What is the output of this pseudocode?

procedure DoSomething(a_1, ... a_n)
 p = a_1
 for i = 2 to n
  temp = p
  for j = 1 to a_i
   p = p * temp

DoSomething(10,2,2,2)

We are getting mixed results. One of us got 10^7, the other 10^27.

I Think I found my error... I keep substituting 10 for p every time, instead of the new value for temp.

EDIT: here's my work:

{10, 2, 2, 2}
p = 10
i = 2 to 4
 temp = p = 10
 j = 1 to 2
  p = 10 * 10 = 10^2
  p = 10^2 * 10 = 10^3
i = 3 to 4
 temp = 10^3
 j = 1 to 2
  p = 10^3 * 10 = 10^4
  p = 10^4 * 10 = 10^5
i = 4 to 4
 temp = 10^5
 j = 1 to 2
  p = 10^5 * 10 = 10^6
  p = 10^6 * 10 = 10^7

10^7

From stackoverflow

• There's a reason folks have called Python "executable pseudocode":

  >>> def doSomething(*args):
  ...     args = list(args);
  ...     p = args.pop(0)
  ...     for i in range(len(args)):
  ...         temp = p
  ...         for j in range(args[i]):
  ...           p *= temp
  ...     return p
  ...
  >>> print doSomething(10,2,2,2)
  1000000000000000000000000000
  

• I entered the program into my TI-89 and got an answer of 1e27 for the value of p.

  t(a)
  Func
    Local i,j,p,tmp
    a[1]->p
    For i,2,dim(a)
      p->tmp
      For j,1,a[i]
        p*tmp->p
      EndFor
    EndFor
    Return p
  EndFunc
  
  t({10,2,2,2})       1.E27
  

• Isn't it ((10^3)^4)^5 = 10 ^ 60 ?

• It's 10^27 as shown by this bit of python code:

  a = [10,2,2,2]
  p = a[0]
  for i in range(1,len(a)):
      temp = p
      for j in range(a[i]):
          p *= temp
  print p
  

  1,000,000,000,000,000,000,000,000,000

  The problems with your code as posted are:

  • in your 10^7 solution, you're always multiplying by 10, not temp (which is increased to the final value of p after the j loop).

  • You're setting temp to arr[i], not p, in your PHP code (which I'll include here so my answer still makes sense after you edited it out of your question :-).

    $arr = array(10, 2, 2, 2);
    $p = $arr[0];
    $temp = 0;
    for($i = 1; $i <= 3; $i++)
    {
        $temp = $arr[$i];
        for($j = 0; $j <= $arr[$i]; $j++)
        {
            $p = $p * $temp;
        }
    }
    echo $p;
    

  TT : Realized this a bit before you posted, but thanks for taking the time to answer.

• In C:

  #include <stdio.h>
  
  double DoSomething(double array[], int count)
  {
    double p, temp;
    int i, j;
  
    p = array[0];
  
    for(i=1;i<count;i++)
    {
      temp = p;
      for(j=0; j<array[i];j++)
      {
        printf("p=%g, temp=%g\n", p, temp); /* useful to see what's going on */
        p = p * temp;
      }
    }
    return p; /* this isn't specified, but I assume it's the procedure output */
  }
  
  double array[4] = {10.0,2.0,2.0,2.0};
  
  int main(void)
  {
    printf("%g\n", DoSomething(array, 4));
    return 0;
  }
  

  And, as others have indicated, 10e27. Note that the above is very verbose from your pseudo code - it could be simplified in many ways.

  I used the Tiny C Compiler - very small, lightweight, and easy to use for simple stuff like this.

• Seems to be a function to calculate

  
  (((a_1^(a_2+1))^(a_3+1))^(a_4+1)...
  

  Thus we get ((10^3)^3)^3 = 10^(3^3) = 10^27

• There is an error in your computation for 10^7, See below. The correct answer is 10^27 {10, 2, 2, 2}

  p = 10
  i = 2 to 4
   temp = p = 10
   j = 1 to 2
    p = 10 * 10 = 10^2
    p = 10^2 * 10 = 10^3
  i = 3 to 4
   temp = 10^3
   j = 1 to 2
    p = 10^3 * 10 = 10^4 -- p=p*temp, p=10^3 and temp=10^3, hence p=10^3 * 10^3.
    p = 10^4 * 10 = 10^5 -- Similarly for other steps.
  i = 4 to 4
   temp = 10^5
   j = 1 to 2
    p = 10^5 * 10 = 10^6
    p = 10^6 * 10 = 10^7
  

  drhorrible : formatting would be helpful. (Use two newlines in the editor to produce one in the final output)

  Charles Duffy : two newlines? ugh! the code-format option (four spaces before each line) looks much better.