trap frogs

a frog jumps around on the number ray and you try to catch it. jumping and catching always alternate. the frog starts at position \(s \in \mathbb{Z}\) and at every move it jumps a distance of \(z \in \mathbb{Z}\) (if \(z>0\), it jumps to the right, others if to the left). \(z\ ) is the same at every jump. catching consists of giving an integer position. you don't know \(z\ ) or \(s\ ). we show that there is a way to always catch the frog.

First, \(a_1 = s\) and \(a_{n+1} = a_n + z = s + n \cdot z\) with \(s,z \in \mathbb{Z}\).

We now choose 

$$h:\mathbb{N} \to \mathbb{Z}^2: h(2^k r) = \left ( (-1)^{k+1} \left \lfloor \frac{k+1}{2} \right \rfloor, (-1)^{\frac{r+1}{2}} \left \lfloor \frac{r+1}{4} \right \rfloor \right ) $$

as the function that assigns (exactly) a number tuple of whole numbers to every natural number. The choice of this function is through the functions \(f(n) = (-1)^n \left \lfloor \frac{n}{2} \right \rfloor\) , the \(\mathbb{N}\) on \(\mathbb{Z}\) and \(g(2^kr) = (k+1, \frac{r+1}{2})\) , which \(\mathbb{N}\) on \(\mathbb{N}^2\) map bijectively, motivated.

We now show the surjectivity of \(h\) ( \(h\) is also injective, but we do not need this property).

Let \((x,y) = (2^{k_1} r_1, 2^{k_2} r_2) \in\mathbb{Z}^2\) . But then

$$h \left ( 2^{2 \cdot 2^{k_1} r_1 - 1} \cdot (4 \cdot 2^{k_2} r_2 - 1) \right ) = (2^{k_1} r_1, 2^{k_2} r_2) = (x,y).$$

Hence: \(\forall (s,z) \in \mathbb{Z}^2 \, \exists \, m \in \mathbb{N}\) with \(h(m) = (x_m,y_m) = (s, z)\) .

If, for example, the move is at \(n = 88\), we calculate \(h(88)=(2,3)\) and choose \(2 + 88 \cdot 3 = 266\) as the position.

Then after exactly \(m\) moves with \(x_m + m \cdot y_m = s + m \cdot z = a_m\) the choice falls on the frog.

Besides \(h\), many other functions are possible, such as the Cantorian pairing function or a bijective spiral.

A simple implementation in JavaScript follows:

See the Pen catch the frog by David Vielhuber (@vielhuber) on CodePen.