I have a haskell assignment in which i have to create a function lastDigit x y of 2 arguments that calculates the sum of all [x^x | (0..x)], mine is too slow and i need to speed it up. Anyone has any ideas??
list :: Integral x=>x->[x]
list 0 = []
list x = list(div x 10) ++ [(mod x 10)]
sqrall :: Integer->[Integer]
sqrall x y = [mod (mod x 10^y)^x 10^y | x <- [1..x]]
lastDigits :: Integer -> Int -> [Integer]
lastDigits x y = drop (length((list(sum (sqrall x y))))-y) (list(sum (sqrall x)))
The main reason this will take too long is because you calculate the entire number of x^x, which scales super exponentially. This means that even for very small x, it will still take a considerable amount of time.
The point is however that you do not need to calculate the entire number. Indeed, you can make use of the fact that x×y mod n = (x mod n) × (y mod n) mod n. For example Haskell's arithmoi package makes use of this [src]:
powMod :: (Integral a, Integral b) => a -> b -> a -> a
powMod x y m
| m <= 0 = error "powModInt: non-positive modulo"
| y < 0 = error "powModInt: negative exponent"
| otherwise = f (x `rem` m) y 1 `mod` m
where
f _ 0 acc = acc
f b e acc = f (b * b `rem` m) (e `quot` 2)
(if odd e then (b * acc `rem` m) else acc)
We can make a specific version for modulo 10 with:
pow10 :: Integral i => i -> i
pow10 x = go x x
where go 0 _ = 1
go i j | odd i = rec * j `mod` 10
| otherwise = rec
where rec = go (div i 2) ((j*j) `mod` 10)
This then matches x^x `mod` 10, except that we do not need to calculate the entire number:
Prelude> map pow10 [1 .. 20]
[1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9,0]
Prelude> [x^x `mod` 10 | x <- [1..20]]
[1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9,0]
Now that we have that, we can also calculate the the sum of the two last digits with integers that range to at most 18:
sum10 :: Int -> Int -> Int
sum10 x y = (x + y) `mod` 10
we thus can calculate the last digit with:
import Data.List(foldl')
lastdigit :: Int -> Int
lastdigit x = foldl' sum10 0 (map pow10 [0 .. x])
For example for x = 26, we get:
Prelude Data.List> lastdigit 26
4
Prelude Data.List> sum [ x^x | x <- [0 .. 26] ]
6246292385799360560872647730684286774
I keep it as an exercise to generalize the above to calculate it for the last y digits. As long as y is relatively small, this will be efficient, since then the numbers never take huge amounts of memory. Furthermore if the numbers have an upper bound, addition, multiplication, etc. are done in constant time. If you however use an Integer, then the numbers can be arbitrary large, and thus operations like addition are not constant.
I'm a beginner to functional languages, and I'm trying to get the whole thing down in Haskell. Here's a quick-and-dirty function that finds all the factors of a number:
factors :: (Integral a) => a -> [a]
factors x = filter (\z -> x `mod` z == 0) [2..x `div` 2]
Works fine, but I found it to be unbearably slow for large numbers. So I made myself a better one:
factorcalc :: (Integral a) => a -> a -> [a] -> [a]
factorcalc x y z
| y `elem` z = sort z
| x `mod` y == 0 = factorcalc x (y+1) (z ++ [y] ++ [(x `div` y)])
| otherwise = factorcalc x (y+1) z
But here's my problem: Even though the code works, and can cut literally hours off the execution time of my programs, it's hideous!
It reeks of ugly imperative thinking: It constantly updates a counter and a data structure in a loop until it finishes. Since you can't change state in purely functional programming, I cheated by holding the data in the parameters, which the function simply passes to itself over and over again.
I may be wrong, but there simply must be a better way of doing the same thing...
Note that the original question asked for all the factors, not for only the prime factors. There being many fewer prime factors, they can probably be found more quickly. Perhaps that's what the OQ wanted. Perhaps not. But let's solve the original problem and put the "fun" back in "functional"!
Some observations:
The two functions don't produce the same output---if x is a perfect square, the second function includes the square root twice.
The first function enumerates checks a number of potential factors proportional to the size of x; the second function checks only proportional to the square root of x, then stops (with the bug noted above).
The first function (factors) allocates a list of all integers from 2 to n div 2, where the second function never allocates a list but instead visits fewer integers one at a time in a parameter. I ran the optimizer with -O and looked at the output with -ddump-simpl, and GHC just isn't smart enough to optimize away those allocations.
factorcalc is tail-recursive, which means it compiles into a tight machine-code loop; filter is not and does not.
Some experiments show that the square root is the killer:
Here's a sample function that produces the factors of x from z down to 2:
factors_from x 1 = []
factors_from x z
| x `mod` z == 0 = z : factors_from x (z-1)
| otherwise = factors_from x (z-1)
factors'' x = factors_from x (x `div` 2)
It's a bit faster because it doesn't allocate, but it's still not tail-recursive.
Here's a tail-recursive version that is more faithful to the original:
factors_from' x 1 l = l
factors_from' x z l
| x `mod` z == 0 = factors_from' x (z-1) (z:l)
| otherwise = factors_from' x (z-1) l
factors''' x = factors_from x (x `div` 2)
This is still slower than factorcalc because it enumerates all the integers from 2 to x div 2, whereas factorcalc stops at the square root.
Armed with this knowledge, we can now create a more functional version of factorcalc which replicates both its speed and its bug:
factors'''' x = sort $ uncurry (++) $ unzip $ takeWhile (uncurry (<=)) $
[ (z, x `div` z) | z <- [2..x], x `mod` z == 0 ]
I didn't time it exactly, but given 100 million as an input, both it and factorcalc terminate instantaneously, where the others all take a number of seconds.
How and why the function works is left as an exercise for the reader :-)
ADDENDUM: OK, to mitigate the eyeball bleeding, here's a slightly saner version (and without the bug):
saneFactors x = sort $ concat $ takeWhile small $
[ pair z | z <- [2..], x `mod` z == 0 ]
where pair z = if z * z == x then [z] else [z, x `div` z]
small [z, z'] = z < z'
small [z] = True
Okay, take a deep breath. It'll be all right.
First of all, why is your first attempt slow? How is it spending its time?
Can you think of a recursive definition for the prime factorization that doesn't have that property?
(Hint.)
Firstly, although factorcalc is "ugly", you could add a wrapper function factors' x = factorscalc x 2 [], add a comment, and move on.
If you want to make a 'beautiful' factors fast, you need to find out why it is slow. Looking at your two functions, factors walks the list about n/2 elements long, but factorcalc stops after around sqrt n iterations.
Here is another factors that also stops after about sqrt n iterations, but uses a fold instead of explicit iteration. It also breaks the problem into three parts: finding the factors (factor); stopping at the square root of x (small) and then computing pairs of factors (factorize):
factors' :: (Integral a) => a -> [a]
factors' x = sort (foldl factorize [] (takeWhile small (filter factor [2..])))
where
factor z = x `mod` z == 0
small z = z <= (x `div` z)
factorize acc z = z : (if z == y then acc else y : acc)
where y = x `div` z
This is marginally faster than factorscalc on my machine. You can fuse factor and factorize and it is about twice as fast as factorscalc.
The Profiling and Optimization chapter of Real World Haskell is a good guide to the GHC suite's performance tools for tackling tougher performance problems.
By the way, I have a minor style nitpick with factorscalc: it is much more efficient to prepend single elements to the front of a list O(1) than it is to append to the end of a list of length n O(n). The lists of factors are typically small, so it is not such a big deal, but factorcalc should probably be something like:
factorcalc :: (Integral a) => a -> a -> [a] -> [a]
factorcalc x y z
| y `elem` z = sort z
| x `mod` y == 0 = factorcalc x (y+1) (y : (x `div` y) : z)
| otherwise = factorcalc x (y+1) z
Since you can't change state in purely
functional programming, I cheated by
holding the data in the parameters,
which the function simply passes to
itself over and over again.
Actually, this is not cheating; this is a—no, make that the—standard technique! That sort of parameter is usually known as an "accumulator," and it's generally hidden within a helper function that does the actual recursion after being set up by the function you're calling.
A common case is when you're doing list operations that depend on the previous data in the list. The two problems you need to solve are, where do you get the data about previous iterations, and how do you deal with the fact that your "working area of interest" for any particular iteration is actually at the tail of the result list you're building. For both of these, the accumulator comes to the rescue. For example, to generate a list where each element is the sum of all of the elements of the input list up to that point:
sums :: Num a => [a] -> [a]
sums inp = helper inp []
where
helper [] acc = reverse acc
helper (x:xs) [] = helper xs [x]
helper (x:xs) acc#(h:_) = helper xs (x+h : acc)
Note that we flip the direction of the accumulator, so we can operate on the head of that, which is much more efficient (as Dominic mentions), and then we just reverse the final output.
By the way, I found reading The Little Schemer to be a useful introduction and offer good practice in thinking recursively.
This seemed like an interesting problem, and I hadn't coded any real Haskell in a while, so I gave it a crack. I've run both it and Norman's factors'''' against the same values, and it feels like mine's faster, though they're both so close that it's hard to tell.
factors :: Int -> [Int]
factors n = firstFactors ++ reverse [ n `div` i | i <- firstFactors ]
where
firstFactors = filter (\i -> n `mod` i == 0) (takeWhile ( \i -> i * i <= n ) [2..n])
Factors can be paired up into those that are greater than sqrt n, and those that are less than or equal to (for simplicity's sake, the exact square root, if n is a perfect square, falls into this category. So if we just take the ones that are less than or equal to, we can calculate the others later by doing div n i. They'll be in reverse order, so we can either reverse firstFactors first or reverse the result later. It doesn't really matter.
This is my "functional" approach to the problem. ("Functional" in quotes, because I'd approach this problem the same way even in non-functional languages, but maybe that's because I've been tainted by Haskell.)
{-# LANGUAGE PatternGuards #-}
factors :: (Integral a) => a -> [a]
factors = multiplyFactors . primeFactors primes 0 [] . abs where
multiplyFactors [] = [1]
multiplyFactors ((p, n) : factors) =
[ pn * x
| pn <- take (succ n) $ iterate (* p) 1
, x <- multiplyFactors factors ]
primeFactors _ _ _ 0 = error "Can't factor 0"
primeFactors (p:primes) n list x
| (x', 0) <- x `divMod` p
= primeFactors (p:primes) (succ n) list x'
primeFactors _ 0 list 1 = list
primeFactors (_:primes) 0 list x = primeFactors primes 0 list x
primeFactors (p:primes) n list x
= primeFactors primes 0 ((p, n) : list) x
primes = sieve [2..]
sieve (p:xs) = p : sieve [x | x <- xs, x `mod` p /= 0]
primes is the naive Sieve of Eratothenes. There's better, but this is the shortest method.
sieve [2..]
=> 2 : sieve [x | x <- [3..], x `mod` 2 /= 0]
=> 2 : 3 : sieve [x | x <- [4..], x `mod` 2 /= 0, x `mod` 3 /= 0]
=> 2 : 3 : sieve [x | x <- [5..], x `mod` 2 /= 0, x `mod` 3 /= 0]
=> 2 : 3 : 5 : ...
primeFactors is the simple repeated trial-division algorithm: it walks through the list of primes, and tries dividing the given number by each, recording the factors as it goes.
primeFactors (2:_) 0 [] 50
=> primeFactors (2:_) 1 [] 25
=> primeFactors (3:_) 0 [(2, 1)] 25
=> primeFactors (5:_) 0 [(2, 1)] 25
=> primeFactors (5:_) 1 [(2, 1)] 5
=> primeFactors (5:_) 2 [(2, 1)] 1
=> primeFactors _ 0 [(5, 2), (2, 1)] 1
=> [(5, 2), (2, 1)]
multiplyPrimes takes a list of primes and powers, and explodes it back out to a full list of factors.
multiplyPrimes [(5, 2), (2, 1)]
=> [ pn * x
| pn <- take (succ 2) $ iterate (* 5) 1
, x <- multiplyPrimes [(2, 1)] ]
=> [ pn * x | pn <- [1, 5, 25], x <- [1, 2] ]
=> [1, 2, 5, 10, 25, 50]
factors just strings these two functions together, along with an abs to prevent infinite recursion in case the input is negative.
I don't know much about Haskell, but somehow I think this link is appropriate:
http://www.willamette.edu/~fruehr/haskell/evolution.html
Edit: I'm not entirely sure why people are so aggressive about the downvoting on this. The original poster's real problem was that the code was ugly; while it's funny, the point of the linked article is, to some extent, that advanced Haskell code is, in fact, ugly; the more you learn, the uglier your code gets, to some extent. The point of this answer was to point out to the OP that apparently, the ugliness of the code that he was lamenting is not uncommon.
Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
I want to see all the different ways you can come up with, for a factorial subroutine, or program. The hope is that anyone can come here and see if they might want to learn a new language.
Ideas:
Procedural
Functional
Object Oriented
One liners
Obfuscated
Oddball
Bad Code
Polyglot
Basically I want to see an example, of different ways of writing an algorithm, and what they would look like in different languages.
Please limit it to one example per entry.
I will allow you to have more than one example per answer, if you are trying to highlight a specific style, language, or just a well thought out idea that lends itself to being in one post.
The only real requirement is it must find the factorial of a given argument, in all languages represented.
Be Creative!
Recommended Guideline:
# Language Name: Optional Style type
- Optional bullet points
Code Goes Here
Other informational text goes here
I will ocasionally go along and edit any answer that does not have decent formatting.
Polyglot: 5 languages, all using bignums
So, I wrote a polyglot which works in the three languages I often write in, as well as one from my other answer to this question and one I just learned today. It's a standalone program, which reads a single line containing a nonnegative integer and prints a single line containing its factorial. Bignums are used in all languages, so the maximum computable factorial depends only on your computer's resources.
Perl: uses built-in bignum package. Run with perl FILENAME.
Haskell: uses built-in bignums. Run with runhugs FILENAME or your favorite compiler's equivalent.
C++: requires GMP for bignum support. To compile with g++, use g++ -lgmpxx -lgmp -x c++ FILENAME to link against the right libraries. After compiling, run ./a.out. Or use your favorite compiler's equivalent.
brainf*ck: I wrote some bignum support in this post. Using Muller's classic distribution, compile with bf < FILENAME > EXECUTABLE. Make the output executable and run it. Or use your favorite distribution.
Whitespace: uses built-in bignum support. Run with wspace FILENAME.
Edit: added Whitespace as a fifth language. Incidentally, do not wrap the code with <code> tags; it breaks the Whitespace. Also, the code looks much nicer in fixed-width.
char //# b=0+0{- |0*/; #>>>>,----------[>>>>,--------
#define a/*#--]>>>>++<<<<<<<<[>++++++[<------>-]<-<<<
#Perl ><><><> <> <> <<]>>>>[[>>+<<-]>>[<<+>+>-]<->
#C++ --><><> <><><>< > < > < +<[>>>>+<<<-<[-]]>[-]
#Haskell >>]>[-<<<<<[<<<<]>>>>[[>>+<<-]>>[<<+>+>-]>>]
#Whitespace >>>>[-[>+<-]+>>>>]<<<<[<<<<]<<<<[<<<<
#brainf*ck > < ]>>>>>[>>>[>>>>]>>>>[>>>>]<<<<[[>>>>*/
exp; ;//;#+<<<<-]<<<<]>>>>+<<<<<<<[<<<<][.POLYGLOT^5.
#include <gmpxx.h>//]>>>>-[>>>[>>>>]>>>>[>>>>]<<<<[>>
#define eval int main()//>+<<<-]>>>[<<<+>>+>->
#include <iostream>//<]<-[>>+<<[-]]<<[<<<<]>>>>[>[>>>
#define print std::cout << // > <+<-]>[<<+>+>-]<<[>>>
#define z std::cin>>//<< +<<<-]>>>[<<<+>>+>-]<->+++++
#define c/*++++[-<[-[>>>>+<<<<-]]>>>>[<<<<+>>>>-]<<*/
#define abs int $n //>< <]<[>>+<<<<[-]>>[<<+>>-]]>>]<
#define uc mpz_class fact(int $n){/*<<<[<<<<]<<<[<<
use bignum;sub#<<]>>>>-]>>>>]>>>[>[-]>>>]<<<<[>>+<<-]
z{$_[0+0]=readline(*STDIN);}sub fact{my($n)=shift;#>>
#[<<+>+>-]<->+<[>-<[-]]>[-<<-<<<<[>>+<<-]>>[<<+>+>+*/
uc;if($n==0){return 1;}return $n*fact($n-1); }//;#
eval{abs;z($n);print fact($n);print("\n")/*2;};#-]<->
'+<[>-<[-]]>]<<[<<<<]<<<<-[>>+<<-]>>[<<+>+>-]+<[>-+++
-}-- <[-]]>[-<<++++++++++<<<<-[>>+<<-]>>[<<+>+>-++
fact 0 = 1 -- ><><><>< > <><>< ]+<[>-<[-]]>]<<[<<+ +
fact n=n*fact(n-1){-<<]>>>>[[>>+<<-]>>[<<+>+++>+-}
main=do{n<-readLn;print(fact n)}-- +>-]<->+<[>>>>+<<+
{-x<-<[-]]>[-]>>]>]>>>[>>>>]<<<<[>+++++++[<+++++++>-]
<--.<<<<]+written+by+++A+Rex+++2009+.';#+++x-}--x*/;}
lolcode:
sorry I couldn't resist xD
HAI
CAN HAS STDIO?
I HAS A VAR
I HAS A INT
I HAS A CHEEZBURGER
I HAS A FACTORIALNUM
IM IN YR LOOP
UP VAR!!1
TIEMZD INT!![CHEEZBURGER]
UP FACTORIALNUM!!1
IZ VAR BIGGER THAN FACTORIALNUM? GTFO
IM OUTTA YR LOOP
U SEEZ INT
KTHXBYE
This is one of the faster algorithms, up to 170!. It fails inexplicably beyond 170!, and it's relatively slow for small factorials, but for factorials between 80 and 170 it's blazingly fast compared to many algorithms.
curl http://www.google.com/search?q=170!
There's also an online interface, try it out now!
Let me know if you find a bug, or faster implementation for large factorials.
EDIT:
This algorithm is slightly slower, but gives results beyond 170:
curl http://www58.wolframalpha.com/input/?i=171!
It also simplifies them into various other representations.
C++: Template Metaprogramming
Uses the classic enum hack.
template<unsigned int n>
struct factorial {
enum { result = n * factorial<n - 1>::result };
};
template<>
struct factorial<0> {
enum { result = 1 };
};
Usage.
const unsigned int x = factorial<4>::result;
Factorial is calculated completely at compile time based on the template parameter n. Therefore, factorial<4>::result is a constant once the compiler has done its work.
Whitespace
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It was hard to get it to show here properly, but now I tried copying it from the preview and it works. You need to input the number and press enter.
I find the following implementations just hilarious:
The Evolution of a Haskell Programmer
Evolution of a Python programmer
Enjoy!
C# Lookup:
Nothing to calculate really, just look it up. To extend it,add another 8 numbers to the table and 64 bit integers are at at their limit. Beyond that, a BigNum class is called for.
public static int Factorial(int f)
{
if (f<0 || f>12)
{
throw new ArgumentException("Out of range for integer factorial");
}
int [] fact={1,1,2,6,24,120,720,5040,40320,362880,3628800,
39916800,479001600};
return fact[f];
}
Lazy K
Your pure functional programming nightmares come true!
The only Esoteric Turing-complete Programming Language that has:
A purely functional foundation, core, and libraries---in fact, here's the complete API: S K I
No lambdas even!
No numbers or lists needed or allowed
No explicit recursion but yet, allows recursion
A simple infinite lazy stream-based I/O mechanism
Here's the Factorial code in all its parenthetical glory:
K(SII(S(K(S(S(KS)(S(K(S(KS)))(S(K(S(KK)))(S(K(S(K(S(K(S(K(S(SI(K(S(K(S(S(KS)K)I))
(S(S(KS)K)(SII(S(S(KS)K)I))))))))K))))))(S(K(S(K(S(SI(K(S(K(S(SI(K(S(K(S(S(KS)K)I))
(S(S(KS)K)(SII(S(S(KS)K)I))(S(S(KS)K))(S(SII)I(S(S(KS)K)I))))))))K)))))))
(S(S(KS)K)(K(S(S(KS)K)))))))))(K(S(K(S(S(KS)K)))K))))(SII))II)
Features:
No subtraction or conditionals
Prints all factorials (if you wait long enough)
Uses a second layer of Church numerals to convert the Nth factorial to N! asterisks followed by a newline
Uses the Y combinator for recursion
In case you are interested in trying to understand it, here is the Scheme source code to run through the Lazier compiler:
(lazy-def '(fac input)
'((Y (lambda (f n a) ((lambda (b) ((cons 10) ((b (cons 42)) (f (1+ n) b))))
(* a n)))) 1 1))
(for suitable definitions of Y, cons, 1, 10, 42, 1+, and *).
EDIT:
Lazy K Factorial in Decimal
(10KB of gibberish or else I would paste it). For example, at the Unix prompt:
$ echo "4" | ./lazy facdec.lazy
24
$ echo "5" | ./lazy facdec.lazy
120
Rather slow for numbers above, say, 5.
The code is sort of bloated because we have to include library code for all of our own primitives (code written in Hazy, a lambda calculus interpreter and LC-to-Lazy K compiler written in Haskell).
XSLT 1.0
The input file, factorial.xml:
<?xml version="1.0"?>
<?xml-stylesheet href="factorial.xsl" type="text/xsl" ?>
<n>
20
</n>
The XSLT file, factorial.xsl:
<?xml version="1.0"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:msxsl="urn:schemas-microsoft-com:xslt" >
<xsl:output method="text"/>
<!-- 0! = 1 -->
<xsl:template match="text()[. = 0]">
1
</xsl:template>
<!-- n! = (n-1)! * n-->
<xsl:template match="text()[. > 0]">
<xsl:variable name="x">
<xsl:apply-templates select="msxsl:node-set( . - 1 )/text()"/>
</xsl:variable>
<xsl:value-of select="$x * ."/>
</xsl:template>
<!-- Calculate n! -->
<xsl:template match="/n">
<xsl:apply-templates select="text()"/>
</xsl:template>
</xsl:stylesheet>
Save both files in the same directory and open factorial.xml in IE.
Python: Functional, One-liner
factorial = lambda n: reduce(lambda x,y: x*y, range(1, n+1), 1)
NOTE:
It supports big integers. Example:
print factorial(100)
93326215443944152681699238856266700490715968264381621468592963895217599993229915\
608941463976156518286253697920827223758251185210916864000000000000000000000000
It does not work for n < 0.
APL (oddball/one-liner):
×/⍳X
⍳X expands X into an array of the integers 1..X
×/ multiplies every element in the array
Or with the built-in operator:
!X
Source: http://www.webber-labs.com/mpl/lectures/ppt-slides/01.ppt
Perl6
sub factorial ($n) { [*] 1..$n }
I hardly know about Perl6. But I guess this [*] operator is same as Haskell's product.
This code runs on Pugs, and maybe Parrot (I didn't check it.)
Edit
This code also works.
sub postfix:<!> ($n) { [*] 1..$n }
# This function(?) call like below ... It looks like mathematical notation.
say 10!;
x86-64 Assembly: Procedural
You can call this from C (only tested with GCC on linux amd64).
Assembly was assembled with nasm.
section .text
global factorial
; factorial in x86-64 - n is passed in via RDI register
; takes a 64-bit unsigned integer
; returns a 64-bit unsigned integer in RAX register
; C declaration in GCC:
; extern unsigned long long factorial(unsigned long long n);
factorial:
enter 0,0
; n is placed in rdi by caller
mov rax, 1 ; factorial = 1
mov rcx, 2 ; i = 2
loopstart:
cmp rcx, rdi
ja loopend
mul rcx ; factorial *= i
inc rcx
jmp loopstart
loopend:
leave
ret
Recursively in Inform 7
(it reminds you of COBOL because it's for writing text adventures; proportional font is deliberate):
To decide what number is the factorial of (n - a number):
if n is zero, decide on one;
otherwise decide on the factorial of (n minus one) times n.
If you want to actually call this function ("phrase") from a game you need to define an action and grammar rule:
"The factorial game" [this must be the first line of the source]
There is a room. [there has to be at least one!]
Factorialing is an action applying to a number.
Understand "factorial [a number]" as factorialing.
Carry out factorialing:
Let n be the factorial of the number understood;
Say "It's [n]".
C#: LINQ
public static int factorial(int n)
{
return (Enumerable.Range(1, n).Aggregate(1, (previous, value) => previous * value));
}
Erlang: tail recursive
fac(0) -> 1;
fac(N) when N > 0 -> fac(N, 1).
fac(1, R) -> R;
fac(N, R) -> fac(N - 1, R * N).
Haskell:
ones = 1 : ones
integers = head ones : zipWith (+) integers (tail ones)
factorials = head integers : zipWith (*) factorials (tail integers)
Brainf*ck
+++++
>+<[[->>>>+<<<<]>>>>[-<<<<+>>+>>]<<<<>[->>+<<]<>>>[-<[->>+<<]>>[-<<+<+>>>]<]<[-]><<<-]
Written by Michael Reitzenstein.
BASIC: old school
10 HOME
20 INPUT N
30 LET ANS = 1
40 FOR I = 1 TO N
50 ANS = ANS * I
60 NEXT I
70 PRINT ANS
Batch (NT):
#echo off
set n=%1
set result=1
for /l %%i in (%n%, -1, 1) do (
set /a result=result * %%i
)
echo %result%
Usage:
C:>factorial.bat 15
F#: Functional
Straight forward:
let rec fact x =
if x < 0 then failwith "Invalid value."
elif x = 0 then 1
else x * fact (x - 1)
Getting fancy:
let fact x = [1 .. x] |> List.fold_left ( * ) 1
Recursive Prolog
fac(0,1).
fac(N,X) :- N1 is N -1, fac(N1, T), X is N * T.
Tail Recursive Prolog
fac(0,N,N).
fac(X,N,T) :- A is N * X, X1 is X - 1, fac(X1,A,T).
fac(N,T) :- fac(N,1,T).
ruby recursive
(factorial=Hash.new{|h,k|k*h[k-1]})[1]=1
usage:
factorial[5]
=> 120
Scheme
Here is a simple recursive definition:
(define (factorial x)
(if (= x 0) 1
(* x (factorial (- x 1)))))
In Scheme tail-recursive functions use constant stack space. Here is a version of factorial that is tail-recursive:
(define factorial
(letrec ((fact (lambda (x accum)
(if (= x 0) accum
(fact (- x 1) (* accum x))))))
(lambda (x)
(fact x 1))))
Oddball examples? What about using the gamma function! Since, Gamma n = (n-1)!.
OCaml: Using Gamma
let rec gamma z =
let pi = 4.0 *. atan 1.0 in
if z < 0.5 then
pi /. ((sin (pi*.z)) *. (gamma (1.0 -. z)))
else
let consts = [| 0.99999999999980993; 676.5203681218851; -1259.1392167224028;
771.32342877765313; -176.61502916214059; 12.507343278686905;
-0.13857109526572012; 9.9843695780195716e-6; 1.5056327351493116e-7;
|]
in
let z = z -. 1.0 in
let results = Array.fold_right
(fun x y -> x +. y)
(Array.mapi
(fun i x -> if i = 0 then x else x /. (z+.(float i)))
consts
)
0.0
in
let x = z +. (float (Array.length consts)) -. 1.5 in
let final = (sqrt (2.0*.pi)) *.
(x ** (z+.0.5)) *.
(exp (-.x)) *. result
in
final
let factorial_gamma n = int_of_float (gamma (float (n+1)))
Freshman Haskell programmer
fac n = if n == 0
then 1
else n * fac (n-1)
Sophomore Haskell programmer, at MIT
(studied Scheme as a freshman)
fac = (\(n) ->
(if ((==) n 0)
then 1
else ((*) n (fac ((-) n 1)))))
Junior Haskell programmer
(beginning Peano player)
fac 0 = 1
fac (n+1) = (n+1) * fac n
Another junior Haskell programmer
(read that n+k patterns are “a disgusting part of Haskell” [1]
and joined the “Ban n+k patterns”-movement [2])
fac 0 = 1
fac n = n * fac (n-1)
Senior Haskell programmer
(voted for Nixon Buchanan Bush — “leans right”)
fac n = foldr (*) 1 [1..n]
Another senior Haskell programmer
(voted for McGovern Biafra Nader — “leans left”)
fac n = foldl (*) 1 [1..n]
Yet another senior Haskell programmer
(leaned so far right he came back left again!)
-- using foldr to simulate foldl
fac n = foldr (\x g n -> g (x*n)) id [1..n] 1
Memoizing Haskell programmer
(takes Ginkgo Biloba daily)
facs = scanl (*) 1 [1..]
fac n = facs !! n
Pointless (ahem) “Points-free” Haskell programmer
(studied at Oxford)
fac = foldr (*) 1 . enumFromTo 1
Iterative Haskell programmer
(former Pascal programmer)
fac n = result (for init next done)
where init = (0,1)
next (i,m) = (i+1, m * (i+1))
done (i,_) = i==n
result (_,m) = m
for i n d = until d n i
Iterative one-liner Haskell programmer
(former APL and C programmer)
fac n = snd (until ((>n) . fst) (\(i,m) -> (i+1, i*m)) (1,1))
Accumulating Haskell programmer
(building up to a quick climax)
facAcc a 0 = a
facAcc a n = facAcc (n*a) (n-1)
fac = facAcc 1
Continuation-passing Haskell programmer
(raised RABBITS in early years, then moved to New Jersey)
facCps k 0 = k 1
facCps k n = facCps (k . (n *)) (n-1)
fac = facCps id
Boy Scout Haskell programmer
(likes tying knots; always “reverent,” he
belongs to the Church of the Least Fixed-Point [8])
y f = f (y f)
fac = y (\f n -> if (n==0) then 1 else n * f (n-1))
Combinatory Haskell programmer
(eschews variables, if not obfuscation;
all this currying’s just a phase, though it seldom hinders)
s f g x = f x (g x)
k x y = x
b f g x = f (g x)
c f g x = f x g
y f = f (y f)
cond p f g x = if p x then f x else g x
fac = y (b (cond ((==) 0) (k 1)) (b (s (*)) (c b pred)))
List-encoding Haskell programmer
(prefers to count in unary)
arb = () -- "undefined" is also a good RHS, as is "arb" :)
listenc n = replicate n arb
listprj f = length . f . listenc
listprod xs ys = [ i (x,y) | x<-xs, y<-ys ]
where i _ = arb
facl [] = listenc 1
facl n#(_:pred) = listprod n (facl pred)
fac = listprj facl
Interpretive Haskell programmer
(never “met a language” he didn't like)
-- a dynamically-typed term language
data Term = Occ Var
| Use Prim
| Lit Integer
| App Term Term
| Abs Var Term
| Rec Var Term
type Var = String
type Prim = String
-- a domain of values, including functions
data Value = Num Integer
| Bool Bool
| Fun (Value -> Value)
instance Show Value where
show (Num n) = show n
show (Bool b) = show b
show (Fun _) = ""
prjFun (Fun f) = f
prjFun _ = error "bad function value"
prjNum (Num n) = n
prjNum _ = error "bad numeric value"
prjBool (Bool b) = b
prjBool _ = error "bad boolean value"
binOp inj f = Fun (\i -> (Fun (\j -> inj (f (prjNum i) (prjNum j)))))
-- environments mapping variables to values
type Env = [(Var, Value)]
getval x env = case lookup x env of
Just v -> v
Nothing -> error ("no value for " ++ x)
-- an environment-based evaluation function
eval env (Occ x) = getval x env
eval env (Use c) = getval c prims
eval env (Lit k) = Num k
eval env (App m n) = prjFun (eval env m) (eval env n)
eval env (Abs x m) = Fun (\v -> eval ((x,v) : env) m)
eval env (Rec x m) = f where f = eval ((x,f) : env) m
-- a (fixed) "environment" of language primitives
times = binOp Num (*)
minus = binOp Num (-)
equal = binOp Bool (==)
cond = Fun (\b -> Fun (\x -> Fun (\y -> if (prjBool b) then x else y)))
prims = [ ("*", times), ("-", minus), ("==", equal), ("if", cond) ]
-- a term representing factorial and a "wrapper" for evaluation
facTerm = Rec "f" (Abs "n"
(App (App (App (Use "if")
(App (App (Use "==") (Occ "n")) (Lit 0))) (Lit 1))
(App (App (Use "*") (Occ "n"))
(App (Occ "f")
(App (App (Use "-") (Occ "n")) (Lit 1))))))
fac n = prjNum (eval [] (App facTerm (Lit n)))
Static Haskell programmer
(he does it with class, he’s got that fundep Jones!
After Thomas Hallgren’s “Fun with Functional Dependencies” [7])
-- static Peano constructors and numerals
data Zero
data Succ n
type One = Succ Zero
type Two = Succ One
type Three = Succ Two
type Four = Succ Three
-- dynamic representatives for static Peanos
zero = undefined :: Zero
one = undefined :: One
two = undefined :: Two
three = undefined :: Three
four = undefined :: Four
-- addition, a la Prolog
class Add a b c | a b -> c where
add :: a -> b -> c
instance Add Zero b b
instance Add a b c => Add (Succ a) b (Succ c)
-- multiplication, a la Prolog
class Mul a b c | a b -> c where
mul :: a -> b -> c
instance Mul Zero b Zero
instance (Mul a b c, Add b c d) => Mul (Succ a) b d
-- factorial, a la Prolog
class Fac a b | a -> b where
fac :: a -> b
instance Fac Zero One
instance (Fac n k, Mul (Succ n) k m) => Fac (Succ n) m
-- try, for "instance" (sorry):
--
-- :t fac four
Beginning graduate Haskell programmer
(graduate education tends to liberate one from petty concerns
about, e.g., the efficiency of hardware-based integers)
-- the natural numbers, a la Peano
data Nat = Zero | Succ Nat
-- iteration and some applications
iter z s Zero = z
iter z s (Succ n) = s (iter z s n)
plus n = iter n Succ
mult n = iter Zero (plus n)
-- primitive recursion
primrec z s Zero = z
primrec z s (Succ n) = s n (primrec z s n)
-- two versions of factorial
fac = snd . iter (one, one) (\(a,b) -> (Succ a, mult a b))
fac' = primrec one (mult . Succ)
-- for convenience and testing (try e.g. "fac five")
int = iter 0 (1+)
instance Show Nat where
show = show . int
(zero : one : two : three : four : five : _) = iterate Succ Zero
Origamist Haskell programmer
(always starts out with the “basic Bird fold”)
-- (curried, list) fold and an application
fold c n [] = n
fold c n (x:xs) = c x (fold c n xs)
prod = fold (*) 1
-- (curried, boolean-based, list) unfold and an application
unfold p f g x =
if p x
then []
else f x : unfold p f g (g x)
downfrom = unfold (==0) id pred
-- hylomorphisms, as-is or "unfolded" (ouch! sorry ...)
refold c n p f g = fold c n . unfold p f g
refold' c n p f g x =
if p x
then n
else c (f x) (refold' c n p f g (g x))
-- several versions of factorial, all (extensionally) equivalent
fac = prod . downfrom
fac' = refold (*) 1 (==0) id pred
fac'' = refold' (*) 1 (==0) id pred
Cartesianally-inclined Haskell programmer
(prefers Greek food, avoids the spicy Indian stuff;
inspired by Lex Augusteijn’s “Sorting Morphisms” [3])
-- (product-based, list) catamorphisms and an application
cata (n,c) [] = n
cata (n,c) (x:xs) = c (x, cata (n,c) xs)
mult = uncurry (*)
prod = cata (1, mult)
-- (co-product-based, list) anamorphisms and an application
ana f = either (const []) (cons . pair (id, ana f)) . f
cons = uncurry (:)
downfrom = ana uncount
uncount 0 = Left ()
uncount n = Right (n, n-1)
-- two variations on list hylomorphisms
hylo f g = cata g . ana f
hylo' f (n,c) = either (const n) (c . pair (id, hylo' f (c,n))) . f
pair (f,g) (x,y) = (f x, g y)
-- several versions of factorial, all (extensionally) equivalent
fac = prod . downfrom
fac' = hylo uncount (1, mult)
fac'' = hylo' uncount (1, mult)
Ph.D. Haskell programmer
(ate so many bananas that his eyes bugged out, now he needs new lenses!)
-- explicit type recursion based on functors
newtype Mu f = Mu (f (Mu f)) deriving Show
in x = Mu x
out (Mu x) = x
-- cata- and ana-morphisms, now for *arbitrary* (regular) base functors
cata phi = phi . fmap (cata phi) . out
ana psi = in . fmap (ana psi) . psi
-- base functor and data type for natural numbers,
-- using a curried elimination operator
data N b = Zero | Succ b deriving Show
instance Functor N where
fmap f = nelim Zero (Succ . f)
nelim z s Zero = z
nelim z s (Succ n) = s n
type Nat = Mu N
-- conversion to internal numbers, conveniences and applications
int = cata (nelim 0 (1+))
instance Show Nat where
show = show . int
zero = in Zero
suck = in . Succ -- pardon my "French" (Prelude conflict)
plus n = cata (nelim n suck )
mult n = cata (nelim zero (plus n))
-- base functor and data type for lists
data L a b = Nil | Cons a b deriving Show
instance Functor (L a) where
fmap f = lelim Nil (\a b -> Cons a (f b))
lelim n c Nil = n
lelim n c (Cons a b) = c a b
type List a = Mu (L a)
-- conversion to internal lists, conveniences and applications
list = cata (lelim [] (:))
instance Show a => Show (List a) where
show = show . list
prod = cata (lelim (suck zero) mult)
upto = ana (nelim Nil (diag (Cons . suck)) . out)
diag f x = f x x
fac = prod . upto
Post-doc Haskell programmer
(from Uustalu, Vene and Pardo’s “Recursion Schemes from Comonads” [4])
-- explicit type recursion with functors and catamorphisms
newtype Mu f = In (f (Mu f))
unIn (In x) = x
cata phi = phi . fmap (cata phi) . unIn
-- base functor and data type for natural numbers,
-- using locally-defined "eliminators"
data N c = Z | S c
instance Functor N where
fmap g Z = Z
fmap g (S x) = S (g x)
type Nat = Mu N
zero = In Z
suck n = In (S n)
add m = cata phi where
phi Z = m
phi (S f) = suck f
mult m = cata phi where
phi Z = zero
phi (S f) = add m f
-- explicit products and their functorial action
data Prod e c = Pair c e
outl (Pair x y) = x
outr (Pair x y) = y
fork f g x = Pair (f x) (g x)
instance Functor (Prod e) where
fmap g = fork (g . outl) outr
-- comonads, the categorical "opposite" of monads
class Functor n => Comonad n where
extr :: n a -> a
dupl :: n a -> n (n a)
instance Comonad (Prod e) where
extr = outl
dupl = fork id outr
-- generalized catamorphisms, zygomorphisms and paramorphisms
gcata :: (Functor f, Comonad n) =>
(forall a. f (n a) -> n (f a))
-> (f (n c) -> c) -> Mu f -> c
gcata dist phi = extr . cata (fmap phi . dist . fmap dupl)
zygo chi = gcata (fork (fmap outl) (chi . fmap outr))
para :: Functor f => (f (Prod (Mu f) c) -> c) -> Mu f -> c
para = zygo In
-- factorial, the *hard* way!
fac = para phi where
phi Z = suck zero
phi (S (Pair f n)) = mult f (suck n)
-- for convenience and testing
int = cata phi where
phi Z = 0
phi (S f) = 1 + f
instance Show (Mu N) where
show = show . int
Tenured professor
(teaching Haskell to freshmen)
fac n = product [1..n]
D Templates: Functional
template factorial(int n : 1)
{
const factorial = 1;
}
template factorial(int n)
{
const factorial =
n * factorial!(n-1);
}
or
template factorial(int n)
{
static if(n == 1)
const factorial = 1;
else
const factorial =
n * factorial!(n-1);
}
Used like this:
factorial!(5)
Java 1.6: recursive, memoized (for subsequent calls)
private static Map<BigInteger, BigInteger> _results = new HashMap()
public static BigInteger factorial(BigInteger n){
if (0 >= n.compareTo(BigInteger.ONE))
return BigInteger.ONE.max(n);
if (_results.containsKey(n))
return _results.get(n);
BigInteger result = factorial(n.subtract(BigInteger.ONE)).multiply(n);
_results.put(n, result);
return result;
}
PowerShell
function factorial( [int] $n )
{
$result = 1;
if ( $n -gt 1 )
{
$result = $n * ( factorial ( $n - 1 ) )
}
$result
}
Here's a one-liner:
$n..1 | % {$result = 1}{$result *= $_}{$result}
Bash: Recursive
In bash and recursive, but with the added advantage that it deals with each iteration in a new process. The max it can calculate is !20 before overflowing, but you can still run it for big numbers if you don't care about the answer and want your system to fall over ;)
#!/bin/bash
echo $(($1 * `( [[ $1 -gt 1 ]] && ./$0 $(($1 - 1)) ) || echo 1`));