Jtdylktuy

I have found the Gridpatterns page.

One can apply the next algoritm and obtaine the grid pattern "1-2-3".

1. On square grid paper start in the middle.


2. Draw a line 1-unit long.


3. Turn a right angle clockwise.


4. Draw a line 2-unit long.


5. Turn a right angle clockwise.


6. Draw a line 3-unit long.


7. Repeat steps 1-6 four times.

Result is below on the left figure. As you can see we returned to the start point (red).

Edit. After the lesnik's answer. Your positions after step 6 will be four points forming a square (right figure).

Question. Is there a rule by which it can be determined that the pattern "a-b-c" will return (convergence) to the starting point? Here a, b, c are integer.

Looks like for any $a$, $b$ and $c$ following your algorithm you will return to the starting point.

After steps 1-6 you shift from original position by some vector $vec{x_1}$ and turn by 90 degrees. Now you repeat steps 1-6, and shift from your current position by some vector $vec{x_2}$. Which is exactly like $vec{x_1}$, but turned by 90 degrees. And so on. Sum of vectors $vec{x_1}$, $vec{x_2}$, $vec{x_3}$ and $vec{x_4}$ is zero because all of them have the same length, and each one is turned by 90 degrees.

Your positions after step 6 will be four points forming a square.