Related
I'm googling about how to find if a number y is power of x and came across this link
Java
public class Solution {
public boolean isPowerOfThree(int n) {
return (Math.log10(n) / Math.log10(3)) % 1 == 0;
} }
Common pitfalls
This solution is problematic because we start using doubles, which
means we are subject to precision errors. This means, we should never
use == when comparing doubles. That is because the result of
Math.log10(n) / Math.log10(3) could be 5.0000001 or 4.9999999. This
effect can be observed by using the function Math.log() instead of
Math.log10().
In order to fix that, we need to compare the result against an
epsilon.
Java
return (Math.log(n) / Math.log(3) + epsilon) % 1 <= 2 * epsilon;
There I didn't understand return (Math.log(n) / Math.log(3) + epsilon) % 1 <= 2 * epsilon;
What is the meaning of that line?
Why EPSILON is used while comparing floating points?
As the quoted section says, because of floating point imprecisions, you can have two numbers that should be exactly equal (if the calculations that created them were carried out with mathematical exactness), but that are instead slightly different.
When you compare them, you want to account for that slight difference and still treat numbers as equal if they differ only by a small amount, called epsilon.
How to choose an appropriate epsilon, though, is a tricky question and highly dependent on the nature of your calculations. I suppose that for this reason, Java does not include a "standard" epsilon constant (some other languages do).
Because natural logarithm calculates using simple series. https://en.wikipedia.org/wiki/Natural_logarithm
Such series may be implemented in math-coprocessor in CPU. You may express any logarithm from natural using simple proportion without big computation time.
I'm trying to use Junit to test a java program, and I'm not sure how to go about testing for upper-bound violations.
Specifically, I have written a simple program to convert between kilometers and miles.
For example, here is the method for converting from miles to kilometers
public static double mileToKm(double mile){
//1.1170347260596139E308 / 0.621371192 = Double.MAX_VALUE
try{
if (mile < 0 || mile > 1.1170347260596139E308){
throw new IllegalArgumentException();
}
else
return mile / 0.621371192;}
return 0;
}
So, I guess my question is two-fold: First, why is it that I can't conjure up an exception when I try
mileToKm(1.1170347260596139E308 + 1)
in junit? I assume it's a rounding issue, but if that's the case then how can I get the exception thrown?
Second, for the method to convert from km to mile, I want to throw an exception if the parameter is greater than Double.MAX_VALUE. How can I pass such a parameter? I can get the Junit test to pass if I just pass as parameter Double.MAX_VALUE * 10, but I also get a message in the Console (this is all in Eclipse Mars 4.5.1, btw) saying 'MAX = 1.7976931348623157E308'. The parameter has to be a double so it can't be BigDecimal or something like that.
OK, I lied, the question is three-fold. What's up with this:
double value = Double.MAX_VALUE * 0.621371192; //max_value * conversion factor
System.out.println(value);
prints 1.1170347260596138E308, but then these two statements
System.out.println(value / 0.621371192);
System.out.println(Double.MAX_VALUE);
print 1.7976931348623155E308 and 1.7976931348623157E308, respectively. In other words, I would expect these two values to both be equivalent to Double.MAX_VALUE, but the first statement has a 5 right before the E, instead of a 7. How can I fix this? Thanks so much, hope this isn't too prolix.
You're confused about floating point numbers.
Firstly, the number 1.1170347260596139E308 + 1 is not representable using primitives in Java, as doubles have ~16 significant digits, and that addition requires 308 significant digits.
Secondly, float/double operations are not idempotent if you use intermediate storage (and most of the times even without it). Floating point operations lose accuracy, and arithmetic methods that retain accuracy over large computations (think weather models) are sought after in the scientific sector.
Thirdly, there's Double.MAX_VALUE, which represents the largest representable number in a primitive in Java; the only other value X such that X > Double.MAX_VALUE can hold is Double.POSITIVE_INFINITY, and that's not a real number.
When I perform simple math in java with doubles and other number data types, the double values seem to randomly vary a bit from the supposed result, which might be 5,59999999997 or 6,0000000002 or something. When I cast to int, the double value is obviously rounded down to the next whole number. Does this mean the double could be both 5 or 6? Or does that "5,999999999997" still count as 6 though which would be depending on the binary float value? If not, is there a way to let the negative vary be rounded up, but not lower values from 5,5 to 5,999999999996?
I mean, I dont really want to round the value as described in my last sentence. I'd like to always round down to the next whole number, but I don't want to cause an extra decrement due to wrong double math results.
Converting a double to an int always rounds down. You can round to the nearest whole integer via Math.round(double). The double is varying from what you expect because of floating point error.
If you want to round, you can use the round() method.
double d = 6 +/- some small error
long l = Math.round(d);
Or you can add 0.5 for positive numbers
long l = (long) (d + 0.5);
or
long l = (long) (d + (d < 0 ? -0.5 : 0.5));
I'm not sure I understand the question. Usually when you cast a double to int you add 0.5 to have a nice round.
From the Java Language Specification:
The Java programming language uses round toward zero when converting a floating value to an
integer (ยง5.1.3), which acts, in this case, as though the number were truncated, discarding
the mantissa bits. Rounding toward zero chooses at its result the format's value closest to
and no greater in magnitude than the infinitely precise result.
So 5,999999999997 when casted to an int will 5 and 6,0000000002 will be 6. If I understand what you are asking with having negative versions of the values (e.g. -5.97), I fail to see how
Math.round() does not suffice you. -6,0000000002 will be rounded to -6 as will -5,999999999997 and every other value above (but not including) -5.5.
The code review tool I use complains with the below when I start comparing two float values using equality operator. What is the correct way and how to do it? Is there a helper function (commons-*) out there which I can reuse?
Description
Cannot compare floating-point values using the equals (==) operator
Explanation
Comparing floating-point values by using either the equality (==) or inequality (!=) operators is not always accurate because of rounding errors.
Recommendation
Compare the two float values to see if they are close in value.
float a;
float b;
if(a==b)
{
..
}
IBM has a recommendation for comparing two floats, using division rather than subtraction - this makes it easier to select an epsilon that works for all ranges of input.
if (abs(a/b - 1) < epsilon)
As for the value of epsilon, I would use 5.96e-08 as given in this Wikipedia table, or perhaps 2x that value.
It wants you to compare them to within the amount of accuracy you need. For example if you require that the first 4 decimal digits of your floats are equal, then you would use:
if(-0.00001 <= a-b && a-b <= 0.00001)
{
..
}
Or:
if(Math.abs(a-b) < 0.00001){ ... }
Where you add the desired precision to the difference of the two numbers and compare it to twice the desired precision.
Whatever you think is more readable. I prefer the first one myself as it clearly shows the precision you are allowing on both sides.
a = 5.43421 and b = 5.434205 will pass the comparison
private static final float EPSILON = <very small positive number>;
if (Math.abs(a-b) < EPSILON)
...
As floating point offers you variable but uncontrollable precision (that is, you can't set the precision other than when you choose between using double and float), you have to pick your own fixed precision for comparisons.
Note that this isn't a true equivalence operator any more, as it isn't transitive. You can easily get a equals b and b equals c but a not equals c.
Edit: also note that if a is negative and b is a very large positive number, the subtraction can overflow and the result will be negative infinity, but the test will still work, as the absolute value of negative infinity is positive infinity, which will be bigger than EPSILON.
Use commons-lang
org.apache.commons.lang.math.NumberUtils#compare
Also commons-math (in your situation more appropriate solution):
http://commons.apache.org/math/apidocs/org/apache/commons/math/util/MathUtils.html#equals(double, double)
The float type is an approximate value - there's an exponent portion and a value portion with finite accuracy.
For example:
System.out.println((0.6 / 0.2) == 3); // false
The risk is that a tiny rounding error can make a comparison false, when mathematically it should be true.
The workaround is to compare floats allowing a minor difference to still be "equal":
static float e = 0.00000000000001f;
if (Math.abs(a - b) < e)
Apache commons-math to the rescue: MathUtils.(double x, double y, int maxUlps)
Returns true if both arguments are equal or within the range of allowed error (inclusive). Two float numbers are considered equal if there are (maxUlps - 1) (or fewer) floating point numbers between them, i.e. two adjacent floating point numbers are considered equal.
Here's the actual code form the Commons Math implementation:
private static final int SGN_MASK_FLOAT = 0x80000000;
public static boolean equals(float x, float y, int maxUlps) {
int xInt = Float.floatToIntBits(x);
int yInt = Float.floatToIntBits(y);
if (xInt < 0)
xInt = SGN_MASK_FLOAT - xInt;
if (yInt < 0)
yInt = SGN_MASK_FLOAT - yInt;
final boolean isEqual = Math.abs(xInt - yInt) <= maxUlps;
return isEqual && !Float.isNaN(x) && !Float.isNaN(y);
}
This gives you the number of floats that can be represented between your two values at the current scale, which should work better than an absolute epsilon.
I took a stab at this based on the way java implements == for doubles. It converts to the IEEE 754 long integer form first and then does a bitwise compare. Double also provides the static doubleToLongBits() to get the integer form. Using bit fiddling you can 'round' the mantissa of the double by adding 1/2 (one bit) and truncating.
In keeping with supercat's observation, the function first tries a simple == comparison and only rounds if that fails. Here is what I came up with some (hopefully) helpful comments.
I did some limited testing, but can't say I've tried all edge cases. Also, I did not test performance. It shouldn't be too bad.
I just realized that this is essentially the same solution as the one offered by Dmitri. Perhaps a bit more concise.
static public boolean nearlyEqual(double lhs, double rhs){
// This rounds to the 6th mantissa bit from the end. So the numbers must have the same sign and exponent and the mantissas (as integers)
// need to be within 32 of each other (bottom 5 bits of 52 bits can be different).
// To allow 'n' bits of difference create an additive value of 1<<(n-1) and a mask of 0xffffffffffffffffL<<n
// e.g. 4 bits are: additive: 0x10L = 0x1L << 4 and mask: 0xffffffffffffffe0L = 0xffffffffffffffffL << 5
//int bitsToIgnore = 5;
//long additive = 1L << (bitsToIgnore - 1);
//long mask = ~0x0L << bitsToIgnore;
//return ((Double.doubleToLongBits(lhs)+additive) & mask) == ((Double.doubleToLongBits(rhs)+additive) & mask);
return lhs==rhs?true:((Double.doubleToLongBits(lhs)+0x10L) & 0xffffffffffffffe0L) == ((Double.doubleToLongBits(rhs)+0x10L) & 0xffffffffffffffe0L);
}
The following modification handles the change in sign case where the value is on either side of 0.
return lhs==rhs?true:((Double.doubleToLongBits(lhs)+0x10L) & 0x7fffffffffffffe0L) == ((Double.doubleToLongBits(rhs)+0x10L) & 0x7fffffffffffffe0L);
There are many cases where one wants to regard two floating-point numbers as equal only if they are absolutely equivalent, and a "delta" comparison would be wrong. For example, if f is a pure function), and one knows that q=f(x) and y===x, then one should know that q=f(y) without having to compute it. Unfortunately the == has two defects in this regard.
If one value is positive zero and the other is negative zero, they will compare as equal even though they are not necessarily equivalent. For example if f(d)=1/d, a=0 and b=-1*a, then a==b but f(a)!=f(b).
If either value is a NaN, the comparison will always yield false even if one value was assigned directly from the other.
Although there are many cases where checking floating-point numbers for exact equivalence is right and proper, I'm not sure about any cases where the actual behavior of == should be considered preferable. Arguably, all tests for equivalence should be done via a function that actually tests equivalence (e.g. by comparing bitwise forms).
First, a few things to note:
The "Standard" way to do this is to choose an constant epsilon, but constant epsilons do not work correctly for all number ranges.
If you want to use a constant epsilon sqrt(EPSILON) the square root of the epsilon from float.h is a generally considered a good value. (this comes from an infamous "orange book" who's name escapes me at the moment).
Floating point division is going to be slow, so you probably want to avoid it for comparisons even if it behaves like picking an epsilon that is custom made for the numbers' magnitudes.
What do you really want to do? something like this:
Compare how many representable floating point numbers the values differ by.
This code comes from this really great article by Bruce Dawson. The article has been since updated here. The main difference is the old article breaks the strict-aliasing rule. (casting floating pointers to int pointer, dereferencing, writing, casting back). While the C/C++ purist will quickly point out the flaw, in practice this works, and I consider the code more readable. However, the new article uses unions and C/C++ gets to keep its dignity. For brevity I give the code that breaks strict aliasing below.
// Usable AlmostEqual function
bool AlmostEqual2sComplement(float A, float B, int maxUlps)
{
// Make sure maxUlps is non-negative and small enough that the
// default NAN won't compare as equal to anything.
assert(maxUlps > 0 && maxUlps < 4 * 1024 * 1024);
int aInt = *(int*)&A;
// Make aInt lexicographically ordered as a twos-complement int
if (aInt < 0)
aInt = 0x80000000 - aInt;
// Make bInt lexicographically ordered as a twos-complement int
int bInt = *(int*)&B;
if (bInt < 0)
bInt = 0x80000000 - bInt;
int intDiff = abs(aInt - bInt);
if (intDiff <= maxUlps)
return true;
return false;
}
The basic idea in the code above is to first notice that given the IEEE 754 floating point format, {sign-bit, biased-exponent, mantissa}, that the numbers are lexicographically ordered if interpreted as signed magnitude ints. That is the sign bit becomes the sign bit, the and the exponent always completely outranks the mantissa in defining magnitude of the float and because it comes first in determining the magnitude of the number interpreted as an int.
So, we interpret the bit representation of the floating point number as a signed-magnitude int. We then convert the signed-magnitude ints to a two's complement ints by subtracting them from 0x80000000 if the number is negative. Then we just compare the two values as we would any signed two's complement ints, and seeing how many values they differ by. If this amount is less than the threshold you choose for how many representable floats the values may differ by and still be considered equal, then you say that they are "equal." Note that this method correctly lets "equal" numbers differ by larger values for larger magnitude floats, and by smaller values for smaller magnitude floats.
Is it safe if I use comparision like this (a is int, b and c is float/double):
a == b
b == c
It may hear ridiculous, but in my old programing language, sometimes 1 + 2 == 3 is false (because left side returns 2.99999999999...). And, what about this:
Math.sqrt(b) == Math.sqrt(c)
b / 3 == 10 / 3 //In case b = 10, does it return true?
In general, no it is not safe due to the fact that so many decimal numbers cannot be precisely represented as float or double values. The often stated solution is test if the difference between the numbers is less than some "small" value (often denoted by a greek 'epsilon' character in the maths literature).
However - you need to be a bit careful how you do the test. For instance, if you write:
if (Math.abs(a - b) < 0.000001) {
System.err.println("equal");
}
where a and b are supposed to be "the same", you are testing the absolute error. If you do this, you can get into trouble if a and b are (say_ 1,999,999.99 and 2,000,000.00 respectively. The difference between these two numbers is less than the smallest representable value at that scale for a float, and yet it is much bigger than our chosen epsilon.
Arguably, a better approach is to use the relative error; e.g. coded (defensively) as
if (a == b ||
Math.abs(a - b) / Math.max(Math.abs(a), Math.abs(b)) < 0.000001) {
System.err.println("close enough to be equal");
}
But even this is not the complete answer, because it does not take account of the way that certain calculations cause the errors to build up to unmanageable proportions. Take a look at this Wikipedia link for more details.
The bottom line is that dealing with errors in floating point calculations is a lot more difficult than it appears at first glance.
The other point to note is (as others have explained) integer arithmetic behaves very differently to floating point arithmetic in a couple of respects:
integer division will truncate if the result is not integral
integer addition subtraction and multiplication will overflow.
Both of these happen without any warning, either at compile time or at runtime.
You do need to exercise some care.
1.0 + 2.0 == 3.0
is true because integers are exactly representable.
Math.sqrt(b) == Math.sqrt(c)
if b == c.
b / 3.0 == 10.0 / 3.0
if b == 10.0 which is what I think you meant.
The last two examples compare two different instances of the same calculation. When you have different calculations with non representable numbers then exact equality testing fails.
If you are testing the results of a calculation that is subject to floating point approximation then equality testing should be done up to a tolerance.
Do you have any specific real world examples? I think you will find that it is rare to want to test equality with floating point.
b / 3 != 10 / 3 - if b is a floating point variable like b = 10.0f, so b / 3 is 3.3333, while 10 / 3 is integer division, so is equal to 3.
If b == c, then Math.sqrt(b) == Math.sqrt(c) - this is because the sqrt function returns double anyways.
In general, you shouldn't be comparing doubles/floats for equation, because they are floating point numbers so you might get errors. You almost always want to compare them with a given precision, i.e.:
b - c < 0.000001
The safest way to compare a float/double with something else is actually to use see if their difference is a small number.
e.g.
Math.abs(a - b) < EPS
where EPS can be something like 0.0000001.
In this way you make sure that precision errors do not affect your results.
== comparison is not particularly safe for doubles/floats in basically any language. An epsilon comparison method (where you check that the difference between two floats is reasonably small) is your best bet.
For:
Math.sqrt(b) == Math.sqrt(c)
I'm not sure why you wouldn't just compare b and c, but an epsilon comparison would work here too.
For:
b / 3 == 10 / 3
Since 10/3 = 3 because of integer division, this will not necessarily give the results you're looking for. You could use 10.0 / 3, though i'm still not sure why you wouldn't just compare b and 10 (using the epsilon comparison method in either case).
Float wrapper class's compare method can be used to compare two float values.
Float.compare(float1,float2)==0
It will compare integer bits corresponding to each float object.
In java you can compare a float with another float,and a double with another double.And you will get true when comparing a double with a float when their precision is equal
b is float and 10 is integer then if you compare both with your this critearia then it will give false because...
b = flaot (meance ans 3.33333333333333333333333)
10 is integer so that make sence.
If you want a more complete explanation, here there is a good one: http://download.oracle.com/docs/cd/E19957-01/806-3568/ncg_goldberg.html (but it is a little bit long)