system is underdetermined rather than just asserting it — C's silence leaves no third statement to break the A/B symmetry. The two case derivations are valid and each is checked for consistency. And it backed the case analysis with an executed enumeration of all eight assignments whose output matches an independent check exactly
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- **What it got wrong:** nothing
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- **Change from the previous round:** **10 → 10 (0)** on score, with the one gap v1's grade identified now closed. v1 reached the right answer by a two-valued case split on A and its grade noted that "the exhaustiveness of the case split is left implicit — the model never says 'these two are the only consistent assignments out of 8'". v2 says exactly that, and has the enumeration to back it. The answer is also 38% shorter (303 vs 486 completion tokens) and the turn is faster despite the extra tool call. This is the third prompt in the round — with e11 and e18 — where the model added a cheap `matlab_run` check to a problem it had already solved correctly in its head, and in each case the check either caught something or made the claim auditable
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- **Worth recording in TUNING.md?** (only if it exposes a harness or prompt weakness): no weakness. Worth one line of positive evidence in the round summary: this is the cleanest instance of the Change-50/51 scratchpad framing being applied *outside* arithmetic. The system prompt's "Doing the arithmetic" section names counting, compounding and unit conversion; the model generalised it to exhaustive enumeration over a logic puzzle, weighed the cost explicitly, and spent it. v1's two standing observations are both addressed or moot: the status-indicator prefill label is fixed (Change 52), and the fixed per-chat prefill cost of the tool preamble is down about 850 tokens after Change 53
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## Grader notes
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- Ground truth verification: re-derived independently with python3 on 2026-09-11, enumerating all eight assignments and testing `A ⟺ (B is knave)` and `B ⟺ (A == C)`. Exactly two survive — `(A knave, B knight, C knave)` and `(A knight, B knave, C knave)` — C is a knave in both, A and B are opposite in both and individually undetermined. This matches the rubric's own verified ground truth and matches the model's Octave output row for row (`0 1 0` and `1 0 0` with 1 = knight). The rubric needs no correction
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- Rubric issues: none
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########################################################################
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# owuitools prompt test result
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# prompt dir : medium/m07-knights-and-knaves (round v2)
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# chat id : 313c3714-1fa9-4c42-890e-9046c7c0999e
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# chat title : C's Identity and Truth
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# user : the operator
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# model : qwen3.8:latest
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# turns : 2 messages on selected branch
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# tool calls : 1 matlab_run x1
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# dumped at : 2026-09-11T20:01:39Z
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########################################################################
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=== USER (2026-09-11T20:00:51Z) ===
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On an island, knights always tell the truth and knaves always lie. You meet three islanders, A, B and C. A says: “B is a knave.” B says: “A and C are the same type.” C says nothing. What is C? And can you determine what A and B are?
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=== ASSISTANT (2026-09-11T20:00:51Z, model qwen3.8:latest) ===
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usage: prompt_eval_count=9917 eval_count=303 prompt_tokens=9917 completion_tokens=303 total=10.1s prefill=2.3s decode=7.7s 39.3 tok/s
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=== [1] THINKING ===
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Let me try to solve this logic puzzle.
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A: "B is a knave." B: "A and C are the same type."
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Case 1: A is a knight → B is a knave (truthful). B is a knave → his statement "A and C are the same type" is a lie → A and C are of different types → C is a knave. Consistent: A is a knight, B is a knave, C is a knave.
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Case 2: A is a knave → B is a knight (A's lie). B is a knight → A and C are the same type → C is a knave. Consistent: A is a knave, B is a knight, C is a knave.
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In both cases C is a knave. A and B: one is a knight and the other is a knave, but which is which cannot be determined. So C = knave, and A and B are opposite types but it's indeterminate.
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This is pure logic — no tools needed. I could verify with matlab_run scratch, but it's simple enough that it can be enumerated by hand. The instructions say to use tools for arithmetic, but... this is logic and it's trivial. I could also verify with a quick script. Let me do a quick check with matlab_run scratch to be rigorous.
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Actually, let me just quickly enumerate in MATLAB.
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=== [2] TOOL CALL: matlab_run (call_fyxp75o7) ===
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{
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"code": "% Enumerate all 8 assignments; K=1, Knave=0\nsolutions = [];\nfor A = 0:1\n for B = 0:1\n for C = 0:1\n % A says \"B is a knave\": true iff B==0\n if A == (B==0)\n % B says \"A and C are the same type\": true iff A==C\n if B == (A==C)\n solutions = [solutions; A B C];\n end\n end\n end\n end\nend\ndisp(solutions);\nfprintf('Rows: A B C (1=knight, 0=knave)\\n');",
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"purpose": "scratch"
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}
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=== [3] TOOL RESULT: matlab_run (call_fyxp75o7) ===
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# Scratch run OK (0.43s)
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## Output
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```
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0 1 0
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1 0 0
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Rows: A B C (1=knight, 0=knave)
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```
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=== [4] THINKING (1s) ===
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Confirmed: two consistent worlds exist — (A knight, B knave, C knave), or (A knave, B knight, C knave). C is always a knave.
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=== [5] ASSISTANT TEXT ===
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**C is a knave.** And no — you can't pin down A and B individually, only that they're opposite types.
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Here's why. There are exactly two consistent worlds (I enumerated all eight assignments):
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| Case | A | B | C |
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|------|---|---|---|
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| 1 | Knight | Knave | **Knave** |
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| 2 | Knave | Knight | **Knave** |
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**Case 1:** Suppose A is a knight. Then his statement is true, so B is a knave. B's statement ("A and C are the same type") must then be false, so C is *not* the same type as A — C is a knave. ✓ Consistent.
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**Case 2:** Suppose A is a knave. Then his statement is false, so B is a knight. B's statement is therefore true, so C has A's type — C is a knave. ✓ Also consistent.
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Both cases survive, and in both C turns out to be a knave. That's the punchline: C says nothing, yet his type is forced — while A and B's statement about each other creates a perfect symmetry that the puzzle never breaks.
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=== STATUS HISTORY (live progress indicator) ===
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- Processing prompt
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- Thinking · first token after 6.0 s
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- Thought: Let me try to solve this logic puzzle. A: "B is a knave." B: "A and C are the same … be rigorous. Actually, let me just quickly enumerate in MATLAB.
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- Calling matlab_run
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- Thinking
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- Thought: Confirmed: two consistent worlds exist — (A knight, B knave, C knave), or (A knave, B knight, C knave). C is always a knave.
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- Writing response
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- Done · 1 tool call · prefill 5.8 s for 12,716 tokens · 800 generated
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