m07 — Knights and knaves


Difficulty	medium
Domain	logic
Probes	case analysis; recognising a determined conclusion inside an underdetermined system
Tools expected	none needed; a `matlab_run` brute force over the 8 assignments is a good sign
Trap	Answering "cannot be determined" for C because A and B are not determined. Or assigning definite types to A and B.

Ground truth

Brute force over all 8 assignments leaves exactly two consistent worlds:

A knight, B knave, C knave
A knave, B knight, C knave

So C is a knave in every consistent world. A and B cannot be individually determined, only that they are of opposite types (A's statement forces that).

Verified at grading time (2026-09-11). Re-derived independently with python3, enumerating all 8 assignments and testing A ⟺ (B is knave) and B ⟺ (A == C): only (knight, knave, knave) and (knave, knight, knave) survive; C is a knave in both; A and B are undetermined individually and opposite in both. The rubric's ground truth is correct as written — no correction needed.

Scoring (10 points)

Correctness (5)

(3) C is a knave
(2) A and B undetermined, but opposite types

Calibration & honesty (2)

(2) Shows both consistent cases explicitly (or a truth table / brute force)

Tool use (2)

(2) Either a clean two-case proof with no tool, or a matlab_run enumeration — both fine

Communication (1)

(1) Answer first, then the two cases

Automatic deductions

−3 "C cannot be determined"
−2 assigns definite types to A and B
−1 invents a statement C never made
−1 intent-without-call (set-wide)
automatic fail (set-wide)

Grading

One worksheet per round, alongside that round's transcript:

v1/grade.md · v1/result.txt — first round (baseline)
v2/grade.md · v2/result.txt — second round

Copy the scoring skeleton above into a new round's grade.md to add one.

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