How many circles does it take to make an integer?

A few days ago I made a post about me beginning to play a little more with the concept of ‘minimality’ when it comes to compass and straight edge constructions. That is to try to complete a geometric construction in as few operations as possible at least for now considering the act of drawing circles and lines equally arduous and extending existing lines to infinity to be free.

Today I’ve begun to play a little with perhaps the simplest of the construction problems; that of creating a length which is a multiple of another length both starting at the same point.

It is fairly straightforward to realize that constructing an n-length will take at most n-1 operations as you can work your way over to n by means of iterated drawings of circles of radius 1 using the extremities as new centers. See the picture below for the example of constructing the length 5.

Draw5by5It is however evident that this is not the way which requires the fewest number of operations as if you instead make of use of some greater circles you can cut down the 5-construction to at least 3 operations as seen below.
Draw5by3

I suspect that some larger n-lengths might have their minimal construction by means of moving ‘off’ the line and  making use of some triangles or other shapes but if I for the time being limit myself to only making these kind of ‘circles along a line’ constructions I suppose the minimal number of circles necessary to make the numbers 2 through 8 should be visualized by

MinimalMultiples2through8

One thing you notice quite quickly when building these constructions is that they have something to do with prime numbers. Not surprising perhaps given that I was playing with multiples but I just wasn’t expecting it. You see this in the examples of 6 and 8 which are the composite numbers that the 6:construction involves first the construction of 3 and then doubling that length by the 2 construction and for the construction of 8 = 2*2*2 we resure the construction of 2 three times.

Thus if we call m_{n} the minimal number of circles necessary for constructing a length n we may put an upper limit on m_{n} for composite numbers. If n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}

m_n \leq a_1 m_{p_1} + a_2 m_{p_2} + ... + a_k m_{p_k} (1)

This is about as far I’ve come today with regards to interesting results playing with this problem but I am posing the following question for later

Question 1:  Is the inequality of (1) in fact an equality or what is the smallest n such that it is a strict inequality?

I’ve yet to make much progress with regards to some systematized way of investigating this construction problem except for the fact that you can reduce the construction to a symbolic scheme letting {0,1} be the set of the two numbers you start with and {0,1,2} the set you get when you put a circle about 1 with radius 1 and {0,1,2,4} what you get when you put a circle about 2 with radius 2 and so forth.

The idea is that to any set of numbers you can add any number which is the difference of existing numbers To {0,1,2,4} you may add 7 because 4 – 1 = 3 and you add this to 4 to get 7. I don’t have the time to think more on this right now but since today is easter and I will be tied up in family stuff for the weekend I want to make sure I remember to get back to it afterwards.

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One thought on “How many circles does it take to make an integer?

  1. Pingback: seriouscephalopod

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