How to solve Wednesday Frog

Every frog hops three times: on Monday, on Tuesday and on Wednesday.

A hop goes straight up, down, left or right and always ends two squares away: the frog hops over the square beside it and lands just beyond it.

A frog may hop only over water or a stone, never over a lily pad or over any frog’s starting square.

On Monday and Tuesday a frog lands on water, and on Wednesday on a lily pad. Exactly one frog lands on each lily pad.

Every stone must be hopped over exactly once. No frog may land on a stone.

No square is used twice, not even on another day or after the frog has left it: not to start from, to land on or to hop over.

Tap a frog, then the square it should hop to. Tapping a square the chosen frog has landed on takes back that hop and every one after it.

With Crosses on, tapping a square puts a cross on it or takes it off. Crosses are your own notes; the game does not check them.

Nothing is given but the frogs, the lily pads and the stones, and there are always as many lily pads as frogs. Every board has exactly one solution, and it can be reached without guessing. A 6 × 6 needs no more than the only path and the one frog for a lily pad or a stone; an 8 × 8 also needs the squares a frog takes on every path, and a 9 × 9 the paths that leave others stuck as well.

The rules above say it all; here each one is drawn. A solid line is a hop that is allowed, a dashed one with a cross at the end a hop that is not, and a dot a square a frog may hop to. The board will not make a hop that is not allowed: it says in one sentence which rule it breaks.

The simplest move is a frog with only one path. You check it hop by hop — on Monday and Tuesday it must land on unused water, on Wednesday on a lily pad — and sometimes only one of the four directions is left. Such a path is settled, and from then on its squares belong to that frog alone: no one else hops over them or lands on them, so every path of another frog that touches them is out.

Some frog must land on every lily pad, and some frog must hop over every stone. So when only one frog can still reach a lily pad, that frog ends there, and its paths to other lily pads are out. The same goes for a stone: if only one frog can still hop over it, that frog’s path must go over it. A 6 × 6 needs a stone settled this way at least once.

When a frog has several paths, you need not know which one it will take to learn something from them. The squares all its paths use — it lands on them or hops over them, whichever way it goes — are certainly its own, and so nobody else’s: every path of another frog that touches them is out. An 8 × 8 needs a frog looked at this way at least once.

When everything else has gone quiet, one question is left: “what if?” Take one of a frog’s paths and check whether, if the frog went that way, everyone would still have a way out: every other frog some path, every stone a frog to hop over it, every lily pad a frog to land on it. If not, that path is out. This is not guessing: nothing is drawn in to try it out; you only look at what that path takes from the others. A 9 × 9 needs a path ruled out this way at least three times.