Re: functional programming excercise from Mastering Mathematica

• To: mathgroup at smc.vnet.net
• Subject: [mg55290] Re: functional programming excercise from Mastering Mathematica
• From: Peter Pein <petsie at arcor.de>
• Date: Fri, 18 Mar 2005 05:34:18 -0500 (EST)
• References: <d1bflp\$lja\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Torsten Coym wrote:
> I'm quite new to Mathematica and its functional programming capabilities
> so I did some reading in John Gray's "Mastering Mathematica". There is
> an excercise in chapter 6 as follows
>
> Write your own function composeList that works just like the built-in
> operation with the same name, using FoldList. Conversely, write your own
> function foldList that works just like the built-in operation with the
> same name, using ComposeList.
>
> Unfortunately, there is no solution given at the end of the book (or I
> didn't find it). I could figure out a way to do the first task:
>
> composeList[funlist_List, var_] := FoldList[(#2[#1]) &, var, funlist]
>
> but I can't manage the second task...
>
> I know it's rather academic, but ... any help is welcome!
>
> Torsten
>
foldList[f_, x_, {}] := {x};
foldList[f_, x_, lst_List] :=
{x, Sequence @@ foldList[f, f[x, First[lst]], Rest[lst]]};

foldList[f, x, {a, b, c}] == FoldList[f, x, {a, b, c}]

==> True

If you need to handle large Lists (length > 250), you'll reach the
recursion limit.
You could try the procedural approach

Clear@foldList;
foldList[f_, x_, lst_] :=
Module[{n = Length[lst], r},
For[i = 1; r = Table[x, {n + 1}], i <= n, i++,
r[[i + 1]] = f[ r[[i]], lst[[i]] ]
];
r];

or even better the builtin FoldList ;-)
--
Peter Pein
Berlin

```

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