Use even and odd

Try one allowed press and notice what changes. What other kinds of pair could a later press choose?

Answer: No. The number of lit lamps always stays odd. All off would mean zero lit lamps, which is even.

Every press changes the lit count by \( -2,\ 0,\ \text{or }+2\). None changes an odd count into an even one. Starting from nine, every later lit count is odd. Which lamps are lit can change; their count still stays odd.

An odd collection leaves one lamp unpaired. Zero is even: an empty collection has no unpaired lamp.

This checks every legal kind of press, so it covers any finite sequence. A few failed attempts would not prove impossibility. The problem is the card’s promise, not Li Wei’s effort.

Answer: P cannot be cleared. Q can, in a minimum of 14 presses.

P has \(3\times7=21\) lamps and cannot reach zero. Q has \(4\times7=28\). An even number of rows lets every lamp match one above or below, so no lamp is left over.

One press can turn off at most two lit lamps. Clearing 28 therefore needs at least 14 presses. The 14 disjoint pair presses attain that bound, so the plan is shortest. It is legal because any pair is allowed; evenness alone is not the plan.

Answer: No. There will always be one lit lamp in A–B and one in C–D.

The four shown states are all the possibilities. Each controlled pair keeps its odd lit count—exactly one—so the total stays two. If A and B had started on instead, the A–B button would clear them in one press. The actual start and the actual controls both matter.

Passing one necessary test does not prove that a route exists.

Answer: 882 is wrong. Parity alone does not prove 883 correct; exact addition does. The actual total is 883 visits.

The even counts contribute complete pairs. The three odd leftovers make another pair and one extra, so the whole sum is odd. Counting odd addends matters, not counting how many addends there are.

For the exact check, group the counts:

For example, 885 is also odd, but it is not this sum. The quick test rejects an error; the addition establishes the exact answer.

Which feature of the phrase lengths could help you organise a complete search?

Answer: The only mixes, in the order (Ripple, Echo, Sweep), are (2, 1, 4) and (1, 3, 3). The second uses the fewest Sweeps: 3.

Use even and odd

Make an organized list

At least one Ripple and one Sweep need two of the seven phrase spaces, so there can be at most five Echoes. The surviving odd Echo counts are 1, 3 and 5. Test each once: begin by making every remaining phrase a Ripple, then replace enough Ripples with Sweeps to reach 37 beats. Each replacement adds two beats.

The list covers every possible odd Echo count; even counts cannot make 37 beats. Check both accepted rows against every rule:

Both have seven phrases, 37 beats and all three types. Their Sweep counts are four and three, so three is the minimum. Parity narrowed the search; the organized list settled the exact possibilities and the minimum.

What can the rules rule out before you try complete plans?

Answer: Only Card C is reachable. Its shortest plan needs 6 actions: four lamp presses and two token trades. A is impossible by lamp parity; B is impossible by a value kept by every trade.

Divide and conquer

Use even and odd

Find an invariant

An invariant is something the allowed moves never change. Here we choose useful scores; they are not extra rules or money.

The tray’s starting value is \(18\times2+4\times3=48\). B asks for \(12\times2+6\times3=42\), so no legal trade sequence reaches B. C asks for \(12\times2+8\times3=48\). Both 42 and 48 are even: parity alone would miss this difference.

C passes that test, but still needs a legal route.

The four disjoint pair presses leave only E on. The two token trades give:

Every trade has enough tokens, and the lamp and token actions leave the other part alone.

Both bounds apply to the same complete Card C finish:

The six-action construction attains both bounds and meets Card C exactly. Thus A and B are impossible, C is possible, and six is the smallest total for any card.

A different valid token check is to count net trades: six fewer shells requires two more forward trades than reverse trades, giving four more waves. The conserved score packages that same restriction into one test for any target card; it is not the only valid method.

Rowan puts the new challenge card beside the model. Li Wei tests their six-action plan once more. The first visitors will have a fair challenge—and a chance to discover their own explanation.

Explore Find an invariant in another story.