Original Post
I have been wondering this for many years and haven''t seen a proof... it''s probably easy. You can''t draw a 6-pointed star from vertex to vertex (starting and finishing each line at an outside vertex) without lifting your pen off the paper, yet you can draw a 5,7,8,9,10,11, etc. pointed one in this manner. It''s because all integers between 2 and (6-1)/2 divide into 6 with no remainder, you can''t skip 1 vertex (you''d get a hexagon, not a star) and you can''t skip 2 vertices, because 2 divides 6 so you just get a triangle and 3 vertices you didn''t visit. Skipping 3 vertices results in a line. Skipping more than 3 verts is the same as skipping less than 3 in the opposite direction. That''s why it can''t be done.
So is 6 the only integer n >4 with the property that all integers from 2 to (n-1)/2 divide into it? I suspect that it is. Hence 6 is the only pointy star you must draw as 2 polygons (if you start from a vertex at the edge)? Has anyone got a proof or a counter-example? Some stars can be drawn in more than one way... 7 pointed ones can be drawn by skipping 2 or 3 vertices. What''s the pattern of the sequence of how many stars you can draw for each number of points? I skived most of my number theory lectures at uni.
Of course you can draw a 6 pointed star without lifting your pen off the paper if you take a turn between the edge vertices (all nodes are 4-nodes).
When I was at school I wrote a program to draw stars in this manner, so it''s got something to do with programming. It didn''t make a very exciting game though.
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