What is the output of this pseudocode? - 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

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 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

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 5

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

Answer 6

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 7

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.
-Adam