I’d like to thank Jonathan Lewis for taking up my challenge and writing up a proof that there is indeed a formula that solves Cary Millsap‘s string problem for not just circles, but for any regular polygon.
Like Jonathan, I found the problem intriguing, and “wasted” a few hours on a Saturday afternoon discovering the formula.
This blog entry isn’t about the formula or proof — it’s rather about the process I used to discover it.
You see, I didn’t know that such a formula actually existed before I set out — but I was convinced that there might be. After all, it would be neat if there was one, wouldn’t it?
This is a common way that I approach problems, especially in the Oracle programming and tuning world — usually I assume the existence of a solution and then I go about trying to find it. One of the things I’m constantly surprised at is how many people “give up” on these kinds of problems, start guessing, guess wrong, and then decide that such a solution either doesn’t exist, or that they’re not smart enough to find it. This post is an attempt to debunk the idea that finding such solutions requires any special smarts — that today there are so many ways to learn and discover that anyone can do it. It just takes perseverance and a bit of dedication.
I started out on my quest to find the formula for regular polygons by searching Wikipedia for “Regular Polygon”
About half-way down the page, I found the following image:
The little letter a looked exactly like what I was looking for — something that resembled the radius of a circle. Before reading this article I had no idea was it was called — from this article I learned it was called the apothem.
From there, I clicked on the link to the definition of the apothem, trying to see if there was a way to calculate its length somehow and relate it to the perimeter of the polygon.
In the article, a way to calculate the apothem based on the length of the sides and the number of them is presented:
a = s / (2 tan (pi / n))
with the comment that the formula can be used when only the perimeter (p) and the number of sides (n) is known because s = p / n.
From here, the rest becomes relatively simple, as I substituted s = p / n for s in the formula for the apothem.
a = (p / n) / (2 tan (pi / n))
Given a new apothem of length a’ (where a’ = a + h); we can work through the formulas to determine the difference between our new perimeter p’ and the original perimeter p.
We start by isolating the perimeter:
a = p / (2n tan (pi/n))
2an tan (pi/n) = p
p = 2an tan(pi/n)
Looking at our new perimeter (p’), using the new apothem (a’):
p’ = 2a’n tan(pi/n)
And we want p’ – p (the difference between the perimeters)
p’ – p = 2a’n tan(pi/n) – 2an tan(pi/n)
Factor out the 2n tan (pi/n) which is common to both terms on the right:
p’ – p = (2n tan(pi/n)) (a’ – a)
And since we know that a’ simply equals a + h (the “height” we’re raising the string):
p’ – p = (2n tan(pi/n))(a + h – a)
(This was my aha! moment , sorry, bad joke)
p’ – p = (2n tan(pi/n))(h)
p’ – p = 2nh tan(pi/n)
Voila, a solution that only requires the number of sides and the “height”.
I was satisfied that this solution matched Cary’s just by simply trying out the formula with 1,000-sided polygon and seeing how it matched — Jonathan went beyond and showed how the approximation of the tan function using a Taylor series (actually a MacLaurin series) gets you to 2hpi.
Why all this focus on a simple problem?
Mainly because it’s been interesting to see how people approach solving it and — in my mind, more interesting — how they attempt to generalize it in a way that makes it really applicable.
When I’m confronted with a particularly challenging Oracle problem, I generally assume there’s a pretty easy way out — I can’t be the first person to ever have encountered such a problem, and the database developers have probably thought of it too, so I usually start with the “Big 3″ manuals:
The Concepts Guide, the SQL Reference Guide, and the PL/SQL Packages and Types Reference.
Don’t re-invent the wheel, and don’t assume that the problem is unsolvable — while we might not always stand on the shoulders of giants, we should start by assuming our problem isn’t unique, and that the proof or truth is out there…