# 15. Floating-Point Arithmetic: Issues and Limitations

Floating-point numbers are represented in computer hardware as base 2 (binary) fractions. For example, the decimal fraction `0.625` has value 6/10 + 2/100 + 5/1000, and in the same way the binary fraction `0.101` has value 1/2 + 0/4 + 1/8. These two fractions have identical values, the only real difference being that the first is written in base 10 fractional notation, and the second in base 2.

Sayangnya, sebagian besar pecahan desimal tidak dapat direpresentasikan persis dengan pecahan biner. Konsekuensinya adalah bahwa, secara umum, angka pecahan floating-point desimal yang Anda masukkan hanya didekati oleh angka-angka pecahan floating-point biner yang sebenarnya disimpan dalam mesin.

Masalahnya lebih mudah dipahami pada awalnya di basis 10. Pertimbangkan fraksi 1/3. Anda dapat memperkirakannya sebagai pecahan basis 10:

    0.3

atau, lebih baik,

    0.33

atau, lebih baik,

    0.333

dan seterusnya. Tidak peduli berapa banyak digit yang Anda ingin tulis, hasilnya tidak akan pernah benar-benar 1/3, tetapi akan menjadi perkiraan yang semakin baik dari 1/3.

Dengan cara yang sama, tidak peduli berapa banyak digit basis 2 yang ingin Anda gunakan, nilai desimal 0.1 tidak dapat direpresentasikan persis sebagai fraksi basis 2. Dalam basis 2, 1/10 adalah percahan berulang yang tak terhingga

    0.0001100110011001100110011001100110011001100110011...

Berhenti pada jumlah bit yang terbatas, dan Anda mendapatkan perkiraan. Pada kebanyakan mesin saat ini, float diperkirakan menggunakan percahan biner dengan pembilang menggunakan 53 bit pertama dimulai dengan bit paling signifikan dan dengan penyebut sebagai pangkat dua. Dalam kasus 1/10, fraksi biner adalah `3602879701896397 / 2 ** 55` yang dekat dengan tetapi tidak persis sama dengan nilai sebenarnya dari 1/10.

Many users are not aware of the approximation because of the way values are displayed. Python only prints a decimal approximation to the true decimal value of the binary approximation stored by the machine. On most machines, if Python were to print the true decimal value of the binary approximation stored for 0.1, it would have to display:

    >>> 0.1
    0.1000000000000000055511151231257827021181583404541015625

That is more digits than most people find useful, so Python keeps the number of digits manageable by displaying a rounded value instead:

    >>> 1 / 10
    0.1

Hanya ingat, meskipun hasil cetakannya terlihat seperti nilai tepat 1/10, nilai sebenarnya yang disimpan adalah pecahan biner terdekat yang dapat direpresentasikan.

Menariknya, ada banyak angka desimal berbeda yang memiliki pecahan biner perkiraan terdekat yang sama. Misalnya, angka `0.1` dan `0.10000000000000001` dan `0.1000000000000000055511151231257827021181583404541015625` semuanya didekati oleh `3602879701896397 / 2 ** 55`. Karena semua nilai desimal ini memiliki perkiraan yang sama, salah satu dari nilai tersebut dapat ditampilkan sambil tetap mempertahankan invarian lainnya `eval(repr(x)) == x`.

Secara historis, Python prompt dan fungsi bawaan `repr()` akan memilih satu dengan 17 digit signifikan, `0.10000000000000001`. Dimulai dengan Python 3.1, Python (pada kebanyakan sistem) sekarang dapat memilih yang paling pendek dan hanya menampilkan `0.1`.

Note that this is in the very nature of binary floating point: this is not a bug in Python, and it is not a bug in your code either. You'll see the same kind of thing in all languages that support your hardware's floating-point arithmetic (although some languages may not display the difference by default, or in all output modes).

For more pleasant output, you may wish to use string formatting to produce a limited number of significant digits:

    >>> format(math.pi, '.12g')  # give 12 significant digits
    '3.14159265359'

    >>> format(math.pi, '.2f')   # give 2 digits after the point
    '3.14'

    >>> repr(math.pi)
    '3.141592653589793'

Sangat penting untuk menyadari bahwa ini adalah, dalam arti sebenarnya, sebuah ilusi: Anda hanya membulatkan display dari nilai mesin yang sebenarnya.

One illusion may beget another. For example, since 0.1 is not exactly 1/10, summing three values of 0.1 may not yield exactly 0.3, either:

    >>> 0.1 + 0.1 + 0.1 == 0.3
    False

Also, since the 0.1 cannot get any closer to the exact value of 1/10 and 0.3 cannot get any closer to the exact value of 3/10, then pre-rounding with `round()` function cannot help:

    >>> round(0.1, 1) + round(0.1, 1) + round(0.1, 1) == round(0.3, 1)
    False

Though the numbers cannot be made closer to their intended exact values, the `math.isclose()` function can be useful for comparing inexact values:

    >>> math.isclose(0.1 + 0.1 + 0.1, 0.3)
    True

Alternatively, the `round()` function can be used to compare rough approximations:

    >>> round(math.pi, ndigits=2) == round(22 / 7, ndigits=2)
    True

Binary floating-point arithmetic holds many surprises like this. The problem with "0.1" is explained in precise detail below, in the "Representation Error" section. See Examples of Floating Point Problems for a pleasant summary of how binary floating point works and the kinds of problems commonly encountered in practice. Also see The Perils of Floating Point for a more complete account of other common surprises.

As that says near the end, "there are no easy answers." Still, don't be unduly wary of floating point! The errors in Python float operations are inherited from the floating-point hardware, and on most machines are on the order of no more than 1 part in 2**53 per operation. That's more than adequate for most tasks, but you do need to keep in mind that it's not decimal arithmetic and that every float operation can suffer a new rounding error.

Sementara kasus patologis memang ada, untuk sebagian besar penggunaan aritmatika floating-point yang santai Anda akan melihat hasil yang Anda harapkan pada akhirnya jika Anda hanya membulatkan tampilan hasil akhir Anda ke jumlah angka desimal yang Anda harapkan. `str()` biasanya mencukupi, dan untuk kontrol yang lebih baik lihat format `str.format()` penentu format di Format string syntax.

Untuk kasus penggunaan yang memerlukan representasi desimal yang tepat, coba gunakan modul `decimal` yang mengimplementasikan aritmatika desimal yang cocok untuk aplikasi akuntansi dan aplikasi presisi tinggi.

Bentuk lain dari aritmatika yang tepat didukung oleh modul `fractions` yang mengimplementasikan aritmatika berdasarkan bilangan rasional (sehingga angka seperti 1/3 dapat direpresentasikan secara tepat).

If you are a heavy user of floating-point operations you should take a look at the NumPy package and many other packages for mathematical and statistical operations supplied by the SciPy project. See <https://scipy.org>.

Python provides tools that may help on those rare occasions when you really do want to know the exact value of a float. The `float.as_integer_ratio()` method expresses the value of a float as a fraction:

    >>> x = 3.14159
    >>> x.as_integer_ratio()
    (3537115888337719, 1125899906842624)

Since the ratio is exact, it can be used to losslessly recreate the original value:

    >>> x == 3537115888337719 / 1125899906842624
    True

The `float.hex()` method expresses a float in hexadecimal (base 16), again giving the exact value stored by your computer:

    >>> x.hex()
    '0x1.921f9f01b866ep+1'

This precise hexadecimal representation can be used to reconstruct the float value exactly:

    >>> x == float.fromhex('0x1.921f9f01b866ep+1')
    True

Karena representasinya tepat, maka berguna untuk porting nilai secara andal di berbagai versi Python (platform independensi) dan pertukaran data dengan bahasa lain yang mendukung format yang sama (seperti Java dan C99).

Another helpful tool is the `sum()` function which helps mitigate loss-of-precision during summation. It uses extended precision for intermediate rounding steps as values are added onto a running total. That can make a difference in overall accuracy so that the errors do not accumulate to the point where they affect the final total:

    >>> 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 == 1.0
    False
    >>> sum([0.1] * 10) == 1.0
    True

The `math.fsum()` goes further and tracks all of the "lost digits" as values are added onto a running total so that the result has only a single rounding. This is slower than `sum()` but will be more accurate in uncommon cases where large magnitude inputs mostly cancel each other out leaving a final sum near zero:

    >>> arr = [-0.10430216751806065, -266310978.67179024, 143401161448607.16,
    ...        -143401161400469.7, 266262841.31058735, -0.003244936839808227]
    >>> float(sum(map(Fraction, arr)))   # Exact summation with single rounding
    8.042173697819788e-13
    >>> math.fsum(arr)                   # Single rounding
    8.042173697819788e-13
    >>> sum(arr)                         # Multiple roundings in extended precision
    8.042178034628478e-13
    >>> total = 0.0
    >>> for x in arr:
    ...     total += x                   # Multiple roundings in standard precision
    ...
    >>> total                            # Straight addition has no correct digits!
    -0.0051575902860057365

## 15.1. Kesalahan Representasi

Bagian ini menjelaskan contoh "0.1" secara terperinci, dan menunjukkan bagaimana Anda dapat melakukan analisis yang tepat atas kasus-kasus seperti ini sendiri. Diasumsikan terbiasa secara mendasar dengan representasi pecahan floating point biner.

Representation error mengacu pada fakta bahwa beberapa pecahan desimal (sebagian besar, sebenarnya) tidak dapat direpresentasikan persis sebagai pecahan biner (basis 2). Ini adalah alasan utama mengapa Python (atau Perl, C, C++, Java, Fortran, dan banyak lainnya) sering tidak akan menampilkan angka desimal tepat yang Anda harapkan.

Why is that? 1/10 is not exactly representable as a binary fraction. Since at least 2000, almost all machines use IEEE 754 binary floating-point arithmetic, and almost all platforms map Python floats to IEEE 754 binary64 "double precision" values. IEEE 754 binary64 values contain 53 bits of precision, so on input the computer strives to convert 0.1 to the closest fraction it can of the form J/2**N where J is an integer containing exactly 53 bits. Rewriting

    1 / 10 ~= J / (2**N)

sebagai

    J ~= 2**N / 10

and recalling that J has exactly 53 bits (is `>= 2**52` but `< 2**53`), the best value for N is 56:

    >>> 2**52 <=  2**56 // 10  < 2**53
    True

That is, 56 is the only value for N that leaves J with exactly 53 bits. The best possible value for J is then that quotient rounded:

    >>> q, r = divmod(2**56, 10)
    >>> r
    6

Since the remainder is more than half of 10, the best approximation is obtained by rounding up:

    >>> q+1
    7205759403792794

Therefore the best possible approximation to 1/10 in IEEE 754 double precision is:

    7205759403792794 / 2 ** 56

Membagi pembilang dan penyebut dengan dua mengurangi pecahan menjadi:

    3602879701896397 / 2 ** 55

Perhatikan bahwa sejak kami mengumpulkan, ini sebenarnya sedikit lebih besar dari 1/10; jika kita belum mengumpulkan, hasil bagi akan sedikit lebih kecil dari 1/10. Tetapi tidak dapatkah hal itu exactly 1/10!

So the computer never "sees" 1/10: what it sees is the exact fraction given above, the best IEEE 754 double approximation it can get:

    >>> 0.1 * 2 ** 55
    3602879701896397.0

If we multiply that fraction by 10**55, we can see the value out to 55 decimal digits:

    >>> 3602879701896397 * 10 ** 55 // 2 ** 55
    1000000000000000055511151231257827021181583404541015625

meaning that the exact number stored in the computer is equal to the decimal value 0.1000000000000000055511151231257827021181583404541015625. Instead of displaying the full decimal value, many languages (including older versions of Python), round the result to 17 significant digits:

    >>> format(0.1, '.17f')
    '0.10000000000000001'

The `fractions` and `decimal` modules make these calculations easy:

    >>> from decimal import Decimal
    >>> from fractions import Fraction

    >>> Fraction.from_float(0.1)
    Fraction(3602879701896397, 36028797018963968)

    >>> (0.1).as_integer_ratio()
    (3602879701896397, 36028797018963968)

    >>> Decimal.from_float(0.1)
    Decimal('0.1000000000000000055511151231257827021181583404541015625')

    >>> format(Decimal.from_float(0.1), '.17')
    '0.10000000000000001'