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Thread: square rooting a BigDecimal
 05202011, 06:27 PM #1
 05202011, 07:58 PM #2Java Code:
package bigroot; import java.math.BigDecimal; import java.math.BigInteger; //sqrtfunction copied of BigFunctionsClass //From The Java Programmers Guide To numerical Computing (Ronald Mak, 2003) public class BigRoot { /** * Compute the square root of x to a given scale, x >= 0. * Use Newton's algorithm. * @param x the value of x * @param scale the desired scale of the result * @return the result value */ public static BigDecimal sqrt(BigDecimal x, int scale) { // Check that x >= 0. if (x.signum() < 0) { throw new IllegalArgumentException("x < 0"); } // n = x*(10^(2*scale)) BigInteger n = x.movePointRight(scale << 1).toBigInteger(); // The first approximation is the upper half of n. int bits = (n.bitLength() + 1) >> 1; BigInteger ix = n.shiftRight(bits); BigInteger ixPrev; // Loop until the approximations converge // (two successive approximations are equal after rounding). do { ixPrev = ix; // x = (x + n/x)/2 ix = ix.add(n.divide(ix)).shiftRight(1); Thread.yield(); } while (ix.compareTo(ixPrev) != 0); return new BigDecimal(ix, scale); } }
Java Code:package bigroot; import java.math.BigDecimal; /** * Test the BigFunctions by comparing results with class java.lang.Math. */ public class TestBigRoot { private static final int SCALE = 40; private void run() { System.out.println("sqrt 2 = " + Math.sqrt(2)); System.out.println(" = " + BigRoot.sqrt(BigDecimal.valueOf(2), SCALE)); System.out.println("sqrt 26 = " + Math.sqrt(26)); System.out.println(" = " + BigRoot.sqrt(BigDecimal.valueOf(26), SCALE)); } public static void main(String args[]) { TestBigRoot test = new TestBigRoot(); try { test.run(); } catch(Exception ex) { System.out.println("ERROR: " + ex.getMessage()); } } }
 05202011, 08:05 PM #3
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 05202011, 09:28 PM #4
Thanks. I am not going to try to understand it now, but it does a great job of what I want it to.
 05212011, 03:28 PM #5
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The only person who got everything done by Friday was Robinson Crusoe.

 05212011, 03:42 PM #7
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 05212011, 04:32 PM #8My reply #3 (accidentally) described how the method in reply #2 works.No bug ever had to calculate its fitnessfunction.
 05212011, 04:42 PM #9
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 05212011, 09:57 PM #10
I am just making a basic calculator. I read a forum post a while back that said if you are adding money or anything like that you should never use double. You should use BigDecimal.
Were you suggesting I convert the BigDecimal to an int or just use double? I haven't made very much java yet. I just do it in my spare time.
 05212011, 10:16 PM #11
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Are you making a general purpose (basic) calculator, or are you building a financial calculator? If it's the former, ordinary doubles would do fine (unless you want unlimited precision). If it is a financial calculator it depends on the type of calculations you want to perform. So, it's your turn to elaborate on this ;)
kind regards,
JosThe only person who got everything done by Friday was Robinson Crusoe.
 05212011, 10:36 PM #12
 05212011, 10:56 PM #13
Just a warning: you'll need a book then like mentioned in the listing, and Jos/Fubarable are right, working with doubles is almost allways more then needed (and a good, (notPolishnotation) calculator is difficult enough). But it would be a challenge. Tell here when you're done!
No bug ever had to calculate its fitnessfunction.
 05222011, 02:33 AM #14
 05222011, 02:53 AM #15Moderator
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No in polish notation your example would be entered as 5,6,5,*,+,9,1,+, (I think!)
The nice thing is that the order of operations is unambiguously specified without parentheses. The downside is that only Poles can understand it.
The notation is described at Polish notation  Wikipedia, the free encyclopedia  the example I gave seems to be Reverse Polish notation  Wikipedia, the free encyclopedia.
 05222011, 10:03 AM #16
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Yep, also, if the arity of the operators/functions is known and fixed you don't need those irritating parentheses in Polish notation either, e.g. 3*(4+5) is * 3 + 4 5 in Polish notation and is unambiguous, as is the reverse Polish notation 3 4 5 + * It's the variadic functions/operators that mess things up, e.g. (+ 1 2 3) where the + operator takes three arguments. There also is the functional notation (sometimes called 'outer Lisp') where the operators are used as functions, so 3*(4+5) is written as *(3, +(4, 5)). If the comma isn't a function/operator itself it can be left out.
kind regards,
JosThe only person who got everything done by Friday was Robinson Crusoe.
 05222011, 12:13 PM #17
I just can't leave for a moment! The explanations of Jos and PBrockway are perfect, and google also knows something about it. I just mentioned it because it is supposedly easier to program: you don't need the parsing. But indeed, only Polish people seem to understand or like it.
(In effect what you do on a small calculator with just an inbetween result showing is a bit like reverse Polish notation (but cheating by using a memory): just enter arguments followed by an operator all the time.)Last edited by Jodokus; 05222011 at 01:23 PM. Reason: forgot "reverse"
No bug ever had to calculate its fitnessfunction.
 05222011, 12:59 PM #18
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That's not entirely true, all three forms need a bit of parsing, e.g. + *  doesn't make sense for a postfix expression (reverse Polish) if there are no operands available on the data stack; equivalently * 3  doesn't make sense as an infix expression, nor does it as a prefix expression (Polish notation). On the other hand those parsers are minimal compared to the lexical analyzers needed in all three of the expression forms.
kind regards,
JosThe only person who got everything done by Friday was Robinson Crusoe.
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