How to solve Frog Slalom

The frog starts on its own square and ends its route on the lily pad with the flower.

It moves only by leaps, straight along a row or a column. After every landing it turns left or right; the turns need not alternate. It never leaps straight on or back.

In every leap the frog flies over exactly one stone. A leap may be any length, and the stone may lie anywhere between the square the frog takes off from and the square it lands on.

The frog flies over every stone exactly once. It never lands on a stone.

The route never crosses itself or comes back. A square the frog has landed on or flown over is taken, even if it is water or the starting square: no later leap lands on it or flies over it. The frog never flies over the lily pad with the flower: it lands there only with its last leap.

You draw the route from the frog: tap the square the frog should land on, and a leap is added. Tapping a square the frog has landed on takes back that leap and every later one.

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 frog, the stones and the flower. Every board has exactly one solution, and it can be reached without guessing. A 6 × 6 needs no more than the first three techniques: every stone has its own leap, the first leap and the last, and who takes over the route. A 7 × 7 also needs a square the route takes whatever the leap, or a leap that leaves another stone nothing; an 8 × 8 needs the latter at least three times, and sometimes the last technique as well: too early for the flower.

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

In every leap the frog flies over one stone, and over every stone once, so there are as many leaps as stones, and every stone has a leap of its own. That is why you start with the stones: where can the frog fly over this one? Along a row or a column, from water on one side to water on the other, with no other stone on the way. A leap that ends on a square with no way on is out — everywhere but at the flower the frog turns after every landing and flies over another stone. And when a stone is left with one leap, that leap is settled.

Two squares of the route are known from the start: the frog’s square and the lily pad with the flower. Exactly one leap leaves the frog’s square, and exactly one ends on the flower. If only one stone’s leaps reach the flower, the route’s last leap goes over that stone — and the same goes for the first leap and the frog’s square.

At every end of a leap, except the frog’s square and the flower, the route turns: two leaps meet there, and the second runs at right angles to the first, over another stone. When a settled leap ends on a square that only one stone’s leaps can reach at right angles, that stone’s leap must start right there. A 6 × 6 needs this technique or the previous one at least once.

Sometimes it is not yet known where the frog will fly over a stone, but it is known where that leap starts or ends: all the leaps the stone has left share a square. The route certainly passes through that square and turns there, and the leap that takes over the route there must belong to the one stone that can reach it at right angles.

Every stone must be flown over. A leap that would leave some stone without any leap is out, even if it breaks no rule itself. It is easiest to see when all the leaps over a stone use the same square — landing on it or flying over it: that square belongs to that stone. A 7 × 7 needs this technique or the previous one at least twice, and an 8 × 8 this one at least three times.

There is one route, and it runs from the frog to the flower. A leap that would join the frog to the flower before the frog has flown over every stone is out, and so is a leap that would close settled leaps into a loop. This technique rarely decides anything, and only on an 8 × 8, but when it does, nothing else will help.