<div>OK, that helps me a little. If I insert "cons (first word)," then I can rebuild the word:</div> <div> </div> <div>(define (cons-function word)<BR> (cond<BR> [(empty? (rest word)) word]<BR> [else (cons (first word) (cons-function (rest word)))]))</div> <div> </div> <div>(cons-function (list 'A 'T 'W)) = (list 'A 'T 'W)</div> <div> </div> <div>The problem is how to add this to a shrinking function:</div> <div> </div> <div>(define (shrinking-function s word)<BR> (cond<BR> [(empty? (rest word)) empty]<BR> [else (list (append (list s) word) (shrinking-function s (rest word)))]))<BR></div> <div>(This shrinking function isn't exactly what I want, but at least it gives me (list 'X 'A 'T 'W) and (list 'X 'T 'W)).</div> <div> </div> <div>So I append them together: </div> <div> </div> <div>(append (cons-function (list 'A 'T 'W))
(shrinking-function 'X (list 'A 'T 'W)))</div> <div> </div> <div>And I get this:</div> <div> </div> <div>(list (list 'X 'A 'T 'W) (list (list 'X 'T 'W) empty)) (list 'A 'T 'W (list 'X 'A 'T 'W) (list (list 'X 'T 'W) empty))</div> <div> </div> <div>Garbage!</div> <div> </div> <div>As a last resort, I try to combine cons and shrinking functions together:</div> <div> </div> <div>(define (cons+shrinking-function s word)<BR> (cond<BR> [(empty? (rest word)) word]<BR> [else (append (cons (first word) (cons+shrinking-function 'X (rest word))) (list s) (rest word))]))</div> <div> </div> <div>(cons+shrinking-function 'X (list 'A 'T 'W)) = (list 'A 'T 'W 'X 'W 'X 'T 'W)</div> <div> </div> <div>It looks kind of nice, but it is even further from the expected result.</div> <div> </div> <div><BR><B><I>Matthias Felleisen <matthias@ccs.neu.edu></I></B> wrote:</div> <BLOCKQUOTE
class=replbq style="PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #1010ff 2px solid"> <DIV><BR></DIV> <DIV>How about (cons 'A (list 'X 'T)) = (list 'A 'X 'T)</DIV> <DIV><BR></DIV> <DIV><BR></DIV> <DIV><BR></DIV><BR> <DIV>On Mar 28, 2008, at 10:53 AM, Cooke Kelsey wrote:<BR class=Apple-interchange-newline> <BLOCKQUOTE type="cite">(append (list 'A) (list 'X 'T)) = (list 'A 'X 'T)<BR><BR><B><I>Matthias Felleisen <<A href="mailto:matthias@ccs.neu.edu">matthias@ccs.neu.edu</A>></I></B> wrote: <BLOCKQUOTE class=replbq style="PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #1010ff 2px solid"> <DIV><BR></DIV> <DIV>Here is a definition of 'adding in the missing element': </DIV> <DIV><BR></DIV> <DIV>* If a list X misses the symbol 'A (at the front), how do you get a list that has the 'A and all of X? </DIV> <DIV><BR></DIV>
<DIV>* X = (list 'X 'T) and 'A --> I want (list 'A 'X 'T). How do I do that? </DIV> <DIV><BR></DIV> <DIV>-- Matthias</DIV> <DIV><BR></DIV></BLOCKQUOTE> <DIV><BR class=khtml-block-placeholder></DIV> <HR SIZE=1> Be a better friend, newshound, and know-it-all with Yahoo! Mobile. <A href="http://us.rd.yahoo.com/evt=51733/*http://mobile.yahoo.com/;_ylt=Ahu06i62sR8HDtDypao8Wcj9tAcJ">Try it now.</A></BLOCKQUOTE></DIV><BR></BLOCKQUOTE><BR><p> 
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