¡Factoriales y ciclos interminables!

33

Como puede saber, el factorial de un entero positivo nes el producto de todos los enteros positivos que son iguales o menores n.

Por ejemplo :

6! = 6*5*4*3*2*1 = 720
0! = 1

Ahora definiremos una operación especial con un nombre irrelevante como sumFac:

Dado un entero positivo n, sumFac(n)es la suma de los factoriales de los dígitos.

Por ejemplo :

sumFac(132) = 1! + 3! + 2! = 9

Tarea

Su misión, ya sea que elija o no aceptarla, es devolver la secuencia (potencialmente infinita) de las aplicaciones sumFaca un entero dado en la entrada.

Ejemplo: 132 -> 132, 9, 362880, 81369, 403927, ...

¡Pero eso no es todo! De hecho, las aplicaciones de sumFaceventualmente crearán un ciclo. ¡También debes devolver este ciclo!

Si su idioma tiene un factorial incorporado, puede usarlo. No soy exigente con el tipo de retorno, solo tiene que devolver la secuencia de aplicaciones sumFac y el ciclo en un formato comprensible para un humano.

EDITAR: para ayudarlo a visualizar mejor cómo debería verse la salida, copié Leaky Nun's justo debajo:

[132, 9, 362880, 81369, 403927, 367953, 368772, 51128, 40444, 97, 367920, 368649, 404670, 5810, 40442, 75, 5160, 842, 40346, 775, 10200, 6, 720, 5043, 151, 122, 5, 120, 4, 24, 26, 722, 5044, 169, 363601, 1454]

¡Solo necesita detener la secuencia cuando el ciclo está por comenzar por segunda vez!

Pero este es el código de golf, por lo que gana la respuesta más corta en bytes.

Tabla de clasificación

Aquí hay un fragmento de pila para generar una tabla de clasificación regular y una descripción general de los ganadores por idioma.

Mostrar fragmento de código

code-golf  math  number  sequence  factorial 
— Monja permeable
fuente
— Leaky Nun
Welcome to PPCG! This looks like a nice challenge, BTW.
— clismique
@Qwerp-Derp Thank you very much! I tried to be creative ^^
@Zgarb Well it's exactly like the output of Leaky Nun. The sequence of the applications and then it shall end just before the beginning of the second cycle. I'll copy his output in the question so everyone can have a clear understanding. Thanks for pointing that out :)
1
@2501 Hardcoding the value is cheating, but concerning the output formatting you can use any separator you want

Respuestas:

19

Jelly, 6 bytes

D!SµÐĿ
    ÐĿ  Repeat until the results are no longer unique. Collects all intermediate results.
D           Convert from integer to decimal (list of digits)
 !          Factorial (each digit)
  S         Sum

Try it online!

I don't see any other way to make it shorter other than to do as told.

Specs

  • Input: 132 (as command-line argument)
  • Output: [132, 9, 362880, 81369, 403927, 367953, 368772, 51128, 40444, 97, 367920, 368649, 404670, 5810, 40442, 75, 5160, 842, 40346, 775, 10200, 6, 720, 5043, 151, 122, 5, 120, 4, 24, 26, 722, 5044, 169, 363601, 1454]
— Leaky Nun
fuente
I didn't expect to have an answer that short. Nice :)
4
@Antoine It's Jelly :D It's always shorter than I think it'll be ;)
— HyperNeutrino
8
@HyperNeutrino Somehow Dennis will come with an even shorter answer
— Leaky Nun
Somehow, yes. Because Dennis. :P
— HyperNeutrino
So... Which character encoding do you use to come up with 6 bytes for those 6 characters? Isn't Jelly supposed to be UTF-8 encoded, meaning that this program is actually 9 bytes?
— LordOfThePigs
10

Python 2, 88 bytes

import math
f=lambda x,l=[]:l*(x in l)or f(sum(math.factorial(int(i))for i in`x`),l+[x])

Try it online!

— ovs
fuente
As a python fan I'm glad someone answered with it, nice :)
9

05AB1E, 12 bytes

[DˆS!O©¯så#®

Try it online!

Explanation

[               # infinite loop
 Dˆ             # add a copy of current value to the global list (initialized as input)
   S            # split current number to digits
    !O          # calculate factorial of each and sum
      ©         # save a copy in register
       ¯så#     # if the current number is in the global list, exit loop
           ®    # retrieve the value from the register for the next iteration
                # implicitly output the global list
— Emigna
fuente
Short and correct, well played !
Thought I could get rid of the s, was wrong, nice answer.
— Magic Octopus Urn
8

Brachylog, 17 bytes

g:I{tẹḟᵐ+}ᵃ⁾L¬≠Lk

Try it online!

Explanation

g:I{     }ᵃ⁾         Accumulate I (a variable) times, with [Input] as initial input:
    t                  Take the last integer
     ẹḟᵐ+              Compute the sum of the factorial of its digits
            L        The result of the accumulation is L
            L­      Not all elements of L are different
               Lk    Output is L minus the last one (which is the start of the loop)
— Fatalize
fuente
What does I mean?
— Leaky Nun
1
@LeakyNun It's a parameter to ᵃ⁾. ᵃ³ means "accumulate 3 times". ᵃ⁾ means "accumulate as many times as the last element of the input", which in that case is I. Since I is a completely free variable, it will try values for it from 0 to +inf.
— Fatalize
8

Wolfram Language, 62 60 56 bytes

Most@NestWhileList[Tr[IntegerDigits@#!]&,#,UnsameQ,All]&

It is really too bad that Wolfram Language has such abominably long function names. *Sigh*

Explanation:

Most[NestWhileList[Tr[IntegerDigits[#]!]&,#,UnsameQ,All]]&
                      IntegerDigits[#]                     (*Split input into list of digits*)
                                      !                    (*Factorial each element in the list*)
                   Tr[                 ]&                  (*Sum the list together*)
     NestWhileList[                      ,#,UnsameQ,All]   (*Iterate the function over itself, pushing each to a list, until a repeat is detected*)
Most[                                                   ]& (*Remove the last element in the list*)
— Scott Milner
fuente
Nice answer. I don't think this can be improved upon.
— Kelly Lowder
1
@KellyLowder Thanks! I was actually able to save two more bytes by mapping the factorial to the list, then summing it with Tr.
— Scott Milner
1
Nice use of NestWhileList[...,All]!
— Greg Martin
6

JavaScript (ES6), 91 89 bytes

Saved 2 bytes thanks to fəˈnɛtɪk

It turns out to be quite similar to the other JS answer.

f=(n,x=!(a=[n]))=>n?f(n/10|0,x+(F=n=>n?n--*F(n):1)(n%10)):~a.indexOf(x)?a:f(x,!a.push(x))

Show code snippet

— Arnauld
fuente
In your factorial function shouldn't you be able to use n instead of n>1 because 0!=1?
— fəˈnɛtɪk
@fəˈnɛtɪk I don't know what I was thinking here. Thank you!
— Arnauld
5

ClojureScript, 146 109 bytes

#(loop[n[%]](let[f(apply +(for[a(str(last n))](apply *(range 1(int a))))](if(some #{f}n)n(recur(conj n f)))))

Yikes, that is a monstrosity. Someone please help me golf this...

Thanks @cliffroot for shaving off a whopping 37 bytes!

This is an anonymous function, to run the function, you have to do this:

(#(...) {arguments})

TIO doesn't have ClojureScript, so here's a link to a ClojureScript REPL.

Here's a link to a Clojure program which prints the last element in the list from 0 to 1000.

Here's the output for 9999:

[9999 1451520 269 363602 1455 265 842 40346 775 10200 6 720 5043 151 122 5 120 4 24 26 722 5044 169 363601 1454]

I have a strong suspicion that all numbers must eventually settle at 1 or the loop [169 363601 1454].

Ungolfed code:

(defn fact-cycle [n]
  (loop [nums [n]]
    (let [fact-num
          (let [str-n (str (last nums))]
            (apply +
              (for [a (range (count str-n))]
                (apply *
                  (range 1
                    (inc (int (nth str-n a))))))))]
      (if (some #{fact-num} nums) nums
        (recur
          (conj nums fact-num))))))

Explanation coming soon!

— clismique
fuente
Quite long but correct ;) I cannot really help you to golf this sorry ^^
the inner for can be (for[a s](apply *(range 1(-(int a)47)))), can't it?
— cliffroot
and this will allow to get rid of the other let #(loop[n[%]](let[f(apply +(for[a(str(last n))](apply *(range 1(-(int a)47)))))](if(some #{f}n)n(recur(conj n f)))))
— cliffroot
oh, seems you don't even need (- ... 47) in ClojureScript, just int will suffice
— cliffroot
well, (inc(int a)) should do for ClojureScript and (-(int a)47) for Clojure.
— cliffroot
5

Perl 6, 64 bytes

{my@a;$_,{[+] .comb.map:{[*] 2..$_}}...^{$_∈@a||!@a.push: $_}}

Try it

Expanded:

{

  my @a;             # array of values already seen

  $_,                # seed sequence with the input

  {
    [+]              # reduce using &infix:<+>
      .comb          # the digits of $_ (implicit method call)
      .map:          # do the following for each
      {
        [*] 2..$_    # get the factorial of
      }
  }


  ...^               # keep generating values until
                     # (｢^｣ means throw away the last value when done)

  {
      $_ ∈ @a        # is it an elem of @a? (｢∈｣ is shorter than ｢(cont)｣)

    ||               # if it's not

      !              # boolean invert so this returns False
        @a.push: $_  # add the tested value to @a
  }
}

Every line above that has { starts a new bare block lambda with an implicit parameter of $_.

I used [*] 2..$_ instead of [*] 1..$_ purely as a micro optimization.

— Brad Gilbert b2gills
fuente
4

JavaScript, 92 bytes

Thanks @Shaggy for golfing off one byte with includes
Thanks @Neil for golfing off two bytes

Code separated into individual functions 92 bytes

f=(x,a=[])=>a.includes(x)?a:f(k(x),a,a.push(x))
p=y=>y?y*p(y-1):1
k=n=>n?p(n%10)+k(n/10|0):0

Code on one line 92 bytes

f=(x,a=[])=>a.includes(x)?a:f((k=n=>n?(p=y=>y?y*p(y-1):1)(n%10)+k(n/10|0):0)(x),a,a.push(x))

Explanation

Initially call the function with just a single argument, therefore a=[].

If x exists in the array a return a a.includes(x)?a:...

Otherwise, append x to a and pass the factorial digit sum and a to the function (a.push(x),f(k(x),a))

p=y=>y?y*p(y-1):1
k=n=>n?p(n%10)+k(n/10|0):0

Factorial Digit sum performed so that it will not exceed the maximum recursion limit.

List of all possible endpoints: 1, 2, 145, 169, 871, 872, 1454, 40585, 45361, 45362, 363601

Try it online!

— fəˈnɛtɪk
fuente
1
Gah, I was so close! Save a byte with f=(x,a=[])=>a.includes(x)?a:(a.push(x),f(k(x),a))
— Shaggy
Can you not write f(k(x),a,a.push(x))? Also, I think you can write k=n=>n&& to save another byte.
— Neil
4

Haskell, 80 67 bytes

g#n|elem n g=g|h<-g++[n]=h#sum[product[1..read[d]]|d<-show n]
([]#)

Try it online! Usage: ([]#) 132

Edit: Saved 13 bytes with typs from Ørjan Johansen!

— Laikoni
fuente
(1) Test and append n instead of s(same as in ovs's Python answer), then f=([]#). (2) Switch the branches, inline s, and use elem.
— Ørjan Johansen
Switch out your ++ for : also.
1
@Canyon It's the wrong order around for that, it would give the final result reversed. You can almost fix it up afterwards, by prepending an extra n: and changing =g to =[], but it seems to be only a tie.
— Ørjan Johansen
4

Pyth, 9 bytes

.us.!MjNT
.us.!MjNTQ  implicit Q

.u          explained below
       N      current value
      j T     convert to decimal (list of digits)
   .!M        factorial of each digit
  s           sum

Try it online!

This answer uses .u ("Cumulative fixed-point. Apply until a result that has occurred before is found. Return all intermediate results.")

— Leaky Nun
fuente
2

R, 120 Bytes

o=scan()
repeat {
q=sum(factorial(as.double(el(strsplit(as.character(o[length(o)]), "")))))
if(q%in%o)break
o=c(o,q)
}
o
— Neil
fuente
you can do o=scan(), use el() instead of [[1]], and gamma(n+1)=factorial(n) which I believe saves a byte, and I think as.numeric is the same as as.double for integers, which also saves a byte, and you can use toString instead of as.character.
— Giuseppe
@Giuseppe Thanks for the input, updated.
— Neil
2

Java 9 JSHell, 213 bytes

n->{Set<Integer>s=new HashSet<>();
return IntStream.iterate(n,i->(""+i).chars()
.map(x->x<50?1:IntStream.rangeClosed(2,x-48)
.reduce(1,(a,b)->a*b)).sum()).boxed()
.takeWhile(x->s.add(x)).collect(Collectors.toList());}

Try it online!

Note: This solution relies on the string representation of a number having code points in the range 48-57. Works for ASCII, UTF-8, Latin-1, all ISO-8859-* character sets, most code pages. Does not work for EBCDIC. I don't think anyone will deduct points for that. :)

Ungolfed:

Function<Integer, List<Integer>> f =        // function from Integer to List of Integer
n -> {
    Set<Integer> s = new HashSet<>();       // memo of values we've seen
    return IntStream.iterate(n,             // iterate over n, f(n), f(f(n)), etc.
    i -> (""+i).chars()                     // the sumFac function; for all chars
        .map(x -> x < 50? 1 :               // give 1 for 0! or 1!
        IntStream.rangeClosed(2, x-48)      // else produce range 2..d 
        .reduce(1,(a,b)->a*b))              // reduction to get the factorial
        .sum())                             // and sum up the factorii!

                                            // now we have a stream of ints
                                            // from applying sumFac repeatedly
        .boxed()                            // box them into Integers (thanks, Java)
        .takeWhile(x->s.add(x))             // and take them while not in the memo
        .collect(Collectors.toList());      // collect them into a list
}

Notes:

  • The return value of Set::add is very helpful here; returns true iff the item was not in the set
  • I was being sarcastic when I said "Thanks, Java"
  • factorii isn't really a word; I just made that up
— David Conrad
fuente
1
I admit this is original ! Nice job :)
@Antoine I admit, Java isn't the best language to golf in, but it's hardly the craziest thing I've done lately. :) codegolf.stackexchange.com/a/117644/794
— David Conrad
2

Pyth, 22 11 bytes

.usm.!sd+Nk

Try it online!

Lots of credit to Leaky Nun's answer, which introduced me to .u, and helped save a massive 11 bytes of this program.

Explanation:

.usm.!sd+NkQ | ending Q is implicitly added
             | Implicit: Q = eval(input())
.u         Q | Repeat the function with initial value Q until a previous value is found. Return all intermediate values
  s          | Summation
   m.!sd     | For each character 'd' in the string, convert to integer and take the factorial
        +Nk  | Convert function argument to string
— K Zhang
fuente
Pyth has more useful functions than one imagines. See my answer as reference.
— Leaky Nun
@LeakyNun I've rewritten my answer to use .u. I guess I'll need to take a look through the character reference again to see if there are any other useful functions there.
— K Zhang
You can use `N to convert to string instead of +Nk.
— Leaky Nun
@LeakyNun Where the N would be obsolete then, and there comes a 9-byte solution...
— Erik the Outgolfer
1

Axiom, 231 bytes

l(a:NNI):List NNI==(r:List NNI:=[];repeat(r:=cons(a rem 10,r);a:=a quo 10;a=0=>break);r)
g(a:NNI):NNI==reduce(+,[factorial(x) for x in l(a)])
h(a:NNI):List NNI==(r:=[a];repeat(a:=g(a);member?(a,r)=>break;r:=cons(a,r));reverse(r))

not golfed functions and some test

-- convert one NNI in its list of digits
listify(a:NNI):List NNI==
    r:List NNI:=[]
    repeat
        r:=cons(a rem 10,r)
        a:=     a quo 10
        a=0=>break
    r

-- g(1234)=1!+2!+3!+4!
SumfactorialDigits(a:NNI):NNI==reduce(+,[factorial(x) for x in listify(a)])

ListGenerateFromSumFactorialDigits(a:NNI):List NNI==
    r:=[a]
    repeat
       a:=SumfactorialDigits(a)
       member?(a,r)=>break
       r:=cons(a,r)
    reverse(r)

(9) -> h 132
   (9)
   [132, 9, 362880, 81369, 403927, 367953, 368772, 51128, 40444, 97, 367920,
    368649, 404670, 5810, 40442, 75, 5160, 842, 40346, 775, 10200, 6, 720,
    5043, 151, 122, 5, 120, 4, 24, 26, 722, 5044, 169, 363601, 1454]
— RosLuP
fuente
1

Java 7, 220 bytes

String c(int n){String r=n+",",c;for(;!r.matches("^"+(c=(n=d(n))+",")+".*|.*,"+c+".*");r+=c);return r;}int d(int n){int s=0;for(String i:(n+"").split(""))s+=f(new Long(i));return s;}long f(long x){return x<2?1:x*f(x-1);}

Explanation:

String c(int n){                            // Method with integer parameter and String return-type
  String r=n+",",                           //  Result-String (which starts with the input integer + a comma
         c;                                 //  Temp String
  for(;!r.matches(                          //  Loop as long as the result-String doesn't match the following regex:
    "^"+(c=(n=d(n))+",")+".*|.*,"+c+".*");  //    "^i,.*|.*,i,.*" where `i` is the current integer
                                            //   `n=d(n)` calculates the next integer in line
                                            //   `c=(n=d(n))+","` sets the temp String to this integer + a comma
    r+=c                                    //   And append the result-String with this temp String
  );                                        //  End of loop
  return r;                                 //  Return the result-String
}                                           // End of method

int d(int n){                               // Separate method (1) with integer parameter and integer return-type
  int s=0;                                  //  Sum
  for(String i:(n+"").split(""))            //  Loop over the digits of `n`
    s+=f(new Long(i));                      //   And add the factorial of these digits to the sum
                                            //  End of loop (implicit / single-line body)
  return s;                                 //  Return the sum
}                                           // End of separate method (1)

long f(long x){                             // Separate method (2) with long parameter and long return-type (calculates the factorial)
                                            // (NOTE: 2x `long` and the `new Long(i)` is shorter than 2x `int` and `new Integer(i)`, hence long instead of int)
  return x<2?                               //  If `x` is 1:
      1                                     //   return 1
    :                                       //  Else:
      x*f(x-1);                             //   return `x` multiplied by the recursive-call of `x-1`
}                                           // End of method (2)

Test code:

Try it here.

class M{
  String c(int n){String r=n+",",c;for(;!r.matches("^"+(c=(n=d(n))+",")+".*|.*,"+c+".*");r+=c);return r;}int d(int n){int s=0;for(String i:(n+"").split(""))s+=f(new Long(i));return s;}long f(long x){return x<2?1:x*f(x-1);}

  public static void main(String[] a){
    System.out.println(new M().c(132));
  }
}

Output:

132,9,362880,81369,403927,367953,368772,51128,40444,97,367920,368649,404670,5810,40442,75,5160,842,40346,775,10200,6,720,5043,151,122,5,120,4,24,26,722,5044,169,363601,1454,
— Kevin Cruijssen
fuente
1

GolfScript, 44 bytes

~]{..)\;10base{,1\{)*}/}%{+}*.@\+@@?)!}do);`

~                                             eval input
 ]                                            initialization
  {                                   }do     do...
   ..)\;                                          initialization
        10base                                    get digits
              {,1\{)*}/}%                         factorial each
                         {+}*                     sum
                             .@\+@@?)!        while result not found
                                         );   drop last element
                                           `  tostring

Try it online!

The factorial part is from here.

— Leaky Nun
fuente
1

C, 161 bytes

f(a){return a?a*f(a-1):1;}
a(l){return l?f(l%10)+a(l/10):0;}
c,t,o;r(i){for(t=o=i;t=a(t),o=a(a(o)),c=t^o;);for(t=i;t^c;printf("%d ",t),c=c|t^o?c:o,t=a(t),o=a(o));}

See it work online.

— 2501
fuente
1

TI-BASIC, 85 79 64 60 bytes

:Prompt L₁                             //Get input as 1 length list, 4 bytes
:Lbl C                                //create marker for looping, see below, 3 bytes
:int(10fPart(Xseq(10^(~A-1),A,0,log(X //split input into list of digits, 20 bytes
:sum(Ans!→X                           //factorial and sum the list, write to new input, 6 bytes
:If prod(L₁-X                         //Test to see if new element is repeated, see below, 7 bytes
:Then                                 //Part of If statement, 2 bytes
:augment(L₁,{X→L₁                     //Push new input to List 1, 10 bytes
:Goto C                               //Loop back to beginning, 3 bytes
:Else                                 //Part of If statement, 2 bytes
:L₁                                   //Print Answer, 2 bytes

Since this is running on a graphing calculator, there is limited RAM. Try testing it with numbers that loop quickly, like 169.

More Explanation:

:int(10fPart(Xseq(10^(~A-1),A,0,log(X
              seq(10^(~A-1),A,0,log(X //Get a list of powers of 10 for each digit (i.e. 1, 0.1, 0.01, etc.)
             X                        //Multiply by input
       fPart(                         //Remove everything but the decimal
     10                               //Multiply by 10 (move one digit in front of the decimal
:int(                                 //Truncate to an integer

If prod(L₁-X works by subtracting the new element from the old list, then multiplying all the elements of the list together. If the element was already in the list, the product will be 0, a falsey value. Otherwise, the product will be a positive integer, a truthy value.

— Scott Milner
fuente
1

J, 40 31 bytes

Edit: 9 bytes saved using the improvements by FrownyFrog. Thanks!

f=.$:@,~`]@.e.~[:+/@:!10#.inv{:

Original code:

f=.[`($:@,)@.([:-.e.~)[:+/!@("."0&":@{:)

In this case I decided to count the bytes for the verb definition, since otherwise it doesn't work in the interpreter.

Explanation:

                         (        {:) - Takes the last element of the array
                               ":@    - converts it to string
                          "."0&       - converts each character back to integer
                       !@             - finds the factorials
                     +/               - sums them
                   [:                 - cap (there are 2 derived verbs above, we need 3 for a fork)
          (    e.~)                   - check if the result is present in the list    
             -.                       - negates the above check
           [:                         - cap
        @.                            - Agenda conjunction, needed for the recursion
  ($:@,)                              - if the result is not in the list, add it to the list and recur
[`                                    - if the result is in the list, display it and stop

Try it online!

— Galen Ivanov
fuente
1
([:-.e.~) -> (1-e.~)
— FrownyFrog
1
— FrownyFrog
@FrownyFrog Thanks, your code is much better! I tried 10#.inv early while experimenting, but then I wrote it 10&#.inv and it was longer, so I rejected it. Thanks for all your suggestions! I have much to learn :)
— Galen Ivanov
@FrownyFrog Reversing the cases for the agenda is so good, I regret I didn't see it :)
— Galen Ivanov
[:+/!@"."0@":@{: is the same length, so there's no improvement with 10#.inv. Just had to drop the ().
— FrownyFrog
1

Tcl, 143 bytes

proc P {n L\ ""} {proc F n {expr $n?($n)*\[F $n-1]:1}
while {[set n [expr [join [lmap d [split $n ""] {F $d}] +]]] ni $L} {lappend L $n}
set L}

Try it online!

— sergiol
fuente
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