How to display a "double" in all its precision???

Discussion in 'Java' started by CS Imam, Aug 7, 2006.

  1. CS Imam

    CS Imam Guest


    Here is a code fragment that is very simple... but I can't get it to

    public static void main(String[] args)
    for (int i = 1; i <= 30 ; i++)
    double x = Math.pow(2, i);
    x = 1 + 1 / x;
    System.out.printf("For i = %d: %.40f%n", i, x);
    Long.toBinaryString(Double.doubleToLongBits(x)) );

    All this code is supposed to do is print out the fractions 1+1/2,
    1+1/4, 1+1/8, etc. When one prints out the raw bits (see
    doubleToLongBits), the code is clearly working.

    But on the regular printf("For i...etc"), at i=17 and above, the
    numbers get frozen at 16 digits displayed after the decimal point (the
    precision). But it's not really the precision, because the bits ARE
    changing correctly. What gives???


    - not a stunningly gorgeous woman who would marry you if you solve this
    CS Imam, Aug 7, 2006
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  2. CS Imam

    EJP Guest

    There *is no more* precision. A double has 53 bits of binary precision
    which is about 16 decimal digits. The 40 decimal digits you're expecting
    would take 133 bits.

    Maybe you're getting confused between 40 bits of binary precision and 40
    decimal digits?
    EJP, Aug 7, 2006
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  3. CS Imam

    CS Imam Guest

    Thank you for your reply, but here is the output to clarify what I am
    seeing. As you will see, the binary representation shows that the
    double is more than capable of representing the numbers in question
    (1/2^i where i goes from 1 to 30). As you said, the double gives 52
    bits of precision. However, when displaying the number in decimal, Java
    appears to be unable to display it correctly - why does it properly
    display in base 2, but not in base 10?

    I hope I was clearer this time - please pardon me if not! But thanks
    for your help...

    For i = 1: 1.5000000000000000000000000000000000000000

    For i = 2: 1.2500000000000000000000000000000000000000

    For i = 3: 1.1250000000000000000000000000000000000000

    For i = 4: 1.0625000000000000000000000000000000000000

    For i = 5: 1.0312500000000000000000000000000000000000

    For i = 6: 1.0156250000000000000000000000000000000000

    For i = 7: 1.0078125000000000000000000000000000000000

    For i = 8: 1.0039062500000000000000000000000000000000

    For i = 9: 1.0019531250000000000000000000000000000000

    For i = 10: 1.0009765625000000000000000000000000000000

    For i = 11: 1.0004882812500000000000000000000000000000

    For i = 12: 1.0002441406250000000000000000000000000000

    For i = 13: 1.0001220703125000000000000000000000000000

    For i = 14: 1.0000610351562500000000000000000000000000

    For i = 15: 1.0000305175781250000000000000000000000000

    For i = 16: 1.0000152587890625000000000000000000000000

    For i = 17: 1.0000076293945312000000000000000000000000

    For i = 18: 1.0000038146972656000000000000000000000000

    For i = 19: 1.0000019073486328000000000000000000000000

    For i = 20: 1.0000009536743164000000000000000000000000

    For i = 21: 1.0000004768371582000000000000000000000000

    For i = 22: 1.0000002384185790000000000000000000000000

    For i = 23: 1.0000001192092896000000000000000000000000

    For i = 24: 1.0000000596046448000000000000000000000000

    For i = 25: 1.0000000298023224000000000000000000000000

    For i = 26: 1.0000000149011612000000000000000000000000

    For i = 27: 1.0000000074505806000000000000000000000000

    For i = 28: 1.0000000037252903000000000000000000000000

    For i = 29: 1.0000000018626451000000000000000000000000

    For i = 30: 1.0000000009313226000000000000000000000000

    CS Imam, Aug 7, 2006

  4. Double.toString, used implicitly in conversion of x to a String,
    produces the shortest string that, when converted back to double, will
    produce the original number.

    BigDecimal is the easiest way I know to get all the digits:

    System.out.println(new BigDecimal(x));

    Patricia Shanahan, Aug 7, 2006
  5. CS Imam

    EJP Guest

    No, 16 decimal places really *is* the precision. The *binary* bits
    change beyond 16 because the radix is binary not decimal. The 16th
    decimal digit expresses a lot more precision than the 16th binary digit.
    You need to understant that. In fact I don't understand what you are
    expecting to see. The last decimal number printed is 1.0000000009313226
    and the last binary number printed converts precisely back to that.
    Nothing is being lost. You can't get 40 decimal digits out of 53 bits,
    you can only get 16.
    EJP, Aug 7, 2006
  6. For any number that can be expressed as a terminating binary fraction,
    including any number that is representable in Java double, there is a
    unique decimal fraction that is EXACTLY equal to it, not just a
    close-enough approximation. I believe that is the answer the OP is
    looking for.

    (In another message, I suggested getting it via BigDecimal).

    Patricia Shanahan, Aug 7, 2006
  7. CS Imam

    CS Imam Guest

    Thanks again for replying, you and Patricia both. The BigDecimal DID
    work, and that is great to solve the problem at hand.

    But I'm really interested in understanding my... misunderstanding I

    I am not looking for 40 places of decimal precision; I only used
    "%.40d" as an overkill to see "all the numbers". I *am* aware that
    doubles are supposed to give only 15 digits of decimal precison
    approximately. However, what I find puzzling is that in binary, we are
    supposed to get 52 (not 53 as far as I know) bits of precision. So here
    is my misunderstanding: I see the bits changing in binary. And yet when
    they are converted into decimal through the "prints" (and Patricia
    pointed out that the problem is really in "toString"), the decimal
    equivalent is NOT precise. And this does not make sense to me. If the
    underlying raw number in binary IS precise, then the converted decimal
    number should be precise as well, right?

    Actually as far as I know, something IS being lost. The last binary
    number, if you convert it to decimal, should be:


    In binary, the underlying bits are as follows:


    So again, if the underlying bits are precisely expressing some number
    within the 52 bits of accuracy, why does converting it to a decimal
    representation fail?

    I really apologize if I am not seeing something that you are

    thanks, and sorry again.
    CS Imam, Aug 7, 2006
  8. CS Imam

    Chris Uppal Guest

    I think what you may be missing is that there is a /range/ of precise decimal
    numbers which would all have the same representation as a double. So, although
    any given double converts exactly into precisely one arbitrary-precision
    decimal number, that number is not the only one which the double value may be
    "trying" to represent.

    The string representation has to /choose/ one value from the infinite set of
    arbitrary-precision decimal numbers which the double value might be intended
    to represent. One option would be to chose the unique element which was
    exactly equal to the double, but that's not the only possible design. In fact
    (and defensibly, IMO, although it would be nice to have a choice) the element
    which is chosen is the one with the fewest digits -- otherwise, for instance,
    0.1D would print out as:
    which is certainly precise, but is probably not what most programmers (or
    users) would wish to see.

    -- chris
    Chris Uppal, Aug 7, 2006
  9. CS Imam

    hiwa Guest

    This is the funniest post of this summer.
    The OP reminds me of some type of elderly people who stubbonly believe
    in an impossible.
    Thanks for a big laugh.

    Learn and study IEEE 754 bit-array-representation of 64 bit FPN.
    hiwa, Aug 7, 2006
  10. Section 3.2.2 Double of ANSI/IEEE Std 754-1985 begins "A 64-bit double
    format number X is divided as shown in in Fig. 2. The value v of X is
    inferred from its constituent fields thus:" followed by a series of cases.

    The relevant case is "(3) If 0 < e < 2047, then ..." followed by a
    formula for the value of a finite, normalized, non-zero double number.
    Using "^" for exponentiation, and "*" for multiplication, and "." for
    the centered dot, it is equivalent to:

    (-1)^s * 2^(e-1023 *(1 "." f) where s is the sign bit, e is the
    exponent, and f is the fraction.

    I interpreted the base article as requesting a print out, in decimal, of
    that value. My BigDecimal suggestion, which does exactly that, worked
    for the OP, so that seems to be the correct interpretation.

    Why do you consider this to be impossible? Or do you disagree with my
    Although I find IEEE 754 floating point arithmetic both interesting and
    useful, I'm completely missing its humor. I know explanations can
    sometimes kill a joke, but perhaps you could explain what is so funny?
    I've read the standard from cover to cover a couple of times, and reread
    individual sections far more often. I've looked at many, many doubles as
    bit patterns. To me, the OP's request seemed quite reasonable. Indeed,
    it is something I've needed for myself when working on understanding
    some of the subtleties of rounding, so I already had a solution. What am
    I missing?

    Patricia Shanahan, Aug 7, 2006
  11. CS Imam wrote:
    Chris has already responded to your main point.

    I just want to clarify the reason for 53 bits of precision, rather than 52.

    It is a consequence of floating point normalization. Even in non-binary
    systems, there are advantages to avoiding, wherever possible, leading
    zero digits in the mantissa. A "normalized" floating point format is one
    in which the most significant digit of the mantissa is never zero.

    For binary, there is an additional advantage. We know the leading digit
    of the mantissa of a normalized float is a binary digit, and is not
    zero. There is no point spending a bit in a dense format on a binary
    digit that must be one, so it does not appear. Normalization buys us an
    extra bit of precision.

    Suppose you have a normalized double with 52 bit fraction f. The full
    mantissa is 1.f, a 1 before the binary point, followed by the 52 bit
    fraction after the binary point.

    Patricia Shanahan, Aug 7, 2006
  12. CS Imam

    jmcgill Guest

    I've looked for a formula before to get this kind of constraint. That
    is, to answer the question, "how many decimal digits of precision is
    N bits of binary precision for a given floating point model?"

    Maybe this is a question for a discrete math forum.
    jmcgill, Aug 7, 2006
  13. CS Imam

    CS Imam Guest


    Just wanted to thank you for your information. You turned out to be
    right on the money. Those posters who are insisting that there is no
    loss of precision, and that it is impossible to do better, and that
    this is a funny post... you need to go back and read.

    Based on what Patricia wrote, I did some searching on the net, and
    found that this is indeed a difficult problem: what to print in decimal
    format given a binary number. And it also turns out that it was
    addressed in a classic paper by Guy Steele and Jon White. Here is a
    link to the paper:

    "How to Print Floating Point Numbers Accurately"

    You may read section 2 "Properties of Radix Conversion" to understand
    the issues involved. If you don't have time, then read Patricia's
    original answer to me. That was the succinct and correct answer, not
    repeatedly insisting that nothing was being lost. Indeed, LOTS is being
    lost... but on purpose it turns out.

    - my hat is off to Ms. Shanahan
    CS Imam, Aug 7, 2006
  14. The superficial, rough answer is that N bits have 2^N possible values
    (using "^" for exponentiation). M decimal digits have 10^M possible values.

    If 10^M = 2^N then M = log10(2^N) = N*log10(2)

    So the rough answer is to multiply N by log10(2), about 0.301.

    53*log10(2) is about 15.954, almost 16.

    Once you get into actual arithmetic, everything gets more complicated.

    Patricia Shanahan, Aug 7, 2006
  15. CS Imam

    jmcgill Guest

    Is it at least correct to claim that 53 bits of binary precision
    guarantees no less than 15 decimal digits? Even this, no doubt,
    depends on the exponent.
    jmcgill, Aug 7, 2006
  16. I'm not sure I see how the exponent affects it that much, as long as you
    are thinking of significant digits. Of course, if you are thinking of
    digits after the decimal point, even a 20 digit decimal floating point
    system does not guarantee 15 decimal digits.

    Patricia Shanahan, Aug 7, 2006
  17. CS Imam

    jmcgill Guest

    I suspected that larger exponents lead to more granularity in the ranges
    or something like that.

    Also, when I posted to the thread I somehow thought I was posting on a C
    group, not java. I realize java programmers do not generally deal with
    the bitwise evaluation of data.

    I'm wondering all this because in other disciplines, "error bounds" is
    always such an early, and often repeated, focus. Yet I had never seen
    nor been asked for error bounds in IEEE numeric representations.
    jmcgill, Aug 7, 2006
  18. I think you need to distinguish more between absolute and relative effects.

    For example, the absolute difference x-y between two consecutive
    representable numbers, x and y, is a strictly increasing function of the
    exponent. The relative difference (x-y)/x varies within a given value of
    the exponent, but does not increase with exponent.
    I'm neither a Java programmer nor a C programmer. I'm a programmer who
    happens be using Java right now. I may ignore some details when working
    in high level languages, but that does not mean I understand them any
    less than when I'm working in assembly language.
    There is a required accuracy for all the basic operations, specified in
    the IEEE 754 standard, although some implementations confuse matters by
    keeping intermediate results with more accuracy. That tends to reduce
    the need for discussion.

    A good library specification should discuss error bounds for those
    functions whose implementation is allowed some slack. See, for example,
    the Java API documentation for sin:

    "The computed result must be within 1 ulp of the exact result."

    ("ulp" is short for Unit Least Place, a difference of one in the least
    significant bit of the fraction. See the top of the referenced page for
    a more detailed explanation.)

    Beyond the basics, the subject rapidly gets very complex, see
    "numerical analysis" in any good technical bookstore or library.
    Floating point application accuracy is a difficult, but intensely
    studied, subject.

    Patricia Shanahan, Aug 7, 2006
  19. CS Imam

    jmcgill Guest

    If you're starting from scratch, start by truly understanding unsigned
    and signed char, then unsigned and signed two's complement, then single
    precision floating point, and then, with a full comprehension of that,
    the full spec won't be that hard to understand.

    Are there CS programs out there that don't include a computer
    organization class where this stuff gets drilled into your brain?
    jmcgill, Aug 8, 2006
  20. CS Imam

    The_Sage Guest

    Isn't it obvious that your "lack of precision" is due to rounding off and
    quantizing errors -- as would be expected?

    Not all floating point decimal numbers can be exactly represented in
    digit-restricted binary, therefore some truncation or rouding will occur during
    conversion. This reduces precision.

    Remember the math lecture that if all your calculations are done with, say
    10-digits, your final answer will not be accurate to 10-digits due to rounding
    errors? If you want 10-digit accuracy, you need more than 10-digits to work
    with. The more operations you perform on a digit-restricted number, the less
    accurate it becomes due to rounding off. This reduces precision and this is the
    reason for the existence of "guard digits", ie -- 10-digit calculators use
    13-digit calculations internally in order to maintain 10-digit accuracy.

    Most application programmers do not list the internal precision of their math
    routines, -- assuming that they are even aware of such math issues. Usually, if
    you are really serious about your math, you will test the math routines for
    accuracy by performing repetitive loops.


    The Sage


    "All those painted screens erected by man to shut out reality
    -- history, religion, duty, social position --
    all were illusions, mere opium fantasies"
    John Fowles, The French Lieutenant's Woman
    The_Sage, Aug 9, 2006
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