in reply to Re: $a++ allowed by $a-- is not ! why?
in thread $a++ allowed by $a-- is not ! why?

Your definitions of one-to-one and onto are somewhat incorrect.

One-to-one (aka injective) means that F(s) = F(s') only happens if s=s'. In other words no more than one value will map to a given target.

Onto (aka surjective) means that for each t in the target space T there is an s such that F(s)=t.

The necessity that F(s) not have 2 values (which you gave as a definition of being one-to-one) is required for F to be a function. The requirement of "exactly one" that you stated for onto is wrong.

Furthermore any function F that is one-to-one has a unique local inverse function G from its image in T back to S such that G(F(s)) is s for all s in S. For G to truly be the inverse of F, you also need F(G(t)) to be t for all t in T, which requires that F be onto.

Therefore any function which is one-to-one and onto is called invertible (because it has an inverse function).

And the entire discussion about strings comes down to this. The ++ operation is not onto the set of alphanumeric strings, and therefore does not have an inverse from the set of all strings. Therefore there is no natural way to define a general -- operator for strings, so none has been defined.

(Exercise for the interested reader - search the archives of p5p for past discussions of the -- operator on strings. I know that it has been discussed extensively there, and am just too lazy to search for it.)

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Re: $a++ allowed by $a-- is not ! why?
by jonadab (Parson) on Sep 02, 2003 at 01:35 UTC

    One-to-one means that F(s) = F(s') only happens if s=s'.

    Onto means that for each t in the target space T there is an s such that F(s)=t.

    The necessity that F(s) not have 2 values is required for F to be a function.

    I knew I should check. (It's been far too long since I dealt with them separately... mostly what matters is whether a given function is a permutation or not.) Anyway, my point that ++ is not onto and therefore not reversible is still valid, even though I mixed up one-to-one pretty badly.


    $;=sub{$/};@;=map{my($a,$b)=($_,$;);$;=sub{$a.$b->()}} split//,".rekcah lreP rehtona tsuJ";$\=$ ;->();print$/
Re: Re: Re: $a++ allowed by $a-- is not ! why?
by kabel (Chaplain) on Sep 02, 2003 at 08:48 UTC
    (Exercise for the interested reader - search the archives of p5p for past discussions of the -- operator on strings. I know that it has been discussed extensively there, and am just too lazy to search for it.)
    thanks. i think i found the starting point into it. further pointers welcome! :-)
      Yup, that is the discussion that I had in mind...