I need an algorithm in C++ to generate insipid numbers. I know what
the "word" equation says, but I can't seem to get it written correctly
in code.  Been beating my head against this keyboard for 4 days trying
to get it figured out on my own!

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Request for Question Clarification by secret901-ga on 10 Mar 2003 01:26 PST

Hi stonewallwoman,
Could you clarify what "insipid numbers" are?  I've tried to research
what they are before attempting this, but couldn't find anything
helpful :-(
secret901-ga

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Request for Question Clarification by secret901-ga on 10 Mar 2003 01:36 PST

I think I found it: by "insipid numbers," do you mean those numbers
that return a 1 after repeated additions of the squares of its digits?

secret901-ga

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Clarification of Question by stonewallwoman-ga on 10 Mar 2003 08:17 PST

Yes, secret901, your second "clarification request" is correct.
Seperate the digits, square them, add them, and keep doing that in a
loop to get either a repeating sequence the ends in 1 or 58. I got a
long email from another source, but to tell you the truth, that guy
went WAY over my head! Hope you can help me in SIMPLE terms! :^)

Hello Stonewallwoman,

Here is a c++ program that should do what you want. I'll first provide
some comments first to describe how the code works.

The function is_insipid is a little bit tricky but runs pretty simply.
The variable i will generate the sum. In each pass through the loop,
 i = i + (x%10)*(x%10)
basically a running sum of the squares of the digits. Note that x%10
is the remainder of x/10 - a digit. x will be divided by 10 each time
through the loop until it is less than 10. Note that integer x/10 is
zero - that will cause the loop to exit. The function then prints the
new sum and returns that sum to the caller.

The main program is pretty simple. Ask the user for a value, copy it
and pass it to is_insipid to compute the next value. Repeat the loop
until the result is 1 or 58. At that point, print the appropriate
answer and exit the program.

I tested it with the values from the web site
  http://www.stonehill.edu/compsci/CS103/Assignments/assignment3D.htm
(scroll down to the second assignment) and this gets the same answers.

If this is not clear - don't hesitate to use a clarification request
to get more information.

  --Maniac


#include <iostream.h>

int is_insipid(int x)

{
  int i = 0;

  while (x>0)
    {
      i += (x % 10)*(x % 10);
      x = x/10;
    }
  cout << "Now have " << i << "\n";
  return i;
}

int main(int argc,
     char *argv[])

{
  int x;
  int y;

  cout << "Enter a value:";
  cin >> x;
  y = x;
  do
    {
      y = is_insipid(y);
    }
  while ((y != 1) && (y != 58));
  if (y == 1)
    cout << x << " is insipid.\n";
  else
    cout << x << " is NOT insipid.\n";
  return 0;
}

  --Maniac

The answer I received from maniac-ga was consise, easy for my beginner
brain to comprhend, and eaisily translated into terms that my
instructor would accept.

This is an answer to the problem posed here:

http://www.stonehill.edu/compsci/CS103/Assignments/assignment3D.htm

It's in Java, since that's the language I know best. Shouldn't be too
hard to convert it to C++. Also, you can either tackle this using 
recursion or iteration. I've supplied solutions using both techniques.

Cheers.
 

  public static void main(String[] args)
  {
    int n = Integer.parseInt(args[0]);
    System.out.print("The sequence starting at " + n + " is: ");
    if(isInsipidIterative(n))
    {
      System.out.println("\nInspid");
    }
    else
    {
      System.out.println("\nNot Inspid");
    }
  }

  public static boolean isInsipidIterative(int n)
  {
    while(true)
    {
      System.out.print(n + " ");
      if(n == 1)
      {
        return true;
      }
      else if(n == 58)
      {
        return false;
      }
      else
      {
        n = sumDigits(n);
      }
    }
  }

  public static boolean isInsipidRecursive(int n)
  {
    System.out.print(n + " ");
    if(n == 1)
    {
      return true;
    }
    else if(n == 58)
    {
      return false;
    }
    else
    {
      return isInsipidRecursive(sumDigits(n));
    }
  }

  public static int sumDigits(int n)
  {
    int sum = 0;
    while(n > 0)
    {
      int digit = n % 10;
      sum += digit * digit;
      n /= 10;
    }
    return sum;
  }