What is the output of this pseudocode?

Question

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

Answer 1

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

Answer 2

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

Answer 3

Isn't it ((10^3)^4)^5 = 10 ^ 60 ?

Answer 4

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;
  

Answer 5

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.

Answer 6

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

Answer 7

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