Hi. I'm reading denotational semantics article on haskell wikibook and don't understand following statement from the ["Interpretation as Least Fixed Point" chapter][1] chapter:
However, there might be multiple fixed points. For instance, there are several f which fulfill the specification
f = n -> if n == 0 then 1 else f (n + 1)
Of course, when executing such a program, the machine will loop forever on f(1) or f(2) and thus not produce any valuable information about the value of f(1).
What is this f exactly? If it's just another recursive function, like the factorial was before, then it has a little difference from it to me: the above f is defined on n <= 0 and not defined on n > 0. Whereas factorial is defined on other part of values. Thus, both functions will not produce any results on certain values, if such program is executed: for the above f this is f(1), f(2), etc, but for factorial this is f(-1), f(-2), etc (who prevents me from executing factorial program on negative integers, after all?). Both functions can be approximated using corresponding functional g and fixed point iteration scheme, like was explained in the article earlier: x0, g(x0), g(g(x0)), .. where x0 == _|_ . So, essentially this functions looks to me exactly the same as factorial and i don't understand what this example is supposed to show? If the right hand side is functional g' defined as g'(f) = n -> if n == 0 then 1 else f (n + 1) and we try to find it's fixed point, i.e. such f, that f = g'(f) then i can't understand, why would we try to execute f(1) at all? And by the way, these "multiple fixed points of function f", are they just (const 1) for n <= 0 and any other values for n > 0? I.e. this if n <= 0 then 1 else undefined is least defined fixed point, but this one: if n <= 1 then 1 else undefined is also fixed point, just not the least defined, and this one: if n <= 0 then 1 else 3 is also a fixed point. Is this correct? Thanks. [1]: https://en.wikibooks.org/wiki/Haskell/Denotational_semantics#Interpretation_...