A robot starts on a cell in an infinite grid. On the first turn it can move 1 cell horizontally or vertically. On the $n$-th turn ($n>1$) it can move $n$ cells horizontally or vertically, but it cannot revisit cells (еxcept the starting cell). Is the robot able to come back to its starting cell?
While Deusovi has a perfectly valid answer to the question, there are other possibilities.
Firstly, there is a possible loophole that allows for a short solution.
If the robot shuts down as soon as it reaches its starting point, then the final turn does not need to be fully completed. In this case you can do it in $5.5$ turns:
1R, 2D, 3D, 4L, 5U, 3R
where the letters indicate the directions of the steps (Right, Left, Up, Down).
Even without that loophole, it is possible to
solve it in $7$ turns.
1R, 2D, 3L, 4D, 5L, 6U, 7R
Note that this is not a golygon since two moves (1 and 7) form a single side of this hexagon.
Proof this is minimal:
The total distance travelled must be even, because any displacement to the right must be compensated by an equal displacement to the left, and similarly for up and down.
Clearly $4$ turns is not sufficient without retracing steps. In $5$ and $6$ turns the robot would travel a distance of $15$ and $21$, but these are odd numbers so not possible. This means $7$ is minimal.