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 checking number and the quotient (except when these two are equal, giving one divisor only -this happens only when n is a square-)) 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. 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 track of where you are in building up the list. Back in the section on natural numbers, there are exercises to create lists of numbers of varying types, for example: ; list-from-a-to-b: number number -> list-of-number (check-expect (list-from-a-to-b 3 7) (list 3 4 5 6 7)) If you can figure out how to write this function, then you just need to include a conditional to decide whether to add the next number into the list you're creating or not. Todd On Wed, Nov 3, 2010 at 7:18 PM, Luke Jordan <luke.jor...@gmail.com> wrote: 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. Those numbers are the numbers from 1 to input. To think about how to get that list, try solving it by hand. If input is 3, how do you go about it? Does 3 divide 3 with no remainder? Yes, we know that numbers divide themselves with no remainder. How about 2? 1? Try it over with larger numbers, like 6 and 10. What process are you using to determine whether the numbers <= input and > 1 divide input with no remainder, and what happens to them if do? What happens if they do not? When does evaluation cease? When it comes to a termination statement, don't forget that list and append are different. Trying not to say too much, but hope I'm still saying something useful. - Luke On Wed, Nov 3, 2010 at 17:43, Ken Hegeland <hege...@yahoo.com> wrote: I am trying to do the problem 26.1.1 in the book HTDP and I just feel like the more I think about it, the more confused I get. I found a post from 2007 with some tips from Jens Axel Søgaard, the link is http://objectmix.com/scheme/186349-htdp-generative-recursion.html. I understand whats to be done, but Im just unsure how to accomplish the task. I believe that trivially solvable in this case is n=1 and n=prime. For n=1 the solution would be (list 1) for n=prime Im thinking it should be (list 1 n), my function is similar to the one on the link with a bit of different organization. When I read jens' tips near the end I am getting confused. The closest I can get to an answer is, (=(tabulate-div 20)(list 20 10 5 2 2)) >From the advice supplied, I was able to say that the smaller problems that you split the program into are, one to get the largest divisor, and one which gets the smallest divisor using the largest. As far as template I am using what is supplied in the book. Im simply stuck, and would love something to help me out, I will continue to run this problem through my head and see what I can come up with. Thanks in advance for any help. _________________________________________________ For list-related administrative tasks: http://lists.racket-lang.org/listinfo/users _________________________________________________ For list-related administrative tasks: http://lists.racket-lang.org/listinfo/users
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