This is a riddle that a coworker of mine posed to me, I have a solution but I'm curious to see what you all arrive at (I'm more interested in the approach than the answer). The question (potentially poorly posed) is described below:
Imagine an infinite chess/checker board ($\mathbb{Z}\times\mathbb{Z}$) which is entirely empty with the exception of two strips of playing pieces such that each piece has a coordinate of the form $(k,-1)$ or $(l,0)$ $\forall l,k \in \mathbb{Z}$. The only admissible movement of pieces is via 'jumping' similar to checkers with the difference that no diagonal jumps are admissible, only those jumps which augment your coordinate for just one dimension (forward, back, and side to side).
EDIT: Only one piece may be jumped at a time, and the piece that has been jumped is removed from the board.
The question is: What is the furthest a playing piece may move in the second dimension?
This problem is some what tricky to explain but I think this should make sense, let me know if further clarification is required.
The answer is that you can get a piece to row $3$ but not to row $4$.
This is not the same problem as Conway's soldier problem, but only because the starting pieces are restricted to the two rows "below the line." We can show that you cannot get a piece to row $4$ (that is, to any $(m,4)$ spot) in a finite number of jumps, as follows:
Let the eventual ending point of the piece we are trying to propel as far above zero as possible be $(k,s)$; without loss of generality we can take $k=0$ since we could change our coordinate numbering in the horizontal direction to make this so. Let us now assign to every lattice point $(x,y)$ on the board a "distance value" $$ d(x,y;s) = |x| + | s - y | $$ and a "potential" value $$ V(x,y;s) = \varphi^{d(x,y)} $$ with $$ \varphi = \frac{\sqrt{5}-1}{2}$$ and let $V(p;s)$ for available piece $p$ which is at $(x_p,y_p)$ be $V(x_p,y_p;s)$. (Clearly, the potential values assigned will depend on the value of $s$.)
It is easy to verify that jumps going toward the "target" point leave $\sum_p V(p;s)$ unchanged, and that jumps that do not go toward the target, or that overshoot either $x=0$ orf $y=x$ decrease $\sum_p V(p;s)$. It is also trivial to see that $V(0,s;s) = \varphi^0 = 1$.
Now let's look at $$ S_s \equiv \sum_{i=-\infty}^\infty \left( V(i,0) + V(i,-1) \right) $$ $S_s$ is the largest possible starting potential possible with all pieces starting in rows $0$ and $-1$,, so if $S_s <1$ then no sequence of jumps can ever get from the starting position to any position with a piece at $(0,s)$.
$$ S_s = \left[ \varphi^s + 2\sum_{n=1}^\infty \varphi^{s+n}\right]\left[ 1 + \varphi \right] =\varphi^s \left[ 1 + \varphi \right] \left[ 1 + 2\sum_{n=1}^\infty \varphi^{n}\right] $$ where the 2 reflects the need to radiate out in pboth directions from $x=0$ and the $[1+\varphi]$ reflects $1$ for row $0$ and an extra power of $\varphi$ for row $-1$.
The infinite sum is a geometric series sum: $$\sum_{n=1}^\infty \varphi^{n} = \frac{\varphi}{1-\varphi} = \frac{1+\sqrt{5}}{2}$$ Then $$ S_s = \varphi^s \left[ 1 + \varphi \right] \left[ 1 + (1+\sqrt{5})\right] = \varphi^{s-1}\frac {\sqrt{5}-1}{2}\frac{\sqrt{5}+1}{2} (2+\sqrt{5}) = \varphi^{s-1} (2+\sqrt{5})$$ In particular, $$ S_4 = \varphi^3 (2+\sqrt{5}) = (\sqrt{5}-2)(2+\sqrt{5}) = 1 $$
So any solution involving a finite number of moves, which therefore cannot involve all the pieces on rows $0$ and $-1$, must start with a total potential $\sum_p V(p;4)$ of less than $1$, and therefore cannot end with a piece on $(0,4)$ (which would have a potential of exactly $1$).
Having shown we cannot get a piece to row $4$, we now can show that we can get a piece to row $3$, in fact, starting with just five pieces centered in row $0$ and three pieces at $(0,-1), (1,-1), (2,-1)$:
Start by jumping to $(0,1)$. Next jump the piece at $(-2,0)$ twice to end at $(0,2)$. Now jump the two pairs on the right side of the board to get them to $((1,1)$ and $(2,1)$. Finally jump the piece at $(2,1)$ twice so that it ends up at $(0,3)$.