I've run across a little problem when trying to solve a Project Euler problem in Julia. I've basically written a recursive function which produces fractions with increasingly large numerators and denominators. I don't want to post the code for obvious reasons, but the last few fractions are as follows:
1180872205318713601//835002744095575440
2850877693509864481//2015874949414289041
6882627592338442563//4866752642924153522
At that point I get an OverflowError(), presumably because the numerator and/or denominator now exceeds 19 digits. Is there a way of handling 'Big' fractions in Julia (i.e. those with BigInt-type numerators and denominators)?
Addendum:
OK, I've simplified the code and disguised it a bit. If anyone wants to wade through 650 Project Euler problems to try to work out which question it is, good luck to them – there will probably be around 200 better solutions!
function series(limit::Int64, i::Int64=1, n::Rational{Int64}=1//1)
while i <= limit
n = 1 + 1//(1 + 2n)
println(n)
return series(limit, i + 1, n)
end
end
series(50)
If I run the above function with, say, 20 as the argument it runs fine. With 50 I get the OverflowError().
Julia defaults to using machine integers. For more information on this see the FAQ: Why does Julia use native machine integer arithmetic?.
In short: the most efficient integer operations on any modern CPU involves computing on a fixed number of bits. On your machine, that's 64 bits.
julia> 9223372036854775805 + 1
9223372036854775806
julia> 9223372036854775805 + 2
9223372036854775807
julia> 9223372036854775805 + 3
-9223372036854775808
Whoa! What just happened!? That's definitely wrong! It's more obvious if you look at how these numbers are represented in binary:
julia> bitstring(9223372036854775805 + 1)
"0111111111111111111111111111111111111111111111111111111111111110"
julia> bitstring(9223372036854775805 + 2)
"0111111111111111111111111111111111111111111111111111111111111111"
julia> bitstring(9223372036854775805 + 3)
"1000000000000000000000000000000000000000000000000000000000000000"
So you can see that those 63 bits "ran out of space" and rolled over — the 64th bit there is called the "sign bit" and signals a negative number.
There are two potential solutions when you see overflow like this: you can use "checked arithmetic" — like the rational code does — that ensures you don't silently have this problem:
julia> Base.Checked.checked_add(9223372036854775805, 3)
ERROR: OverflowError: 9223372036854775805 + 3 overflowed for type Int64
Or you can use a bigger integer type — like the unbounded BigInt:
julia> big(9223372036854775805) + 3
9223372036854775808
So an easy fix here is to remove your type annotations and dynamically choose your integer types based upon limit:
function series(limit, i=one(limit), n=one(limit)//one(limit))
while i <= limit
n = 1 + 1//(1 + 2n)
println(n)
return series(limit, i + 1, n)
end
end
julia> series(big(50))
#…
1186364911176312505629042874//926285732032534439103474303
4225301286417693889465034354//3299015554385159450361560051
Related
So in my text book there is this example of a recursive function using f#
let rec gcd = function
| (0,n) -> n
| (m,n) -> gcd(n % m,m);;
with this function my text book gives the example by executing:
gcd(36,116);;
and since the m = 36 and not 0 then it ofcourse goes for the second clause like this:
gcd(116 % 36,36)
gcd(8,36)
gcd(36 % 8,8)
gcd(4,8)
gcd(8 % 4,4)
gcd(0,4)
and now hits the first clause stating this entire thing is = 4.
What i don't get is this (%)percentage sign/operator or whatever it is called in this connection. for an instance i don't get how
116 % 36 = 8
I have turned this so many times in my head now and I can't figure how this can turn into 8?
I know this is probably a silly question for those of you who knows this but I would very much appreciate your help the same.
% is a questionable version of modulo, which is the remainder of an integer division.
In the positive, you can think of % as the remainder of the division. See for example Wikipedia on Euclidean Divison. Consider 9 % 4: 4 fits into 9 twice. But two times four is only eight. Thus, there is a remainder of one.
If there are negative operands, % effectively ignores the signs to calculate the remainder and then uses the sign of the dividend as the sign of the result. This corresponds to the remainder of an integer division that rounds to zero, i.e. -2 / 3 = 0.
This is a mathematically unusual definition of division and remainder that has some bad properties. Normally, when calculating modulo n, adding or subtracting n on the input has no effect. Not so for this operator: 2 % 3 is not equal to (2 - 3) % 3.
I usually have the following defined to get useful remainders when there are negative operands:
/// Euclidean remainder, the proper modulo operation
let inline (%!) a b = (a % b + b) % b
So far, this operator was valid for all cases I have encountered where a modulo was needed, while the raw % repeatedly wasn't. For example:
When filling rows and columns from a single index, you could calculate rowNumber = index / nCols and colNumber = index % nCols. But if index and colNumber can be negative, this mapping becomes invalid, while Euclidean division and remainder remain valid.
If you want to normalize an angle to (0, 2pi), angle %! (2. * System.Math.PI) does the job, while the "normal" % might give you a headache.
Because
116 / 36 = 3
116 - (3*36) = 8
Basically, the % operator, known as the modulo operator will divide a number by other and give the rest if it can't divide any longer. Usually, the first time you would use it to understand it would be if you want to see if a number is even or odd by doing something like this in f#
let firstUsageModulo = 55 %2 =0 // false because leaves 1 not 0
When it leaves 8 the first time means that it divided you 116 with 36 and the closest integer was 8 to give.
Just to help you in future with similar problems: in IDEs such as Xamarin Studio and Visual Studio, if you hover the mouse cursor over an operator such as % you should get a tooltip, thus:
Module operator tool tip
Even if you don't understand the tool tip directly, it'll give you something to google.
I just read a post from Quora:
http://www.quora.com/Is-Julia-ready-for-production-use
At the bottom, there's an answer said:
2 ^ 3 ^ 4 = 0
I tried it myself:
julia> 2 ^ 3 ^ 4
0
Personally I don't consider this a bug in the language. We can add parenthesis for clarity, both for Julia and for our human beings:
julia> (2 ^ 3) ^ 4
4096
So far so good; however, this doesn't work:
julia> 2 ^ (3 ^ 4)
0
Since I'm learning, I'd like to know, how Julia evaluate this expression to 0? What's the evaluation precedent?
julia> typeof(2 ^ 3 ^ 4)
Int64
I'm surprised I couldn't find a duplicate question about this on SO yet. I figure I'll answer this slightly differently than the FAQ in the manual since it's a common first question. Oops, I somehow missed: Factorial function works in Python, returns 0 for Julia
Imagine you've been taught addition and multiplication, but never learned any numbers higher than 99. As far as you're concerned, numbers bigger than that simply don't exist. So you learned to carry ones into the tens column, but you don't even know what you'd call the column you'd carry tens into. So you just drop them. As long as your numbers never get bigger than 99, everything will be just fine. Once you go over 99, you wrap back down to 0. So 99+3 ≡ 2 (mod 100). And 52*9 ≡ 68 (mod 100). Any time you do a multiplication with more than two factors of 10, your answer will be zero: 25*32 ≡ 0 (mod 100). Now, after you do each computation, someone could ask you "did you go over 99?" But that takes time to answer… time that could be spent computing your next math problem!
This is effectively how computers natively do arithmetic, except they do it in binary with 64 bits. You can see the individual bits with the bits function:
julia> bits(45)
"0000000000000000000000000000000000000000000000000000000000101101"
As we multiply it by 2, 101101 will shift to the left (just like multiplying by 10 in decimal):
julia> bits(45 * 2)
"0000000000000000000000000000000000000000000000000000000001011010"
julia> bits(45 * 2 * 2)
"0000000000000000000000000000000000000000000000000000000010110100"
julia> bits(45 * 2^58)
"1011010000000000000000000000000000000000000000000000000000000000"
julia> bits(45 * 2^60)
"1101000000000000000000000000000000000000000000000000000000000000"
… until it starts falling off the end. If you multiply more than 64 twos together, the answer will always zero (just like multiplying more than two tens together in the example above). We can ask the computer if it overflowed, but doing so by default for every single computation has some serious performance implications. So in Julia you have to be explicit. You can either ask Julia to check after a specific multiplication:
julia> Base.checked_mul(45, 2^60) # or checked_add for addition
ERROR: OverflowError()
in checked_mul at int.jl:514
Or you can promote one of the arguments to a BigInt:
julia> bin(big(45) * 2^60)
"101101000000000000000000000000000000000000000000000000000000000000"
In your example, you can see that the answer is 1 followed by 81 zeros when you use big integer arithmetic:
julia> bin(big(2) ^ 3 ^ 4)
"1000000000000000000000000000000000000000000000000000000000000000000000000000000000"
For more details, see the FAQ: why does julia use native machine integer arithmetic?
If I enter a value, for example
1234567 ^ 98787878
into Wolfram Alpha it can provide me with a number of details. This includes decimal approximation, total length, last digits etc. How do you evaluate such large numbers? As I understand it a programming language would have to have a special data type in order to store the number, let alone add it to something else. While I can see how one might approach the addition of two very large numbers, I can't see how huge numbers are evaluated.
10^2 could be calculated through repeated addition. However a number such as the example above would require a gigantic loop. Could someone explain how such large numbers are evaluated? Also, how could someone create a custom large datatype to support large numbers in C# for example?
Well it's quite easy and you can have done it yourself
Number of digits can be obtained via logarithm:
since `A^B = 10 ^ (B * log(A, 10))`
we can compute (A = 1234567; B = 98787878) in our case that
`B * log(A, 10) = 98787878 * log(1234567, 10) = 601767807.4709646...`
integer part + 1 (601767807 + 1 = 601767808) is the number of digits
First, say, five, digits can be gotten via logarithm as well;
now we should analyze fractional part of the
B * log(A, 10) = 98787878 * log(1234567, 10) = 601767807.4709646...
f = 0.4709646...
first digits are 10^f (decimal point removed) = 29577...
Last, say, five, digits can be obtained as a corresponding remainder:
last five digits = A^B rem 10^5
A rem 10^5 = 1234567 rem 10^5 = 34567
A^B rem 10^5 = ((A rem 10^5)^B) rem 10^5 = (34567^98787878) rem 10^5 = 45009
last five digits are 45009
You may find BigInteger.ModPow (C#) very useful here
Finally
1234567^98787878 = 29577...45009 (601767808 digits)
There are usually libraries providing a bignum datatype for arbitrarily large integers (eg. mapping digits k*n...(k+1)*n-1, k=0..<some m depending on n and number magnitude> to a machine word of size n redefining arithmetic operations). for c#, you might be interested in BigInteger.
exponentiation can be recursively broken down:
pow(a,2*b) = pow(a,b) * pow(a,b);
pow(a,2*b+1) = pow(a,b) * pow(a,b) * a;
there also are number-theoretic results that have engenedered special algorithms to determine properties of large numbers without actually computing them (to be precise: their full decimal expansion).
To compute how many digits there are, one uses the following expression:
decimal_digits(n) = 1 + floor(log_10(n))
This gives:
decimal_digits(1234567^98787878) = 1 + floor(log_10(1234567^98787878))
= 1 + floor(98787878 * log_10(1234567))
= 1 + floor(98787878 * 6.0915146640862625)
= 1 + floor(601767807.4709647)
= 601767808
The trailing k digits are computed by doing exponentiation mod 10^k, which keeps the intermediate results from ever getting too large.
The approximation will be computed using a (software) floating-point implementation that effectively evaluates a^(98787878 log_a(1234567)) to some fixed precision for some number a that makes the arithmetic work out nicely (typically 2 or e or 10). This also avoids the need to actually work with millions of digits at any point.
There are many libraries for this and the capability is built-in in the case of python. You seem primarily concerned with the size of such numbers and the time it may take to do computations like the exponent in your example. So I'll explain a bit.
Representation
You might use an array to hold all the digits of large numbers. A more efficient way would be to use an array of 32 bit unsigned integers and store "32 bit chunks" of the large number. You can think of these chunks as individual digits in a number system with 2^32 distinct digits or characters. I used an array of bytes to do this on an 8-bit Atari800 back in the day.
Doing math
You can obviously add two such numbers by looping over all the digits and adding elements of one array to the other and keeping track of carries. Once you know how to add, you can write code to do "manual" multiplication by multiplying digits and putting the results in the right place and a lot of addition - but software will do all this fairly quickly. There are faster multiplication algorithms than the one you would use manually on paper as well. Paper multiplication is O(n^2) where other methods are O(n*log(n)). As for the exponent, you can of course multiply by the same number millions of times but each of those multiplications would be using the previously mentioned function for doing multiplication. There are faster ways to do exponentiation that require far fewer multiplies. For example you can compute x^16 by computing (((x^2)^2)^2)^2 which involves only 4 actual (large integer) multiplications.
In practice
It's fun and educational to try writing these functions yourself, but in practice you will want to use an existing library that has been optimized and verified.
I think a part of the answer is in the question itself :) To store these expressions, you can store the base (or mantissa), and exponent separately, like scientific notation goes. Extending to that, you cannot possibly evaluate the expression completely and store such large numbers, although, you can theoretically predict certain properties of the consequent expression. I will take you through each of the properties you talked about:
Decimal approximation: Can be calculated by evaluating simple log values.
Total number of digits for expression a^b, can be calculated by the formula
Digits = floor function (1 + Log10(a^b)), where floor function is the closest integer smaller than the number. For e.g. the number of digits in 10^5 is 6.
Last digits: These can be calculated by the virtue of the fact that the expression of linearly increasing exponents form a arithmetic progression. For e.g. at the units place; 7, 9, 3, 1 is repeated for exponents of 7^x. So, you can calculate that if x%4 is 0, the last digit is 1.
Can someone create a custom datatype for large numbers, I can't say, but I am sure, the number won't be evaluated and stored.
EDIT
So it seems I "underestimated" what varying length numbers meant. I didn't even think about situations where the operands are 100 digits long. In that case, my proposed algorithm is definitely not efficient. I'd probably need an implementation who's complexity depends on the # of digits in each operands as opposed to its numerical value, right?
As suggested below, I will look into the Karatsuba algorithm...
Write the pseudocode of an algorithm that takes in two arbitrary length numbers (provided as strings), and computes the product of these numbers. Use an efficient procedure for multiplication of large numbers of arbitrary length. Analyze the efficiency of your algorithm.
I decided to take the (semi) easy way out and use the Russian Peasant Algorithm. It works like this:
a * b = a/2 * 2b if a is even
a * b = (a-1)/2 * 2b + a if a is odd
My pseudocode is:
rpa(x, y){
if x is 1
return y
if x is even
return rpa(x/2, 2y)
if x is odd
return rpa((x-1)/2, 2y) + y
}
I have 3 questions:
Is this efficient for arbitrary length numbers? I implemented it in C and tried varying length numbers. The run-time in was near-instant in all cases so it's hard to tell empirically...
Can I apply the Master's Theorem to understand the complexity...?
a = # subproblems in recursion = 1 (max 1 recursive call across all states)
n / b = size of each subproblem = n / 1 -> b = 1 (problem doesn't change size...?)
f(n^d) = work done outside recursive calls = 1 -> d = 0 (the addition when a is odd)
a = 1, b^d = 1, a = b^d -> complexity is in n^d*log(n) = log(n)
this makes sense logically since we are halving the problem at each step, right?
What might my professor mean by providing arbitrary length numbers "as strings". Why do that?
Many thanks in advance
What might my professor mean by providing arbitrary length numbers "as strings". Why do that?
This actually change everything about the problem (and make your algorithm incorrect).
It means than 1234 is provided as 1,2,3,4 and you cannot operate directly on the whole number. You need to analyze your algorithm in terms of #additions, #multiplications, #divisions.
You should expect a division to be a bit more expensive than a multiplication, and a multiplication to be lot more expensive than an addition. So a good algorithm try to reduce the number of divisions and multiplications.
Check out the Karatsuba algorithm, (ps don't copy it that's not what your teacher want) is one of the fastest for this specification.
Add 3): Native integers are limited in how large (or small) numbers they can represent (32- or 64-bit integers for example). To represent arbitrary length numbers you can choose strings, because then you are not really limited by this. The problem is then, of course, that your arithmetic units are not really made to add strings ;-)
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);
}