This is an extension of the discussion
A robot is placed on a grid point. At each move the robot must take three steps along the edge of the grid. After each step the robot must turn right. Lengths of each step are $a$, $b$, and $c$ edges. It doesn't matter in which order $a$, $b$, $c$ are, since the trajectory forms a loop. A permutation of $a$, $b$, $c$ amounts to starting somewhere else on the loop or mirroring the loop. We can assume $0<a≤b≤c$. After each move the robot must turn right too. The robot can revisit grid points and edges. After four moves the robot must return to the start grid point and stop.
There are five different combinations $a$, $b$, $c$ for determining the move trajectory:
The first three steps in each figure are highlighted by red color.
All grid points are numbered in the order of passing, if the grid point has already been visited, then its number does not change.
In addition, the color of the grid point changes. If a grid point was visited right-down, right-up, left-down, or left-up (i.e. corner point), then this point is marked by red color. If a grid point has been visited from top to bottom or from bottom to top (i.e. through point), then the point is marked by yellow. Grid points that were visited as corner and through are marked in green (mix point), and points that could be visited as corner and / or through points are marked in blue (central point).
Question 1. Can you write down a relation between the number of different grid points (corner, through, mix and central) for any $a$, $b$, and $c$?
Question 2. Can you write down the minimum number of edges you need to get a closed loop for any $a$, $b$, and $c$?
I am looking for an answer based on graph theory.
My partical answer
CASE A. If $a+b>c$ and ($a=b<c$ or $a<b=c$) then the loop forms something like a Swiss cross in square (figure A). It has $4$ central grid points (blue), $4$ corner grid points (red), $8$ mix grid points (green), and $4(c-3)$ through points (yellow).
CASE B. If $a+b<c$ and $a=b$ then the loop forms something like an apple command key ⌘ (figure B). It has $4$ central grid points (blue), $12$ corner grid points (red), and $4(2a+c-5)$ through points (yellow).
CASE C. If $a+b = c$ then the loop forms something like a windmill (figure C). It has $1$ central grid point (blue), $8$ corner grid points (red), $4$ mix grid points (green), and $4(2a+2b-3)$ if $a \neq 1$, else $4(2b-3)$ through points (yellow).
CASE D. If $a+b > c$ and $a \ne b < c$ then the loop forms something like a flower (figure D). It has $12$ central grid points (blue), $12$ corner grid points (red), $0$ mix grid points (green), and $4(a+b+c-7)$ through points (yellow).
CASE E. The case $a = b = c$ is simplest one (figure F). We have $4$ corner grid points (red), $4(a-1)$ through points (yellow), and the minimum number of edges is $4a$.
