[Jprogramming] bug or feature?

[Roger Hui]

That x f@[&0 y is equivalent to f^:x y follows from the defn  
x (f&n) y <=> (f&n)^:x y , not the other way around.



----- Original Message -----
From: Roger Hui <rhui000 at shaw.ca>
Date: Friday, August 31, 2007 11:10
Subject: Re: [Jprogramming] bug or feature?
To: Programming forum <programming at jsoftware.com>

> > x (f&n) y <=> (f&n)^:x y
> > 
> > but I can't quite see how that follows from what
> > the Vocabulary says:
> > 
> > The phrase x f@[&0 y is equivalent to f^:x y , apply the monad 
> f 
> > x times to y .
> 
> Since  x f@[&0 y  is  f@[&0^:x y according to the defn,
> it comes down to the meaning of  f@[&0 z  .  But:
> 
> f@[&0 z
> z f@[ 0     defn of u&n
> f z[0       defn of @
> f z
> 
> That is, the monad  f@[&0  is the same as the monad f,
> whence f@[&0^:x y is the same as f^:x y.
> 
> 
> 
> ----- Original Message -----
> From: Nollaig MacKenzie <nollaig at amhuinnsuidhe.net>
> Date: Friday, August 31, 2007 10:56
> Subject: Re: [Jprogramming] bug or feature?
> To: Programming forum <programming at jsoftware.com>
> 
> > 
> > On 2007.08.31 16:47:16, you,
> >  the extraordinary Yuri Burger, emitted:
> > 
> > > 
> > > how it works ???
> > > 
> > >   p=:*&0.1
> > >   p 10
> > > 1
> > >   1 p 10
> > > 1
> > >   2 p 10
> > > 0.1
> > >   3 p 10
> > > 0.01
> > >   4 p 10
> > > 0.001
> > >   5 p 10
> > > 0.0001
> > > 
> > > 
> > > 
> > > 
> > >   (,&'-') '|'
> > > |-
> > >   2 (,&'-') '|'
> > > |--
> > >   10 (,&'-') '|'
> > > |----------
> > 
> > x (f&n) y <=> (f&n)^:x y
> > 
> > but I can't quite see how that follows from what
> > the Vocabulary says:
> > 
> > The phrase x f@[&0 y is equivalent to f^:x y , apply the monad 
> f 
> > x times to y .