Fibonacci Code Golf

2019-01-31 08:20发布

站内文章 / 前端开发
65 0 0

问题:

Generate the Fibonacci sequence in the fewest amount of characters possible. Any language is OK, except for one that you define with one operator, f, which prints the Fibonacci numbers.

Starting point: 25 14 characters in Haskell:

f=0:1:zipWith(+)f(tail f) 

f=0:scanl(+)1f

回答1:

RePeNt, 9, 8 chars

1↓[2?+1]

Or 10 chars with printing:

1↓[2?+↓£1]

Run using:

RePeNt "1↓[2?+1]"

RePeNt is a stack based toy language I wrote (and am still improving) in which all operators/functions/blocks/loops use Reverse Polish Notation (RPN).

Command      Explanation                                              Stack
-------      -----------                                              -----

1            Push a 1 onto the stack                                  1
↓            Push last stack value                                    1 1
[            Start a do-while loop                                    1 1
2?           Push a two, then pop the 2 and copy the last 2 stack     1 1 1 1
             items onto the stack
+            Add on the stack                                         1 1 2
↓£           Push last stack value then print it                      1 1 2
1            Push a 1 onto the stack                                  1 1 2 1
]            Pop value (1 in this case), if it is a 0 exit the loop   1 1 2
             otherwise go back to the loop start.

The answer is on the stack which builds itself up like:

1 1
1 1 2
1 1 2 3
1 1 2 3 5

It never terminates (it has the eqivilent of a C#/JAVA do { } while(true) loop) because the sequence will never terminate, but a terminating solution can be written thus:

N_1↓nI{2?+}

which is 12 chars.

I wonder if anyone will ever read this :(



回答2:

18 characters of English..

"Fibonacci Sequence"

ok, I fail. :)



回答3:

13 chars of Golfscript:

2,~{..p@+.}do

Update to explain the operation of the script:

  1. 2, makes an array of [0 1]
  2. ~ puts that array on the stack
  3. So, at the time we run the do, we start the stack off with 0 1 (1 at top of stack)

The do loop:

  1. Each . duplicates the top item of the stack; here, we do this twice (leaving us with 0 1 1 1 on initial run)
  2. p prints the topmost value (leaving us with 0 1 1)
  3. @ rotates the top 3 items in the stack, so that the third-topmost is at the top (1 1 0)
  4. + adds the top 2 items in the stack (leaving 1 1)
  5. . duplicates the top value, so that the do loop can check its truthiness (to determine whether to continue)

Tracing this mentally a couple of loops will be enough to tell you that this does the required addition to generate the Fibonacci sequence values.

Since GolfScript has bignums, there will never be an integer overflow, and so the top-of-stack value at the end of the do loop will never be 0. Thus, the script will run forever.



回答4:

Language: C++ Compiler Errors
Characters: 205

#define t template <int n> struct 
#define u template <> struct f
t g { int v[0]; };
t f { enum { v = f<n-1>::v + f<n-2>::v }; g<v> x;};
u<1> { enum { v = 1 }; };
u<0> { enum { v = 0 }; };
int main() { f<10> x; }


回答5:

Perl 6 - 22 characters:

sub f{1,1...{$^a+$^b}}


回答6:

x86 (C-callable) realmode, 14 bytes.
Input is  n  on stack, returns  Fn  in AX.

59 31 C0 E3 08 89 C3 40 93 01 D8 E2 FB C3


回答7:

Brainfuck, 33 characters:

+.>+.[<[>+>+<<-]>.[<+>-]>[<+>-]<]


回答8:

22 characters with dc:

1[pdd5**v1++2/lxx]dsxx

Invoke with either:

dc -e'1[pdd5**v1++2/lxx]dsxx'

Or:

echo '1[pdd5**v1++2/lxx]dsxx' | dc

Note: not my work, poached from perlmonks.



回答9:

J, 27 characters for a non-recursive function:

f=:3 :'{:}.@(,+/)^:y(0 1x)'

+/ sums over a list.
(,+/) appends the sum of a list to its tail.
}.@(,+/) sums a list, appends an element to its tail, and drops the first element.
}.@(,+/)^:y iterates the above function y times.
}.@(,+/)^:y(0 1x) applies the above function to the list (0,1) (the x makes it an integer).
{:}.@(,+/)^:y(0 1x) takes the last element of the output list of the above.
f=:3 :'{:}.@(,+/)^:y(0 1x)' defines f to be a function on one variable y.



回答10:

For the record:

  • Lua (66 chars): function f(n)if n<2 then return n else return f(n-1)+f(n-2)end end
  • JavaScript (41 chars): function f(n){return n<2?n:f(n-1)+f(n-2)}
  • Java (41 chars): int f(int n){return n<2?n:f(n-1)+f(n-2);}

I am not much adept of super concise languages... :-P

Chris is right, I just took the simple, recursive algorithm. Actually, the linear one is even shorter in Lua (thanks to multiple assignment)! JavaScript isn't so lucky and Java is worse, having to declare vars...

  • Lua (60 chars): function f(n)a=1;b=0;for i=1,n do a,b=b,a+b end return b end
  • JavaScript (60 chars): function f(n){a=1;b=i=0;for(;i++<n;){x=a+b;a=b;b=x}return b}
  • Java (71 chars): int f(int n){int a=1,b=0,i=0;for(;i++<n;){int x=a+b;a=b;b=x;}return b;}

I would write Lua's code with local a,b=1,0 but it is longer, so let's pollute _G! ;-) Idem for JS.

For completeness, here are the terminal recursive versions. Lua's one, using tail call, is as fast as the linear one (but 69 chars, it is the longest!) - need to call them with three params, n,1,0.

  • Lua (69 char, longer!): function f(n,a,b)if n<1 then return b else return f(n-1,b,a+b)end end
  • JavaScript (44 chars): function f(n,a,b){return n<1?b:f(n-1,b,a+b)}
  • Java (52 chars): int f(int n,int a,int b){return n<1?b:f(n-1,b,a+b);}


回答11:

Corrected after comments (thanks Sebastian), it wasn't a sequence solution, so here we go with 42 chars (includes the \n):

def f(a=0,b=1):
 while 1:yield a;a,b=b,a+b

OLD post below

Python, 38 chars.

f=lambda n:n if n<2 else f(n-1)+f(n-2)

Not so short but the most readable in my opinion :P

EDIT: Here is the analytic way (if someone needs to see it in python :-)

f=lambda n:int(.5+(.5+5**.5/2)**n/5**.5)


回答12:

Windows XP (and later versions) batch script. This batch function when given a single argument - amount, generates amount+1 Fibonacci numbers and returns them as a string (BATCH doesn't really have sets) in variable %r% (369 characters, or 347 characters - if we remove indentation):

:f
    set i=0
    set r=1
    set n=1
    set f=0
    :l
        if %n% GTR %~1 goto e
        set f=%f% %r%
        set /A s=%i%+%r%
        set i=%r%
        set r=%s%
        set /A n+=1
        goto l
    :e
    set r=%f%
    exit /B 0

And here's the complete script, to see it in action (just copy-past it into a CMD or BAT file and run it):

@echo off
call :ff 0
call :ff 1
call :ff 2
call :ff 3
call :ff 5
call :ff 10
call :ff 15
call :ff 20
exit /B 0

:ff
    call :f "%~1"
    echo %~1: %r%
    exit /B 0

:f
    set i=0
    set r=1
    set n=1
    set f=0
    :l
        if %n% GTR %~1 goto e
        set f=%f% %r%
        set /A s=%i%+%r%
        set i=%r%
        set r=%s%
        set /A n+=1
        goto l
    :e
    set r=%f%
    exit /B 0


回答13:

Microsoft Batch - 15 characters

Old challenge, but the world must know it is possible:

%1
%0 %1%2 %1 #

Output is to stderr in unary, counting only the # characters. Depending on the host system's space restrictions, it may produce only the first 14 numbers or so.



回答14:

Language: dc, Char count: 20

Shorter dc solution.

dc -e'1df[dsa+plarlbx]dsbx'


回答15:

F#:

(0,1)|>Seq.unfold(fun(a,b)->Some(a,(b,a+b)))

44 Chars



回答16:

Here's my best using scheme, in 45 characters:

(let f((a 0)(b 1))(printf"~a,"b)(f b(+ a b)))


回答17:

MS Excel: 11 characters:

=SUM(A1:A2)

Type 1 in the top 2 cells, then put the above formula in cell A3. Copy the formula down the spreadsheet.

Starts losing accuracy due to floating point rounding on row 74.
Exceeds 10^307 and overflows to a #NUM! error on row 1477.



回答18:

Generate the Fibonacci sequence. sequence SEQUENCE!



回答19:

C#

I'm seeing a lot of answers that don't actually generate the sequence, but instead give you only the fibonacci number at position *n using recursion, which when looped to generate the sequence gets increasingly slower at higher values of n.

using System;
static void Main()
{
  var x = Math.Sqrt(5);
  for (int n = 0; n < 10; n++)
    Console.WriteLine((Math.Pow((1 + x) / 2, n) - Math.Pow((1 - x) / 2, n)) / p) ;
}


回答20:

let rec f l a b =function 0->a::l|1->b::l|n->f (a::l) b (a+b) (n-1) in f [] 1 1;;

80 characters, but truly generates the sequence, in linear time.



回答21:

Ruby (30 characters):

def f(n)n<2?n:f(n-1)+f(n-2)end


回答22:

@Andrea Ambu

An iterative pythonic fibonacci()'s version should look something like that:

def fibonacci(a=0, b=1):
    while True:
        yield b
        a, b = b, a+b


回答23:

Lua - 49 chars

function f(n)return n<2 and n or f(n-1)+f(n-2)end


回答24:

Befunge-93

31 chars

Will output an infinite list of the Fibonacci numbers, from 0 upwards, separated by tabs (could be reduced to 29 chars by deleting 9, in the first row, at the expense of no whitespace between numbers).

Unfortunately, all the Befunge-93 interpreters I've tried seem to overflow after 65k, so the output is only correct until and including 46368 (which is F24).

#v::1p1>01g:.\:01p+9,#
 >     ^

Confirmed to work (with caveat above) with the Befunge-93 interpreter in Javascript and the Visual Befunge Applet Full.

I'm proud to say this is a completely original work (i.e. I did not copy this code from anyone), and it's much shorter than the Befunge solution currently on Rosetta Code.



回答25:

BrainF**k:

>+++++>+>+<[[>]<<[>>+>+<<<-]>>>[<<<+>>>-]<<[>+>+<<-]>>[<<+>>-]<[<]>-]

That'll generate the first 5. To generate more, replace the 5 + at the beginning with more: eg:

>++++++++++++++++++++++>+>+<[[>]<<[>>+>+<<<-]>>>[<<<+>>>-]<<[>+>+<<-]>>[<<+>>-]<[<]>-]


回答26:

Not the shortest, but the fastest at the time of posting. :-)

float f(float n) {
    return (pow(1+sqrt(5.0))/2.0),n) - pow(1+sqrt(5.0))/2.0),n)/sqrt(n));
}


回答27:

33 characters in C:

F(n){return n<2?n:F(n-1)+F(n-2);}


回答28:

Delphi Prism (Delphi for .net)

f:func<int32,int32>:=n->iif(n>1,f(n-1)+f(n-2),n)

49 chars



回答29:

The previous Ruby example won't work w/o either semicolons or newlines, so it's actually 32 chars. Here's the first example to actually output the sequence, not just return the value of a specified index.

Ruby:
53 chars, including newlines:

def f(n);n<2?1:f(n-1)+f(n-2);end
0.upto 20 {|n|p f n}

or if you want function that outputs a usable data structure, 71 chars:

def f(n);n<2?1:f(n-1)+f(n-2);end
def s(n);(0..n).to_a.map {|n| f(n)};end

or accepting command-line args, 70 chars:

def f(n);n<2?1:f(n-1)+f(n-2);end
p (0..$*[0].to_i).to_a.map {|n| f(n)}


回答30:

PDP-11 Assembler (source)

    .globl  start
    .text
start:
    mov $0,(sp)
    mov $27,-(sp)
    jsr pc, lambda
print_r1:
    mov $outbyte,r3
div_loop:
    sxt r0
    div $12,r0
    add $60,r1
    movb    r1,-(r3)
    mov r0,r1
    tst r1
    jne div_loop
    mov $1,r0
    sys 4; outtext; 37
    mov $1,r0
    sys 1
lambda:
    mov 2(sp),r1
    cmp $2,r1
    beq gottwo
    bgt gotone
    sxt r0
    div $2,r0
    tst r1
    beq even
odd:
    mov 2(sp),r1
    dec r1
    sxt r0
    div $2,r0
    mov r0,-(sp)
    jsr pc,lambda
    add $2,sp
    mov r0,r3
    mov r1,r2
    mov r3,r4
    mul r2,r4
    mov r5,r1
    mov r3,r4
    add r2,r4
    mul r2,r4
    add r5,r1
    mul r3,r3
    mov r3,r0
    mul r2,r2
    add r3,r0
    rts pc
even:
    mov 2(sp),r1
    sxt r0
    div $2,r0
    dec r0
    mov r0,-(sp)
    jsr pc,lambda
    add $2,sp
    mov r0,r3
    mov r1,r2
    mov r2,r4
    mul r2,r4
    mov r5,r1
    mov r2,r4
    add r3,r4
    mul r4,r4
    add r5,r1
    mov r2,r4
    add r3,r4
    mul r2,r4
    mov r5,r0
    mul r2,r3
    add r3,r0
    rts pc
gotone:
    mov $1,r0
    mov $1,r1
    rts pc
gottwo:
    mov $1,r0
    mov $2,r1
    rts pc

    .data
outtext:
    .byte 62,63,162,144,40,106,151,142,157,156
    .byte 141,143,143,151,40,156,165,155
    .byte 142,145,162,40,151,163,40
    .byte 60,60,60,60,60
outbyte:
    .byte 12


标签： language-agnostic code-golf fibonacci rosetta-stone
举报

收藏的人(0)

Ta的文章 更多文章
登录 后发表评论
0条评论
还没有人评论过~