Shortest Sierpinski Walk

This graphic is a negative of the other one shown, i.e., white $\rightarrow$ black, black $\rightarrow$ white. The white space indicates the possible routes to walk on.

Classification of Sierpinski carpet: The point $(x,y)$ exists in the Sierpinski carpet if and only if in the base 3 representation $x = 0.x_1 x_2 x_3\ldots$ , $y = 0.y_1 y_2 y_3 \ldots$ ,
1. There is no value of $n$ such that $(x_n, y_n ) = (1,1)$ , OR
2. If there is a value of $n$ such that $(x_n, y_n) = (1,1)$ , then either for $x_N = 0$ for all $N>n$ or $y_N = 0$ for all $N > n$ .

Note: For clarity, we take the "simplified form", where we disallow the eventual repeating decimal $\overline{222\ldots }$ .

Proof of classification: This follows because at the $n$ th iteration, the squares that are removed are those with $(x_n, y_n) = (1,1)$ . The second condition is to add back the left and bottom sides of the square since we only removed the interior.

Claim: The route that we're looking for, is the line connecting the points from $(0,0)$ to $( \frac{1}{3}, \frac{2}{3} )$ and then to $(1,1)$ . (The symmetric route to the point $(\frac{2}{3}, \frac{1}{3} )$ also works.

Proof of minimality: When we remove the first central square, we already see that this path is the shortest possible. More formally, let's separate the square into 2 parts with the line $y = 1 - x$ . Then, the shortest distance out of the lower left half would be from $(0,0)$ to $(\frac{1}{3} , \frac{2}{3} )$ OR from $(0,0)$ to $(\frac{2}{3} , \frac{1}{3} )$ . Similarly, for the upper right half, the shortest distance is from $(1,1)$ to $(\frac{1}{3} , \frac{2}{3} )$ OR $(1,1)$ to $(\frac{2}{3} , \frac{1}{3} )$ . This establishes that the path we chose, is the shortest possible in the first iteration.
We cannot hope to do better, since in subsequent iterations, more area is removed.

Proof of existence: The bigger question is why this path continues to exist throughout all iterations of removing central squares. Let's consider the first half of the path (under the line $y = 1 -x$ ), and the second half is dealt in a similar manner.

We will prove by contradiction that the points lying on $y = 2x$ , $0 \leq x \leq \frac{1}{3}$ lie on the Sierpinski carpet.
Suppose not. Then, there is a point $(x,y)$ with $y = 2x$ and $x_n = y_n = 1$ .
Observe that the fractional part of $3^{n-1} \times 2x$ looks like $0. 2 (2x_{n+1} ) ( 2x_{n+2} ) \ldots \geq \frac{2}{3}$
and the fractional part of $3^{n-1} \times y$ looks like $0. 1 y_{n+1} y_{n+2} \ldots < \frac{2}{3}$ .
Hence, the fractional part of $3^{n-1} (2x-y)$ is non-zero, and so this term cannot be 0.
This contradicts $y = 2x$ .

This graphic is a negative of the other one shown, i.e., white $\rightarrow$ black, black $\rightarrow$ white

Draw a line from one vertex of the unit square to the midpoint of the other side, as shown in graphic. This line passes through the vertex of the largest white [black] square. Hence, if all the other sub-squares are constructed in a like manner, this line also passes through the vertex of the largest white [black] square of each, no matter how small. Homothety, anyone? The distance from the vertex of the unit square to the vertex of the largest white [black] square then works out to

$\dfrac{1\sqrt{5}}{3}$

so that for the total path, $\;$ $a+b+c=2+5+3=10$