16
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It seems that many people would like to have this, so it's now a sequel to this challenge!

Definition: a prime power is a natural number that can be expressed in the form pn where p is a prime and n is a natural number.

Task: Given a prime power pn > 1, return the power n.

Testcases:

input output
9     2
16    4
343   3
2687  1
59049 10

Scoring: This is . Shortest answer in bytes wins.

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  • 2
    \$\begingroup\$ Note: This challenge might be trivial in some golfing languages, but it's not so trivial for some mainstream languages, as well as the language of June 2018, QBasic. \$\endgroup\$ – Erik the Outgolfer Jun 20 '18 at 23:55
  • \$\begingroup\$ Can we output True instead of 1? Alternatively, float instead of ints? \$\endgroup\$ – Jo King Jun 21 '18 at 0:33
  • 1
    \$\begingroup\$ @JoKing yes, yes. \$\endgroup\$ – Leaky Nun Jun 21 '18 at 0:34
  • \$\begingroup\$ @EriktheOutgolfer Challenge accepted :D \$\endgroup\$ – DLosc Jun 22 '18 at 4:46

31 Answers 31

7
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05AB1E, 2 bytes

Òg

Try it online!

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5
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Python 3, 49 bytes

f=lambda n,x=2:n%x and f(n,x+1)or n/x<2or-~f(n/x)

Try it online!

Outputs True instead of 1 (as allowed by OP). Recursive function that repeatedly finds the lowest factor and then calls the function again with the next lowest power until it reaches 1. This is an extension of my answer to the previous question.

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4
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Pyth, 2

Count prime factors:

lP

Online test.

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4
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Python 2, 37 bytes

f=lambda n,i=2:i/n or(n%i<1)+f(n,i+1)

Try it online!

Counts factors. Apparently I wrote the same golf in 2015.

Narrowly beats out the non-recursive

Python 2, 38 bytes

lambda n:sum(n%i<1for i in range(1,n))

Try it online!

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4
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Bash + GNU utilities, 22

  • 2 bytes saved thanks to @H.PWiz and @Cowsquack
factor|tr -cd \ |wc -c

Try it online!

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  • 1
    \$\begingroup\$ Does factor|sed s/\ //|wc -w work? \$\endgroup\$ – H.PWiz Jun 21 '18 at 10:32
  • 1
    \$\begingroup\$ What about factor|tr -cd \ |wc -c? \$\endgroup\$ – Kritixi Lithos Jun 21 '18 at 11:10
3
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dc, 50 41 bytes

1si[dli1+dsi%0<X]dsXx[dli/dli<Y]sYdli<Yzp

Try it online!

Takes input from the top of the stack (in TIO, put the input in the header to load it onto the stack before execution). Outputs to stdout.

Explanation

Registers used:

i: the current trial divisor, while X is running. Later, the divisor we've found.

X: the macro dli1+dsi%0<X, which has the effect "increment i, then check the modulus with the value on the stack (which will be the original input). If it's not zero, repeat".

Y: the macro dli/dli<Y, which has the effect "Add to the stack a copy of the current top of the stack, divided by i. Repeat until i is reached."

Full program:

1si                 Initialize i
[dli1+dsi%0<X]dsXx  Define and run X
[dli/dli<Y]sY       Define Y
dli<Y               Run Y, but only if needed (if the input wasn't just i)
z                   The stack is i^n, i^(n-1), ... ,i, so print the stack depth
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  • \$\begingroup\$ I found a much better solution! Editing now... \$\endgroup\$ – Sophia Lechner Jun 21 '18 at 1:13
3
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face, 86 bytes

(%d@)\$*,c'$,io>Av"[""mN*c?*m1*mp*m%*s1"$pN1p:~+p1p%%Np?%~:=/NNp+?1?-%N1?%=p%'$i?w1'%>

Hooray, longer than Java!

Try it online!

I am particularly fond of the trick of using the return value of sscanf. Normally the return value would be discarded, but here it will always be 1, because we're always reading a single number as input. We can take advantage of this by assigning its return value to the variable 1, saving the 2 bytes that would otherwise be required to assign 1 to 1 explicitly.

(%d@)

\$*,c'$,io>  ( setup - assign $ to "%d", * to a number, o to stdout )
Av"[""mN*    ( set " to input and allocate space for N for int conversion )
c?*          ( calloc ?, starting it at zero - this will be the output )
m1*          ( allocate variable "1", which gets the value 1 eventually )
mp*m%*       ( p is the prime, % will be used to store N mod p )

s1"$pN       ( scan " into N with $ as format; also assigns 1 to 1 )

1p:~         ( begin loop, starting p at 1 )
  +p1p       ( increment p )
  %%Np       ( set % to N mod p )
?%~          ( repeat if the result is nonzero, so that we reach the factor )

:=           ( another loop to repeatedly divide N by p )
  /NNp       ( divide N by p in-place )
  +?1?       ( increment the counter )
  -%N1       ( reuse % as a temp variable to store N-1 )
?%=          ( repeat while N-1 is not 0 -- i.e. break when N = 1 )

p%'$i?       ( sprintf ? into ', reusing the input format string )
w1'%>        ( write to stdout )
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3
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Attache and Wolfram Language (Mathematica) polyglot, 10 bytes

PrimeOmega

Try Attache online! Try Mathematica online!

Simply a builtin for computing the number of prime factors N has.

Explanation

Since N = pk, Ω(N) = Ω(pk) = k, the desired result.

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2
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Java 8, 59 bytes

A lambda from int to int.

x->{int f=1,c=0;while(x%++f>0);for(;x>1;c++)x/=f;return c;}

Try It Online

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2
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J, 4 bytes

#@q:

q: gives the list of prime factors, # gives the length of the list.

Try it online!

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2
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R, 37 bytes

length(numbers::primeFactors(scan()))

Try it online!

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  • 1
    \$\begingroup\$ sum(x|1) is nearly always shorter than length(x) \$\endgroup\$ – Giuseppe Jun 21 '18 at 15:56
2
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Stax, \$\require{cancel}\xcancel 4 3\$ bytes

|f%

Run and debug it

Length of prime factorization.

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  • 5
    \$\begingroup\$ Ahh.. you're breaking the crossed out 4 is still regular 4 ;( meme. ;p (It was getting old anyway though.. So well done I guess) \$\endgroup\$ – Kevin Cruijssen Jun 21 '18 at 7:47
  • 1
    \$\begingroup\$ \$\text{Yay, MathJax abuse!}\$ But remember to put the cross before the actual bytecount otherwiae the leaderboard snippet may not be able to recognize it. \$\endgroup\$ – user202729 Jun 22 '18 at 8:53
2
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MATL, 3 bytes

Yfz

Try it online!

Explanation:

     % Implicit input: 59049
Yf   % Factorize input [3, 3, 3, 3, 3, 3, 3, 3, 3, 3]
  z  % Number of non-zero elements: 10
     % Implicit output
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2
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Jelly, 3 2 bytes

Æḍ

Try it online!

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2
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Whitespace, 141 bytes

[S S S N
_Push_0][S N
S _Duplicate_0][T   N
T   T   _Read_STDIN_as_number][T    T   T   _Retrieve][S S S T  N
_Push_1][N
S S N
_Create_Label_LOOP_1][S S S T   N
_Push_1][T  S S S _Add][S N
S _Duplicate][S T   S S T   S N
_Copy_2nd_input][S N
T   _Swap_top_two][T    S T T   _Modulo][N
T   S S N
_If_0_Jump_to_Label_BREAK_1][N
S N
N
_Jump_to_Label_LOOP_1][N
S S S N
_Create_Label_BREAK_1][S S S N
_Push_0][S T    S S T   S N
_Copy_2nd_input][N
S S T   N
_Create_Label_LOOP_2][S N
S _Duplicate_input][S S S T N
_Push_1][T  S S T   _Subtract][N
T   S S S N
_If_0_Jump_to_Label_BREAK_2][S N
T   _Swap_top_two][S S S T  N
_Push_1][T  S S S _Add][S N
T   _Swap_top_two][S T  S S T   S N
Copy_2nd_factor][T  S T S _Integer_divide][N
S N
T   N
_Jump_to_Label_LOOP_2][N
S S S S N
_Create_Label_BREAK_2][S N
N
_Discard_top][T N
S T _Print_as_number]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Try it online (with raw spaces, tabs and new-lines only).

Explanation in pseudo-code:

Integer n = STDIN as input
Integer f = 1
Start LOOP_1:
  f = f + 1
  if(n modulo-f == 0)
    Call function BREAK_1
  Go to next iteration of LOOP_1

function BREAK_1:
  Integer r = 0
  Start LOOP_2:
    if(n == 1)
      Call function BREAK_2
    r = r + 1
    n = n integer-divided by f
    Go to next iteration of LOOP_2

function BREAK_2:
  Print r as number to STDOUT
  Program stops with an error: Exit not defined

Example run: input = 9

Command    Explanation                    Stack           Heap    STDIN   STDOUT   STDERR

SSSN       Push 0                         [0]
SNS        Duplicate top (0)              [0,0]
TNTT       Read STDIN as number           [0]             {0:9}   9
TTT        Retrieve                       [9]             {0:9}
SSSTN      Push 1                         [9,1]           {0:9}
NSSN       Create Label_LOOP_1            [9,1]           {0:9}
 SSSTN     Push 1                         [9,1,1]         {0:9}
 TSSS      Add top two (1+1)              [9,2]           {0:9}
 SNS       Duplicate top (2)              [9,2,2]         {0:9}
 STSSTSN   Copy 2nd from top              [9,2,2,9]       {0:9}
 SNT       Swap top two                   [9,2,9,2]       {0:9}
 TSTT      Modulo top two (9%2)           [9,2,1]         {0:9}
 NTSSN     If 0: Jump to Label_BREAK_1    [9,2]           {0:9}
 NSNN      Jump to Label_LOOP_1           [9,2]           {0:9}

 SSSTN     Push 1                         [9,2,1]         {0:9}
 TSSS      Add top two (2+1)              [9,3]           {0:9}
 SNS       Duplicate top (3)              [9,3,3]         {0:9}
 STSSTSN   Copy 2nd                       [9,3,3,9]       {0:9}
 SNT       Swap top two                   [9,3,9,3]       {0:9}
 TSTT      Modulo top two (9%3)           [9,3,0]         {0:9}
 NTSSN     If 0: Jump to Label_BREAK_1    [9,3]           {0:9}
NSSSN      Create Label_BREAK_1           [9,3]           {0:9}
SSSN       Push 0                         [9,3,0]         {0:9}
STSSTSN    Copy 2nd from top              [9,3,0,9]       {0:9}
NSSTN      Create Label_LOOP_2            [9,3,0,9]       {0:9}
 SNS       Duplicate top (9)              [9,3,0,9,9]     {0:9}
 SSSTN     Push 1                         [9,3,0,9,9,1]   {0:9}
 TSST      Subtract top two (9-1)         [9,3,0,9,8]     {0:9}
 NTSSSN    If 0: Jump to Label_BREAK_2    [9,3,0,9]       {0:9}
 SNT       Swap top two                   [9,3,9,0]       {0:9}
 SSSTN     Push 1                         [9,3,9,0,1]     {0:9}
 TSSS      Add top two (0+1)              [9,3,9,1]       {0:9}
 SNT       Swap top two                   [9,3,1,9]       {0:9}
 STSSTSN   Copy 2nd from top              [9,3,1,9,3]     {0:9}
 TSTS      Integer-divide top two (9/3)   [9,3,1,3]       {0:9}
 NSNTN     Jump to Label_LOOP_2           [9,3,1,3]       {0:9}

 SNS       Duplicate top (3)              [9,3,1,3,3]     {0:9}
 SSSTN     Push 1                         [9,3,1,3,3,1]   {0:9}
 TSST      Subtract top two (3-1)         [9,3,1,3,2]     {0:9}
 NTSSSN    If 0: Jump to Label_BREAK_2    [9,3,1,3]       {0:9}
 SNT       Swap top two                   [9,3,3,1]       {0:9}
 SSSTN     Push 1                         [9,3,3,1,1]     {0:9}
 TSSS      Add top two (1+1)              [9,3,3,2]       {0:9}
 SNT       Swap top two                   [9,3,2,3]       {0:9}
 STSSTSN   Copy 2nd from top              [9,3,2,3,3]     {0:9}
 TSTS      Integer-divide top two (3/3)   [9,3,2,1]       {0:9}
 NSNTN     Jump to Label_LOOP_2           [9,3,2,1]       {0:9}

 SNS       Duplicate top (1)              [9,3,2,1,1]     {0:9}
 SSSTN     Push 1                         [9,3,2,1,1,1]   {0:9}
 TSST      Subtract top two (1-1)         [9,3,2,1,0]     {0:9}
 NTSSSN    If 0: Jump to Label_BREAK_2    [9,3,2,1]       {0:9}
NSSSSN     Create Label_BREAK_2           [9,3,2,1]       {0:9}
 SNN       Discard top                    [9,3,2]         {0:9}
 TNST      Print as integer               [9,3]           {0:9}           2
                                                                                    error

Program stops with an error: No exit found.

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2
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Brachylog, 2 bytes

ḋl

Try it online!

Explanation

ḋ        Prime decomposition
 l       Length
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1
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Python 2, 62 bytes

def f(n,p=2,i=0):
	while n%p:p+=1
	while n>p**i:i+=1
	return i

Try it online!

Nothing fancy here.

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  • 1
    \$\begingroup\$ You can save three bytes by making it a full program: Try it online! \$\endgroup\$ – dylnan Jun 21 '18 at 0:23
1
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Japt, 3 bytes

k l

Try it online!

Explanation:

k l
k     Get the prime factors of the input
  l   Return the length
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1
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Actually, 2 bytes

ol

Try it online!

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1
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Haskell, 27 bytes

f n=sum$(0^).mod n<$>[2..n]

Try it online!

Counts factors. Compare:

Haskell, 28 bytes

f n=sum[1|0<-mod n<$>[2..n]]

Try it online!

Haskell, 28 bytes

f n=sum[0^mod n i|i<-[2..n]]

Try it online!

Haskell, 30 bytes

f n=sum[1|i<-[2..n],mod n i<1]

Try it online!

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1
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Octave, 18 bytes

@(x)nnz(factor(x))

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Does what it says on the tin: Number of non-zero elements in the prime factorization of the input.

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1
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Cjam, 5 bytes

rimf,

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Explanation:

ri      take the input and convert it to an int
  mf    factors the input
    ,   take the length of the list

Builtins are great!

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  • \$\begingroup\$ Submissions must be programs or functions by default, and we don't consider this a function. Both rimf, (full program) and {mf,} (function) would be valid. \$\endgroup\$ – Dennis Jun 21 '18 at 20:49
  • \$\begingroup\$ @Dennis Yeah, I think I'm kind of confused on that. I also looked at allowed stardard io before and wondered about what I should actually submit... I actually wanted to ask a question on meta about that. But you confirmed that, so thanks! \$\endgroup\$ – Chromium Jun 22 '18 at 3:19
1
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QBasic, 51 bytes

INPUT x
p=2
WHILE x/p>x\p
p=p+1
WEND
?LOG(x)/LOG(p)

Uses the same algorithm as the "Recover the prime" solution to find the base, then uses rules of logarithms to get the exponent: \$log(p^n) = n \cdot log(p)\$.

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0
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Gaia, 2 bytes

ḍl

Try it online!

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0
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JavaScript (ES6), 37 bytes

f=(n,k=2)=>n%k?n>1&&f(n,k+1):1+f(n/k)

Try it online!

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0
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Perl 6, 36 bytes

{round .log/log (2..*).first: $_%%*}

Looks for the first factor (2..*).first: $_%%*, then from there calculates the approximate value (logs won't get it exact) and rounds it.

Try it online!

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0
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Pari/GP, 8 bytes

bigomega

Try it online!

bigomega(x): number of prime divisors of x, counted with multiplicity.


Pari/GP, 14 bytes

n->numdiv(n)-1

Try it online!

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0
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Racket, 31 bytes

(car(cdr(perfect-power(read))))

Try it online!

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0
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Perl 6, 18 bytes

{+grep($_%%*,^$_)}

Try it online!

Anonymous code block that gets a list of factors and coerces it to a number.

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0
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JavaScript (Node.js), 29 bytes

f=(n,k=n)=>--k&&!(n%k)+f(n,k)

Try it online! Note: Stack overflows for larger inputs.

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