# How can a 16-sided non-self-intersecting polygon be drawn on a 4-by-4 grid?

Let there be a square 4-by-4 grid of points in the plane.

How can a 16-sided non-self-intersecting polygon be drawn on a 4-by-4 grid if the points are the vertices of the polygon?

.

(Don't count reflections/rotations as different polygons.)

• Can we draw diagonals?
– Carl
Apr 18 '16 at 0:02
• Is it considered to be an intersection if the perimeter touches the same point twice, but does not cross? (For instance, it comes in from the bottom, leaves left, does something else, comes in from the top, and leave right.) Apr 18 '16 at 17:02
• @ Passage - There must be exactly two line segments meeting at a point whenever a point is used to draw the polygon. . . . . . . . . . . . . . . . @ Carl - Yes. Apr 18 '16 at 19:50

For example it could be something like this:

• Congratulations - you posted 2 seconds ahead of me :) . Apr 18 '16 at 0:15
• @Lawrence Probably ascii drawing is a little bit slower ;) Apr 18 '16 at 0:18
• Heh. I made a mistake, corrected it, then checked it again. +1 on yours, though :) . Apr 18 '16 at 0:28

x-x x-x
| |/  |
x x x-x
\   \
x-x x x
|  /| |
x-x x-x

• haha, that's innovative xD Apr 18 '16 at 3:13