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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.)

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  • $\begingroup$ Can we draw diagonals? $\endgroup$
    – Carl
    Apr 18, 2016 at 0:02
  • $\begingroup$ 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.) $\endgroup$
    – Passage
    Apr 18, 2016 at 17:02
  • $\begingroup$ @ Passage - There must be exactly two line segments meeting at a point whenever a point is used to draw the polygon. . . . . . . . . . . . . . . . @ Carl - Yes. $\endgroup$ Apr 18, 2016 at 19:50

2 Answers 2

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For example it could be something like this:

16 sided polygon

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  • 2
    $\begingroup$ Congratulations - you posted 2 seconds ahead of me :) . $\endgroup$
    – Lawrence
    Apr 18, 2016 at 0:15
  • $\begingroup$ @Lawrence Probably ascii drawing is a little bit slower ;) $\endgroup$
    – kamenf
    Apr 18, 2016 at 0:18
  • $\begingroup$ Heh. I made a mistake, corrected it, then checked it again. +1 on yours, though :) . $\endgroup$
    – Lawrence
    Apr 18, 2016 at 0:28
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Please excuse the ascii art.

x-x x-x
| |/  |
x x x-x
 \   \
x-x x x
|  /| |
x-x x-x
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  • $\begingroup$ haha, that's innovative xD $\endgroup$
    – ABcDexter
    Apr 18, 2016 at 3:13

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