Trouble with float on C [duplicate]
1/2 does integer division, its result is 0, change either or both operand to double, e.g: 1.0/2
square-root
Is there a short-hand for nth root of x in Python?
nth root of x is x^(1/n), so you can do 9**(1/2) to find the 2nd root of 9, for example. In general, you can compute the nth root of x as: x**(1/n) Note: In Python 2, you had to do 1/float(n) or 1.0/n so that the result would be a float rather than an int. … Read more
How can I find the Square Root of a Java BigInteger?
Just for fun: public static BigInteger sqrt(BigInteger x) { BigInteger div = BigInteger.ZERO.setBit(x.bitLength()/2); BigInteger div2 = div; // Loop until we hit the same value twice in a row, or wind // up alternating. for(;;) { BigInteger y = div.add(x.divide(div)).shiftRight(1); if (y.equals(div) || y.equals(div2)) return y; div2 = div; div = y; } }
How is the square root function implemented? [closed]
Simple implementation using Binary Search with C++ double root(double n){ // Max and min are used to take into account numbers less than 1 double lo = min(1, n), hi = max(1, n), mid; // Update the bounds to be off the target by a factor of 10 while(100 * lo * lo < n) … Read more
Looking for an efficient integer square root algorithm for ARM Thumb2
Integer Square Roots by Jack W. Crenshaw could be useful as another reference. The C Snippets Archive also has an integer square root implementation. This one goes beyond just the integer result, and calculates extra fractional (fixed-point) bits of the answer. (Update: unfortunately, the C snippets archive is now defunct. The link points to the … Read more
Square root of BigDecimal in Java
I’ve used this and it works quite well. Here’s an example of how the algorithm works at a high level. Edit: I was curious to see just how accurate this was as defined below. Here is the sqrt(2) from an official source: (first 200 digits) 1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157273501384623091229702492483605585073721264412149709993583141322266592750559275579995050115278206057147 and here it is using the approach I outline … Read more
Writing your own square root function
The following computes floor(sqrt(N)) for N > 0: x = 2^ceil(numbits(N)/2) loop: y = floor((x + floor(N/x))/2) if y >= x return x x = y This is a version of Newton’s method given in Crandall & Pomerance, “Prime Numbers: A Computational Perspective”. The reason you should use this version is that people who know … Read more
John Carmack’s Unusual Fast Inverse Square Root (Quake III)
FYI. Carmack didn’t write it. Terje Mathisen and Gary Tarolli both take partial (and very modest) credit for it, as well as crediting some other sources. How the mythical constant was derived is something of a mystery. To quote Gary Tarolli: Which actually is doing a floating point computation in integer – it took a … Read more
Program that solves a simple math equation [closed]
See also the ScriptEngine. import java.awt.*; import java.awt.event.*; import javax.script.ScriptEngine; import javax.script.ScriptEngineManager; import javax.script.ScriptException; import javax.swing.*; class EvaluateString { public static void main(String[] args) { Runnable r = new Runnable() { @Override public void run() { JPanel gui = new JPanel(new BorderLayout(5,5)); final JTextField input = new JTextField( “Math.pow(2.6,22)+ Math.pow(3.9,15)”,19); final JTextField output = new … Read more
Square root of number python
Use the math module instead of the cmath module; the latter is for complex numbers only: >>> import math >>> print math.sqrt(25) 5.0 For what it’s worth, the cmath result is correct, if you expected a complex number. You could take just the .real component of the result, but since there is a regular floating … Read more