<html><body style="word-wrap: break-word; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space; "> <br><div><html>On Mar 28, 2008, at 1:24 PM, Cooke Kelsey wrote:</html><br class="Apple-interchange-newline"><blockquote type="cite"><div>Oh right, in order to include all the words I need more than an append. Here's a function that adds a letter to ALL the words:</div> <div><br>(define (add-first-letter first-letter low)<br> (cond <br> [(empty? (rest low)) (list (cons first-letter low))]<br> [else (cons (cons first-letter (list (first low))) (add-first-letter first-letter (rest low)))]))<br></div> <div>;example:</div>(add-first-letter 'X (list 'A 'T 'E 'Y 'W 'Z)) <div>;result:</div> <div>(list (list 'X 'A) (list 'X 'T) (list 'X 'E) (list 'X 'Y) (list 'X 'W) (list 'X 'Z))</div></blockquote><div><br></div><div><br></div><div>So if you apply add-first-letter to (first word) and the result of (insert-everywhere/in-one-word (rest word)), you get back all but one of the word you want. Which brings me to the second question: </div><div><br></div><br><blockquote type="cite"> <div> <span class="Apple-style-span" style="-webkit-text-stroke-width: -1; ">That was a lot more complicated than I thought. I guess that should be a helper function. </span></div> <div> </div> <div>However, when I try to make a main function (insert-in-single-word) that calls add-first-letter, I end up going around in the same circle as before.</div></blockquote><div><br></div><div><br></div><div>Because you failed to read and answer my second question. -- Matthias</div><div><br></div><div><br></div><div><br></div><div><br></div><br><blockquote type="cite"> <div><br><b><i>Matthias Felleisen <<a href="mailto:matthias@ccs.neu.edu">matthias@ccs.neu.edu</a>></i></b> wrote:</div> <blockquote class="replbq" style="PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #1010ff 2px solid"><br> <div>On Mar 28, 2008, at 12:34 PM, Cooke Kelsey wrote:<br class="Apple-interchange-newline"> <blockquote type="cite"> <div>OK, here's a simple function that adds a letter to a word:</div> <div> </div> <div>(define (add-first-letter first-letter word)<br> (append (list first-letter) word))</div></blockquote> <div><br></div> <div><br></div> <div>That's not what I asked you to _design_. Please read the suggestion carefully. And again, (append (list foo) bar) is equal to (cons foo bar). Why do yu keep using append? </div> <div><br></div> <div><br></div><br> <blockquote type="cite"> <div>(add-first-letter 'A (list 'X 'T 'E))=(list 'A 'X 'T 'E')<br></div> <div>So my question is, where does (list 'A 'T 'E) come from? </div></blockquote> <div><br></div> <div>It is your sample input to the function. It says "Given" in that column. </div> <div><br></div><br> <blockquote type="cite"> <div>In your table below, you say "the recursion: "X" "TE" --> "XTE" "TXE" "TEX". Actually I have never been able to create such a recursion. That's the big mystery to me. </div></blockquote> <div><br></div> <div><br></div> <div>The purpose statement of the function tells you what the recursive call should produce. So when you code, you may assume that the recursive call works properly.</div> <div><br></div> <div><br></div> <blockquote type="cite"> <div></div> <div><br></div> <div>I just read Dave Yrueta's encouraging note, and he says I need to focus on your hint about adding 'A to the list, "in the simplest and most natural way." Well, all I can think of is to append it, like I did above. But there is no recursive element, and I cannot call it or plug it into the main recursive function. It is floating out in space.</div> <div><br><b><i>Matthias Felleisen <<a href="mailto:matthias@ccs.neu.edu">matthias@ccs.neu.edu</a>></i></b> wrote:</div> <blockquote class="replbq" style="PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #1010ff 2px solid"> <div><br></div> <div>Okay you have gotten to the point where you're jumping even faster. Slogan: </div> <div><br></div> <div> furiously fast programming leads to nowhere. </div> <div><br></div> <div>;; --- </div> <div><br></div> <div>All I was getting at is a simple point: </div> <div><br></div> <div>Your recursive calls in insert-everywhere/one-word produced almost all the work in the recursion. Indeed, it always produces all but one word, except that the first letter from the original word is missing. Example: </div> <div><br></div> <div> given: the recursion in the final result I want plus </div> <div> 'X "ATE" "X" "TE" --> "XTE" "TXE" "TEX" "AXTE" "ATXE" "ATXE" "ATEX" "XATE" </div> <div><br></div> <div>Given the recursive result (the list of words that are too short by the first letter in the orginal word) can you design a function that adds this letter (no matter what it is) to the front of all these words you have gotten back? </div> <div><br></div> <div>DONE? </div> <div><br></div> <div>After that: what is missing? </div> <div><br></div> <div>-- Matthias</div> <div><br></div> <div><br></div> <div><br></div> <div><br></div><br> <div>On Mar 28, 2008, at 11:51 AM, Cooke Kelsey wrote:<br class="Apple-interchange-newline"> <blockquote type="cite"> <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 <<a href="mailto:matthias@ccs.neu.edu">matthias@ccs.neu.edu</a>></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! 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