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The number 6

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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. "Most people think, great God will come from the sky, take away everything, and make everybody feel high" - Bob Marley

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> So is 6 the only integer n >4 with the property that all
> integers from 2 to (n-1)/2 divide into it?

I think so. Every bigger number must also be divisible by 2 and 3, so the next is 12, then 18, 24, 30, ... and I think it''s safe to say that for each of these numbers has a number relatively prime to it that is less than half it. Wouldn''t like to prove it though.

> What''s the pattern of the sequence of how many stars you can
> draw for each number of points?

I think the formula is the whole number part of

(n + 1) / 2 - 2

giving a table of

3 4 5 6 7 8 9 10 11...
0 0 1 1 2 2 3 3 4...

If you want stars which are all in one piece, i.e. that can be drawn in one go as you describe, you need stars where the number of polygon corners you count for each star edge is relatively prime to the polygon size. For polygons with a prime number of corners this is all of them, but for others it''s not. The series looks like this:

3 4 5 6 7 8 9 10 11...
0 0 1 0 2 1 2 1 4...

With zero such stars possible for a hexagon, as expected. There''s not going to be a predictable pattern or formula for this series, just as there''s not one for the number of primes or the nth prime.

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Google for "Euler function". The special property of 6 is that Phi(6)=2 (1 and 5 are the only numbers below 6 that have no common factors with 6).

If a number''s factorization is n = P1^i1 * P2^i2 * ... * Pk^ik, then
Phi(n) = P1^(i1-1)*(P1-1) * P2^(i2-1)*(P2-1) * ... * Pk^(ik-1)*(Pk-1)
From that, it is very easy to prove that no number larger than 6 will have that property.

The sequence that johnb refers to at the end is just (Phi(n)-2)/2. the -2 is because 1 and (n-1) are not good steps to draw a star (you would draw a polygon), and the /2 is because steping k or (n-k) gives the same result.

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Cheers folks. So it was another of Euler's discoveries then?

My explanation wasn't that great, I should have said that for n pointed stars you need to find a number k such that 2k, ... , (n-1)k aren't congruent to zero modulo n.

Skipping 2 then 3 vertices when drawing a 6-pointed star is cheating in my book! (EDIT: and I just drew it, and it produces a 5 pointed star anyway!)

"Most people think, great God will come from the sky, take away everything, and make everybody feel high" - Bob Marley

[edited by - Paradigm Shifter on January 29, 2003 5:22:38 AM]

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