Conway's soldiers cannot reach the fifth row

Idea. Use a potential function with the golden ratio so that jumps toward the target keep the total unchanged and all other jumps decrease it.

Weights. Let $\sigma=\frac{\sqrt5-1}2\approx0.618$, so $\sigma^2=1-\sigma$. The weight of the point $P=(x,y)$ is $\sigma^{d(P)}$ with $d(P)=|x|+|5-y|$. The target $T=(0,5)$ has weight $1$.

Initial weight. Using $\sum_{k\ge0}\sigma^k=\frac1{1-\sigma}=\frac1{\sigma^2}$ and $\sum_{x\in\mathbb Z}\sigma^{|x|}=1+\frac{2\sigma}{1-\sigma}=1+\frac2\sigma$, for $y\le0$,
$$W_0=\sigma^5\cdot\frac1{\sigma^2}\left(1+\frac2\sigma\right)=\sigma^3+2\sigma^2=1.$$

A move does not increase the weight. A move takes three consecutive points $P,Q,R$ on a line and replaces soldiers on $P,Q$ by one on $R$. Along a line, the taxicab distance to $T$ is convex, so the triple $(d(P),d(Q),d(R))$ has one of the forms:
- $(n+2,n+1,n)$: change $\sigma^n-\sigma^{n+1}-\sigma^{n+2}=\sigma^n(1-\sigma-\sigma^2)=0$;
- $(n,n+1,n+2)$: change $<0$;
- $(n+1,n,n+1)$: change $<0$.
So the total weight never increases.

Conclusion. After any finite sequence of moves the weight is at most $W_0=1$. Suppose a soldier reached $T$. It contributes weight $1$; but infinitely many other soldiers remain, each with positive weight, so the total would exceed $1$. Contradiction. ∎

Remark. The golden ratio is exactly the value making jumps toward the target neutral, which is why row $4$ can be reached but row $5$ cannot.

Lesson. Weight/potential functions: choose weights that make the dangerous move neutral and every other move a decrease.