Self-Similar Geometric Recurrences (Fractal Length/Perimeter)

Many "fractal" konkur problems (Koch snowflake, Cantor-style lengths, Sierpiński perimeters) hide a very simple engine: a self-similar substitution rule applied to every piece of the figure, over and over. If you spot the rule, the answer is just a geometric sequence — no calculus, no figure-counting.

The substitution rule

Fix attention on one straight segment of length $L$. A single iteration is described by two numbers:

• $k$ = how many segments each segment is replaced by,
• $r$ = the length scale factor of each new segment ($0 < r < 1$, usually $r=\tfrac{1}{3}$ or $\tfrac{1}{2}$).

So one iteration sends

$L \;\longrightarrow\; k\,(rL),$

meaning the length of that piece is multiplied by the constant ratio $q = kr$. Because every segment obeys the same rule, the total length $P_n$ of the whole figure obeys the same multiplier:

$P_{n+1} = (kr)\,P_n.$

Solving the recurrence

This is a first-order linear recurrence with constant ratio $q = kr$, so it is purely geometric:

$\boxed{\,P_n = (kr)^{\,n}\,P_0\,}$

where $P_0$ is the length/perimeter of the starting figure (step $0$). Watch the indexing: "after $n$ steps" means multiply $n$ times. Equivalently, tracking the number of segments $N_n = k^{n} N_0$ and the length of each $\ell_n = r^{n}\ell_0$ separately gives the same product $P_n = N_n\,\ell_n = (kr)^n P_0$.

The behavior as $n\to\infty$ is decided by $q=kr$:

How to apply it

1. Look at a single segment after one step. Count $k$ (new segments) and read off $r$ (scale).
2. Form the ratio $q = kr$.
3. Identify $P_0$ = perimeter of the initial figure (don't forget it has several sides).
4. Plug into $P_n = (kr)^n P_0$, being careful whether the question counts from step $0$ or step $1$.

Mini-example: Koch curve from one segment

Koch replaces the middle third of a segment by the two sides of a small equilateral triangle, so one segment becomes $k=4$ segments each scaled by $r=\tfrac{1}{3}$. Thus $q = 4\cdot\tfrac{1}{3} = \tfrac{4}{3}$. Starting from a single segment of length $L=P_0$, after $n$ steps the length is

$P_n = \left(\tfrac{4}{3}\right)^{n} L.$

For the full Koch snowflake the seed is an equilateral triangle, so $P_0 = 3s$ for side $s$, and the perimeter after $n$ steps is $P_n = 3s\left(\tfrac{4}{3}\right)^{n}$ — which diverges, encircling a finite area with an infinite boundary.

Exam trick: The whole problem reduces to the ratio $q=kr$ of one segment — never try to count all the little edges in the final figure. And mind the step count: the figure "obtained at the $n$-th step" has been transformed $n$ times, so the exponent is $n$, not $n-1$ (and not $n+1$).