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Today's Topics:
1. Re: ($) operator (Daniel Fischer)
2. [fschwi...@localhost: Re: [Haskell-beginners] ($) operator]
(Frank Schwidom)
3. Re: ($) operator (Brent Yorgey)
4. Re: ($) operator (David Morse)
5. Re: ($) operator (Jan Jakubuv)
6. Re: ($) operator (Frank Schwidom)
7. Re: ($) operator (Brent Yorgey)
8. Re: ($) operator (Dave Bayer)
9. Calculating Time Complexity (Matthew J. Williams)
10. Re: Calculating Time Complexity (Krzysztof Skrz?tnicki)
----------------------------------------------------------------------
Message: 1
Date: Sat, 24 Jan 2009 21:09:31 +0100
From: Daniel Fischer <daniel.is.fisc...@web.de>
Subject: Re: [Haskell-beginners] ($) operator
To: John Hartnup <john.hart...@gmail.com>, beginners@haskell.org
Message-ID: <200901242109.31199.daniel.is.fisc...@web.de>
Content-Type: text/plain; charset="iso-8859-1"
Am Samstag, 24. Januar 2009 20:37 schrieb John Hartnup:
> Hi.
>
> I'm working through Real World Haskell, and although it's going well
> (I just finished the exercise to write a glob matcher without using a
> regex library, and I'm pleased as punch), I keep seeing the ($)
> operator, and I'm not sure I understand its use. If the book explains
> it, I've been unable to find it.
>
> Empirically, it seems like:
> a $ b c d e f
> .. is equivalent to ..
> a (b c d e f)
Indeed. ($) is the function application operator,
f $ x = f x
, so
a $ b c d e f
is equivalent to
a (b c d e f)
and
a $ b c $ d e f
to
a (b c (d e f)).
>
> But is that its only purpose? To placate the LISP haters by removing
> parentheses?
More or less. Although the idea was more to enhance readability than placating
LISP haters :)
It's also handy to pass to some higher order functions, although offhand I
can't think of any situation where you couldn't pass id instead of ($).
>
> (1 +) 2 does the same thing as (1 +) $ 2, and has the same type.
>
> Am I missing something?
>
> Thanks,
> John
------------------------------
Message: 2
Date: Sat, 24 Jan 2009 22:14:02 +0000
From: Frank Schwidom <schwi...@gmx.net>
Subject: [fschwi...@localhost: Re: [Haskell-beginners] ($) operator]
To: beginners@haskell.org
Message-ID: <20090124221402.gb4...@bigbox>
Content-Type: text/plain; charset=us-ascii
----- Forwarded message from Frank Schwidom <fschwi...@localhost> -----
Date: Sat, 24 Jan 2009 21:40:54 +0000
From: Frank Schwidom <fschwi...@localhost>
To: John Hartnup <john.hart...@gmail.com>
Subject: Re: [Haskell-beginners] ($) operator
Message-ID: <20090124214053.ga4...@bigbox>
References: <6c1ba2fc0901241137x242cf4b4w2398896af45ae...@mail.gmail.com>
MIME-Version: 1.0
Content-Type: text/plain; charset=us-ascii
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In-Reply-To: <6c1ba2fc0901241137x242cf4b4w2398896af45ae...@mail.gmail.com>
User-Agent: Mutt/1.5.16 (2007-06-09)
Status: RO
On Sat, Jan 24, 2009 at 07:37:29PM +0000, John Hartnup wrote:
> Hi.
>
> I'm working through Real World Haskell, and although it's going well
> (I just finished the exercise to write a glob matcher without using a
> regex library, and I'm pleased as punch), I keep seeing the ($)
> operator, and I'm not sure I understand its use. If the book explains
> it, I've been unable to find it.
>
> Empirically, it seems like:
> a $ b c d e f
> .. is equivalent to ..
> a (b c d e f)
>
> But is that it's only purpose? To placate the LISP haters by removing
> parentheses?
>
> (1 +) 2 does the same thing as (1 +) $ 2, and has the same type.
>
> Am I missing something?
>
> Thanks,
> John
>
>
> --
> "There is no way to peace; peace is the way"
> _______________________________________________
> Beginners mailing list
> Beginners@haskell.org
> http://www.haskell.org/mailman/listinfo/beginners
maybe can this answer your question:
A:
((==) True) (1 == 1) -- won't work: ((==) True) 1 == 1
=>
((==) True) ((==) 1 1) -- won't work: ((==) True) (==) 1 1
=>
((==) True) $ (==) 1 1
=>
((==) True) $ 1 == 1
Regards
----- End forwarded message -----
------------------------------
Message: 3
Date: Sat, 24 Jan 2009 17:52:37 -0500
From: Brent Yorgey <byor...@seas.upenn.edu>
Subject: Re: [Haskell-beginners] ($) operator
To: beginners@haskell.org
Message-ID: <20090124225237.ga13...@seas.upenn.edu>
Content-Type: text/plain; charset=us-ascii
On Sat, Jan 24, 2009 at 07:37:29PM +0000, John Hartnup wrote:
> Hi.
>
> I'm working through Real World Haskell, and although it's going well
> (I just finished the exercise to write a glob matcher without using a
> regex library, and I'm pleased as punch), I keep seeing the ($)
> operator, and I'm not sure I understand its use. If the book explains
> it, I've been unable to find it.
>
> Empirically, it seems like:
> a $ b c d e f
> .. is equivalent to ..
> a (b c d e f)
>
> But is that it's only purpose? To placate the LISP haters by removing
> parentheses?
>
> (1 +) 2 does the same thing as (1 +) $ 2, and has the same type.
>
> Am I missing something?
More generally, in a language with first-class and higher-order
functions like Haskell, it's simply useful to have a function which
performs function application. For example, one place that ($) comes
in handy is when applying each function in a list of functions to a
single input value. Using ($) you can just write something like
map ($5) [(+1), (^2), abs]
which evaluates to [6,25,5].
-Brent
------------------------------
Message: 4
Date: Sun, 25 Jan 2009 00:25:29 -0500
From: David Morse <dcmo...@gmail.com>
Subject: Re: [Haskell-beginners] ($) operator
To: beginners@haskell.org
Message-ID:
<11f192740901242125l22a7d8f1i9ccf0bd9dbcf3...@mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
They definitely dropped the ball on introducing '$'. I think you just
hit the seam between two of the authors, and it fell through the
cracks.
------------------------------
Message: 5
Date: Sun, 25 Jan 2009 12:45:03 +0000
From: Jan Jakubuv <jaku...@gmail.com>
Subject: Re: [Haskell-beginners] ($) operator
To: John Hartnup <john.hart...@gmail.com>
Cc: beginners@haskell.org
Message-ID:
<d4a7b1730901250445t35b10e42wbb1a446f0d6d...@mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
2009/1/24 John Hartnup <john.hart...@gmail.com>:
>
> But is that it's only purpose? To placate the LISP haters by removing
> parentheses?
>
Also one important difference is that the standard application
operator (space) associates to the left while $ associates to the
right. It says how missing parenthesis are filled in.
For example
(a b c d e) stands for ((((a b) c) d) e)
while
(a $ b $ c $ d $ e) stands for (a (b (c (d e))))
Sometimes you need the former behavior sometimes the later. The $ is
useful for example when an argument to an unary function is itself a
function application. For example you can write
filter (<4) $ map (+1) $ [100,99,1,2,3,4]
In this case you can think of $ to be like a "unix pipe".
Sincerely,
jan.
------------------------------
Message: 6
Date: Sun, 25 Jan 2009 14:25:03 +0000
From: Frank Schwidom <schwi...@gmx.net>
Subject: Re: [Haskell-beginners] ($) operator
To: beginners@haskell.org
Message-ID: <20090125142503.ga4...@bigbox>
Content-Type: text/plain; charset=us-ascii
On Sat, Jan 24, 2009 at 07:37:29PM +0000, John Hartnup wrote:
> Hi.
>
> I'm working through Real World Haskell, and although it's going well
> (I just finished the exercise to write a glob matcher without using a
> regex library, and I'm pleased as punch), I keep seeing the ($)
> operator, and I'm not sure I understand its use. If the book explains
> it, I've been unable to find it.
>
> Empirically, it seems like:
> a $ b c d e f
> .. is equivalent to ..
> a (b c d e f)
>
> But is that it's only purpose? To placate the LISP haters by removing
> parentheses?
>
> (1 +) 2 does the same thing as (1 +) $ 2, and has the same type.
>
> Am I missing something?
>
> Thanks,
> John
I made some Examples for better understanding:
-- creating an $ like function:
f g x = (g x)
-- some tests comparing $ with f:
5 == ($ 2) (+ 3)
5 == (`f` 2) (+ 3)
5 == (\x -> (($) x 2)) (+ 3)
5 == (\x -> (f x 2)) (+ 3)
Regards
------------------------------
Message: 7
Date: Sun, 25 Jan 2009 09:12:06 -0500
From: Brent Yorgey <byor...@seas.upenn.edu>
Subject: Re: [Haskell-beginners] ($) operator
To: John Hartnup <john.hart...@gmail.com>
Cc: beginners@haskell.org
Message-ID: <20090125141206.ga18...@seas.upenn.edu>
Content-Type: text/plain; charset=us-ascii
On Sun, Jan 25, 2009 at 05:34:35AM +0000, John Hartnup wrote:
> 2009/1/24 Brent Yorgey <byor...@seas.upenn.edu>:
>
> >
> > map ($5) [(+1), (^2), abs]
> >
> > which evaluates to [6,25,5].
> >
>
> Just to check I understand correctly - you're using ($) to 'promote'
> an expression into a function, in order that it may be used in higher
> order functions?
In this case, I'm using ($) to *make* a higher-order function---($5)
is a function which takes its argument (which must be a function) and
applies it to 5. (I'm assuming you're familiar with 'operator
sections', if you aren't I can explain that too.)
-Brent
------------------------------
Message: 8
Date: Sun, 25 Jan 2009 09:13:41 -0500
From: Dave Bayer <ba...@cpw.math.columbia.edu>
Subject: Re: [Haskell-beginners] ($) operator
To: beginners@haskell.org
Message-ID: <f79252ae-c2ce-4a89-bcd7-7fbeeebad...@math.columbia.edu>
Content-Type: text/plain; charset=US-ASCII; format=flowed; delsp=yes
2009/1/24 John Hartnup <john.hart...@gmail.com>:
> But is that it's only purpose? To placate the LISP haters by removing
> parentheses?
On Jan 25, 2009, at 12:25 AM, David Morse wrote:
> They definitely dropped the ball on introducing '$'. I think you just
> hit the seam between two of the authors, and it fell through the
> cracks.
Haskell isn't Cobol, but in comparison with the lambda calculus, it
flows a bit better like a natural language. '$' helps with this.
For example, in the line
> mempty = en $ en []
...one clearly doesn't need to use $, parenthesizing (en []) has the
same meaning. However, reading left-to-right, the $ coaxes one into
dropping top-down into a new view of the line, partially abstracting
out or forgetting what came before, without having to maintain a
mental stack and brace oneself for the closing ).
I've read books on numeracy and vision, that present the same
conclusion: As our brain evolved new systems for seeing or counting,
the old systems didn't get thrown out, they got buried by the new
layers, but they're still wired in. Part of our reaction to a line of
lisp that ends in seven )))))) is no different than a pigeon or a
horse seeing the same number of objects; we're not sure. Then for us a
new linguistic layer kicks in, "I can see two groups of three, that's
even" and we rise above the animals, but our single conscious core got
hijacked to sort this out.
People take the ergonomics of cockpit design very seriously, to reduce
the cognitive load on pilots in emergencies. On Apollo 13, they nearly
blew it by flipping a switch the wrong way. I fly a lot, so I'm very
grateful that the "purists" that gravitate to math and computer
science aren't instead in ergonomic design.
I actually love lisp, or more precisely Scheme. I came back to Haskell
for strong typing and class abstractions; Haskell more directly
supports making a theory about what one is coding. In order to use
Scheme, I had to defang it of its parentheses. I didn't believe that
any of the web proposals for this saw actual use, so I rolled my own
preprocessor, using indentation to indicate outline structure, and
using '|' roughly as Haskell uses '$'. There are other issues one
inevitably faces, e.g. missing starts to outline levels, but I came to
love '$' as Haskell uses it.
Some people develop a point-free style in Haskell, e.g.
> blah foo . blah . blah foo foo . blah
and different people write code that piles up uses of '$' e.g.
> blah foo $ blah $ blah foo foo $ blah
With decent syntax coloring the second line can be given as light a
touch as the first.
Ultimately, '$' helps mitigate a left / right tension in Haskell that
exists in all languages, and in algebra itself.
In algebra, "f g" can mean either "f then g" or "g then f". It means
"g then f" in any situation crippled by the ancestry of the function
composition "f(g(x))", and it means "f then g" any time one has the
free choice of easier ergonomics in a culture that reads left to right.
Quick, ten times, work out out the permutation cycle product
> (1 2 3) (2 3 4 5) (3 4 5 6) (4 5 6 2) (5 6 1 2) (1 2 3) (2 3 4 5)
> (3 4 5 6) (4 5 6 2) (5 6 1 2)
under each assumption for the order of operations. One wraps around
through parentheses either way, but using the "f then g" convention,
all other scanning is consistently left to right, not flitting back
and forth like a deranged water bug.
Now try composing a complex set of type signatures for Haskell
functions. Do you feel like a deranged water bug? The arrows go the
wrong way, don't they! You'd be playing a nice game of dominoes,
canceling off neighbors and so forth, if the arrows went the other way.
So Haskell doesn't have its left / right tension completely worked
out, but I'd say '$' isn't part of the problem, it's part of the
solution.
------------------------------
Message: 9
Date: Mon, 26 Jan 2009 00:04:48 +0000
From: "Matthew J. Williams" <matthewjwillia...@googlemail.com>
Subject: [Haskell-beginners] Calculating Time Complexity
To: beginners@haskell.org
Message-ID: <497cfe10.0c53300a.065a.ffff8...@mx.google.com>
Content-Type: text/plain; charset="us-ascii"; format=flowed
Dear all,
In general terms, how would one calculate the time complexity of a
given algorithm? Feel free to make use of my pseudo code in your answer:
/* input:
2-D array A of size n by n
output: a number max */
Max := 0
For i := 1 to n
sum := 0
For j := 1 to n
sum := sum + A[i][j]
End for
If sum > max then max := sum
End for
Output max
Sincerely
Matthew J. Williams
------------------------------
Message: 10
Date: Mon, 26 Jan 2009 01:51:51 +0100
From: Krzysztof Skrz?tnicki <gte...@gmail.com>
Subject: Re: [Haskell-beginners] Calculating Time Complexity
To: "Matthew J. Williams" <matthewjwillia...@googlemail.com>
Cc: beginners@haskell.org
Message-ID:
<220e47b40901251651j1af2c6afteeb2e10b52f19...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
On Mon, Jan 26, 2009 at 01:04, Matthew J. Williams <
matthewjwillia...@googlemail.com> wrote:
> Dear all,
>
> In general terms, how would one calculate the time complexity of a given
> algorithm?
I would count most significant operations. The result is Theta(n^2) (see
http://en.wikipedia.org/wiki/Big_O_notation for definition of Big Theta
notation).
Feel free to make use of my pseudo code in your answer:
>
> /* input:
> 2-D array A of size n by n
> output: a number max */
> Max := 0
> For i := 1 to n
> sum := 0
> For j := 1 to n
> sum := sum + A[i][j]
> End for
> If sum > max then max := sum
> End for
> Output max
>
>
Christopher SkrzÄtnicki
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