Connexes: fonction phi (n) itérée .

Votre défi est de calculer la fonction phi itérée:

f(n) = number of iterations of φ for n to reach 1.

Où φest la fonction totiente d'Euler .

OEIS connexe .

En voici le graphique:

Règles:

Votre objectif est de sortir f(n)de n=2à n=100.

C'est le code-golf, donc le code le plus court l'emporte.

Voici les valeurs que vous pouvez vérifier:

1, 2, 2, 3, 2, 3, 3, 3, 3, 4, 3, 4, 3, 4, 4, 5, 3, 4, 4, 4, 4, 5, 4, 5, 4, 4, 4, 5, 4, 5, 5, 5, 5, 5, 4, 5, 4, 5, 5, 6, 4, 5, 5, 5, 5, 6, 5, 5, 5, 6, 5, 6, 4, 6, 5, 5, 5, 6, 5, 6, 5, 5, 6, 6, 5, 6, 6, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 6, 5, 6, 7, 5, 7, 5, 6, 6, 7, 5, 6, 6, 6, 6, 6, 6, 7, 5, 6, 6

Haskell , 53 52 octets

Merci nimi d'avoir sauvé 1 octet!

f<$>[2..100]
f 1=0
f n=1+f(sum[1|1<-gcd n<$>[1..n]])

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sum[1|1<-gcd n<$>[1..n]]donne φ(n)(Tiré de flawr , merci!)

f est une fonction récursive qui calcule 1+φ(n) si n ne l'est pas 1et renvoie 0si elle l' nest 1, car il n'y a plus d'itérations à effectuer pour atteindre1

f<$>[2..100]Crée enfin une liste def appliqués à chaque élément de[2..100]

Haskell , 70 69 68 octets

La fonction (\n->sum[1|1<-gcd n<$>[1..n]]) est la fonction totiente, que nous appliquons à plusieurs reprises dans la fonction anonyme. Merci @laikoni pour -1 octet!

EDIT: Je viens de découvrir que @xnor a utilisé cette fonction de totient exacte dans un défi précédent .

length.fst.span(>1).iterate(\n->sum[1|1<-gcd n<$>[1..n]])<$>[2..100]

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MATL , 16 15 octets

99:Q"@`_Zptq}x@

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Explication

99:       % Push [1 2 ... 99]
Q         % Add 1 element-wise: gives [2 3 ... 100]
"         % For each k in that array
  @       %   Push k
  `       %   Do...while
    _Zp   %     Euler's totient function
     tq   %     Duplicate, subtract 1. This is the loop condition
  }       %   Finally (execute on loop exit)
  x       %     Delete
  @       %     Push latest k
          %   End (implicit)
          % End (implicit)
          % Display stack (implicit)

Ancienne version, 16 octets

99:Qt"t_Zp]v&X<q

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Explication

99:       % Push [1 2 ... 99]
Q         % Add 1 element-wise: gives [1 2 ... 100]
t"        % Duplicate. For each (i.e. do the following 100 times)
  t       %   Duplicate
  _Zp     %   Euler's totient function, element-wise
]         % End
v         % Concatenate vertically. Gives a 100×100 matrix
&X<       % Row index of the first minimizing entry for each column.
          % The minimum is guaranteed to be 1, because the number of
          % iterations is more than sufficient.
q         % Subtract 1. Display stack (implicit)

Gelée , 12 11 10 9 octets

³ḊÆṪÐĿ>1S

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-1 octet grâce à HyperNeutrino!

-1 octet merci à M. Xcoder!

-1 octet grâce à Dennis

Comment ça fonctionne

³ḊÆṪÐĿ>1S - Main link. No arguments
³         - Yield 100
 Ḋ        - Dequeue. Creates the list [2, 3 ... 99, 100]
    ÐĿ    - For each element, while the results of the next function
          - aren't unique, collect the results...
  ÆṪ      -   Next function: Totient
      >1  - Greater than one?
        S - Sum the columns

Comme cela a été fait par Dennis, je (naturellement) ne sais pas pourquoi cela fonctionne, juste que cela fonctionne.

APL (Dyalog) , 50 29 25 octets

Regardez-moi, pas de totient intégré!

4 octets enregistrés grâce à @ H.PWiz

{⍵=1:0⋄1+∇+/1=⍵∨⍳⍵}¨1+⍳99

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Comment?

Apparemment, j'ai d'abord opté pour la formule totiente la plus longue (et la plus difficile). Voir l'historique des révisions.

⍳⍵- 1àn

⍵∨ - gcd avec n

1= - égal à 1?

+/ - résume-les tous

Ceci est le totient. Tout le reste est enveloppant pour le comptage ( 1+∇) et l'application sur la plage 2..100( ¨1+⍳99).

Mathematica, 44 octets

(i=-1;#+1//.x_:>EulerPhi[++i;x];i)&~Array~99

-10 octets de @Misha Lavrov
-2 octets de @ user202729

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J REPL, 23 octets

<:#@(5&p:^:a:)"0|2+i.99

Je n'ai pas vérifié, mais cela fonctionne probablement en J régulier si vous le définissez comme un nom (j'ai joué au golf sur mon téléphone sur le REPL).

Intégrés, yo.

Je dirais qu'il y a au moins 2-3 octets à raser (off-by-one à cause de la façon dont a:fonctionne, devoir utiliser |comme noop, etc.).

JavaScript (ES6), 115 ... 104 99 octets

Le codage en dur peut être plus court, mais essayons une approche purement mathématique.

f=n=>n>97?6:(P=(n,s=0)=>k--?P(n,s+(C=(a,b)=>b?C(b,a%b):a<2)(n,k)):s>1?1+P(k=s):1)(k=n+2)+' '+f(-~n)

console.log(f())

Python 2 , 80 octets

n,_=r=2,0;exec'r+=r[sum(k/n*k%n>n-2for k in range(n*n))]+1,;print r[n];n+=1;'*99

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Python 2 , 82 octets

l=0,1
exec"n=len(l);p=2\nwhile n%p:p+=1\nl+=l[p-1]+l[n/p]-n%4%3/2,;print l[n];"*99

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Il utilise les observations qui:

• f(a*b) = f(a) + f(b) - 1, sauf que -1est omis si aet bsont tous deux pairs
• f(p) = f(p-1) + 1quand pest premier, avecf(2)=1

These imply that if n has prime factorization n = 2**a * 3**b * 5**c * 7**d * 11**e * ..., then f(n) = max(a,1) + b + 2*c + 2*d + 3*e + ..., where each p>2 in the factorization contributes f(p-1).

I'm not sure if these continue to hold past n=100, but if they do, they give a way to define and calculate f without using φ.

Bubblegum, 49 bytes

00000000: 5d88 0511 0020 0003 ab2c 024e ff64 e8a3  ].... ...,.N.d..
00000010: 379f 956b f05d 206c 0545 7274 743a b876  7..k.] l.Ertt:.v
00000020: 2267 27f9 9f4d 9b9d fc85 e7e6 994d 6eb0  "g'..M.......Mn.
00000030: 2b                                       +

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PowerShell, 110 bytes

$a=,0*101;2..100|%{$i=$_;for($z=$j=0;++$j-lt$i;$z+=$k-eq1){for($k=$j;$j%$k-or$i%$k;$k--){}};($a[$i]=$a[$z]+1)}

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Mathematical approach.

Actually, looking through it, very similar to the C answer, but developed independently. Creates an array of 0s, loops from 2 to 100, then calculates phi using the gcd formulation. The part in parens at the end both saves the result into $a for the next go-round, and places a copy on the pipeline, which results in the implicit output.

PowerShell, 112 bytes

"122323333434344534444545444545555545455645555655565646555656556656665656565656656757566756666667566"-split'(.)'

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Hard-coded. Ho-hum. Shorter than I could get a mathematical approach by about 10-15 bytes.

Python 2, 83 bytes

n=2
exec"print len(bin(n))-3+n%2-~n%9/8-(0x951a5fddc040419d4005<<19>>n&1);n+=1;"*99

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Combines a heuristic estimate with a hardcoded constant that corrects each estimate as either -0 or -1.

Husk, 10 17 bytes

mö←LU¡Sȯṁε⌋ḣtḣ100

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Edit: +7 bytes for actually mapping the function over the range that was asked for, before it was only the function computing A003434.

Explanation

The following computes A003434:

←LU¡S(ṁ(ε⌋))ḣ -- takes a number as input, for example: 39
   ¡          -- iterate the following function on the input: [39,24,8,4,2,1,1,1..]
    S(     )ḣ --   with itself (x) and the range [1..x]..
      ṁ(  )   --   ..map and sum the following
        ε⌋    --     0 if gcd not 1 else 1
  U           -- longest unique prefix: [39,24,8,4,2,1]
 L            -- length: 6
←             -- decrement: 5

The m(....)ḣ100 part just map that function over the range [2..100], not sure how I missed that part before :S

PHP, 98 bytes

1,2,<?=join(',',str_split(unpack('H*','##3444E4DEEDEEUUEEVEUVUVVFUVVUfVfVVVVVegWVgVffgV')[1]))?>,6

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I packed all digits into a binary string. After unpacking it, converting it to a an array and then mergin the array again, i only had to prepend the 1,2 and append the 6 as those wouldnt fit or caused a control code to appear.

Perl 6, 47 bytes

map {($_,{+grep 1==* gcd $_,^$_}...1)-1},2..200

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05AB1E, 11 bytes

тL¦ε[DNs#sÕ

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Explanation

тL¦           # push range [2 ... 100]
   ε          # apply to each
    [         # start a loop
     D        # duplicate the current number
      N       # push the loop iteration counter
       s      # swap one copy of the current number to the top of the stack
        #     # if true, break the loop
         s    # swap the second copy of the current number to the top of the stack
          Õ   # calculate eulers totient

C, 112 bytes

a[101];f(i,j,k,t){for(a[i=1]=0;i++<100;printf("%d ",a[i]=a[t]+1))for(t=j=0;++j<i;t+=k==1)for(k=j;j%k||i%k;k--);}

Ungolfed:

a[101];
f(i,j,k,t){
    for(a[1]=0,i=2;i<=100;i++) {   // initialize
        for(t=j=0;++j<i;t+=k==1)   // count gcd(i, j) == 1 (t = phi(i))
            for(k=j;j%k||i%k;k--); // calculate k = gcd(i, j)
        printf("%d ",a[i]=a[t]+1); // print and store results
    }
}

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Alumin, 87 bytes

hhhhhdadtqdhcpkkmzyhqkhwzydqhhwdrdhhhwrysrshhwqdrybpkshehhhwrysrarhcpkksyhaydhehycpkkmr

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Explanation

hhhhhdadt      CONSTANT 100

RANGE FROM 100 to 0
q
  dhc
p

REMOVE 0 AND 1
kk

OVER EACH ELEMENT...
m
  zyh
  q
    kh
    wzyd
    q
      DUPLICATE TOP TWO ELEMENTS...
      hhwdrdhhhwrysrshhw
      GCD...
      qdryb
    p
    ks
    he
    hhhw
    ry
    s
    rarhc
  p
  IS IT ONE? IF SO TERMINATE (FIXPOINT)
  kksyhaydhehyc
p
kk
m
REVERSE THE VALUES
r

Pyth, 38 bytes (not competitive)

.e-+1sl+1kb_jC"Éõ4ÕYHø\\uÊáÛ÷â¿"3

Try it on the Pyth Herokuapp, because it doesn't work on TIO for whatever reason.

I have no doubt the explicit Pyth solution is smaller, but I wanted to see how small I could get the code by compressing the sequence, and learn Pyth I guess. This uses the fact that an upper bound of the sequence is log2(n)+1.

Explanation

.e-+1sl+1kb_jC"Éõ4ÕYHø\\uÊáÛ÷â¿"3
             C"Éõ4ÕYHø\\uÊáÛ÷â¿"   interpret string as base 256 integer
            j                   3  convert to array of base 3 digits
           _                       invert sequence (original had leading 0s)
.e                                 map with enumeration (k=index, b=element)
       +1k                                   k+1
     sl                            floor(log(   ))
   +1                                             +1
  -       b                                         -b

I got the compressed string via Ci_.e+1-sl+1ksb"122323333434344534444545444545555545455645555655565646555656556656665656565656656757566756666667566"3, which is just the opposite of the code above with a few type conversions.

Ohm v2, 41 bytes

“ ‽W3>€þΣÌιZ§Á HgüυH§u·β}Bā€ΣNπáÂUõÚ,3“8B

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Literally completely hardcoded... I actually took the sequence above, stripped everything that wasn't a number, interpreted it as base 8, then turned it into Ohm's built-in base 255 number representation. That's what the quotes do. Then, the program simply turns that into base 8 again.