Computing x^n in C++ without Math Library

[ineedhelpnow]

Program should compute the power x^2

#include<iostream>
int main() {
double x;
int n;
double pow=1;

cout<<"Please enter x:";
cin>>x;
cout<<"Please enter n:";
cin>>n;

[B]LOOP FOR X^N[/B]
pow=pow*x

cout<<"The result is: "<<pow<<endl;
return 0;
}

as you can see, there is no math library included and i need to be able to perform the operation x^n. i figured everything out except the main part which is the loop. do i need to use a do loop?

 

[Nono713]

$$x^0 = 1$$
$$x^1 = x = x^0 \cdot x = x^{1 - 1} \cdot x$$
$$x^2 = (x) \cdot x = x^1 \cdot x = x^{2 - 1} \cdot x$$
$$x^3 = ((x) \cdot x) \cdot x = x^2 \cdot x = x^{3 - 1} \cdot x$$
$$\vdots$$
$$x^n = x^{n - 1} \cdot x$$

See the pattern? So you can use a for-loop, like this:

1. set xn = 1, and assert that x^0 = xn, which is correct, so far so good
2. for i = 1 to n, set xn = xn * x, and assert that at the end of the ith iteration, xn = x^i. this holds because then xn = x^(i - 1) * x = x^i, as the previous iteration computed x^(i - 1) and put it into xn
3. once the loop terminates, xn = x^n, and you are done

Can you implement that with a for loop (or other)? Does it work?

PS: there is a more efficient way of computing x^n that takes only around log(n) multiplications instead of n. Can you work it out? (if you can't, don't worry about it, it is a bit advanced).

 

[ineedhelpnow]

it looks like he didnt give initialize i for us so i don't think we can add anything to the code from besides what he gave us. is that the only loop that would work? if so, i guess i have no choice but to add i.

im not sure how to do it with log(n). i suck at programming. ill die with anything advanced. (Giggle)

 

[Nono713]

it looks like he didnt give initialize i for us so i don't think we can add anything to the code from besides what he gave us. is that the only loop that would work? if so, i guess i have no choice but to add i.

im not sure how to do it with log(n). i suck at programming. ill die with anything advanced. (Giggle)

... you could add a variable i, yes. Also note that you aren't actually using the loop variable in the loop (it's just used to get something that runs n times) so you can get by with a while loop, perhaps as:

while (n--)
{
// stuff
}

which will run exactly n times.

 

[ineedhelpnow]

oh i see. but why n--?

 

[Nono713]

oh i see. but why n--?

The idea is to decrease n until it reaches zero, doing one iteration every time you decrease it. You can rewrite it as n = n - 1 if you prefer.

 

[ineedhelpnow]

but how do you know you want to decrease. what if n is 1 and next you want it to be 2. or next you want it to be 7.

 

[Evgeny.Makarov]

Another version of the loop is

for (int i = 0; i < n; i++)
  // body of the loop

 

[ineedhelpnow]

yes that one is simpler but like i was telling Bac, since my instructor didnt initialize the variable i, I am not sure if he wants us to or if he wants us to use a different loop.

 

[Nono713]

yes that one is simpler but like i was telling Bac, since my instructor didnt initialize the variable i, I am not sure if he wants us to or if he wants us to use a different loop.

Maybe try asking him? I mean, I get that he doesn't want you to completely change the program, but it seems to me that adding a for loop does not detract from the spirit of the question, which is: implement the naive algorithm to compute x^n.

 

[ineedhelpnow]

yeah its kind of difficult getting in touch with him and i don't know if ill have time to see him before the test but i think ill send him an email now. ill stick with the for loop. its simpler. i just wish he was sometimes more clear about what in the world he wants :) thank you guys

 

[Evgeny.Makarov]

The code I wrote both declares and initializes [m]i[/m] in the first line. Note that it says [m]int i = 0[/m] and not [m]i = 0[/m]. In the latter case, [m]i[/m] would have to be declared previously. As it is, no changes outside of the loop are necessary. On the other hand, you need a loop counter; otherwise you don't know how many iterations you've completed.

 

[ineedhelpnow]

for (int i = 0; i < n; i++)
pow=pow*x;

is that right?

 

[Evgeny.Makarov]

for (int i = 0; i < n; i++)
pow=pow*x;

is that right?

Yes, that's right.

 

[ineedhelpnow]

thanks Bac. thanks EM.

 

[ineedhelpnow]

i think ill just do it your way. and if the teacher says anything ill tell him to (Swearing) i reeeaaaally appreciate your help mark to take the time and to actually show me how to exactly do the code. thank you for showing me how to do it in a MORE CLEAR MANNER than my instructor would. you're the best programmer i know around