[plt-scheme] again it is newbie with ex 11.4.6 N[<=20]

> Develop the function
>
> ;; tabulate-f-up-to-20 : N [<= 20]  ->  N

1. Note the typo. This should be ... -> list-of-N or even ... -> 
list-of-pair-n-x-n . Sorry.

> (define (tabulate-f-up-to-20 n-below-20) ...)
> which tabulates the values of f for natural numbers less than 20. 
> Specifically, it consumes a natural number n less than or equal to 20 
> and produces a list of posns, each of which has the shape (make-posn n 
> (f n)) for some n between 0 and n (inclusively)

2. But read the text, which is a long purpose statement. Given n <= 20, 
you want the values for (f n), (f (add1 n)) ... all the way up to (f 
20).

3. Now read your expected answer and compare.

-- Matthias



On Jan 26, 2006, at 5:12 AM, arnuld wrote:

> hello everybody,
>
>                                  well, it's me again. i know what are 
> you thinking, this newbie is really totally
> brain-less. but i have spent my 6 months of financial-sums to order 
> HtDP from USA and i want to solve 100% of exercises given in the book 
> at any cost even if i need to spend 8 hrs. in front my AMD64. Also Mr. 
> Matthias is very helpful regarding problems faced by newbies.
>
> anyway here is the original exercise from the book:
>
> --- Exercise 11.4.6.   In exercises 11.2.2, 11.4.4, and 11.4.5, we 
> developed functions that tabulate the mathematical function f in 
> various ranges. In both cases, the final function produced a list of 
> posns that was ordered in descending order. That is, an expression 
> like (tabulate-f 3) yields the list
> (cons (make-posn 3 2.4)
>   (cons (
> make-posn 2 3.4)
>     (cons (make-posn 1 3.6)
>       (cons (make-posn 0 3.0)
>         empty))))
>
> If we prefer a list of posns in ascending order, we must look at a 
> different data collection, natural numbers up to a certain point in 
> the chain:
>
> A natural number [<= 20] (N[<=20]) is either
> 	1 	
> 20 or
> 	2 	
> (sub1 n) if n is a natural number [ <= 20].
>
> Of course, in high school, we refer to N[<=-1] as the negative 
> integers.
>
> Develop the function
> ;; tabulate-f-up-to-20 : N [<=
>  20]  ->  N
> (define (tabulate-f-up-to-20 n-below-20) ...)
>
>  which tabulates the values of f for natural numbers less than 20. 
> Specifically, it consumes a natural number n less than or equal to 20 
> and produces a list of posns, each of which has the shape (make-posn n 
> (f n)) for some n between 0 and n (inclusively). ----
>
> i have developed the solution but according to the data-definition it 
> is wrong (i know it) but i am not able to solve it, even i have solved 
> all the previous problems of sec-11.
>
> may you help me out?
>
> ------------------------ start 
> ----------------------------------------------------------
> ;; A natural number [<=20] is either:
> ;;               1. 20, or
> ;;               2. (sub1 n) where nis a natural number[<=20]
>
> ;; in this case natural numbers end upto 0, because beyond zero starts 
> the series of -ve integers..
> ;; which we do not refer to as natural numbers in this case.
>
> ;; tabulate-f-upto-20 : N[<=20] -> list
> ;; to pruduce a list of posns in ascending order which has shape 
> (make-posn n (f n)) for N[<=20] and upto n only.
> ;; (define (tabulate-f-upto-20 n)..)
>
>
> #| TEMPLATE
> (define (tabulate-f-upto-20 n)
>    (cond
>     [(= n 20)...]               
>     [else....(tabulate-f-upto-20 (add1 n))...]))
>
> ;; (= n 20) it is because we want the list in ascending order
> ;; 2nd it exactly matches the data-definition
>
>  (tabulate-f-upto-20 20)
> ;; expected answer
> empty
>
> (tabulate-f20 5)
> ;; expected answer
> (cons (make-posn 0 (f 0))
>       (cons (make-posn 1 (f 1))
>             (cons (make-posn 2 (f 2))
>                   (cons (make-posn 3 (f 3))
>                         (cons (make-posn 4 (f 4)) (cons (make-posn 5 
> (f 5)) empty)))))) |#
>
>
> (define (tabulate-f-upto-20 n)
>   (cond
>     [(= n 20) empty]               
>     [else (cons (make-posn n (f n)) (tabulate-f-upto-20 (add1 n)))]))
>
>
> (define (f x)
>   (+ (* 3 (* x x))
>      (+ (* -6 x)
>         -1)))
>
> ;; tests
> (tabulate-f-upto-20 20)
> ;; expected answer
> 'empty
>
> (tabulate-f-upto-20 5)
> ;; expected answer
> (cons (make-posn 0 (f 0))
>       (cons (make-posn 1 (f 1))
>             (cons (make-posn 2 (f 2))
>                   (cons (make-posn 3 (f 3))
>                         (cons (make-posn 4 (f 4)) (cons (make-posn 5 
> (f 5)) empty))))))
>
>  --------------------------- end 
> -------------------------------------------------------
> OUPUT it produces:
>
> 'empty
> -- in 2nd test-case it produces all posn from 5-20 not from 0-5 as 
> problem wants.--
>
>
>  --
> "the great 
> intellectuals"_________________________________________________
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