Modify a given pseudocode such that it contains no loops

Question

Consider the pseudocode:

read n (non-zero natural number)
x <- 1
y <- n
d <- 2
while x < y
{
    if n % d = 0
    { 
        x <- d
        y <- [n / d]

    }

    d <- d + 1

}

if x = y
{
    write 'D', x
}

else
{
    write 'N'
}

I have to modify this pseudocode such that there are no loops in it, so I have to get rid of that while loop at the top. I went through some examples, namely the numbers {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 100} and the code resulted in showing N for the numbers {2, 3, 5, 6, 7, 8} and for {1, 4, 9, 100} it showed D followed by their respective square roots ({1, 2, 3, 10} respectively).

So I concluded that the code outputs D only when n is a perfect square, and then it shows its square root. For the numbers that are not perfect squares, it outputs N.

That means I have to change the above pseudocode such that it checks if the number n is a perfect square or not. But how can I do that without using ANY loops? Especially since this is pseudocode, so I don't have a function like sqrt(n). I got this exercise from a source that usually has simple problems, so it must be something simple that I just don't see, nothing complicated. But I don't see any way of using the given variables, or creating new ones to check if the given number n is a perfect square without any loops.

I disagree that you don't have `sqrt(n)` in pseudocode. I would actually claim that you have whatever operation you want. So your analysis makes a lot of sense and I think the intended solution is indeed to just take the square root and check if it's integer. But of course it's hard to be sure that that is what the asker inteded. — Vincent van der Weele, May 04 '20 at 10:59
you can check a number is sqrt by calling function or using for loops.... you can not ignore both... — GolamMazid Sajib, May 04 '20 at 11:22
The problem amounts to writing some code in an unspecified language. Without details about what the language is, and what counts as a loop, it can't really be answered. Maybe `if exists(x natural number) such that x*x == n then write 'D', x else write 'N'`. Or with recursion, or with a builtin integer square root... — Paul Hankin, May 04 '20 at 11:26

Chema · Answer 1 · 2020-07-16T11:50:18.530

One approach would be replace the while expression with a recursive function.

As an example

read n (non-zero natural number)
  // recursive function loop, inside of method read
  loop(num,x,y,d)
    if x < y
      if num % d = 0
        loop(num, d, n / d, d + 1) // recursive function call      
      else 
        loop(num, x, y, d + 1)  // recursive function call      
    else  // exit of the loop function
      if x = y
        return d - 1  // it is a perfect square
      else
        return -1      // it is not a perfet square
        
  // recursive function call      
  res = loop(n,1,n,2)  
  if res = -1
    write 'N'
  else    
    write res

written in Scala

object PerfetSquare {
  def read(n: Int): Int = {
    @tailrec
    def loop(num: Int,x: Int,y: Int, d: Int): Int = {
      if(x < y) {
        if(num % d == 0) {
          loop(num, d, n / d, d + 1)
        } else {
          loop(num, x, y, d + 1)
        }
      } else if (x == y){
        d - 1  // it is a perfect square
      } else {
        -1     // it is not a perfect square
      }
    }
    loop(n,1,n,2)
  }

  def main(args: Array[String]): Unit = {
   val res = read(64)
    if(res == -1) println("N")
    else println(s"D $res")

   val res2 = read(65)
    if(res2 == -1) println("N")
    else println(s"D $res2")
  }
}

output

D 8
N

Modify a given pseudocode such that it contains no loops

1 Answers1