Essay · December 2025

Escaping the Killer by Always Getting Closer

Aadesh Tikhe · PDF · Download · Desmos

A rule that improves things at every step can still never finish. The rule here: get closer to the target, every single step.

The setup

You stand at point $B$. A killer starts at point $A$ and will kill you the moment he reaches $B$. He is dumb but perfectly obedient, and you get to program him with one rule: after every step, he must be closer to you than he was before. You also choose how much closer he gets each step, and in which direction he moves.

The natural belief is that this rule guarantees your death: if he keeps getting closer, how can he not arrive? But that belief hides an assumption: that getting closer also limits how much useless movement is possible. So the real question is: can you force him to walk an enormous total distance, even an infinite one, while he gets closer at every step?

The exploit

“Distance to $B$ must decrease” only says that each step lands inside the circle around $B$ through his current position. It limits his radial motion and says nothing about tangential motion. He can circle almost forever while the radius shrinks by tiny amounts. Displacement is the straight line from $A$ to $B$; total distance is the sum of every step, and a spiral can make the second as large as you like compared with the first.

Figure 1. The killer follows $r(\theta)=R-\dfrac{m\theta}{2\pi}$ with $R=6$. His distance to you only ever shrinks, yet for small $m$ the distance walked explodes. Shrink $m$ and press replay.

The Archimedean spiral

Take $r(\theta)=R-\dfrac{m\theta}{2\pi}$. As $\theta$ increases the radius strictly decreases, so the rule is never broken. He needs $N=R/m$ full turns to reach the centre, and each turn adds about one circumference of walking for only $m$ of real progress.

The arc length from $\theta=0$ to $\theta=2\pi R/m$ is

$$L=\int_0^{2\pi R/m}\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta =\int_0^{2\pi R/m}\sqrt{\Bigl(R-\frac{m\theta}{2\pi}\Bigr)^2+\Bigl(\frac{m}{2\pi}\Bigr)^2}\,d\theta.$$

Substitute $u=R-\frac{m\theta}{2\pi}$, so $d\theta=-\frac{2\pi}{m}\,du$ and the limits run from $u=R$ to $u=0$. With $a=\frac{m}{2\pi}$,

$$L=\frac{2\pi}{m}\int_0^{R}\sqrt{u^2+a^2}\,du =\frac{2\pi}{m}\left[\frac{u}{2}\sqrt{u^2+a^2}+\frac{a^2}{2}\ln\Bigl|u+\sqrt{u^2+a^2}\Bigr|\right]_0^R.$$

Evaluating and simplifying gives the plug-and-play formula

$$\boxed{L=\frac{\pi R^2}{m}\sqrt{1+\Bigl(\frac{m}{2\pi R}\Bigr)^2}+\frac{m}{4\pi}\ln\!\left(\frac{2\pi R}{m}\Bigl(1+\sqrt{1+\Bigl(\frac{m}{2\pi R}\Bigr)^2}\Bigr)\right)}$$

Taking the limit

As $m\to0^+$ the dominant term $\pi R^2/m$ grows without bound: $$\lim_{m\to0^+}\frac{\pi R^2}{m}=\infty.$$ So the killer can be made to walk an arbitrarily long distance (in the limit, an infinite one) while always getting closer. The discrete picture is the harmonic series: if step $n$ moves him $1/n$ along the path while his distance to you still falls, then $\sum 1/n$ diverges.

Figure 2. $L$ against $m$ for $R=6$ on log–log axes; the line has slope $-1$. Hollow points are the rows of the table below.
Rmm / 2πR√(1 + (m/2πR)²)L
610.026531.000352113.4811
60.80.021221.000225141.6929
60.50.013261.000088226.4144
60.20.005311.000014565.5890
60.10.002651.0000041131.0301
60.050.001331.0000012261.9778
60.0010.000031.000000113097.3365

Why this matters

The original paper

The essay as typeset in December 2025.

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