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Summing. I'm still living in imperative land.
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Chui Tey  
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 More options Sep 24 2007, 12:21 am
Newsgroups: comp.lang.prolog
From: Chui Tey <t...@cognoware.com>
Date: Sun, 23 Sep 2007 21:21:27 -0700
Local: Mon, Sep 24 2007 12:21 am
Subject: Summing. I'm still living in imperative land.
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Hi all,

I've got a program here:

% different sized nuts in the toolbox
t(1).
t(2).
t(4).

% build a list of nuts
ts([]).
ts([t(H)|Ts]) :- ts(Ts), t(H).

% get sum of nut sizes after
% arranging nuts as a palindrome
go(SumOfH):-
  length(Ts, 3),
  ts(Ts),
  reverse(Ts, Ts),
  findall(H, (t(H), memberchk(t(H), Ts)), Hs),
  sumlist(Hs, SumOfH),
  display(Ts).

The question is: is there a simpler way to do this without memberchk?
memberchk seems awfully redundant.


 
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Markus Triska  
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 More options Sep 24 2007, 1:36 am
Newsgroups: comp.lang.prolog
From: Markus Triska <tri...@logic.at>
Date: Mon, 24 Sep 2007 07:36:17 +0200
Local: Mon, Sep 24 2007 1:36 am
Subject: Re: Summing. I'm still living in imperative land.
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Chui Tey <t...@cognoware.com> writes:
> ts([t(H)|Ts]) :- ts(Ts), t(H).

Why carry the t/1 functor with you? Just collect the sizes, like:

  ts([T|Ts]) :- ts(Ts), t(T).

>   display(Ts).

You want a relation sum_nuts/2 between a sum S and a list of
palindromic nut sizes summing to S. If you define just that, the
toplevel will handle display for you.

--
comp.lang.prolog FAQ: http://www.logic.at/prolog/faq/


 
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Chui Tey  
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 More options Sep 24 2007, 2:18 am
Newsgroups: comp.lang.prolog
From: Chui Tey <t...@cognoware.com>
Date: Sun, 23 Sep 2007 23:18:46 -0700
Local: Mon, Sep 24 2007 2:18 am
Subject: Re: Summing. I'm still living in imperative land.
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On Sep 24, 3:36 pm, Markus Triska <tri...@logic.at> wrote:

> Chui Tey <t...@cognoware.com> writes:
> > ts([t(H)|Ts]) :- ts(Ts), t(H).

> Why carry the t/1 functor with you? Just collect the sizes, like:

>   ts([T|Ts]) :- ts(Ts), t(T).

I was trying to do this (Python equivalent, apologies):

  # height, thickness, size of the head
  nuts = [(5, 10, 2),
          (8, 10, 3), ...]

  nuts_sublist = foo(nuts) #
  total_height = sum([nut[0] for nut in nuts_sublist])
  total_thickness = sum([nut[1] for nut in nuts_sublist])
  etc...

I don't want to rebuild nuts_sublist to calculate each statistic. It's
hard learning to talk another language!


 
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Markus Triska  
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 More options Sep 24 2007, 2:54 am
Newsgroups: comp.lang.prolog
From: Markus Triska <tri...@logic.at>
Date: Mon, 24 Sep 2007 08:54:53 +0200
Local: Mon, Sep 24 2007 2:54 am
Subject: Re: Summing. I'm still living in imperative land.
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Chui Tey <t...@cognoware.com> writes:
>   total_height = sum([nut[0] for nut in nuts_sublist])

Suppose Sub = [t(5,10,2),t(8,10,3)], then:

   ?- maplist(arg(1), $Sub, Hs), sumlist(Hs, TotalH).
   %@ Hs = [5, 8],
   %@ TotalH = 13

>   total_thickness = sum([nut[1] for nut in nuts_sublist])

   ?- maplist(arg(2), $Sub, Ts), sumlist(Ts, TotalT).
   %@ Ts = [10, 10],
   %@ TotalT = 20

--
comp.lang.prolog FAQ: http://www.logic.at/prolog/faq/


 
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Chui Tey  
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 More options Sep 24 2007, 6:51 am
Newsgroups: comp.lang.prolog
From: Chui Tey <t...@cognoware.com>
Date: Mon, 24 Sep 2007 03:51:00 -0700
Subject: Re: Summing. I'm still living in imperative land.
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>    ?- maplist(arg(2), $Sub, Ts), sumlist(Ts, TotalT).
>    %@ Ts = [10, 10],
>    %@ TotalT = 20

Thanks! arg/3 is exactly what I was looking for!

 
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