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'
