If you've solved the permutations problem, then in my opinion, you have the
ability to solve any problem in the book. Sounds like you're just having a
weird block. Don't worry -- my guess is it happens to everyone!
The question Wooks put to you in one of his previous posts -- how can you
preserv
On Nov 7, 2010, at 1:29 AM, Ken Hegeland wrote:
> I felt very dumb trying to figure this problem out.
You shouldn't feel dumb. I suspect that this problem needs to go because it
doesn't drive home the points I had had in mind when I wrote it. Sorry
-- Matthias
_
I am able to define a generative function for tabulate the divisors of 20, so
hopefully tomorrow I can figure out how to abstract it, Im sure I might have a
bit more trouble with the final step I need to figure out for abstraction. I
just wanted to say thanks to wooks and anyone else, I felt ver
Yes, I have completed all the previous challenges, I have completed most 3
times, and some 2 times. For the most part with all the previous challenges I
never had too much trouble. The only other problems that gave me a decent
amount of trouble were the permutation function, and computing natura
Hi Ken --
Just a quick question to make sure we are all on the "same page," as it
were: have you completed all exercises in prior sections?
Dave
On Sat, Nov 6, 2010 at 2:24 PM, Ken Hegeland wrote:
> Something that looks like this:
>
> (define(tabulate1 n i)
> (cond
> [(> i n)empty]
> [(=(remai
Something that looks like this:
(define(tabulate1 n i)
(cond
[(> i n)empty]
[(=(remainder n i)0)(cons i(tabulate1 n(add1 i)))]
[else(tabulat1 n(add1 i))]))
(define(tabulate n)
(tabulate1 n 1))
I
feel like I understand how it works recursively, just the generative
recursion method makes me stru
An alternative approach.
1. Write a program called tabulate-20 that returns all the factors of 20
starting from 1.
2. Abstract tabulate-20 so that it works for numbers other than 20.
_
For list-related
I tried to get you to approach this as though nobody had told you about
generative recursion.
In which case you would/should have tried the tried and tested formula you
know, which should have led you to
(tabulate-div 20) and the simple recursion that follows is (tabulate-div 19).
You have
I have been trying to take a break for the weekend and I am going to try to go
back at it with a fresh perspective Monday.
One
thought I did have though, you said tabulate-div means we want all the
divisors of 20, so the next step would be tab-div(sub1 n), or 19, Im
thinking that would mean a
Forget about the term generative recursion for a second.
I've coined a phrase that for me encapsulates much of the approach of the
Design Recipe - I call it Delta Programming.
Why - well you have a result you would like, you have your input and you have
in your template a bunch of expressions
#x27;; lukejor...@gmail.com
>
> *Cc:* users@racket-lang.org
> *Subject:* Re: [racket] Generative recursion
>
> Actually there is one more way. We only have to check numbers 1 up to the
> integer-sqrt of n. For each check whose remainder is 0, we immediately have
> two divisors, the ch
0 12:06
To: 'Todd O'Bryan'; lukejor...@gmail.com
Cc: users@racket-lang.org
Subject: Re: [racket] Generative recursion
Actually there is one more way. We only have to check numbers 1 up to the
integer-sqrt of n. For each check whose remainder is 0, we immediately have
two divisors, th
-))
Jos
_
From: users-boun...@racket-lang.org [mailto:users-boun...@racket-lang.org]
On Behalf Of Todd O'Bryan
Sent: 05 November 2010 10:31
To: lukejor...@gmail.com
Cc: users@racket-lang.org
Subject: Re: [racket] Generative recursion
There are really two ways of doing this problem
There are really two ways of doing this problem. The way I'd probably use is
to make a list of all the *possible* divisors and then use the filter
function to pull out the actual divisors.
The way you're probably thinking of requires a helper function for the
recursion, because you need to keep tr
I found implementing this trickier than grasping the solution as well.
Stick with it. I don't see that you need any functions related prime
numbers. Perhaps if input is prime that is a trivial case, but try to focus
on what the output is: A list of numbers that can evenly divide the input.
Thos
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