Text version of this lesson
Even and odd counts decide questions that counting one by one never could.
Key takeaway: If every move changes a count by two, that count keeps its parity forever — and that alone can settle a puzzle.
Question 1
Which lights are ON?
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Why: Even counts turn a half-filled column into an instruction: exactly one more light, and you know where.
Hint: Row A already has an even count, so its third bulb has to stay off.
Question 2
Which lights are ON?
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Why: Odd rows plus a total of three leaves no room for a row of three — one per row, one per column.
Hint: Three rows, three lights, and every row odd: each row gets exactly one.
Question 3
Which lights are ON?
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Why: The same parity argument scales: one light per row, one per column, and the last one has nowhere else to go.
Hint: Four rows and four lights with every row odd means one light per row.
Question 4
Which lights are ON?
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Why: Odd rows force one light each; even columns force those lights to pair up. Both rules together fix the board.
Hint: Every row is odd and there are only four lights, so each row holds exactly one.
Question 5
Pressing a light flips it and its right-hand neighbour. Light 6 is neighboured by light 1.
Leave lights 1 and 4 ON, and the rest OFF.
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Why: Each press flips exactly two lights, so the count of lit lights can never become odd. Any reachable target has an even number ON.
Hint: Every press changes the number of lit lights by two or by nothing, so that number is always even.
Question 6
Pressing a light flips it and its right-hand neighbour. Light 5 is neighboured by light 1.
Leave lights 2 and 5 ON, and the rest OFF.
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Why: The invariant does not care about the ring size: two lights per press keeps the lit count even, always.
Hint: The ring has an odd length, but each press still flips exactly two lights.