C#•3y ago

Calculating square root of any number using a lower and upper bound that guesses towards solution.

how? ;-;

30 Replies

What have you tried?

CattoOP•3y ago

The square root of x is calculated iterative using a lower and upper bound that represent guesses towards the solution.  The lower bound is the last known value that the true square root must be above and the upper bound is the last know value that the true square root must be below

The square root of x is calculated iterative using a lower and upper bound that represent guesses towards the solution.  The lower bound is the last known value that the true square root must be above and the upper bound is the last know value that the true square root must be below

Nothing, because im trying to figure out how would i achieve this: like i geniunenly dont get what lower or upper bound its asking for? and how id used it to find square root of any number

Yawnder•3y ago

Ok. Let's say I tell you "give me the square root of 483269867. What initial bound do you think could mean / be?

CattoOP•3y ago

well the next closeset square number that has a square root number close to that? so for example square root of 50, would have square root of 49 the next closest, which is 7^2 right?

Yawnder•3y ago

Not quite. You're assuming you know about 7^2 What I was expecting is "the square root has to be between 0 and 483269867" Makes sense?

CattoOP•3y ago

mhmmmm yes i guess

Yawnder•3y ago

Well, not 483269867. 483269867 + 1. So basically, you know that it's above 0, and you know it's bellow V + 1.

CattoOP•3y ago

hmm okay, so maybe between 483269867 - 1 and 483269867 + 1? that -1 is lower ound and +1 upper

Yawnder•3y ago

No, it doesn't work for the lower bound.

CattoOP•3y ago

smh

Yawnder•3y ago

The square root is lower than V - 1

CattoOP•3y ago

Ohhh okay true true

Yawnder•3y ago

Let's compute the square root of 4.5 using the method, it will be simpler. You know it's between 0 and 5.5 (excluded) What you then do is you take a number between those, you square it. (do it (manually), I'm waiting)

CattoOP•3y ago

wait why 5.5 and not like 5?

Yawnder•3y ago

(Ideally, you would take the mid-point, not random) Any number higher by 0.25 would do I think.

CattoOP•3y ago

ah ok fair enough, btw i think im onto something, so i then do 0 + 5.5 and divide it by 2, and then i square that and see how close i am to 4.5?

Yawnder•3y ago

So yes, the bound can start at 0 and V + 0.25 Yup. If you're above, you move the upper bound to what you tried. If you're bellow, you move the lower bound to what you tried.

CattoOP•3y ago

Holy crap, wait wait so oooooo

Yawnder•3y ago

And you do it as long as Math.Abs(v1 - v2) > tolerance. (don't use these variable names, it's just for illustrative purposes)

CattoOP•3y ago

mhmmmm i seee, so if its above 4.5.... then i try 0 + what i just tried / 2 and thne squared right?

Yawnder•3y ago

Yup

CattoOP•3y ago

Holy moly you a genius, id have never thought of that lol i couldnt find solutions to this anywhere as well

Yawnder•3y ago

decimal target = 547M;
decimal tolerance = 0.00001M;
decimal lowerBound = 0;
decimal upperBound = target + 0.25M;

decimal currentAttempt;
decimal currentValue;

do
{
    currentAttempt = (lowerBound + upperBound) / 2M;
    currentValue = currentAttempt * currentAttempt;

    if (currentValue < target)
    {
        lowerBound = currentAttempt;
    }
    else if (currentValue > target)
    {
        upperBound = currentAttempt;
    }
}
while (Math.Abs(target - currentValue) > tolerance);

Console.WriteLine(currentAttempt);

decimal target = 547M;
decimal tolerance = 0.00001M;
decimal lowerBound = 0;
decimal upperBound = target + 0.25M;

decimal currentAttempt;
decimal currentValue;

do
{
    currentAttempt = (lowerBound + upperBound) / 2M;
    currentValue = currentAttempt * currentAttempt;

    if (currentValue < target)
    {
        lowerBound = currentAttempt;
    }
    else if (currentValue > target)
    {
        upperBound = currentAttempt;
    }
}
while (Math.Abs(target - currentValue) > tolerance);

Console.WriteLine(currentAttempt);

CattoOP•3y ago

Oh holy crap you even made the code in like 10 seconds!

MODiX•3y ago

Yawnder#7904

REPL Result: Success

decimal target = 547M;
decimal tolerance = 0.00001M;
decimal lowerBound = 0;
decimal upperBound = target + 0.25M;

decimal currentAttempt;
decimal currentValue;

do
{
    currentAttempt = (lowerBound + upperBound) / 2M;
    currentValue = currentAttempt * currentAttempt;

    if (currentValue < target)
    {
        lowerBound = currentAttempt;
    }
    else if (currentValue > target)
    {
        upperBound = currentAttempt;
    }
}
while (Math.Abs(target - currentValue) > tolerance);

Console.WriteLine(currentAttempt);

decimal target = 547M;
decimal tolerance = 0.00001M;
decimal lowerBound = 0;
decimal upperBound = target + 0.25M;

decimal currentAttempt;
decimal currentValue;

do
{
    currentAttempt = (lowerBound + upperBound) / 2M;
    currentValue = currentAttempt * currentAttempt;

    if (currentValue < target)
    {
        lowerBound = currentAttempt;
    }
    else if (currentValue > target)
    {
        upperBound = currentAttempt;
    }
}
while (Math.Abs(target - currentValue) > tolerance);

Console.WriteLine(currentAttempt);

Console Output

23.388031024485826492309570312

23.388031024485826492309570312

Compile: 664.938ms | Execution: 36.923ms | React with ❌ to remove this embed.

CattoOP•3y ago

that is hella impressive, btw what does the M stand for tho?

Yawnder•3y ago

To indicate that the number preceding it is a decimal. https://learn.microsoft.com/en-us/dotnet/csharp/language-reference/builtin-types/floating-point-numeric-types#real-literals

CattoOP•3y ago

Ahhh i see that code is so much simpler than i thought it would have been ngl i thought i needed like tons of math

Yawnder•3y ago

There probably are more efficient ways to compute that, but that's how I would do it based on the question you asked.

CattoOP•3y ago

hey i wouldnt have thought of anything lol, so that will 100% do, especially when i couldnt find any solutions to this problem online

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Calculating square root of any number using a lower and upper bound that guesses towards solution.

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