###

Question:

# The Challenge

Write a program that acts as a Fractran interpreter. The shortest interpreter by character count, in any language, is the winner. Your program must take two inputs: The fractran program to be executed, and the input integer n. The program may be in any form that is convenient for your program - for example, a list of 2-tuples, or a flat list. The output must be a single integer, being the value of the register at the end of execution.

## Fractran

Fractran is a trivial esoteric language invented by John Conway. A fractran program consists of a list of positive fractions and an initial state n. The interpreter maintains a program counter, initially pointing to the first fraction in the list. Fractran programs are executed in the following fashion:

- Check if the product of the current state and the fraction currently under the program counter is an integer. If it is, multiply the current state by the current fraction and reset the program counter to the beginning of the list.
- Advance the program counter. If the end of the list is reached, halt, otherwise return to step 1.

For details on how and why Fractran works, see the esolang entry and this entry on good math/bad math.

## Test Vectors

**Program:** [(3, 2)]**Input:** 72 (2^{3}3^{2})**Output:** 243 (3^{5})

**Program:** [(3, 2)]**Input:** 1296 (2^{4}3^{4})**Output:** 6561 (3^{8})

**Program:** [(455, 33), (11, 13), (1, 11), (3, 7), (11, 2), (1, 3)]**Input:** 72 (2^{3}3^{2})**Output:** 15625 (5^{6})

**Bonus test vector:**

Your submission does not need to execute this last program correctly to be an acceptable answer. But kudos if it does!

**Program:** [(455, 33), (11, 13), (1, 11), (3, 7), (11, 2), (1, 3)]**Input:** 60466176 (2^{10}3^{10})**Output:** 7888609052210118054117285652827862296732064351090230047702789306640625 (5^{100})

## Submissions & Scoring

Programs are ranked strictly by length in characters - shortest is best. Feel free to submit both a nicely laid out and documented and a 'minified' version of your code, so people can see what's going on.

**The language 'J' is not admissible. This is because there's already a well-known solution in J on one of the linked pages.** If you're a J fan, sorry!

As an extra bonus, however, anyone who can provide a working fractran interpreter *in* fractran will receive a 500 reputation point bonus. In the unlikely event of multiple self-hosting interpreters, the one with the shortest number of fractions will receive the bounty.

## Winners

The official winner, after submitting a self-hosting fractran solution comprising 1779 fractions, is Jesse Beder's solution. Practically speaking, the solution is too slow to execute even 1+1, however.

Incredibly, this has since been beaten by another fractran solution - Amadaeus's solution in only 84 fractions! It is capable of executing the first two test cases in a matter of seconds when running on my reference Python solution. It uses a novel encoding method for the fractions, which is also worth a close look.

Honorable mentions to:

- Stephen Canon's solution, in 165 characters of x86 assembly (28 bytes of machine code)
- Jordan's solution in 52 characters of ruby - which handles long integers
- Useless's solution in 87 characters of Python, which, although not the shortest Python solution, is one of the few solutions that isn't recursive, and hence handles harder programs with ease. It's also very readable.

###

Solution:1

# Fractran - 1779 fractions

**(Edit: fixed)**

(I hope people are still following this thread, because this took a while!)

It appears SO won't let me post something as long as this, so I posted the Fractran source here.

Input is specified as follows:

First, we encode a fraction `m/n = p_0^a0... p_k^ak`

by:

- Start with 1. Then, for each
`ai`

: - Multiply by
`p_2i^ai`

if`ai > 0`

- Multiply by
`p_2i+1^{-ai}`

if`a_i < 0`

This way, we encode any fraction as a positive integer. Now, given a progoram (sequence of encoded fractions F0, F1, ...), we encode that by

`p_0^F0 p1^F1 ... `

Finally, input to the interpreter is given by:

`2^(program) 3^(input) 5 `

where `program`

and `input`

are encoded as above. For example, in the first test problem, `3/2`

gets encoded to `15`

, so the program gets encoded to `2^15`

; and `108`

gets encoded to `500`

. So, we pass

`2^{2^15} 3^500 5 `

to the program. The output, then is of the form

`2^(program) 3^(output) `

so in the first example, it'll be

`2^{2^15} 3^3125 `

## How does it work?

I wrote a meta-language that compiles down to Fractran. It allows for functions (simple Fractran and sequences of other functions), and a `while`

loop and `if`

statement (for convenience!). The code for that can be found here.

If you want to compile that code down to Fractran yourself, my (C++) program can be found here [tar.gz]. In a stunning display of dogfooding (and showing off), I used my C++ YAML parser yaml-cpp, so you'd have to download and link with that. For both yaml-cpp and the "compiler", you'll need CMake for cross-platform makefile generating.

The usage of this program is:

`./fracc interpreter.frp `

The it reads the name of a function from standard input, and writes the corresponding "pseudo-Fraction" (I'll explain that in a second) to standard output. So to compile the interpreter (the Interpret function), you could run

`echo "Interpret" | ./fracc interpreter.frp > interpret `

The output ("pseudo-Fractran") will be a sequence of lines, each with a string of space-separated digits. A line corresponds to a fraction: if the `n`

th digit in the line is `an`

, then the fraction is the product of `p_n^an`

.

It's very easy to convert this to Fractran, but if you're lazy, you can use to-fractions.py. [**Note**: earlier I had a C++ program, and I had carelessly ignored integer overflow. I translated it to Python to avoid this.]

**Note about input**: if you want to test out a different function this way, the convention is always the same. It has a number of parameters (usually the comment above the function explains this) in pseudo-Fractran, so give it what it wants, plus a `1`

on the very next slot (so in ordinary Fractran, multiply once by the first prime that it won't use). This is a "signal" bit to the function to start going.

## However,

I don't recommend actually trying to run the Fractran interpreter (alas). I tested many of its components, and, for example, the function `IncrementPrimes`

, which takes a pair of primes and returns the next two primes, takes about **8 minutes** to run, using my silly C++ interpreter (no need to post that :). Plus, it goes (at least) quadratically in the number of function calls - doubling the number of function calls makes it take at least four times as long (more if there are `while`

loops or `if`

statements). So I'm guessing that running the interpreter will take at least days, if not years :(

So how do I know it works? Well, of course I'm not 100% certain, but I'm pretty close. First of all, I tested many, many of its components, and in particular, I tested all of the elements of the meta-language (sequences of functions and `if`

and `while`

statements) very thoroughly.

Also, the meta-language is easy to translate into your favorite language, and even easier to translate to C++, since all parameters of functions are passed by reference. If you're feeling lazy again, you can download my translation here [tar.gz] (there's no makefile; it's just two .cpp files, so directly calling gcc is fine).

So you can compare the two interpreters, run the C++ version (it also takes input/output in pseudo-Fractran), check that that works, and then convince yourself that the meta-language works too.

## Or!

If you're feeling inspired, and *really* want to see this interpreter interpreted, you can write a "clever" Fractran interpreter based around the type of Fractran output that we get. The output is very structured - sequences of functions are implemented using signals, so if you somehow cache where the interpreter was, you could jump there immediately if nothing important changed. This, I think, would dramatically speed up the program (perhaps cutting down running time by one or more powers).

But, I'm not really sure how to do this, and I'm happy with what's done, so I'll leave it as an exercise for the reader.

###

Solution:2

# Fractran: 84 fractions

`FTEVAL = [197*103/(2^11*101), 101/103, 103*127/(2*101), 101/103, 109/101, 2*23/(197*109), 109/23, 29/109,197*41*47/(31*59), 11^10*53/(127*197), 197/53, 37/197, 7^10*43/(11^10*37), 37/43, 59/(37*47), 59/47, 41*61/59, 31*67/(41*61), 61/67, 7*67/(127*61), 61/67,101/71, 73/(127^9*29), 79/(127^2*73), 83/(127*73), 89/(2*29), 163/29, 127^11*89/79, 337/83, 2*59/89, 71/61, 7*173/(127*163), 163/173, 337*167/163, 347/(31*337), 337/347, 151/337, 1/71,19*179/(3*7*193), 193/179, 157/(7*193), 17*181/193, 7*211/(19*181), 181/211, 193/181, 157/193, 223/(7*157), 157/223, 281*283/239, 3*257*269/(7*241), 241/269, 263/241, 7*271/(257*263), 263/271, 281/263, 241/(17*281), 1/281, 307/(7*283), 283/307, 293/283, 71*131/107, 193/(131*151), 227/(19*157), 71*311/227, 233/(151*167*311), 151*311/229, 7*317/(19*229), 229/317, 239*331/217, 71*313/157, 239*251/(151*167*313), 239*251/(151*313), 149/(251*293), 107/(293*331), 137/199, 2^100*13^100*353/(5^100*137), 2*13*353/(5*137), 137/353, 349/137, 107/349, 5^100*359/(13^100*149), 5*359/(13*149), 149/359, 199/149] `

This is written entirely by hand. I did make up a pseudo language to be able to express things more clearly, but I did not write a compiler and opted to write optimized Fractran code directly.

FTEVAL takes input `3^initial_state * 5^encoded_program * 199`

, produces intermediate results `3^interpreted_program_state * 199`

, and completely halts at a number divisible by `233`

.

The interpreted program is embeded as a list of base 10 digits inside a single base 11 number, using the digit "a" to mark the boundary except at the very end. The addition program [3/2] is encoded as

`int("3a2", 11) = 475. `

The multiplication program [455/33, 11/13, 1/11, 3/7, 11/2, 1/3] is encoded as

`int("1a3a11a2a3a7a1a11a11a13a455a33", 11) = 3079784207925154324249736405657 `

which is a truly large number.

The first test vector finished in **less than one second**, produced the desired result after 4545 iterations and halted after 6172 iterations. Here is the complete output.

Unfortunately, sage segfaulted when I tried the second test vector (but I think it'll work under Nick's implementation using prime exponent vectors).

The space here is really too small to explain everything. But here is my pseudocode. I will write up my process in a couple of days, hopefully.

`# Notations: # %p # designates the exponent of prime factor p that divides the # current state. # mov x y # directly translates to the fraction y/x; its meaning: test if x divides # the current state, if so divide the current state by x and multiply it by # y. In effect, the prime exponents of x and y are exchanged. A Fractran # program only comprises of these instructions; when one is executable, the # program continues from the beginning. # dec x => mov x, 1 # wipes out all factors of x # inc x => mov 1, x # this form is here for the sake of clarity, it usually appears in a # loop's entry statement and is merged as such when compiled # sub n ret m {...} # conceptually represents a Fractran sub-program, which will execute just # like a normal Fractran program, that is, the sub-program's statements # execute when triggered and loop back. The sub-program only exits when none of # its statement is executable, in which occasion we multiply the program's # state by m. We can also explicitly break out early on some conditions. # It is also possible to enter a sub-prorgram via multiple entry points and # we must take care to avoiding this kind of behavior (except in case where # it is desirable). # entry point 101: return 29 # Divide %2 modulo 11: # - quotient is modified in-place # - remainder goes to %127 sub 101 ret 101 { mov 2^11, 197 } sub 101 ret 109 { mov 2, 127 } sub 109 ret 29 { mov 197, 2 } # entry point 59: return 61 # Multiply %127 by 10^%31 then add result to %7, # also multiply %31 by 10 in-place. sub 59 ret 41*61 { mov 31, 197*41 sub 197 ret 37 { mov 127, 11^10 } sub 37 { mov 11^10, 7^10 } } sub 61 ret 61 { mov 41, 31 } sub 61 ret 61 { mov 127, 7 } # the case where %31==0 # entry point 71: return 151 if success, 151*167 if reached last value # Pop the interpreted program stack (at %2) to %7. sub 71 { # call sub 101 inc 101 # if remainder >= 9: mov 29*127^9, 73 # if remainder == 11, goto 79 mov 73*127^2, 79 # else: # if remainder == 10, goto 83 mov 73*127, 83 # else: # if quotient >= 1: goto 89 mov 29*2, 89 # else: goto 163 mov 29, 163 # 79: restore remainder to original value, then goto 89 mov 79, 127^11*89 # 83: reached a border marker, ret mov 83, 337 # 89: the default loop branch # restore quotient to original value, call 59 and loop when that rets mov 2*89, 59 mov 61, 71 # 163: reached stack bottom, # ret with the halt signal sub 163 ret 337*167 { mov 127, 7 } # 337: clean up %31 before ret sub 337 ret 151 { dec 31 } } # entry point 193, return 157 # Divide %3 modulo %7: # - quotient goes to %17 # - remainder goes to %19 sub 193 ret 17*181 { mov 3*7, 19 } mov 7*193, 157 sub 181 ret 193 { mov 19, 7 } mov 193, 157 sub 157 ret 157 { dec 7 } # entry point 239: return 293 # Multiply %17 by %7, result goes to %3 mov 239, 281*283 sub 241 { mov 7, 3*257 } sub 263 ret 281 { mov 257, 7 } mov 281*17, 241 sub 283 ret 293 { dec 7 } # entry point 107: return 149 if success, 233 if stack empty # Pop the stack to try execute each fraction sub 107 { # pop the stack inc 71*131 # 151: popped a value # call divmod %3 %7 mov 131*151, 193 # if remainder > 0: mov 19*157, 227 # pop and throw away the numerator mov 227, 71*311 # if stack is empty: halt! mov 151*167*311, 233 # else: call 239 to multiply back the program state and gave loop signal mov 151*311, 229 sub 229 ret 239*331 { mov 19, 7 } # else: (remainder == 0) # pop the numerator mov 157, 71*313 # clear the stack empty signal if present # call 239 to update program state and gave ret signal mov 151*167*313, 239*251 mov 151*313, 239*251 # after program state is updated # 313: ret mov 293*251, 149 # 331: loop mov 293*331, 107 } # main sub 199 { # copy the stack from %5 to %2 and %13 sub 137 ret 137 { mov 5^100, 2^100*13^100 } sub 137 ret 349 { mov 5, 2*13 } # call sub 107 mov 349, 107 # if a statement executed, restore stack and loop sub 149 ret 149 { mov 13^100, 5^100 } sub 149 ret 199 { mov 13, 5 } } `

###

Solution:3

**x86_64 assembly**, 165 characters (28 bytes of machine code).

State is passed in %rdi, Program (pointer to null-terminated array of fractions) is in %rsi. Results are returned in %rax per the usual C-style calling conventions. Using non-standard calling conventions or Intel syntax (this is AT&T syntax) would drop a few more characters, but I'm lazy; someone else can do that. An instruction or two can almost certainly be saved by re-arranging the control flow, if someone wants to do that, feel free.

Intermediate computations (state*numerator) can be up to 128 bits wide, but only 64 bit state is supported.

`_fractran: 0: mov %rsi, %rcx // set aside pointer to beginning of list 1: mov (%rcx), %rax // load numerator test %rax, %rax // check for null-termination of array jz 9f // if zero, exit mul %rdi mov 8(%rcx), %r8 // load denominator div %r8 test %rdx, %rdx // check remainder of division cmovz %rax, %rdi // if zero, keep result jz 0b // and jump back to program start add $16, %rcx // otherwise, advance to next instruction jmp 1b 9: mov %rdi, %rax // copy result for return ret `

Delete comments, extraneous whitespace, and the verbose label `_fractran`

for minimized version.

###

Solution:4

## Ruby, ~~58~~ ~~57~~ ~~56~~ ~~53~~ 52 characters

This is my first-ever code golf entry, so please be gentle.

`def f(n,c)d,e=c.find{|i,j|n%j<1};d ?f(n*d/e,c):n end `

Usage:

`irb> f 108, [[455, 33], [11, 13], [1,11], [3,7], [11,2], [1,3]] => 15625 irb> f 60466176, [[455, 33], [11, 13], [1, 11], [3, 7], [11, 2], [1, 3]] => 7888609052210118054117285652827862296732064351090230047702789306640625 `

Pretty version (252):

`def fractran(instruction, program) numerator, denominator = program.find do |numerator, denominator| instruction % denominator < 1 end if numerator fractran(instruction * numerator / denominator, program) else instruction end end `

## Ruby, ~~53~~ 52 using Rational

Inspired by gnibbler's solution I was able to get a solution using Rational down to ~~53~~ 52 characters. ~~Still one longer than the (less elegant) solution above.~~

`def f(n,c)c.map{|i|return f(n*i,c)if i*n%1==0};n end `

Usage:

`irb> require 'rational' irb> f 60466176, [Rational(455, 33), Rational(11, 13), Rational(1, 11), Rational(3, 7), Rational(11, 2), Rational(1, 3)] => Rational(7888609052210118054117285652827862296732064351090230047702789306640625, 1) `

(A `to_i`

call for prettier output would add 5 more characters.)

###

Solution:5

# Golfscript - 32

{:^{1=1$\%!}?.1={~@\/*^f}{}if}:f ; 108 [[3 2]] f p # 243 ; 1296 [[3 2]] f p # 6561 ; 108 [[455 33][11 13][1 11][3 7][11 2][1 3]] f p # 15625 ; 60466176 [[455 33][11 13][1 11][3 7][11 2][1 3]] f p # 7888609052210118054117285652827862296732064351090230047702789306640625

###

Solution:6

## Haskell, 102 characters

`import List import Ratio l&n=maybe n((&)l.numerator.(n%1*).(!!)l)$findIndex((==)1.denominator.(n%1*))l `

$ ghci Prelude> :m List Ratio Prelude List Ratio> let l&n=maybe n((&)l.numerator.(n%1*).(!!)l)$findIndex((==)1.denominator.(n%1*))l Prelude List Ratio> [3%2]&108 243 Prelude List Ratio> [3%2]&1296 6561 Prelude List Ratio> [455%33,11%13,1%11,3%7,11%2,1%3]&108 15625

88 with relaxed restrictions on the input/output format.

`import List import Ratio l&n=maybe n((&)l.(*)n.(!!)l)$findIndex((==)1.denominator.(*)n)l `

Prelude List Ratio> let l&n=maybe n((&)l.(*)n.(!!)l)$findIndex((==)1.denominator Prelude List Ratio> [455%33,11%13,1%11,3%7,11%2,1%3]&108 15625 % 1

###

Solution:7

## Python, ~~83~~ ~~82~~ ~~81~~ ~~72~~ 70 characters.

It's convenient to have input as fractions.Fraction objects. Same idea as in Ruby solution.

`def f(n,c):d=[x for x in c if x*n%1==0];return d and f(n*d[0],c) or n # Test code: from fractions import Fraction as fr assert f(108, [fr(3, 2)]) == 243 assert f(1296, [fr(3, 2)]) == 6561 assert f(108, [fr(455, 33), fr(11, 13), fr(1, 11), fr(3, 7), fr(11, 2), fr(1, 3)]) == 15625 assert f(60466176, [fr(455, 33), fr(11, 13), fr(1, 11), fr(3, 7), fr(11, 2), fr(1, 3)]) == 7888609052210118054117285652827862296732064351090230047702789306640625 `

###

Solution:8

## C, ~~159~~ ~~153~~ ~~151~~ ~~131~~ ~~111~~ 110 characters

`v[99],c,d;main(){for(;scanf("%d",v+c++););while(d++,v[++d]) *v%v[d]?0:(*v=*v/v[d]*v[d-1],d=0);printf("%d",*v);} `

$ cc f.c $ echo 108 3 2 . | ./a.out; echo 243 $ echo 1296 3 2 . | ./a.out; echo 6561 $ echo 108 455 33 11 13 1 11 3 7 11 2 1 3 . | ./a.out; echo 15625

###

Solution:9

# Python - 53

Improvement thanks to Paul

`f=lambda n,c:next((f(n*x,c)for x in c if x*n%1==0),n) `

testcases

`from fractions import Fraction as fr assert f(108, [fr(3, 2)]) == 243 assert f(1296, [fr(3, 2)]) == 6561 assert f(108, [fr(455, 33), fr(11, 13), fr(1, 11), fr(3, 7), fr(11, 2), fr(1, 3)]) == 15625 assert f(60466176, [fr(455, 33), fr(11, 13), fr(1, 11), fr(3, 7), fr(11, 2), fr(1, 3)]) == 7888609052210118054117285652827862296732064351090230047702789306640625 `

**Python - 54 Without using Fraction**

`f=lambda n,c:next((f(n*i/j,c)for i,j in c if n%j<1),n) `

**Python - 55**

This one is somewhat theoretical. The first two cases run ok, but the other two fail from recursion depth. Maybe someone can get it to work with a generator expression

`f=lambda n,c:([f(n*i/j,c)for i,j in c if n%j<1]+[n])[0] `

Here's one possibility, but grows to 65 even without including the import

`from itertools import chain f=lambda n,c:(chain((f(n*i/j,c)for i,j in c if n%j<1),[n])).next() `

###

Solution:10

**F#: 80 chars**

`let rec f p=function|x,(e,d)::t->f p (if e*x%d=0I then(e*x/d,p)else(x,t))|x,_->x `

Here's an expanded version using `match pattern with |cases`

instead of `function`

:

`//program' is the current remainder of the program //program is the full program let rec run program (input,remainingProgram) = match input, remainingProgram with | x, (e,d)::rest -> if e*x%d = 0I then //suffix I --> bigint run program (e*x/d, program) //reset the program counter else run program (x, rest) //advance the program | x, _ -> x //no more program left -> output the state `

Test code:

`let runtests() = [ f p1 (108I,p1) = 243I f p1 (1296I,p1) = 6561I f p2 (108I,p2) = 15625I f p2 (60466176I,p2) = pown 5I 100] `

And result (tested in F# interactive):

`> runtests();; val it : bool list = [true; true; true; true] `

**Edit** let's have some more fun with this, and calculate some primes (see linked page in the starting post). I've written a new function `g`

that yields the intermediate values of the state.

`//calculate the first n primes with fractran let primes n = let ispow2 n = let rec aux p = function | n when n = 1I -> Some p | n when n%2I = 0I -> aux (p+1) (n/2I) | _ -> None aux 0 n let pp = [(17I,91I);(78I,85I);(19I,51I);(23I,38I);(29I,33I);(77I,29I);(95I,23I); (77I,19I);(1I,17I);(11I,13I);(13I,11I);(15I,14I);(15I,2I);(55I,1I)] let rec g p (x,pp) = seq { match x,pp with |x,(e,d)::t -> yield x yield! g p (if e*x%d=0I then (e*x/d,p) else (x,t)) |x,_ -> yield x } g pp (2I,pp) |> Seq.choose ispow2 |> Seq.distinct |> Seq.skip 1 //1 is not prime |> Seq.take n |> Seq.to_list `

Takes a whopping 4.7 seconds to cough up the first 10 prime numbers:

`> primes 10;; Real: 00:00:04.741, CPU: 00:00:04.005, GC gen0: 334, gen1: 0, gen2: 0 val it : int list = [2; 3; 5; 7; 11; 13; 17; 19; 23; 29] `

This is, without doubt, the most bizarre and slow prime number generator I've ever written. I'm not sure whether that's a good thing or a bad thing.

###

Solution:11

A Javascript one: **99 characters**. No bonus vector :(

`function g(n,p,q,i,c){i=0;while(q=p[i],c=n*q[0],(c%q[1]?++i:(n=c/q[1],i=0))<p.length){};return n;};`

Input is in the format `[[a,b],[c,d]]`

. I took advantage of Javascript's lenience: instead of doing `var x=0, y=0;`

, you can add as many parameters as you like. It doesn't matter whether you actually pass them or not, since they default to `null`

.

Pretty version:

function g(n,p) { var q, c, i=0; while(i < p.length) { q = p[i]; c = n * q[0]; if(c % q[1] != 0) { ++i; } else { n = c % q[1]; i = 0; } } return n; };

###

Solution:12

# Python, ~~110~~ ~~103~~ ~~95~~ 87 characters

## frc.py

`def f(n,c): d=c while len(d): if n%d[1]:d=d[2:] else:n=d[0]*n/d[1];d=c return n `

## test.py

(shows how to drive it)

`from frc import f def test(): """ >>> f(108, [3,2]) 243 >>> f(1296, [3,2]) 6561 >>> f(108, [455,33,11,13,1,11,3,7,11,2,1,3]) 15625 >>> f(60466176, [455, 33,11, 13,1, 11,3, 7,11, 2,1, 3]) 7888609052210118054117285652827862296732064351090230047702789306640625L """ pass import doctest doctest.testmod() `

###

Solution:13

**C#:**

Tidy version:

`using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace Test { class Program { static void Main(string[] args) { int ip = 1; decimal reg = Convert.ToInt32(args[0]); while (true) { if (ip+1 > args.Length) { break; } decimal curfrac = Convert.ToDecimal(args[ip]) / Convert.ToDecimal(args[ip+1]); if ((curfrac * reg) % 1 == 0) { ip = 1; reg = curfrac * reg; } else { ip += 2; } } Console.WriteLine(reg); Console.ReadKey(true); } } } `

Cut down version weighing in at **201 chars** (without the namespace declarations or any of that, just the single using statement (not system) and the Main function):

`using System;namespace T{using b=Convert;static class P{static void Main(string[] a){int i=1;var c=b.ToDecimal(a[0]);while(i+1<=a.Length){var f=b.ToDecimal(a[i])/b.ToDecimal(a[i+1]);if((f*c)%1==0){i=1;c*=f;}else{i+=2;}}Console.Write(c);}}} `

Examples (input is through command line arguments):

`input: 108 3 2 output: 243.00 input: 1296 3 2 output: 6561.0000 input: 108 455 33 11 13 1 11 3 7 11 2 1 3 output: 45045.000000000000000000000000 `

###

Solution:14

**Groovy**, ~~136~~ ~~117~~ 107 characters.

Call as groovy fractal.groovy [input state] [program vector as list of numbers]

`a=args.collect{it as int} int c=a[0] for(i=1;i<a.size;i+=2) if(c%a[i+1]==0){c=c/a[i+1]*a[i];i=-1} println c `

Sample

`bash$ groovy fractal.groovy 108 455 33 11 13 1 11 3 7 11 2 1 3 Output: 15625 `

###

Solution:15

## Perl, ~~84~~ 82 char

Uses standard input.

`@P=<>=~/\d+/g;$_=<>; ($a,$%)=@P[$i++,$i++],$_*$a%$%or$i=0,$_*=$a/$%while$i<@P; print `

Takes 110 chars to pass the bonus test:

`use Math'BigInt blcm;@P=<>=~/\d+/g;$_=blcm<>; ($%,$=)=@P[$i++,$i++],$_*$%%$=or$i=0,($_*=$%)/=$=while$i<@P;print `

###

Solution:16

Haskell: ~~116~~ 109 characters

`f p x[]=x f p x((n,d):b)|x*n`mod`d==0=f p(x*n`div`d)p|True=f p x b main=do{p<-readLn;x<-readLn;print$f p x p} `

This ended up as somewhat of a knockoff of Dario's entry.

###

Solution:17

## Scheme: 326

I thought a Scheme submission was needed, for parity. I also just wanted the excuse to play with it. (Excuse my rudimentary knowledge, I'm sure this could be optimized, and I am open to suggestions!)

`#lang scheme (define fractran_interpreter (lambda (state pc program) (cond ((eq? pc (length program)) (print state)) ((integer? (* state (list-ref program pc))) (fractran_interpreter (* state (list-ref program pc)) 0 program)) (else (fractran_interpreter state (+ pc 1) program))))) `

**Tests:**

`(fractran_interpreter 108 0 '(3/2))`

`(fractran_interpreter 60466176 0 '(455/33 11/13 1/11 3/7 11/2 1/3))`

I get the bonus vector! *(using Dr. Scheme, allocating 256 mb)*

###

Solution:18

**Lua:**

Tidy code:

`a=arg; ip=2; reg=a[1]; while a[ip] do curfrac = a[ip] / a[ip+1]; if (curfrac * reg) % 1 == 0 then ip=2; reg = curfrac * reg else ip=ip+2 end end print(reg) `

Compact code weighing in at **98 chars** (reduction suggested by Scoregraphic on my other answer, and more suggested by gwell):

`a=arg i=2 r=a[1]while a[i]do c=a[i]/a[i+1]v=c*r if v%1==0 then i=2 r=v else i=i+2 end end print(r) `

Run from the command line, supplying the base number first then the series of fractions presented as numbers with space separation, like the following:

`C:\Users\--------\Desktop>fractran.lua 108 3 2 243 C:\Users\--------\Desktop>fractran.lua 1296 3 2 6561 C:\Users\--------\Desktop>fractran.lua 108 455 33 11 13 1 11 3 7 11 2 1 3 15625 `

(manually typed some of that in because it's a pain to get stuff out of the command line, though that is the results returned)

Does NOT handle the bonus vector sadly :(

###

Solution:19

## Reference implementation in Python

This implementation operates on prime factorizations.

First, it decodes a list of fraction tuples by encoding the numerator and denominator as a list of (idx, value) tuples, where idx is the number of the prime (2 is prime 0, 3 is prime 1, and so forth).

The current state is a list of exponents for each prime, by index. Executing an instruction requires first iterating over the denominator, checking if the indexed state element is at least the specified value, then, if it matches, decrementing state elements specified in the denominator, and incrementing those specified in the numerator.

This approach is about 5 times the speed of doing arithmetic operations on large integers in Python, and is a lot easier to debug!

A further optimisation is provided by constructing an array mapping each prime index (variable) to the first time it is checked for in the denominator of a fraction, then using that to construct a 'jump_map', consisting of the next instruction to execute for each instruction in the program.

`def primes(): """Generates an infinite sequence of primes using the Sieve of Erathsones.""" D = {} q = 2 idx = 0 while True: if q not in D: yield idx, q idx += 1 D[q * q] = [q] else: for p in D[q]: D.setdefault(p + q, []).append(p) del D[q] q += 1 def factorize(num, sign = 1): """Factorizes a number, returning a list of (prime index, exponent) tuples.""" ret = [] for idx, p in primes(): count = 0 while num % p == 0: num //= p count += 1 if count > 0: ret.append((idx, count * sign)) if num == 1: return tuple(ret) def decode(program): """Decodes a program expressed as a list of fractions by factorizing it.""" return [(factorize(n), factorize(d)) for n, d in program] def check(state, denom): """Checks if the program has at least the specified exponents for each prime.""" for p, val in denom: if state[p] < val: return False return True def update_state(state, num, denom): """Checks the program's state and updates it according to an instruction.""" if check(state, denom): for p, val in denom: state[p] -= val for p, val in num: state[p] += val return True else: return False def format_state(state): return dict((i, v) for i, v in enumerate(state) if v != 0) def make_usage_map(program, maxidx): firstref = [len(program)] * maxidx for i, (num, denom) in enumerate(program): for idx, value in denom: if firstref[idx] == len(program): firstref[idx] = i return firstref def make_jump_map(program, firstref): jump_map = [] for i, (num, denom) in enumerate(program): if num: jump_map.append(min(min(firstref[idx] for idx, val in num), i)) else: jump_map.append(i) return jump_map def fractran(program, input, debug_when=None): """Executes a Fractran program and returns the state at the end.""" maxidx = max(z[0] for instr in program for part in instr for z in part) + 1 state = [0]*maxidx if isinstance(input, (int, long)): input = factorize(input) for prime, val in input: state[prime] = val firstref = make_usage_map(program, maxidx) jump_map = make_jump_map(program, firstref) pc = 0 length = len(program) while pc < length: num, denom = program[pc] if update_state(state, num, denom): if num: pc = jump_map[pc] if debug_when and debug_when(state): print format_state(state) else: pc += 1 return format_state(state) `

###

Solution:20

## Perl 6: 77 Characters (experimental)

`sub f(@p,$n is copy){ loop {my$s=first {!($n%(1/$_))},@p or return $n;$n*=$s}} `

Newline is optional. Call as:

`say f([3/2], 1296).Int; say f([455/33, 11/13, 1/11, 3/7, 11/2, 1/3], 60466176).Int; `

Readable version:

`sub Fractran (@program, $state is copy) { loop { if my $instruction = first @program: -> $inst { $state % (1 / $inst) == 0 } { $state *= $instruction; } else { return $state.Int; } } } `

### Notes:

The colon notation

`first @program: pointy-sub`

doesn't work on current implementations; first BLOCK, @program has to be used instead.Rakudo appears to have a buggy

`Rat`

giving incorrect results. Current Niecza runs all of the test programs correctly and quickly, including the "bonus" fraction.

###

Solution:21

## Haskell, 142 characters

Without any additional libraries and full I/O.

`t n f=case f of{(a,b):f'->if mod n b == 0then(\r->r:(t r f))$a*n`div`b else t n f';_->[]} main=readLn>>=(\f->readLn>>=(\n->print$last$t n f)) `

###

Solution:22

## Java, ~~200~~ ~~192~~ 179 characters

I think everyone knows that Java would not have the shortest implementation, but I wanted to see how it would compare. It solves the trivial examples, but not the bonus one.

Here is the minimized version:

`class F{public static void main(String[]a){long p=new Long(a[0]);for(int i=1;i<a.length;){long n=p*new Long(a[i++]),d=new Long(a[i++]);if(n%d<1){p=n/d;i=1;}}System.out.print(p);}} `

java -cp . F 108 455 33 11 13 1 11 3 7 11 2 1 3

`15625 `

java -cp . F 1296 3 2

`6561 `

Here is the cleaned-up version:

`public class Fractran { public static void main(String[] args) { long product = new Long(args[0]); for (int index = 1; index < args.length;) { long numerator = product * new Long(args[index++]); long denominator = new Long(args[index++]); if (numerator % denominator < 1) { product = numerator / denominator; index = 1; } // if } // for System.out.print(product); } } `

###

Solution:23

## Scheme 73 characters

My first attempt, at doing this with completely standard R^{5}RS Scheme, came in at 104 characters:

(define(f p n)(let l((q p)(n n))(if(null? q)n(let((a(* n(car q))))(if(integer? a)(l p a)(l(cdr q)n))))))

Running against a few items in the test vector:

> (f '(3/2) 1296) 6561 > (f '(455/33 11/13 1/11 3/7 11/2 1/3) 60466176) 7888609052210118054117285652827862296732064351090230047702789306640625

If you assume that `Î»`

is bound to `lambda`

and `let/cc`

is defined (as they are in PLT Scheme; see below for definitions for running this in Schemes that don't define those), then I can adapt Jordan's second Ruby solution to Scheme, which comes out to 73 characters (note that the argument order is the reverse of my first solution, but the same as Jordan's; in this version, that saves one character).:

(define(f n p)(let/cc r(map(Î»(i)(if(integer?(* n i))(r(f(* n i)p))))p)n))

If I don't have `Î»`

and `let/cc`

predefined, then this one comes in at 111 characters (88 if the fairly common `call/cc`

abbreviation is defined):

(define(f n p)(call-with-current-continuation(lambda(r)(map(lambda(i)(if(integer?(* n i))(r(f(* n i)p))))p)n)))

Definitions of `Î»`

and `let/cc`

:

(define-syntax Î» (syntax-rules () ((_ . body) (lambda . body))) (define-syntax let/cc (syntax-rules () ((_ var . body) (call-with-current-continuation (lambda (var) . body)))))

###

Solution:24

## A bit late... dc 84 chars

Just for fun a `dc`

solution (OpenBSD)

`[ss1sp]s1[Rlp1+sp]s2?l1xz2/sz[z2/ds_:bl_:az0<L]dsLx 1[;als*lp;b~0=1e2lpdlz!<L]dsLxlsp `

It handles all the cases:

`$ dc fractran.dc 455 33 11 13 1 11 3 7 11 2 1 3 60466176 7888609052210118054117285652827862296732064351090230047702789306640625 `

###

Solution:25

I can't leave comments yet but here's a "slightly" shorter version of RCIX's C# version (i believe it's 7 chars shorter)

`using System;namespace T{static class P{static void Main(string[] a){int i=1;Func<string,decimal> d=Convert.ToDecimal;var c=d(a[0]);while(i+1<=a.Length){var f=d(a[i])/d(a[++i]);if((f*c)%1==0){i=1;c*=f;}else i++;}Console.Write(c);}}} `

which uses

`Func<string,decimal> d=Convert.ToDecimal `

and calls `d();`

instead of

`using b=Convert; `

and repeatedly calling `b.ToDecimal();`

.

I also removed an unnecessary pair of curly braces around the else statement to gain 1 char :).

I also replaced the `a[i+1]`

with `a[++i]`

and in the following else body i replaced `i+=2`

with `i++`

to gain another char :P

**Note:If u also have question or solution just comment us below or mail us on toontricks1994@gmail.com**

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