I have the same set of data and am running the same code, but sometimes I get different results at the 19th decimal place and beyond. Although this is not a great concern to me for numbers less than 0.0001, it makes me wonder whether 19th decimal place is Raku's limit of precision?
Word 104 differ:
0.04948872986571077 19 chars
0.04948872986571079 19 chars
Word 105 differ:
0.004052062278212545 20 chars
0.0040520622782125445 21 chars
TL;DR See the doc's outstanding Numerics page.
(I had forgotten about that page before I wrote the following answer. Consider this answer at best a brief summary of a few aspects of that page.)
There are two aspects to this. Internal precision and printing precision.
100% internal precision until RAM is exhausted
Raku supports arbitrary precision number types. Quoting Wikipedia's relevant page:
digits of precision are limited only by the available memory of the host system
You can direct Raku to use one of its arbitrary precision types.[1] If you do so it will retain 100% precision until it runs out of RAM.
Arbitrary precision type
Corresponding type checking[2]
Example of value of that type
Int
my Int $foo ...
66174449004242214902112876935633591964790957800362273
FatRat
my FatRat $foo ...
66174449004242214902112876935633591964790957800362273 / 13234889800848443102075932929798260216894990083844716
Thus you can get arbitrary internal precision for integers and fractions (including arbitrary precision decimals).
Limited internal precision
If you do not direct Raku to use an arbitrary precision number type it will do its best but may ultimately switch to limited precision. For example, Raku will give up on 100% precision if a formula you use calculates a Rat and the number's denominator exceeds 64 bits.[1]
Raku's fall back limited precision number type is Num:
On most platforms, [a Num is] an IEEE 754 64-bit floating point numbers, aka "double precision".
Quoting the Wikipedia page for that standard:
Floating point is used ... when a wider range is needed ... even if at the cost of precision.
The 53-bit significand precision gives from 15 to 17 significant decimal digits precision (2−53 ≈ 1.11 × 10−16).
Printing precision
Separate from internal precision is stringification of numbers.
(It was at this stage that I remembered the doc page on Numerics linked at the start of this answer.)
Quoting Printing rationals:
Keep in mind that output routines like say or put ... may choose to display a Num as an Int or a Rat number. For a more definitive string to output, use the raku method or [for a rational number] .nude
Footnotes
[1] You control the type of a numeric expression via the types of individual numbers in the expression, and the types of the results of numeric operations, which in turn depend on the types of the numbers. Examples:
1 + 2 is 3, an Int, because both 1 and 2 are Ints, and a + b is an Int if both a and b are Ints;
1 / 2 is not an Int even though both 1 and 2 are individually Ints, but is instead 1/2 aka 0.5, a Rat.
1 + 4 / 2 will print out as 3, but the 3 is internally a Rat, not an Int, due to Numeric infectiousness.
[2] All that enforcement does is generate a run-time error if you try to assign or bind a value that is not of the numeric type you've specified as the variable's type constraint. Enforcement doesn't mean that Raku will convert numbers for you. You have to write your formulae to ensure the result you get is what you want.[1] You can use coercion -- but coercion cannot regain precision that's already been lost.
When I save a large number in R as an object the wrong number is saved? Why is that?
options("scipen"=100, "digits"=4)
num <- 201912030032451613
num
#> [1] 201912030032451616
Created on 2019-12-12 by the reprex package (v0.2.1.9000)
As #Roland says, this is a floating point issue (the Wikipedia page on floating point numbers is as good as anything). Unpacking it a bit though, R has specific integer format but it is limited to 32 bit integers:
> str(-2147483647L)
int -2147483647
> str(2147483647L)
int 2147483647
> str(21474836470L)
num 21474836470
Warning message:
non-integer value 21474836470L qualified with L; using numeric value
So, when R gets your number it is storing it as a floating point number not an integer. Floating point numbers are limited in how much precision they can store and typically only have about 17 significant digits. Because your number has more significant digits than that there is loss of precision. Losing precision in the smallest digits doesn't usually matter for computer arithmetic, but if your big number is a key of some kind (or a date stamp) then you are in more trouble. The bit64 package is designed with this kind of use case in mind, or you could import it as a string, depending on what you want to do.
round(Number, Precision) ->
Power = math:pow(10, Precision),
round(Pi * Power) / Power.
I can't for the life of me figure out how this function actually works.
First, you have a mistake in the function. It should be:
round(Number, Precision) ->
Power = math:pow(10, Precision),
round(Number * Power) / Power.
With Number = 10.23456 and Precision = 2, the line:
Power = math:pow(10, Precision)
results in:
Power = 10 * 10 = 100
And Number * Power is equal to:
10.23456 * 100
which is:
1023.456
Then, round(1023.456) is equal to:
1023
Dividing that number by Power, or 100, gives you:
10.23
The algorithm you are using works like this:
Move the number of decimal places you are interested in keeping to the left of the decimal point. If you want to keep one decimal place, you multiply the number by 10 (=> 102.3456); if you want to keep two decimal places, you multiply the number by 100 (=> 1023.456); if you want to keep 3 decimal places, you multiply the number by 1000 (=> 10234.56); etc.
Get rid of everything to the right of the decimal point using round().
Move the number of decimal places you wanted to keep back to the right of the decimal point.
So, if you have the number 10.23456 and you want to keep 1 decimal place, if you follow the steps in the algorithm you get:
102.3456
102
10.2
With the number 10.23456 and wanting to keep 2 decimal places, if you follow the steps in the algorithm you get:
1023.456
1023
10.23
With the number 10.23456 and wanting to keep 3 decimal places, if you follow the steps in the algorithm you get:
10234.56
10234
10.234
Theorem:
The required number of digits (in base t) to represent the positive integer S in base t is ⟦logtS⟧+1 (⟦.⟧: floor function).
I wondered, what is the required number of digits (in base 2) to represent the maximum positive double (floating point) number in computer. I have 64-bit OS and 32-bit R on it. Hence, I did:
.Machine$double.xmax # 1.797693e+308
typeof(.Machine$double.xmax) # double
floor(log(.Machine$double.xmax, 2))+1 # 1025
.Machine$integer.max # 2147483647
class(.Machine$integer.max) # integer
floor(log(.Machine$integer.max, 2))+1 # 31; (1 bit for sign bit)
So, the theory is OK for integers.
(1) But what about the double equivalent of the theorem? I.e., what is the required number of digits (in base t) to represent the double in base t?
(2) This may be difficult with real numbers with decimals. So, perhaps, one may know the equivalent of the theorem for decimalless reals (that is ">2147483647").
In particular, where does the 1025 above come from?
(3) Would I get 63 if I used 64-bit OS and 64-bit R for the following?
floor(log(.Machine$integer.max, 2))+1 # 63??; (1 bit for sign bit??)
Ad 3) I don't know about doubles but the integer internal representation is still 32 bits even on 64 bit systems. If you want to go bigger you need to use some sort of library for that for example 'bit64'
You will get more detailed information with help(double) and help(integer)
I'm trying to learn C and have come across the inability to work with REALLY big numbers (i.e., 100 digits, 1000 digits, etc.). I am aware that there exist libraries to do this, but I want to attempt to implement it myself.
I just want to know if anyone has or can provide a very detailed, dumbed down explanation of arbitrary-precision arithmetic.
It's all a matter of adequate storage and algorithms to treat numbers as smaller parts. Let's assume you have a compiler in which an int can only be 0 through 99 and you want to handle numbers up to 999999 (we'll only worry about positive numbers here to keep it simple).
You do that by giving each number three ints and using the same rules you (should have) learned back in primary school for addition, subtraction and the other basic operations.
In an arbitrary precision library, there's no fixed limit on the number of base types used to represent our numbers, just whatever memory can hold.
Addition for example: 123456 + 78:
12 34 56
78
-- -- --
12 35 34
Working from the least significant end:
initial carry = 0.
56 + 78 + 0 carry = 134 = 34 with 1 carry
34 + 00 + 1 carry = 35 = 35 with 0 carry
12 + 00 + 0 carry = 12 = 12 with 0 carry
This is, in fact, how addition generally works at the bit level inside your CPU.
Subtraction is similar (using subtraction of the base type and borrow instead of carry), multiplication can be done with repeated additions (very slow) or cross-products (faster) and division is trickier but can be done by shifting and subtraction of the numbers involved (the long division you would have learned as a kid).
I've actually written libraries to do this sort of stuff using the maximum powers of ten that can be fit into an integer when squared (to prevent overflow when multiplying two ints together, such as a 16-bit int being limited to 0 through 99 to generate 9,801 (<32,768) when squared, or 32-bit int using 0 through 9,999 to generate 99,980,001 (<2,147,483,648)) which greatly eased the algorithms.
Some tricks to watch out for.
1/ When adding or multiplying numbers, pre-allocate the maximum space needed then reduce later if you find it's too much. For example, adding two 100-"digit" (where digit is an int) numbers will never give you more than 101 digits. Multiply a 12-digit number by a 3 digit number will never generate more than 15 digits (add the digit counts).
2/ For added speed, normalise (reduce the storage required for) the numbers only if absolutely necessary - my library had this as a separate call so the user can decide between speed and storage concerns.
3/ Addition of a positive and negative number is subtraction, and subtracting a negative number is the same as adding the equivalent positive. You can save quite a bit of code by having the add and subtract methods call each other after adjusting signs.
4/ Avoid subtracting big numbers from small ones since you invariably end up with numbers like:
10
11-
-- -- -- --
99 99 99 99 (and you still have a borrow).
Instead, subtract 10 from 11, then negate it:
11
10-
--
1 (then negate to get -1).
Here are the comments (turned into text) from one of the libraries I had to do this for. The code itself is, unfortunately, copyrighted, but you may be able to pick out enough information to handle the four basic operations. Assume in the following that -a and -b represent negative numbers and a and b are zero or positive numbers.
For addition, if signs are different, use subtraction of the negation:
-a + b becomes b - a
a + -b becomes a - b
For subtraction, if signs are different, use addition of the negation:
a - -b becomes a + b
-a - b becomes -(a + b)
Also special handling to ensure we're subtracting small numbers from large:
small - big becomes -(big - small)
Multiplication uses entry-level math as follows:
475(a) x 32(b) = 475 x (30 + 2)
= 475 x 30 + 475 x 2
= 4750 x 3 + 475 x 2
= 4750 + 4750 + 4750 + 475 + 475
The way in which this is achieved involves extracting each of the digits of 32 one at a time (backwards) then using add to calculate a value to be added to the result (initially zero).
ShiftLeft and ShiftRight operations are used to quickly multiply or divide a LongInt by the wrap value (10 for "real" math). In the example above, we add 475 to zero 2 times (the last digit of 32) to get 950 (result = 0 + 950 = 950).
Then we left shift 475 to get 4750 and right shift 32 to get 3. Add 4750 to zero 3 times to get 14250 then add to result of 950 to get 15200.
Left shift 4750 to get 47500, right shift 3 to get 0. Since the right shifted 32 is now zero, we're finished and, in fact 475 x 32 does equal 15200.
Division is also tricky but based on early arithmetic (the "gazinta" method for "goes into"). Consider the following long division for 12345 / 27:
457
+-------
27 | 12345 27 is larger than 1 or 12 so we first use 123.
108 27 goes into 123 4 times, 4 x 27 = 108, 123 - 108 = 15.
---
154 Bring down 4.
135 27 goes into 154 5 times, 5 x 27 = 135, 154 - 135 = 19.
---
195 Bring down 5.
189 27 goes into 195 7 times, 7 x 27 = 189, 195 - 189 = 6.
---
6 Nothing more to bring down, so stop.
Therefore 12345 / 27 is 457 with remainder 6. Verify:
457 x 27 + 6
= 12339 + 6
= 12345
This is implemented by using a draw-down variable (initially zero) to bring down the segments of 12345 one at a time until it's greater or equal to 27.
Then we simply subtract 27 from that until we get below 27 - the number of subtractions is the segment added to the top line.
When there are no more segments to bring down, we have our result.
Keep in mind these are pretty basic algorithms. There are far better ways to do complex arithmetic if your numbers are going to be particularly large. You can look into something like GNU Multiple Precision Arithmetic Library - it's substantially better and faster than my own libraries.
It does have the rather unfortunate misfeature in that it will simply exit if it runs out of memory (a rather fatal flaw for a general purpose library in my opinion) but, if you can look past that, it's pretty good at what it does.
If you cannot use it for licensing reasons (or because you don't want your application just exiting for no apparent reason), you could at least get the algorithms from there for integrating into your own code.
I've also found that the bods over at MPIR (a fork of GMP) are more amenable to discussions on potential changes - they seem a more developer-friendly bunch.
While re-inventing the wheel is extremely good for your personal edification and learning, its also an extremely large task. I don't want to dissuade you as its an important exercise and one that I've done myself, but you should be aware that there are subtle and complex issues at work that larger packages address.
For example, multiplication. Naively, you might think of the 'schoolboy' method, i.e. write one number above the other, then do long multiplication as you learned in school. example:
123
x 34
-----
492
+ 3690
---------
4182
but this method is extremely slow (O(n^2), n being the number of digits). Instead, modern bignum packages use either a discrete Fourier transform or a Numeric transform to turn this into an essentially O(n ln(n)) operation.
And this is just for integers. When you get into more complicated functions on some type of real representation of number (log, sqrt, exp, etc.) things get even more complicated.
If you'd like some theoretical background, I highly recommend reading the first chapter of Yap's book, "Fundamental Problems of Algorithmic Algebra". As already mentioned, the gmp bignum library is an excellent library. For real numbers, I've used MPFR and liked it.
Don't reinvent the wheel: it might turn out to be square!
Use a third party library, such as GNU MP, that is tried and tested.
You do it in basically the same way you do with pencil and paper...
The number is to be represented in a buffer (array) able to take on an arbitrary size (which means using malloc and realloc) as needed
you implement basic arithmetic as much as possible using language supported structures, and deal with carries and moving the radix-point manually
you scour numeric analysis texts to find efficient arguments for dealing by more complex function
you only implement as much as you need.
Typically you will use as you basic unit of computation
bytes containing with 0-99 or 0-255
16 bit words contaning wither 0-9999 or 0--65536
32 bit words containing...
...
as dictated by your architecture.
The choice of binary or decimal base depends on you desires for maximum space efficiency, human readability, and the presence of absence of Binary Coded Decimal (BCD) math support on your chip.
You can do it with high school level of mathematics. Though more advanced algorithms are used in reality. So for example to add two 1024-byte numbers :
unsigned char first[1024], second[1024], result[1025];
unsigned char carry = 0;
unsigned int sum = 0;
for(size_t i = 0; i < 1024; i++)
{
sum = first[i] + second[i] + carry;
carry = sum - 255;
}
result will have to be bigger by one place in case of addition to take care of maximum values. Look at this :
9
+
9
----
18
TTMath is a great library if you want to learn. It is built using C++. The above example was silly one, but this is how addition and subtraction is done in general!
A good reference about the subject is Computational complexity of mathematical operations. It tells you how much space is required for each operation you want to implement. For example, If you have two N-digit numbers, then you need 2N digits to store the result of multiplication.
As Mitch said, it is by far not an easy task to implement! I recommend you take a look at TTMath if you know C++.
One of the ultimate references (IMHO) is Knuth's TAOCP Volume II. It explains lots of algorithms for representing numbers and arithmetic operations on these representations.
#Book{Knuth:taocp:2,
author = {Knuth, Donald E.},
title = {The Art of Computer Programming},
volume = {2: Seminumerical Algorithms, second edition},
year = {1981},
publisher = {\Range{Addison}{Wesley}},
isbn = {0-201-03822-6},
}
Assuming that you wish to write a big integer code yourself, this can be surprisingly simple to do, spoken as someone who did it recently (though in MATLAB.) Here are a few of the tricks I used:
I stored each individual decimal digit as a double number. This makes many operations simple, especially output. While it does take up more storage than you might wish, memory is cheap here, and it makes multiplication very efficient if you can convolve a pair of vectors efficiently. Alternatively, you can store several decimal digits in a double, but beware then that convolution to do the multiplication can cause numerical problems on very large numbers.
Store a sign bit separately.
Addition of two numbers is mainly a matter of adding the digits, then check for a carry at each step.
Multiplication of a pair of numbers is best done as convolution followed by a carry step, at least if you have a fast convolution code on tap.
Even when you store the numbers as a string of individual decimal digits, division (also mod/rem ops) can be done to gain roughly 13 decimal digits at a time in the result. This is much more efficient than a divide that works on only 1 decimal digit at a time.
To compute an integer power of an integer, compute the binary representation of the exponent. Then use repeated squaring operations to compute the powers as needed.
Many operations (factoring, primality tests, etc.) will benefit from a powermod operation. That is, when you compute mod(a^p,N), reduce the result mod N at each step of the exponentiation where p has been expressed in a binary form. Do not compute a^p first, and then try to reduce it mod N.
Here's a simple ( naive ) example I did in PHP.
I implemented "Add" and "Multiply" and used that for an exponent example.
http://adevsoft.com/simple-php-arbitrary-precision-integer-big-num-example/
Code snip
// Add two big integers
function ba($a, $b)
{
if( $a === "0" ) return $b;
else if( $b === "0") return $a;
$aa = str_split(strrev(strlen($a)>1?ltrim($a,"0"):$a), 9);
$bb = str_split(strrev(strlen($b)>1?ltrim($b,"0"):$b), 9);
$rr = Array();
$maxC = max(Array(count($aa), count($bb)));
$aa = array_pad(array_map("strrev", $aa),$maxC+1,"0");
$bb = array_pad(array_map("strrev", $bb),$maxC+1,"0");
for( $i=0; $i<=$maxC; $i++ )
{
$t = str_pad((string) ($aa[$i] + $bb[$i]), 9, "0", STR_PAD_LEFT);
if( strlen($t) > 9 )
{
$aa[$i+1] = ba($aa[$i+1], substr($t,0,1));
$t = substr($t, 1);
}
array_unshift($rr, $t);
}
return implode($rr);
}