4 ms·

I did not know this! :) Thanks for sharing. Another thing I heard, but have not been able to prove (and is not in the Wiki article), is that if you record whet

by hawkice 12y ago

I did not know this! :) Thanks for sharing. Another thing I heard, but have not been able to prove (and is not in the Wiki article), is that if you record whether you take the Left or Right segments in the algorithm, you can construct a continued fraction for any real number as int(N) + 1 / (1 (L -> -, R -> +) (number of times you took that direction before "turning around") / 1 (L -> -, R -> +) (#direction) ... ))). My inability to prove may be primarily because I never really got the flavor of proofs about continued/infinite fractions -- considering the source I'd say it's likely true (or I misremembered). You don't happen to know where I'd look for hints on the proof there, do you? :)

robinhouston 12y ago
Yes, this is explained in http://en.wikipedia.org/wiki/Stern–Brocot_tree http://en.wikipedia.org/wiki/Stern–Brocot_tree Such a beautiful circle of ideas!
sheetjs 12y ago
I'm sure it's hiding somewhere in Concrete Mathematics (knuth/graham/patashnik), and I'm sure there are more elegant proofs, but it's not too hard to work out this proof yourself. Here is a hint: Suppose you were at a point in the mediant algorithm where the two terms were a/b and c/d. Taking the left segment transforms this to [a/b, (a+c)/(b+d)] and taking the right segment transforms this to [(a+c)/(b+d), c/d]. Represent this as a matrix of the form [[a,b],[c,d]]. In this form, what does "take the left segment" look like in terms of a matrix operation? what does "take the right segment" look like? Can you find any useful properties (for example, how do you repeat) A convenient root matrix is [[0,1],[1,0]]