CODEWITHSUNDEEP
functional-programmingschemeracket
Nico Bellic
Nico Bellic

Reputation: 363

Loops and iterations in scheme.

I was going over some problems for my discrete math class, and read an exercise that caught my attention (note: this is not homework. I'm just purely curious).

Question: How many times is the "print" statement executed for the following program segment? (i, j, k, m are integers):

for i := 1 to 20 do
    for j := 1 to i do
        for k := 1 to j do
            for m := 1 to k do
                print (i * j) + (k * m)

I tried to do it in python, but it was boring because I got it done in a few seconds. So for fun I tried to do it with DrRacket using scheme as the language. However, after reading documentation on loops, I can't seem to find reference for loops of this kind. So, using this specific example (or I guess a general one with an indefinite number of loops), how can this problem be solved?

Upvotes: 1

Views: 655

Answers (2)

Óscar López
Óscar López

Reputation: 235984

A loop like this:

for i := 1 to 20 do
    for j := 1 to i do
        print [i, j]

Is equivalent to this in Racket, assuming that a loop range that looks like this: 1 to 20 includes all the numbers from 1 up to (and including) 20.

#lang racket

(for* ((i (in-range 1 21))
       (j (in-range 1 (add1 i))))
  (displayln (list i j)))

Notice that the above is not limited to one nested loop, you can declare as many nested loops and iteration variables as needed.

Upvotes: 6

Will Ness
Will Ness

Reputation: 71070

MIT Scheme:

(do ((cnt 0) (i 1 (+ i 1))) ((> i 20) cnt)
   (do ((j 1 (+ 1 j))) ((> j i)) 
     (do ((k 1 (+ 1 k))) ((> k j))
       (do ((m 1 (+ 1 m))) ((> m k)) 
         (set! cnt (+ 1 cnt))))))
;Value: 8855

do is in R5RS.

But this is really a math problem, not a programming problem.

Upvotes: 0

Related Questions