Permutation parity — the ladder
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The crossings that will not come out even
Draw a rearrangement as strings from one row of pegs to another and count where they cross. The count depends on how the strings are drawn; whether it is odd or even does not, and that single bit is what makes determinants exist and a sliding puzzle unsolvable.
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The only bit that survives
A shuffle can be called even or odd, and the label behaves under composition. Ask whether some cleverer label — a number out of three, or out of four — could behave the same way, and the answer is that nothing else can — one bit is exactly what a permutation gives up.
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The puzzle that is exactly half solvable
A sliding puzzle sold with two tiles swapped is not a hard puzzle; it is an impossible one, and the proof is a quantity that no slide can change. The same argument, run three times at once, says that one arrangement of a scrambled cube in twelve is reachable.
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When the label may be a matrix
A permutation carries exactly one bit into any commutative target, and commutativity is the restriction doing all the work. Drop it — let the label be a matrix — and what survives is a short finite table, computed here from traces and checked for orthogonality over every pair of rows.