CODEWITHSUNDEEP
c++algorithmmathsquare-root
faustborne
faustborne

Reputation: 21

Finding a square root through iteration in c++

I have a homework assignment due and I am having trouble with my loops. I first have to find the next highest and lowest square roots through loops which I already can do. Next my assignment tells me that I need to get an approximation of the square root which I do by averaging the next highest and lowest square roots of the integer. Then I must ask the user for the number of decimal places of accuracy they want. Here is a quote from the assignment:

A count-controlled loop should then be constructed; it will execute once for each of the desired decimal positions; in the example, this loop will execute four times (once each for the tenths, hundredths, thousandths, and ten-thousandths decimal positions). Use a counter such as decimalPosition to keep track of which pass the loop is on.

This is where I have trouble, I am using a while loop based on the number of decimal positions the user has entered but my loop doesn't complete the loop. I am new to programming so please forgive me if this is a truly simple. Here is my while code:

for (int decimalPosition = 1; decimalPosition <= decimal; decimalPosition++)
{
    while (baseRoot*baseRoot > num)
    {
        baseRoot = baseRoot - (pow((.1),decimalPosition));
        cout << fixed << setprecision(decimal) << baseRoot << endl;
    }

}

here is my output so far

Enter a number you wish to know the square root of: 8
Enter the number of decimal places of accuracy you want: 7
Find the square root of 8 to 7 decimal places: 
2.6000000
2.7000000
2.8000000
2.9000000
2.9000000 square root of 8.0000000

Upvotes: 1

Views: 5111

Answers (2)

Joe Liu
Joe Liu

Reputation: 61

To make your loop works, you can add a statement

baseRoot = baseRoot + (pow((.1),decimalPosition));

after the while loop because your need to make sure baseRoot is bigger than the answer before each iteration. Like this:

for (int decimalPosition = 1; decimalPosition <= decimal; decimalPosition++)
{
    while (baseRoot*baseRoot > num)
    {
        baseRoot = baseRoot - (pow((.1),decimalPosition));
        cout << fixed << setprecision(decimal) << baseRoot << endl;
    }
    baseRoot = baseRoot + (pow((.1),decimalPosition));
}

Now you can get the answer 2.8284271.

By the way, there's another efficient way called bisection method (similar to binary search) to solve this kind of problem (related to Monotonic function) without too much math:

double mySqrt(double x, double epsilon) {
    double left = 0, right = x;
    while (right - left > epsilon) {
        double mid = (left + right) / 2;
        if (mid * mid > x) {
            right = mid;
        } else {
            left = mid;
        }
    }
    return left;
}

It's simple, stupid :)

Upvotes: 0

duffymo
duffymo

Reputation: 308733

It's called Newton's method, and its convergence is quadratic. That should help you figure it out.

PS - The Babylonians discovered it first, but Newton gets credit for it.

Upvotes: 4

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