23
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Given a positive integer number \$n\$ output its perfect radical.

Definition

A perfect radical \$r\$ of a positive integer \$n\$ is the lowest integer root of \$n\$ of any index \$i\$:

$$r = \sqrt[i]{n}$$

where \$r\$ is an integer.

In other words \$i\$ is the maximum exponent such that \$r\$ raised to \$i\$ is \$n\$:

$$n = r^i$$

This is OEIS A052410.

Special cases

For \$n = 1\$ we don't really care about the degree \$i\$ as we are asked to return \$r\$ in this challenge.

  • Just take \$r=1\$ for \$n=1\$.
  • Since there is an OEIS for this and it starts from 1 you don't have to handle \$n=0\$.

Note

A positive integer \$n\$ is expressed in the form \$100...000\$ if we convert it to base \$r\$ For example the perfect radical of \$128\$ is \$2\$ which is \$1000000\$ in base \$2\$, a \$1\$ followed by \$i -1\$ \$0\$s.

Input: a positive integer. You don't not have to handle inputs not supported by your language (obviously, abusing this is a standard loophole.)

Output: the perfect radical of that number.

You may instead choose to take a positive integer \$n\$ and output the radicals of the first \$n\$ positive integers, or to output the infinite list of radicals.

Test cases

This is a list of all numbers \$n \le 10000\$ where \$n \ne r\$ (expect for \$n = 1\$, included as an edge case, included also some cases where r==n for completeness sake ) :

[n, r]
[1, 1],
[2,2],
[3,3],
[4, 2],
[5,5],
[6,6],
[7,7],
[8, 2],
[9, 3],
[10,10],
[16, 2],
[25, 5],
[27, 3],
[32, 2],
[36, 6],
[49, 7],
[64, 2],
[81, 3],
[100, 10],
[121, 11],
[125, 5],
[128, 2],
[144, 12],
[169, 13],
[196, 14],
[216, 6],
[225, 15],
[243, 3],
[256, 2],
[289, 17],
[324, 18],
[343, 7],
[361, 19],
[400, 20],
[441, 21],
[484, 22],
[512, 2],
[529, 23],
[576, 24],
[625, 5],
[676, 26],
[729, 3],
[784, 28],
[841, 29],
[900, 30],
[961, 31],
[1000, 10],
[1024, 2],
[1089, 33],
[1156, 34],
[1225, 35],
[1296, 6],
[1331, 11],
[1369, 37],
[1444, 38],
[1521, 39],
[1600, 40],
[1681, 41],
[1728, 12],
[1764, 42],
[1849, 43],
[1936, 44],
[2025, 45],
[2048, 2],
[2116, 46],
[2187, 3],
[2197, 13],
[2209, 47],
[2304, 48],
[2401, 7],
[2500, 50],
[2601, 51],
[2704, 52],
[2744, 14],
[2809, 53],
[2916, 54],
[3025, 55],
[3125, 5],
[3136, 56],
[3249, 57],
[3364, 58],
[3375, 15],
[3481, 59],
[3600, 60],
[3721, 61],
[3844, 62],
[3969, 63],
[4096, 2],
[4225, 65],
[4356, 66],
[4489, 67],
[4624, 68],
[4761, 69],
[4900, 70],
[4913, 17],
[5041, 71],
[5184, 72],
[5329, 73],
[5476, 74],
[5625, 75],
[5776, 76],
[5832, 18],
[5929, 77],
[6084, 78],
[6241, 79],
[6400, 80],
[6561, 3],
[6724, 82],
[6859, 19],
[6889, 83],
[7056, 84],
[7225, 85],
[7396, 86],
[7569, 87],
[7744, 88],
[7776, 6],
[7921, 89],
[8000, 20],
[8100, 90],
[8192, 2],
[8281, 91],
[8464, 92],
[8649, 93],
[8836, 94],
[9025, 95],
[9216, 96],
[9261, 21],
[9409, 97],
[9604, 98],
[9801, 99],
[10000, 10]

Rules

  • Input/output can be given by any convenient method.
  • You can print it to STDOUT, return it as a function result or error message/s.
  • Either a full program or a function are acceptable.
  • Standard loopholes are forbidden.
  • This is so all usual golfing rules apply, and the shortest code (in bytes) wins.

Sandbox

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9
  • 2
    \$\begingroup\$ I believe this is A052410 \$\endgroup\$
    – Giuseppe
    Dec 1, 2020 at 21:36
  • 1
    \$\begingroup\$ Suggest adding \$0 \to 0\$ and \$1 \to 1\$ to the testcases. \$\endgroup\$
    – Noodle9
    Dec 1, 2020 at 22:31
  • 1
    \$\begingroup\$ I have edited the question: since there is an OEIS and it starts from 1 you don't have to handle n=0, I'll add a test for n=1 \$\endgroup\$
    – AZTECCO
    Dec 1, 2020 at 23:11
  • 3
    \$\begingroup\$ If we don't need to handle one it should say "positive" rather than "non-negative". \$\endgroup\$
    – Jonathan Allan
    Dec 2, 2020 at 3:45
  • 2
    \$\begingroup\$ Suggest adding a test case where r == n \$\endgroup\$
    – Stef
    Dec 2, 2020 at 9:20

27 Answers 27

Reset to default
11
\$\begingroup\$

J, 14 bytes

(%+./)&.(_&q:)

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(%+./)&.(_&q:)
      &.(_&q:) number to prime exponents
(%+./)         divide them by their GCD
      &.(_&q:) prime exponents to number
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1
  • \$\begingroup\$ Beautiful use of under. \$\endgroup\$
    – Jonah
    Dec 1, 2020 at 23:22
8
\$\begingroup\$

Jelly, 10 bytes

ÆEgƒ0:@ƊÆẸ

Try it online!

ÆE:g/$ÆẸ errors given 1.

ÆE            Take the exponents of the input's prime factorization.
     :@Ɗ      Divide each exponent by
  gƒ0         the exponents' GCD (or 0 in the case that there are none).
        ÆẸ    Let the result be the exponents of the output's prime factorization.
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3
  • 1
    \$\begingroup\$ Unfortunately, this errors for 1: Try it online! \$\endgroup\$
    – caird coinheringaahin g ♦
    Dec 1, 2020 at 23:27
  • 2
    \$\begingroup\$ As impressive as this answer is (and I‘ve dropped a +1 because it is a nice answer), I will never understand how an answer twice the length of mine got almost twice as many upvotes. Guess it‘ll stay a mystery \$\endgroup\$
    – caird coinheringaahin g ♦
    Dec 6, 2020 at 3:56
  • 1
    \$\begingroup\$ @cairdcoinheringaahing Indeed... \$\endgroup\$
    – Unrelated String
    Dec 6, 2020 at 21:31
8
\$\begingroup\$

Brachylog, 6 bytes

1|~^hℕ

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1|~^hℕ  with the implicit input n
1       input and output is 1
 |      or
  ~^    find two numbers [r, i] so that r^i = n
    h   return r
     ℕ  to limit the search space: r must be positive

Search tries lowest i first, so we get the maximum r for free.

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4
  • \$\begingroup\$ I spent the last ten minutes trying to get this working with ≜... I have too little faith in CLP(FD)! \$\endgroup\$
    – Unrelated String
    Dec 1, 2020 at 21:41
  • 1
    \$\begingroup\$ @UnrelatedString I had ℕᵐ≜h and a >. first, but just before posting tried deleting stuff in true codegolf fashion and it kept working. :-) \$\endgroup\$
    – xash
    Dec 1, 2020 at 21:45
  • 1
    \$\begingroup\$ This outputs 0 for 1 (should output 1): Try it online! \$\endgroup\$
    – caird coinheringaahin g ♦
    Dec 1, 2020 at 23:30
  • 1
    \$\begingroup\$ @cairdcoinheringaahing Ah, I only checked 0 -> 0 earlier. Pesky special cases, increasing bytes by 50%. :-) Thanks! \$\endgroup\$
    – xash
    Dec 1, 2020 at 23:45
8
\$\begingroup\$

Python 3, 55 \$\cdots\$ 59 57 bytes

Added 7 bytes to fix an error kindly pointed by user.
Saved 3 bytes thanks to user!!!

lambda n:{r**i:r for i in range(n)for r in range(n+1)}[n]

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2
  • \$\begingroup\$ 59 bytes if you're fine with being wasteful \$\endgroup\$
    – user
    Dec 1, 2020 at 22:32
  • 4
    \$\begingroup\$ @user Wasteful's fine by me if it saves bytes - thanks! :D \$\endgroup\$
    – Noodle9
    Dec 1, 2020 at 22:37
7
\$\begingroup\$

Husk, 8 bytes

VȯεΣ`B¹ḣ

Try it online!

V           # index of first truthy element of
 ȯ          # applying 3 functions to
       ḣ    # 1...input
    `B¹     # convert input to this base
   Σ        # sum of digits
  ε         # is at most 1
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7
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Jelly, 8 6 5 bytes

bR§i1

Try it online!

Uses the fact that n in base r has the format 100...000, meaning that the sum of the digits only equals 1 in base r

-1 byte (indirectly) thanks to Dominic van Essen's answer, make sure to give them an upvote

How it works

bR§i1 - Main link. Takes n on the left
 R    - [1, 2, 3, ..., n]
b     - Convert n to each base 1, 2, 3, ..., n
  §   - Sum of the digits of each
   i1 - First index of 1
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2
  • 1
    \$\begingroup\$ And here I thought nothing would come of the aside about base conversion! \$\endgroup\$
    – Unrelated String
    Dec 1, 2020 at 21:20
  • 1
    \$\begingroup\$ @UnrelatedString I had the same thought, before remembering a trick from one of Jonathan Allan's old answers for checking numbers in the form 100...00 \$\endgroup\$
    – caird coinheringaahin g ♦
    Dec 1, 2020 at 21:22
7
\$\begingroup\$

05AB1E, 8 11 8 bytes

-3 thanks to @ovs!

L¦BíCXkÌ

Try it online!

I am trying to somehow implement a log function to check whether a number matches the regex 10*, but that is too mathematical for me...

Wait, how?

L              # Push all numbers natural numbers up to input          [1, 2, 3 ... I]
 ¦             # What is that 1 doing there? Remove it!                [2, 3, 4 ... I]
  B            # Convert the input to each of the bases e.g input: 9   [1001, 100, 21...]
   í           # Reverse each string                                   [1001, 001, 12...]
    C          # Convert each from binary to decimal                   [9, 1, 4...]              (How though! Can someone explain?)
     Xk        # Get first index of 1                                  1
       Ì       # Add 2                                                 3
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3
  • 3
    \$\begingroup\$ This hangs for input of 0 and returns 2 for input of 1. \$\endgroup\$
    – Giuseppe
    Dec 1, 2020 at 22:44
  • 1
    \$\begingroup\$ i,} can be ≠i for -1. \$\endgroup\$
    – Kevin Cruijssen
    Dec 2, 2020 at 9:30
  • 1
    \$\begingroup\$ i,}∞ can be L for -3 ;) (For n=1 Xk returns -1, and -1+2=1) \$\endgroup\$
    – ovs
    Dec 2, 2020 at 10:36
6
\$\begingroup\$

k4, 26 24 21 18 bytes

-2 bytes by ignoring n=0 case

-3 bytes by applying @caird coinheringaahing's logic

-3 bytes by simplifying/combining operations

{(x{+/y\:x}'!x)?1}

Benefits from list ? value returning the count of the list if the value isn't present in it, and from convenient weirdness with the n=0 and n=1 edge cases.

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5
\$\begingroup\$

Retina 0.8.2, 60 bytes

.+
$*
(?<=(?=((?=((1*)(?=\5\3+$)1)(\2*$))\4)*1$)^(..+)).*

1

Try it online! Link includes test cases. Explanation:

.+
$*

Convert to unary.

(?<=(?=...$)^(..+)).*

Delete the earliest suffix leaving behind the smallest prefix \$r\$ (captured into \5) such that the \$n\$ matches the following:

((?=((1*)(?=\5\3+$)1)(\2*$))\4)*1

Find \$k\$ \3 such that \$n-r\$ is divisible by \$k\$, but also \$n\$ is divisible by \$k+1\$ \2. Apparently this can only be satisfied by \$n=r(k+1)\$, but I can't find the answer where this is proved. \$(r-1)(k+1)\$ is then subtracted from \$n\$, resulting in \$k+1\$. This is then repeated until \$n\$ is reduced to \$1\$, which is matched at the end.

1

Convert to decimal.

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5
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R, 47 bytes

n=scan();match(n,sapply(0:n,"^",1:n))%/%n-(n<2)

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Struggled for ages trying to beat Giuseppe's answer, only to be totally outgolfed (seconds before posting) by Robin Ryder's comment (now an answer)...

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1
  • 1
    \$\begingroup\$ It's the journey that matters, not the byte count at the end. \$\endgroup\$
    – user
    Dec 1, 2020 at 23:37
5
\$\begingroup\$

R, 37 33 bytes

-4 bytes thanks to Dominic van Essen

match(T,!log(n<-scan(),1:n)%%1,1)

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A different (and shorter) approach than the one used in Giuseppe's and Dominic van Essen's R answers.

Finds the first integer k such that log(n,k) is an integer, or returns 1 if there is no such k (which corresponds to the special case n=1).

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5
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JavaScript (ES7), 36 bytes

Recursively looks for the highest \$i\le n\$ such that \$k=n^{1/i}\$ is an integer. Then returns this \$k\$.

f=(n,i=n)=>(k=n**(1/i))%1?f(n,i-1):k

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JavaScript (ES7), 37 bytes

A slightly longer version that performs more recursive calls but is not subject to rounding errors.

n=>(g=k=>k**i-n?g(k-1||i--|n):k)(i=n)

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1
  • \$\begingroup\$ I like to read your answers too much. Using %1 to check being floating point is nice. \$\endgroup\$
    – Soner from The Ottoman Empire
    Dec 4, 2020 at 4:10
5
\$\begingroup\$

Python 3.8, 53 51 bytes (thanks @user for pointing out extra spaces)

def r(n):
    i=n
    while(a:=n**(1/i))%1:i-=1
    return a

Try it online!

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4
  • 1
    \$\begingroup\$ Nice answer! You left a couple spaces in, so it's really 51 bytes. \$\endgroup\$
    – user
    Dec 2, 2020 at 19:51
  • \$\begingroup\$ @user Thanks! Took me some time to perfect it, and in the end, I forgot about removing the extra spaces \$\endgroup\$
    – nobody
    Dec 2, 2020 at 19:53
  • \$\begingroup\$ It's a pity this isn't allowed \$\endgroup\$
    – user
    Dec 2, 2020 at 20:03
  • \$\begingroup\$ @user True, I spent some time trying to make something similar work, but I couldn't. \$\endgroup\$
    – nobody
    Dec 2, 2020 at 20:08
4
\$\begingroup\$

R, 49 44 bytes

n=scan();which(outer(1:n,n:1,"^")==n,T)[1,1]

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Thanks to Dominic van Essen for pointing out a bug.

n <- scan()				# read input
arr <- outer(0:n,1:n,"^")		# create the array of powers (0^(1:n), 1^(1:n), ... n^(1:n))
arr <- t(arr)				# transpose, so the array is ((0:n)^1, (0:n)^2, ... (0:n)^n)
ind <- which(arr==n,T)			# get 1-based array indices where arr == n. So they are a matirx of rows of [i+1,r+1] pairs, sorted in increasing order of r
ind[1,2]-1				# extract the appropriate r.
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8
  • 1
    \$\begingroup\$ Nice, but what about 0 and 1? \$\endgroup\$
    – Dominic van Essen
    Dec 1, 2020 at 21:38
  • \$\begingroup\$ @DominicvanEssen easily fixed by starting the second 0:n at 1 instead. \$\endgroup\$
    – Giuseppe
    Dec 1, 2020 at 21:39
  • 1
    \$\begingroup\$ 37 bytes \$\endgroup\$
    – Robin Ryder
    Dec 1, 2020 at 23:21
  • 2
    \$\begingroup\$ @RobinRyder ...and, along the same lines, 33 bytes \$\endgroup\$
    – Dominic van Essen
    Dec 1, 2020 at 23:34
  • 1
    \$\begingroup\$ @RobinRyder you should post that as your own! \$\endgroup\$
    – Giuseppe
    Dec 2, 2020 at 0:11
4
\$\begingroup\$

Factor, 49 bytes

[ dup [1,b] 2dup '[ _ swap _ n^v member? ] find ]

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Slow for larger inputs because it tries to evaluate a large power.

[
  dup [1,b] 2dup       ! ( n 1..n n 1..n )
  '[                   ! Put stack items in the `_`s in the quotation
    _(n) swap _(1..n)  !   ( elt -- n elt 1..n )
    n^v member?        !   Test if n appears in elt^(1..n)
  ] find               ! Find the first number in 1..n that satisfies the above
]
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4
\$\begingroup\$

05AB1E, 6 bytes

Inspired by SunnyMoon's answer.

LÀ.ΔBR

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L         # push the range [1, 2, ..., n]
 À        # rotate the 1 to the back: [2, 3, ..., n, 1]
  .Δ      # find the first integer where ...
    B     #   the input converted to that base
     R    #   reversed
          #   implicit: is equal to 1 as a number

05AB1E, 8 bytes

LÀ.Δ.n.ï

Try it online! There was a bug with .ï, which has recently been fixed, but the interpreter on TIO is not up to date.

L         # push the range [1, 2, ..., n]
 À        # rotate the 1 to the back: [2, 3, ..., n, 1]
  .Δ      # find the first integer where ...
    .n    #   the logarithm of the input in that base
      .ï  #   is an integer
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3
\$\begingroup\$

Japt -g, 10 bytes

I feel like I'm missing a trick here.

õ ï æ@¥Xrp

Try it

õ ï æ@¥Xrp     :Implicit input of integer U
õ              :Range [1,U]
  ï            :Cartesian product with itself
    æ          :Get first pair that returns true
     @         :When passed through the following function as X
      ¥        :  Test U for equality with
       Xr      :  X reduced by
         p     :    Exponentiation
               :Implicit output of first element of that pair
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3
\$\begingroup\$

Stax, 10 bytes

┌Pèó~JRå▲ï

Run and debug it

Explanation

R{xs|E|+1=}j
R            range 1..n
 {        }j get first number i where:
  xs|E       input(x) in base i digits
      |+     summed
        1=   equals 1
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3
\$\begingroup\$

05AB1E, 12 (10) bytes

ÓDā<Ør0š¿÷mP

Port of @UnrelatedString's Jelly answer.

The 0š shouldn't be necessary, but unfortunately there is a bug in 05AB1E for ¿ with empty lists.

Try it online or verify all test cases.

Explanation:

Ó             # Get the prime exponents of the (implicit) input-integer
 D            # Duplicate this list of exponents
  ā           # Push a list in the range [1, length] (without popping the list itself)
   <          # Decrease each by one to make the range [0, length)
    Ø         # Get the n'th prime for each of these indices
     r        # Reverse the three lists on the stack
      0š      # Prepend a 0 (work-around for `¿` bug with empty lists)
        ¿     # Pop and get the greatest common divisor (gcd) of this list
         ÷    # Integer-divide all values in the list by this gcd
              # (we use integer-division due to another bug that isn't on TIO yet,
              #  as well as to get an integer output, instead of float)
          m   # Take the primes we created earlier to the power of these values
           P  # And take the product of that
              # (after which it is output implicitly)
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3
\$\begingroup\$

C (gcc), 65 60 58 bytes

Saved 5 bytes thanks to Sisyphus!!!

p;i;r;f(n){for(r=0;r++<n;)for(p=i=1;i++<n;p*=r)n=p-n?n:r;}

Try it online!

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2
  • \$\begingroup\$ You can write n=r instead of return r, if you don't mind some undefined behavior. \$\endgroup\$
    – Sisyphus
    Dec 2, 2020 at 1:56
  • \$\begingroup\$ @Sisyphus I would normally avoid the dreaded UB like the plague, but if it'll save some bytes here: bring it on! Thanks! :D \$\endgroup\$
    – Noodle9
    Dec 2, 2020 at 8:40
3
\$\begingroup\$

Wolfram Language (Mathematica), 27 bytes

Most@*Internal`PerfectPower

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-20 bytes from att

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4
  • 1
    \$\begingroup\$ 40 bytes \$\endgroup\$
    – att
    Dec 2, 2020 at 3:03
  • \$\begingroup\$ 36 \$\endgroup\$
    – att
    Dec 2, 2020 at 7:38
  • \$\begingroup\$ 31 bytes \$\endgroup\$
    – att
    May 13, 2022 at 23:01
  • \$\begingroup\$ Or Most@*Internal`PerfectPower for 27 bytes, with list-wrapped output. \$\endgroup\$
    – att
    May 13, 2022 at 23:21
2
\$\begingroup\$

Octave, 33 bytes

@(n)[~,j]=find((t=1:n)'.^t'==n,1)

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How it works

@(n)                               % anonymous function with input n
               (t=1:n)             % let t = [1, 2, ..., n] (row vector)
                       .^          % element-wise power with broadcast...
                      '            % of t transposed...
                         t         % raised to t. Gives n×n matrix of powers
                          '==n     % test each entry for equality with n
    [~,j]=find(               ,1)  % col index of the first true entry (in linear order)
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2
\$\begingroup\$

Scala, 50 bytes

Back to 50 bytes because n=0 doesn't have to be handled anymore!

n=>1.to(n)find(r=>1.to(n)exists(n==math.pow(r,_)))

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2
\$\begingroup\$

Charcoal, 19 16 bytes

ＮθＩ⊕⌕Ｅθ⎇ιΣ↨θ⊕ιθ¹

Try it online! Link is to verbose version of code. Edit: Now back to a reformulation of my original answer. Works by converting n to each base 1..n and finding the first 1-indexed value with a digit sum of 1. Conveniently this automatically works for an input of 0 (the resulting list is empty, so the 1-indexed position is 0), so the only edge case is base 1 as Charcoal cannot convert to unary, but the digit sum is always n in base 1 anyway. Explanation:

Ｎθ                  Input `n` as a number
      θ             `n`
     Ｅ              Map over implicit range
        ι           Current value
       ⎇      θ     If zero then `n` else
          ↨θ        `n` converted to base
            ⊕ι      Incremented value
         Σ          Sum of digits
    ⌕          ¹    Find first occurrence of literal `1`
   ⊕                Increment (convert to 1-indexing)
  Ｉ                 Cast to string
                    Implicitly print
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2
  • \$\begingroup\$ Since there is an OEIS , and it starts from 1 you don't have to handle n=0 \$\endgroup\$
    – AZTECCO
    Dec 1, 2020 at 23:00
  • 1
    \$\begingroup\$ @AZTECCO As it happens my latest approach works for 0 without any special-casing! \$\endgroup\$
    – Neil
    Dec 3, 2020 at 10:49
2
\$\begingroup\$

Perl 5 -p, 34 bytes

++$\until grep"@F"==$\**$_,1..$_}{

Try it online!

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1
\$\begingroup\$

Python 3, 55 bytes

f=lambda n,r=1,i=1:r*(r**i==n)or f(n,r+(i>n),i>n or-~i)

Try it online!

My first golf in over a year! It's a bit longer than this answer, but doesn't use that nasty floating point. As many good code golf answers do, this hits the recursion limit pretty soon.

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2
  • \$\begingroup\$ place f = lambda it is better to write` lambda` and put f= in the header. tio.run/… \$\endgroup\$
    – Danis
    Dec 23, 2020 at 6:01
  • 2
    \$\begingroup\$ I don't think that's allowed in this case unfortunately, since I use f recursively in the program \$\endgroup\$
    – ArBo
    Dec 23, 2020 at 9:22
1
\$\begingroup\$

Python 3, 74 bytes

lambda n,r=round:r(n**[1/i for i in range(1,n+1)if r(n**(1/i))**i==n][-1])

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