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*To*: categories@mta.ca (Categories List)*Subject*: categories: Re: Real midpoints*From*: Peter Selinger <selinger@math.lsa.umich.edu>*Date*: Sun, 16 Jan 2000 20:23:36 -0500 (EST)*In-Reply-To*: <199912261845.NAA19441@saul.cis.upenn.edu> from "Peter Freyd" at Dec 26, 1999 01:45:08 PM*Sender*: cat-dist@mta.ca

Dusko Pavlovic wrote on Jan 15: > peter's bipointed approach, however, induces a *redundant* > representation. this allows him to extract the algebraic operations > coinductively. Sorry for lagging a few weeks behind in this discussion. Something has been bothering me about Peter's definition of the midpoint operation (Dec 26). How can it be coinductive, but the resulting function is not computable? After mulling it over, I realized that Peter's definition is *not* coinductive: He gives an equation which has a unique fixpoint, but it is not a coinductive fixpoint. I believe that Peter's representation has not enough redundancy to define the algebraic operations in a purely coinductive manner. Note however that Peter really only wanted to define the algebraic structure from the final coalgebra structure on I = [-1,1]; he did not claim to do it purely coinductively. Recall Peter's definition of the midpoint operation (where h:I-->I is the "halfing" map h(x) = x|0). I have corrected some typos in lines 4+5: > Let g be the endo-function on I x I defined recursively by: > > g<x,y> = if dx = T and dy = T then <x,y> else > if dx = T and uy = B then (h x h) (g<ux,dy>) else > if ux = B and dy = T then (h x h) (g<dx,uy>) else > if ux = B and uy = B then <x,y>. > > The values of g lie in the first and third quadrants, that is, those > points such that either dx = dy = T or ux = uy = B. The two maps > > g d x d g u x u > I x I --> I x I --> I x I and I x I --> I x I --> I x I > > give a coalgebra structure on I x I. The midpoint operation may be > defined as the induced map to the final coalgebra. I think Vaughan has already pointed out that this is not well-defined, because the four cases in the definition of g are not mutually disjoint and the right-hand sides do not match; thus let us assume we write "dx != T" instead of "ux = B", and similarly for y. Note that the resulting map is neither continuous nor monotone; thus, we are definitely forced to work with bipointed sets, not posets, for this construction. My point is that the definition of g is neither recursive nor co-recursive. The definition of g is merely expressed as a fixpoint equation, and Peter has set up things so that there is indeed a unique solution to this equation. However, the reason the fixpoint is unique is because the "h" part is contracting, and not because the equation is co-recursive. If we write g as a pair <g1, g2>: I x I --> I, we can separate the equation for g into two equations about g1 and g2. A priori, these two equations are mutually dependent, but interestingly, it turns out that they are independent, so we need not worry about simultaneous fixpoints. The equation for g1 can be written as s x s dist I x I -------> (I + I) x (I + I) ------> IxI + IxI + IxI + IxI | | (1) | g1 | pi1 + g1 + g1 + pi1 | | [h-,h,h,h+] I <---------------------------------- I + I + I + I (add downward arrow heads), where s: I --> I v I --> I + I is the composition of the coalgebra morphism F for I with one of the two obvious maps I v I --> I + I (the one that sends the midpoint to the second component, say). "dist" is distributivity, and h- and h+ are the maps defined by h-(x) = x|-1 and h+(x) = x|1 (these have explicit definitions similar to that of the halfing map h). The equation for g2 is similar. One obtains it by replacing pi1 with pi2. For this diagram to be considered a co-recursive definition of g1, it would have to be equivalent to a diagram of the form G I x I ---------------> (I x I) v (I x I) | | (2) | g1 | g1 v g1 | | F I ---------------------> I v I for some suitable coalgebra G on I x I. This, however, is not the case, unless (in a suitable sense) G is just as complex as g1 is. For instance, the first bit (trit?) of G(x,y) is the same as that of g(x,y), which is the same as the first bit of midpoint(x,y). But the first bit of the midpoint cannot be computed with finite lookahead (consider x = 1/3 = +-+-+-... and y = -1/3 = -+-+-+...). Thus, the first bit of g or G can't be computed either, and no matter through how many levels of co-recursion we go, we never arrive at an elementary function. Thus, there is no chance of defining the midpoint operation from elementary functions through a finite chain of co-recursions. It ought to be possible to formalize this argument. Intuitively, note that in diagram (2), g1 appears in something like a "tail recursive" position, whereas in diagram (1) it does not. Thus, if G is computable with finite lookahead (as a function on pairs of streams), and g1 is defined like in (2), then g1 is also computable with finite lookahead: namely, G will tell us the first trit, plus a continuation, and the tail recursive call will compute the rest. Thus computable functions on streams are closed under co-recursion (as they should). Finally, Peter's definition of the midpoint function utilizes the function g and then lifts it through one level of co-recursion. It is, however, possible, to define the midpoint function directly without going through this g: It is the unique morphism m: I x I --> I that makes the following diagram commute: s x s dist I x I -------> (I + I) x (I + I) ------> IxI + IxI + IxI + IxI | | (1) | m | m + m + m + m | | [h-,h,h,h+] I <---------------------------------- I + I + I + I. Or, if you prefer, it is the unique solution of m<x,y> = if dx = T and dy = T then h+ (m<ux,uy>) else if dx = T and dy != T then h (m<ux,dy>) else if dx != T and dy = T then h (m<dx,uy>) else if dx != T and dy != T then h- (m<dx,dy>) This definition, of course, is not co-recursive either. If I am not mistaken, it is more or less what Vaughan defined in his mail on Dec 24, except he separated the case for dx = T into two cases ux != B and (dx=T and ux=B) (and similarly for y). Best wishes, -- Peter Selinger

**Follow-Ups**:**categories: Re: Real midpoints***From:*Dusko Pavlovic <dusko@kestrel.edu>

**References**:**categories: Real midpoints***From:*Peter Freyd <pjf@saul.cis.upenn.edu>

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