Suppose I want to calculate 2^10000 using Math.pow(). For which my code is-
long num = (long)Math.pow(2, 10000);
System.out.println(num);
Here I'm getting output as 9223372036854775807 which I guess is the limit of long. But if I write like this-
double n = Math.pow(2, 10000);
System.out.println(n);
the o/p is Infinity which is a bit weird. Anyone please help me with this.
This will work, but be aware that BigInteger is not very efficient and I don't consider it suitable for production work for math problems.
public static void main( String[] args ) {
BigInteger two = new BigInteger( "2" );
BigInteger twoToTenThousand = two.pow( 10000 );
System.out.println( twoToTenThousand );
}
2^10000 overflows the limit of both longs and doubles. However, in Java, overflowing in longs and doubles are treated differently. If you add 1 to the largest possible value in a long, you wrap around and the result is the smallest possible value for a long, equal to -2^63. On the other hand, if you add 1 to the largest possible value in a double, you get Double.POSITIVE_INFINITY. See https://docs.oracle.com/javase/8/docs/api/java/lang/Long.html and https://docs.oracle.com/javase/8/docs/api/java/lang/Double.html for more information.
i am working on a decoder, and i need to take 1020 to the 275th power, and mod that by 1073, but when i do try that, it wont print out the correct answer
double decoded = (1020^275)%1073;
That is the code i am trying, but it will print out 751, and its supposed to print 4, anyone have any tips?
^ is the xor operator; however, 1020 to the 275th power is a HUGE number and will overflow even if using double or long, so it should be represented by BigInteger:
System.out.println(BigInteger.valueOf(1020).pow(275).mod(BigInteger.valueOf(1073)));
Output:
4
Note: I used BigInteger#mod, but BigInteger#remainder will return the same value in your case (as you're only dealing with non-negative values).
You can use BigInteger.pow() followed by BigInteger.mod().
However the BigInteger class specifically includes in operation for your task: BigInteger.modPow():
System.out.println(BigInteger.valueOf(1020).modPow(BigInteger.valueOf(275), BigInteger.valueOf(1073)));
gives 4.
Try this out:
BigDecimal remainder = new BigDecimal(1020).pow(275).remainder(new BigDecimal(1073));
System.out.println(remainder);
After applying remainder, it can be converted to long as its max value will be 1072(in this case):
remainder.longValueExact();
double is bounded and has precision limitations, try:
System.out.println(Math.pow(1020, 275)); // Infinity
Usually, you should use BigDecimal or BigInteger to manipulate big numbers,
BigDecimal bigDecimal = new BigDecimal(1020);
bigDecimal = bigDecimal.pow(275);
bigDecimal = bigDecimal.remainder(new BigDecimal(1073));
System.out.println(bigDecimal); //4
Xor is differ from power function. 2 ^ 3, The output of an XOR gate is true only when exactly one of its inputs is true. If both of an XOR gate's inputs are false, or if both of its inputs are true, then the output of the XOR gate is false.
double maximum value = 1.7976931348623157E308
Actually (1020^275) = 2.3176467070363862212063591722218e+827 which greater than double maximum value.
Go with BigDecimal.
System.out.println(new BigDecimal(1020.0).pow(275).remainder(new BigDecimal(1073)));
public class doublePrecision {
public static void main(String[] args) {
double total = 0;
total += 5.6;
total += 5.8;
System.out.println(total);
}
}
The above code prints:
11.399999999999
How would I get this to just print (or be able to use it as) 11.4?
As others have mentioned, you'll probably want to use the BigDecimal class, if you want to have an exact representation of 11.4.
Now, a little explanation into why this is happening:
The float and double primitive types in Java are floating point numbers, where the number is stored as a binary representation of a fraction and a exponent.
More specifically, a double-precision floating point value such as the double type is a 64-bit value, where:
1 bit denotes the sign (positive or negative).
11 bits for the exponent.
52 bits for the significant digits (the fractional part as a binary).
These parts are combined to produce a double representation of a value.
(Source: Wikipedia: Double precision)
For a detailed description of how floating point values are handled in Java, see the Section 4.2.3: Floating-Point Types, Formats, and Values of the Java Language Specification.
The byte, char, int, long types are fixed-point numbers, which are exact representions of numbers. Unlike fixed point numbers, floating point numbers will some times (safe to assume "most of the time") not be able to return an exact representation of a number. This is the reason why you end up with 11.399999999999 as the result of 5.6 + 5.8.
When requiring a value that is exact, such as 1.5 or 150.1005, you'll want to use one of the fixed-point types, which will be able to represent the number exactly.
As has been mentioned several times already, Java has a BigDecimal class which will handle very large numbers and very small numbers.
From the Java API Reference for the BigDecimal class:
Immutable,
arbitrary-precision signed decimal
numbers. A BigDecimal consists of an
arbitrary precision integer unscaled
value and a 32-bit integer scale. If
zero or positive, the scale is the
number of digits to the right of the
decimal point. If negative, the
unscaled value of the number is
multiplied by ten to the power of the
negation of the scale. The value of
the number represented by the
BigDecimal is therefore (unscaledValue
× 10^-scale).
There has been many questions on Stack Overflow relating to the matter of floating point numbers and its precision. Here is a list of related questions that may be of interest:
Why do I see a double variable initialized to some value like 21.4 as 21.399999618530273?
How to print really big numbers in C++
How is floating point stored? When does it matter?
Use Float or Decimal for Accounting Application Dollar Amount?
If you really want to get down to the nitty gritty details of floating point numbers, take a look at What Every Computer Scientist Should Know About Floating-Point Arithmetic.
When you input a double number, for example, 33.33333333333333, the value you get is actually the closest representable double-precision value, which is exactly:
33.3333333333333285963817615993320941925048828125
Dividing that by 100 gives:
0.333333333333333285963817615993320941925048828125
which also isn't representable as a double-precision number, so again it is rounded to the nearest representable value, which is exactly:
0.3333333333333332593184650249895639717578887939453125
When you print this value out, it gets rounded yet again to 17 decimal digits, giving:
0.33333333333333326
If you just want to process values as fractions, you can create a Fraction class which holds a numerator and denominator field.
Write methods for add, subtract, multiply and divide as well as a toDouble method. This way you can avoid floats during calculations.
EDIT: Quick implementation,
public class Fraction {
private int numerator;
private int denominator;
public Fraction(int n, int d){
numerator = n;
denominator = d;
}
public double toDouble(){
return ((double)numerator)/((double)denominator);
}
public static Fraction add(Fraction a, Fraction b){
if(a.denominator != b.denominator){
double aTop = b.denominator * a.numerator;
double bTop = a.denominator * b.numerator;
return new Fraction(aTop + bTop, a.denominator * b.denominator);
}
else{
return new Fraction(a.numerator + b.numerator, a.denominator);
}
}
public static Fraction divide(Fraction a, Fraction b){
return new Fraction(a.numerator * b.denominator, a.denominator * b.numerator);
}
public static Fraction multiply(Fraction a, Fraction b){
return new Fraction(a.numerator * b.numerator, a.denominator * b.denominator);
}
public static Fraction subtract(Fraction a, Fraction b){
if(a.denominator != b.denominator){
double aTop = b.denominator * a.numerator;
double bTop = a.denominator * b.numerator;
return new Fraction(aTop-bTop, a.denominator*b.denominator);
}
else{
return new Fraction(a.numerator - b.numerator, a.denominator);
}
}
}
Observe that you'd have the same problem if you used limited-precision decimal arithmetic, and wanted to deal with 1/3: 0.333333333 * 3 is 0.999999999, not 1.00000000.
Unfortunately, 5.6, 5.8 and 11.4 just aren't round numbers in binary, because they involve fifths. So the float representation of them isn't exact, just as 0.3333 isn't exactly 1/3.
If all the numbers you use are non-recurring decimals, and you want exact results, use BigDecimal. Or as others have said, if your values are like money in the sense that they're all a multiple of 0.01, or 0.001, or something, then multiply everything by a fixed power of 10 and use int or long (addition and subtraction are trivial: watch out for multiplication).
However, if you are happy with binary for the calculation, but you just want to print things out in a slightly friendlier format, try java.util.Formatter or String.format. In the format string specify a precision less than the full precision of a double. To 10 significant figures, say, 11.399999999999 is 11.4, so the result will be almost as accurate and more human-readable in cases where the binary result is very close to a value requiring only a few decimal places.
The precision to specify depends a bit on how much maths you've done with your numbers - in general the more you do, the more error will accumulate, but some algorithms accumulate it much faster than others (they're called "unstable" as opposed to "stable" with respect to rounding errors). If all you're doing is adding a few values, then I'd guess that dropping just one decimal place of precision will sort things out. Experiment.
You may want to look into using java's java.math.BigDecimal class if you really need precision math. Here is a good article from Oracle/Sun on the case for BigDecimal. While you can never represent 1/3 as someone mentioned, you can have the power to decide exactly how precise you want the result to be. setScale() is your friend.. :)
Ok, because I have way too much time on my hands at the moment here is a code example that relates to your question:
import java.math.BigDecimal;
/**
* Created by a wonderful programmer known as:
* Vincent Stoessel
* xaymaca#gmail.com
* on Mar 17, 2010 at 11:05:16 PM
*/
public class BigUp {
public static void main(String[] args) {
BigDecimal first, second, result ;
first = new BigDecimal("33.33333333333333") ;
second = new BigDecimal("100") ;
result = first.divide(second);
System.out.println("result is " + result);
//will print : result is 0.3333333333333333
}
}
and to plug my new favorite language, Groovy, here is a neater example of the same thing:
import java.math.BigDecimal
def first = new BigDecimal("33.33333333333333")
def second = new BigDecimal("100")
println "result is " + first/second // will print: result is 0.33333333333333
Pretty sure you could've made that into a three line example. :)
If you want exact precision, use BigDecimal. Otherwise, you can use ints multiplied by 10 ^ whatever precision you want.
As others have noted, not all decimal values can be represented as binary since decimal is based on powers of 10 and binary is based on powers of two.
If precision matters, use BigDecimal, but if you just want friendly output:
System.out.printf("%.2f\n", total);
Will give you:
11.40
You're running up against the precision limitation of type double.
Java.Math has some arbitrary-precision arithmetic facilities.
You can't, because 7.3 doesn't have a finite representation in binary. The closest you can get is 2054767329987789/2**48 = 7.3+1/1407374883553280.
Take a look at http://docs.python.org/tutorial/floatingpoint.html for a further explanation. (It's on the Python website, but Java and C++ have the same "problem".)
The solution depends on what exactly your problem is:
If it's that you just don't like seeing all those noise digits, then fix your string formatting. Don't display more than 15 significant digits (or 7 for float).
If it's that the inexactness of your numbers is breaking things like "if" statements, then you should write if (abs(x - 7.3) < TOLERANCE) instead of if (x == 7.3).
If you're working with money, then what you probably really want is decimal fixed point. Store an integer number of cents or whatever the smallest unit of your currency is.
(VERY UNLIKELY) If you need more than 53 significant bits (15-16 significant digits) of precision, then use a high-precision floating-point type, like BigDecimal.
private void getRound() {
// this is very simple and interesting
double a = 5, b = 3, c;
c = a / b;
System.out.println(" round val is " + c);
// round val is : 1.6666666666666667
// if you want to only two precision point with double we
// can use formate option in String
// which takes 2 parameters one is formte specifier which
// shows dicimal places another double value
String s = String.format("%.2f", c);
double val = Double.parseDouble(s);
System.out.println(" val is :" + val);
// now out put will be : val is :1.67
}
Use java.math.BigDecimal
Doubles are binary fractions internally, so they sometimes cannot represent decimal fractions to the exact decimal.
/*
0.8 1.2
0.7 1.3
0.7000000000000002 2.3
0.7999999999999998 4.2
*/
double adjust = fToInt + 1.0 - orgV;
// The following two lines works for me.
String s = String.format("%.2f", adjust);
double val = Double.parseDouble(s);
System.out.println(val); // output: 0.8, 0.7, 0.7, 0.8
Doubles are approximations of the decimal numbers in your Java source. You're seeing the consequence of the mismatch between the double (which is a binary-coded value) and your source (which is decimal-coded).
Java's producing the closest binary approximation. You can use the java.text.DecimalFormat to display a better-looking decimal value.
Short answer: Always use BigDecimal and make sure you are using the constructor with String argument, not the double one.
Back to your example, the following code will print 11.4, as you wish.
public class doublePrecision {
public static void main(String[] args) {
BigDecimal total = new BigDecimal("0");
total = total.add(new BigDecimal("5.6"));
total = total.add(new BigDecimal("5.8"));
System.out.println(total);
}
}
Multiply everything by 100 and store it in a long as cents.
Computers store numbers in binary and can't actually represent numbers such as 33.333333333 or 100.0 exactly. This is one of the tricky things about using doubles. You will have to just round the answer before showing it to a user. Luckily in most applications, you don't need that many decimal places anyhow.
Floating point numbers differ from real numbers in that for any given floating point number there is a next higher floating point number. Same as integers. There's no integer between 1 and 2.
There's no way to represent 1/3 as a float. There's a float below it and there's a float above it, and there's a certain distance between them. And 1/3 is in that space.
Apfloat for Java claims to work with arbitrary precision floating point numbers, but I've never used it. Probably worth a look.
http://www.apfloat.org/apfloat_java/
A similar question was asked here before
Java floating point high precision library
Use a BigDecimal. It even lets you specify rounding rules (like ROUND_HALF_EVEN, which will minimize statistical error by rounding to the even neighbor if both are the same distance; i.e. both 1.5 and 2.5 round to 2).
Why not use the round() method from Math class?
// The number of 0s determines how many digits you want after the floating point
// (here one digit)
total = (double)Math.round(total * 10) / 10;
System.out.println(total); // prints 11.4
Check out BigDecimal, it handles problems dealing with floating point arithmetic like that.
The new call would look like this:
term[number].coefficient.add(co);
Use setScale() to set the number of decimal place precision to be used.
If you have no choice other than using double values, can use the below code.
public static double sumDouble(double value1, double value2) {
double sum = 0.0;
String value1Str = Double.toString(value1);
int decimalIndex = value1Str.indexOf(".");
int value1Precision = 0;
if (decimalIndex != -1) {
value1Precision = (value1Str.length() - 1) - decimalIndex;
}
String value2Str = Double.toString(value2);
decimalIndex = value2Str.indexOf(".");
int value2Precision = 0;
if (decimalIndex != -1) {
value2Precision = (value2Str.length() - 1) - decimalIndex;
}
int maxPrecision = value1Precision > value2Precision ? value1Precision : value2Precision;
sum = value1 + value2;
String s = String.format("%." + maxPrecision + "f", sum);
sum = Double.parseDouble(s);
return sum;
}
You can Do the Following!
System.out.println(String.format("%.12f", total));
if you change the decimal value here %.12f
So far I understand it as main goal to get correct double from wrong double.
Look for my solution how to get correct value from "approximate" wrong value - if it is real floating point it rounds last digit - counted from all digits - counting before dot and try to keep max possible digits after dot - hope that it is enough precision for most cases:
public static double roundError(double value) {
BigDecimal valueBigDecimal = new BigDecimal(Double.toString(value));
String valueString = valueBigDecimal.toPlainString();
if (!valueString.contains(".")) return value;
String[] valueArray = valueString.split("[.]");
int places = 16;
places -= valueArray[0].length();
if ("56789".contains("" + valueArray[0].charAt(valueArray[0].length() - 1))) places--;
//System.out.println("Rounding " + value + "(" + valueString + ") to " + places + " places");
return valueBigDecimal.setScale(places, RoundingMode.HALF_UP).doubleValue();
}
I know it is long code, sure not best, maybe someone can fix it to be more elegant. Anyway it is working, see examples:
roundError(5.6+5.8) = 11.399999999999999 = 11.4
roundError(0.4-0.3) = 0.10000000000000003 = 0.1
roundError(37235.137567000005) = 37235.137567
roundError(1/3) 0.3333333333333333 = 0.333333333333333
roundError(3723513756.7000005) = 3.7235137567E9 (3723513756.7)
roundError(3723513756123.7000005) = 3.7235137561237E12 (3723513756123.7)
roundError(372351375612.7000005) = 3.723513756127E11 (372351375612.7)
roundError(1.7976931348623157) = 1.797693134862316
Do not waste your efford using BigDecimal. In 99.99999% cases you don't need it. java double type is of cource approximate but in almost all cases, it is sufficiently precise. Mind that your have an error at 14th significant digit. This is really negligible!
To get nice output use:
System.out.printf("%.2f\n", total);
I'm playing with numbers in Java, and want to see how big a number I can make. It is my understanding that BigInteger can hold a number of infinite size, so long as my computer has enough Memory to hold such a number, correct?
My problem is that BigInteger.pow accepts only an int, not another BigInteger, which means I can only use a number up to 2,147,483,647 as the exponent. Is it possible to use the BigInteger class as such?
BigInteger.pow(BigInteger)
Thanks.
You can write your own, using repeated squaring:
BigInteger pow(BigInteger base, BigInteger exponent) {
BigInteger result = BigInteger.ONE;
while (exponent.signum() > 0) {
if (exponent.testBit(0)) result = result.multiply(base);
base = base.multiply(base);
exponent = exponent.shiftRight(1);
}
return result;
}
might not work for negative bases or exponents.
You can only do this in Java by modular arithmetic, meaning you can do a a^b mod c, where a,b,c are BigInteger numbers.
This is done using:
BigInteger modPow(BigInteger exponent, BigInteger m)
Read the BigInteger.modPow documentation here.
The underlying implementation of BigInteger is limited to (2^31-1) * 32-bit values. which is almost 2^36 bits. You will need 8 GB of memory to store it and many times this to perform any operation on it like toString().
BTW: You will never be able to read such a number. If you tried to print it out it would take a life time to read it.
Please be sure to read the previous answers and comments and understand why this should not be attempted on a production level application. The following is a working solution that can be used for testing purposes only:
Exponent greater than or equal to 0
BigInteger pow(BigInteger base, BigInteger exponent) {
BigInteger result = BigInteger.ONE;
for (BigInteger i = BigInteger.ZERO; i.compareTo(exponent) != 0; i = i.add(BigInteger.ONE)) {
result = result.multiply(base);
}
return result;
}
This will work for both positive and negative bases. You might want to handle 0 to the power of 0 according to your needs, since that's technically undefined.
Exponent can be both positive or negative
BigDecimal allIntegersPow(BigInteger base, BigInteger exponent) {
if (BigInteger.ZERO.compareTo(exponent) > 0) {
return BigDecimal.ONE.divide(new BigDecimal(pow(base, exponent.negate())), 2, RoundingMode.HALF_UP);
}
return new BigDecimal(pow(base, exponent));
}
This re-uses the first method to return a BigDecimal with 2 decimal places, you can define the scale and rounding mode as per your needs.
Again, you should not do this in a real-life, production-level system.
java wont let you do BigInteger.Pow(BigInteger) but you can just put it to the max integer in a loop and see where a ArithmeticException is thrown or some other error due to running out of memory.
2^2,147,483,647 has at least 500000000 digit, in fact computing pow is NPC problem, [Pow is NPC in the length of input, 2 input (m,n) which they can be coded in O(logm + logn) and can take upto nlog(m) (at last the answer takes n log(m) space) which is not polynomial relation between input and computation size], there are some simple problems which are not easy in fact for example sqrt(2) is some kind of them, you can't specify true precision (all precisions), i.e BigDecimal says can compute all precisions but it can't (in fact) because no one solved this up to now.
I can suggest you make use of
BigInteger modPow(BigInteger exponent, BigInteger m)
Suppose you have BigInteger X, and BigInteger Y and you want to calculate BigInteger Z = X^Y.
Get a large Prime P >>>> X^Y and do Z = X.modPow(Y,P);
For anyone who stumbles upon this from the Groovy side of things, it is totally possible to pass a BigInteger to BigInteger.pow().
groovy> def a = 3G.pow(10G)
groovy> println a
groovy> println a.class
59049
class java.math.BigInteger
http://docs.groovy-lang.org/2.4.3/html/groovy-jdk/java/math/BigInteger.html#power%28java.math.BigInteger%29
Just use .intValue()
If your BigInteger is named BigValue2, then it would be BigValue2.intValue()
So to answer your question, it's
BigValue1.pow(BigValue2.intValue())