Moving from the Stern-Brocot tree to the tree of surreals is done by a function f closely related to the Minkowski Question Mark function. In fact, when mapping rationals in the range (1,2) to their corresponding entries in the surreal tree in (0,1), we get f(x) = ?(x-1). The answer to the question "What function plays the role of lg in the Stern-Brocot tree?" where lg is log_2, is g(x)=f^-1(lg(f(x))) and doesn't seem to have a simpler form. ------- Original message -------- The section [of ONAG] on the game "contorted fractions" explores the area in great depth, but there's another way of looking at continued fractions with surreal numbers that I haven't seen mentioned anywhere. The shift operator acting on the Stern-Brocot tree deletes the root (x=1) and the left branch (x<1). This is equivalent to subtracting 1 until the number is less than or equal to 1. When we find ourselves at the root, we halt; if we're in the left branch, we jump to the mirror position in the right branch (1/x) and continue. We can do the same operation on the tree of surreals. Subtracting 1 will get you to the LR branch--that is, first go right, then left (0<x<1). To get to the root or left branch (x<=0), we have to apply the "pseudo-lg" function: p-lg(x) = { x * 2^k - k - 1 where 1/2^k <= x < 1/2^{k-1} This is just a linear interpolation between powers of 1/2. Finally, if we're at the root (x=0) we halt; if we're in the left branch, we jump to the mirror image in the right branch (-x). If we use lg instead of p-lg, cf's typically involve the Lambert W-function, and the omega constant plays the role of the golden mean. What function plays the role of lg in the Stern-Brocot tree? -- Mike Stay staym@clear.net.nz http://www.cs.auckland.ac.nz/~msta039