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### Re: Re: \$a++ allowed by \$a-- is not ! why?

by kabel (Chaplain)
 on Sep 01, 2003 at 16:35 UTC ( #288160=note: print w/replies, xml ) Need Help??

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

sorry, i did not mean that permutation ;) it was just meant to be a visual description of the commutitativity law.

Modern algebra would say that although ++ is one-to-one, it is not onto.
why? it is a mapping from N to N (if i understand you correctly).
-- OTOH (as it is currently defined) is theoretically a permutation over the set of all finite numbers ...
the association is not clear to me. why do you think -- is a permutation? (what is a permutation for you?).

i think we make too much fuss about nothing. perhaps we should continue this by mail?
• Comment on Re: Re: \$a++ allowed by \$a-- is not ! why?

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Re: \$a++ allowed by \$a-- is not ! why?
by jonadab (Parson) on Sep 01, 2003 at 18:33 UTC
Modern algebra would say that although ++ is one-to-one, it is not onto.
why? it is a mapping from N to N (if i understand you correctly).

Given a function F from set S to set T, F is said to be one-to-one if F(s) is equal to a unique t in T for all s in S, and F is said to be onto if for all t in T there exists exactly one s in S such that F(s)=t. In our case both S and T are the same set, the set of all finite numbers and alphanumeric strings. (I believe (Inf)++ is also defined, but that can be considered a special case.)

++ as it is defined in Perl is one-to-one because each possible number or alphanumeric string has a unique successor, but it is not onto because it is not true that each number or alphanumeric string has a unique predecessor.

why do you think -- is a permutation?

Because, its domain and range are the same set (specifically, the set of all finite numbers; again, (Inf)-- is also defined but can be considered a special case, a polymorphism if you will) and it is both one-to-one and onto.

You can also consider ++ to be a permutation over the set of all finite numbers, if you consider the string magic to be a polymorphism, but ++ is definitely not a permutation over the set of alphanumeric strings, because it is not onto.

\$;=sub{\$/};@;=map{my(\$a,\$b)=(\$_,\$;);\$;=sub{\$a.\$b->()}} split//,".rekcah lreP rehtona tsuJ";\$\=\$ ;->();print\$/
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.)

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\$/
(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! :-)

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