[plt-scheme] creating nested lists in DrScheme

 From: Aniket Karmarkar (akarmarkar at mail.bradley.edu) Date: Fri May 7 23:55:43 EDT 2010 Previous message: [plt-scheme] creating nested lists in DrScheme Next message: [plt-scheme] creating nested lists in DrScheme Messages sorted by: [date] [thread] [subject] [author]

```That's what this function does:
(define (equal L1 L2)
(cond
((eq? (caar L1) (car L2)) (car L1))
(else (equal (cdr L1) L2))
)
)

This function find the sublist which match. I had it before but I forgot to
include it. So it returns (b c d). I tried the map function but you need the
arguments to have same number in both arguments. :(

On Fri, May 7, 2010 at 10:39 PM, Stephen Bloch <sbloch at adelphi.edu> wrote:

>
> On May 7, 2010, at 10:42 PM, Aniket Karmarkar wrote:
>
> This is what I am trying to do.
> Write the function partialpaths, of two arguments. The first argument is a
> list of lists, representing a graph(adjacency list representation of a
> graph). The second argument is a partial path in the reverse order, for
> example the path "from a to b to c" is represented by the list (c b a). The
> function should return a list of sublists containing all possible expansions
> of this partial path by one node. For example: (partialpaths '( (a b c) (b c
> d) (c d a) (d)) '(b a)) should return: ( (c b a) (d b a))
> I get the (b c d) part now I need to attach the cdr of this to b a multiple
> times so I try putting it in a loop.
>
> Here's what I have:
> (define nod ())
> (define ma '('()))
> (define (partialpaths L1 L2)
>   (set! nod (equal L1 L2))
>   (do()
>     ;exit test
>     ((null? (cdr nod)))
>     (list ma (attach(cdr nod)L2))
>     (set! nod (cdr nod))
>   )
>   ma
> )
>
>
> Yigg.  Are the loop, the mutation, and the global variables really
> necessary?  With all due respect, this doesn't look like Scheme code to me
> :-)
>
> (define (attach List1 List2)
>   (list (car List1) List2)
> )
>
>
> This almost certainly doesn't do what you think it does.
>
> Seems to me what you want to do is extract the first (i.e. last) element of
> the path, find the row of the table that starts with it, and cons each of
> the other elements of that row onto the path.
>
> So off the top of my head, I would define a variable equal to (car path),
> then another equal to the row (if any) in the adjacency matrix whose car
> matches it (this may require a helper function), then map over the cdr of
> the row a function which conses its argument onto the path.  I did this in
> about six lines (not counting test cases) with no loops, no mutation, no
> global variables, and only one recursion (in the helper function).
>
>
>
> Stephen Bloch
> sbloch at adelphi.edu
>
>
>
>
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