Circular primes

Question

Given a number determine if it is a circular prime.

A circular prime is a prime where if you rotate the digits you still get prime numbers. 1193 is circular prime because 1931 and 9311 and 3119 are all primes.

Input:          Output:
1193            circular
11939           circular
1456            nothing
193939          circular
1111111         nothing
7               circular
23              nothing

Shortest code wins.

Not a big deal, but they're sort of disjoint. codegolf.stackexchange.com/tags/code-challenge/info "A code challenge is a competition for creative ways to solve a programming puzzle for an objective criterion other than code size." — Cruncher, Nov 12, 2013 at 19:36
-1 You stated: "Shortest code wins" but awarded the check to a submission that was 4 times as long as the shortest submitted! — DavidC, Nov 14, 2013 at 19:50
@DavidCarraher Some problems on this site require specific output. This is one. Deal with it and stop whining. "Oh, poor me, my output 99,98,97... is equivalent to the lyrics of '99 bottles of beer on the wall' but it wasn't an accepted answer!" — boothby, Nov 14, 2013 at 22:13
@Boothby. Bad example. "99 bottles of beer on the wall" is all about crafting a specific output, the lyrics of a song. The present example is advertised a test of circularity of a number: "Given a number determine if it is a circular prime." Example output was given. But it was not stated (nor should it have been, in my view) that that specific format is required). — DavidC, Nov 14, 2013 at 22:21

rationalis · Accepted Answer · 2014-08-14 08:38:05Z

8

J, 47 chars (w/ formatted output)

('nothing';'circular'){~*/1&p:".(|."0 1~i.&#)":

J, 23 chars (boolean only)

*/1&p:".(|."0 1~i.&#)":

": is a string conversion; i.&# produces a range of integers [0,len), and |."0 1~ is a rotation of the string by each successive integer of the range.

". converts the list of rotations back into numbers, 1&p: converts the list of numbers into booleans (i.e. "prime?" predicate), and */ is a multiply reduce over the booleans (i.e. and).

edited Aug 14, 2014 at 8:38

answered Nov 13, 2013 at 5:29

rationalis

4962 silver badges8 bronze badges

\$\begingroup\$ Wow I really like this :) nicely done \$\endgroup\$
– Quillion
Nov 13, 2013 at 13:56
\$\begingroup\$ Add an {'nothing';'circular' and beat the Python submission :) \$\endgroup\$
– marinus
Nov 14, 2013 at 23:46
1

\$\begingroup\$ @marinus Done, though it really hurts my sense of aesthetics... The interface between the human and the program is always the least fun to code. \$\endgroup\$
– rationalis
Nov 15, 2013 at 4:03
\$\begingroup\$ Months after the fact, I finally just realized I could replace [:i.# with i.&#, which was bugging me at the time... L'esprit_de_l'escalier. Take that, GolfScript! \$\endgroup\$
– rationalis
Aug 14, 2014 at 8:41

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Daniel Lubarov · Accepted Answer · 2013-11-14 19:47:02Z

7

Python, 106 chars

p=input();n=`p`
for i in n:
 n=n[1:]+n[0];e=2
 while`e`!=n:p*=int(n)%e;e+=1
print'cniortchuilnagr'[p<1::2]

edited Nov 14, 2013 at 19:47

answered Nov 13, 2013 at 0:19

Daniel Lubarov

3211 silver badge9 bronze badges

1

\$\begingroup\$ I see what you did there with the answer :) very clever \$\endgroup\$
– Quillion
Nov 13, 2013 at 15:29
2

\$\begingroup\$ Oh man! I just did mine in Python and I came up with "cniortchuilnagr" independently. \$\endgroup\$
– danmcardle
Nov 14, 2013 at 3:32
1

\$\begingroup\$ Whoa! I got the idea from codegolf.stackexchange.com/a/406/9395 \$\endgroup\$
– Daniel Lubarov
Nov 14, 2013 at 3:38
3

\$\begingroup\$ You can make this both shorter and correct if you use for i in n... otherwise, this will fail to rule out certain primes with more than 10 digits. \$\endgroup\$
– boothby
Nov 14, 2013 at 19:02

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Community · Accepted Answer · 2020-06-17 09:04:33Z

7

Mathematica 62 61

f@s_ := And @@ PrimeQ[FromDigits@StringRotateLeft[s, #] & /@ Range@9]

Usage

f["1193"]

True

f["11939"]

True

f["1456"]

False

f["193939"]

True

f["1111111"]

False

f["7"]

True

f["23"]

False

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 13, 2013 at 2:24

DavidC

25.3k2 gold badges51 silver badges105 bronze badges

1

\$\begingroup\$ You can use FromDigits instead of ToExpression. \$\endgroup\$
– alephalpha
Nov 13, 2013 at 2:27
\$\begingroup\$ I had no idea that FromDigits could handle strings! \$\endgroup\$
– DavidC
Nov 13, 2013 at 2:34

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Marcks Thomas · Accepted Answer · 2013-11-13 13:09:02Z

5

Mathematica, 75 64

And@@PrimeQ[FromDigits/@NestList[RotateLeft,IntegerDigits@#,9]]&

edited Nov 13, 2013 at 13:09

answered Nov 12, 2013 at 23:05

Marcks Thomas

1614 bronze badges

\$\begingroup\$ You can save a few characters by using RotateRight or RotateLeft to rotate the digits. \$\endgroup\$
– DavidC
Nov 13, 2013 at 0:45
\$\begingroup\$ @DavidCarraher: Thanks, it saves quite a few, actually. How at one point I must have thought my elaborate Part trick would be shorter, I find most puzzling, but even now, you're still a good three characters ahead. \$\endgroup\$
– Marcks Thomas
Nov 13, 2013 at 13:13

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Doorknob · Accepted Answer · 2013-11-15 13:17:28Z

5

Ruby, 110 109 chars

require'mathn'
x=gets.chop.split''
puts (0..x.size).all?{|n|x.rotate(n).join.to_i.prime?}?:circular:'nothing'

edited Nov 15, 2013 at 13:17

answered Nov 12, 2013 at 21:55

Doorknob

71.3k19 gold badges141 silver badges382 bronze badges

1

\$\begingroup\$ You can shorten 1 character by using Symbol instead of String for :circular. \$\endgroup\$
– manatwork
Nov 15, 2013 at 9:31

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Community · Accepted Answer · 2020-06-17 09:04:33Z

4

GolfScript, 47 characters

:s,,{)s/(+~:P{(.P\%}do(},!"nothing
circular"n/=

Input must be given on STDIN without trailing newline (online example).

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 15, 2013 at 8:53

Howard

23.4k2 gold badges43 silver badges81 bronze badges

\$\begingroup\$ It's beautiful :) \$\endgroup\$
– Quillion
Nov 15, 2013 at 15:04

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plannapus · Accepted Answer · 2013-11-13 08:56:52Z

3

R, 139

a=scan()
p=0
A=nchar(a)
S=substr
for(i in 1:A){b=as.integer(paste0(S(a,i+1,A),S(a,1,i)));p=p+!sum(!b%%1:b)<3}
c('nothing','circular')[!p+1]

answered Nov 13, 2013 at 8:56

plannapus

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Community · Accepted Answer · 2020-06-17 09:04:33Z

3

Python3.3, 160C

def m():
 s=input()
 r=range
 for i in r(len(s)):
  a=int(s)
  if all(a%i for i in r(2,a)):return 0
  s=s[1:]+s[:1]
 return 1 
print("cniortchuilnagr"[m()::2])

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 14, 2013 at 3:31

danmcardle

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protist · Accepted Answer · 2013-11-25 22:08:21Z

3

APL, 53

'nothing' 'circular'[1+^/{⌊/⍵|⍨1↓⍳⍵-1}¨⍎¨{⍵⌽∆}¨⍳⍴∆←⍞]

answered Nov 25, 2013 at 22:08

protist

6103 silver badges7 bronze badges

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a spaghetto · Accepted Answer · 2015-12-23 23:17:20Z

3

Smalltalk, 195 chars

[:x|r:=[:n|(n\\10)*(10 raisedTo:n log floor)+(n//10)].s:=OrderedCollection with:x.[(m:=r value:s last)=x]whileFalse:[s add:m].^(s allSatisfy:[:each|each isPrime])ifTrue:['circular']ifFalse:['nothing']]

formatted

[:x|
  r:=[:n|(n\\10)*(10 raisedTo:n log floor)+(n//10)].
  s:=OrderedCollection with:x.
  [(m:=r value:s last)=x]whileFalse:[s add:m].
  ^(s allSatisfy:[:each|each isPrime])ifTrue:['circular']ifFalse:['nothing']]

edited Dec 23, 2015 at 23:17

a spaghetto

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answered Nov 20, 2013 at 7:51

Paul Richter

7905 silver badges6 bronze badges

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Shaggy · Accepted Answer · 2020-10-10 21:05:02Z

Japt, 23 bytes

Would obviously be a hell of a lot shorter without the unnecessarily strict output requirement, which accounts for more than half the score here.

ÆìéX jÃ×g`Í(ˆg circÔ³`¸

Try it (includes all test cases)

ÆìéX jÃ×g`...`¸     :Implicit input of integer U
Æ                   :Map each X in the range [0,U)
 ì                  :   Convert U to digit array
  éX                :   Rotate right X times and convert back to integer
     j              :  Is prime?
      Ã             :End map
       ×            :Reduce by multiplication
        g           :Index into (0-based)
         `...`      :  Compressed string "circular nothing"
              ¸     :  Split on spaces

Razetime · Accepted Answer · 2020-10-11 09:37:58Z

2

Husk, ²² 19 bytes

?¨ÿċ◊¨¨ø±¨ΛoṗdSṀṙŀd

Try it online!

-3 bytes from Jo King. (simplified using Ṁ)

edited Oct 11, 2020 at 9:37

answered Oct 10, 2020 at 16:35

Razetime

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\$\begingroup\$ You can simplify that last bit of logic for 19 bytes \$\endgroup\$
– Jo King ♦
Oct 11, 2020 at 8:43

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SunnyMoon · Accepted Answer · 2020-10-11 13:49:28Z

05AB1E, 7 6 bytes

^{-1 thanks to @ovs!}

gL._pP

Outputs 1 if a circular prime, and 0 if not! :D

Try it online!

This is actually my first effortful code answer in this site! Yey!

How?

g         # Get the number of digits in the input number.
 L        # Push a list of all numbers from 1 to the length.
  ._      # For each number 'n' in the list, rotate the input to to the left n times and push a list of all outcomes.
    p     # Is each rotation a prime? If prime, convert that number to 1, else convert it to 0.
     P    # Push the product of all 1's or 0's (if a rotation is not prime, the product should be zero).
          # Print the product automatically.

05AB1E, 17 16 bytes

^{Again -1 thanks to @ovs!}

gL._pP“‹ë“si“ÒŽ“

This code follows the output standard suggested by the test cases. I wrote this just in case my previous answer is invalid. If it is not, please count 6B as my score!

Try it online!

How again?

gL._pP              # Works just like the previous answer.
      “‹ë“          # Push the string "nothing" from the 05AB1E dictionary.
          s         # Bring back the number output again.
           i        # If the number is 1...
            “ÒŽ“    # Push the string "circular" again from the 05AB1E dictionary.

Thanks to @Kevin Cruijssen's tip on compressing strings! :)

Edit: Wow, I noticed that both of my answers are currently the shortest (during the time of writing). That is kind of ironic for a first effortful answer. That means I won!

My interpretation of the spec was the same as yours until I looked through the solutions. Pretty sure now that we have to output the words in the spec and the comments on the challenge would seem to back that up. — Shaggy, Oct 10, 2020 at 19:36
@Shaggy Yea, but a Mathematica answer claims that it outputs "True" and "False" for output. — SunnyMoon, Oct 10, 2020 at 19:49
If you rotate the input by the input length, you still get the input. This means gL works instead of g<Ý. And for this the builtin ā (length range) exists. — ovs, Oct 11, 2020 at 7:47

Stack Exchange Network

Circular primes

13 Answers 13

J, 47 chars (w/ formatted output)

J, 23 chars (boolean only)

Python, 106 chars

Mathematica 62 61

Mathematica, 75 64

Ruby, 110 109 chars

GolfScript, 47 characters

R, 139

Python3.3, 160C

APL, 53

Smalltalk, 195 chars

Japt, 23 bytes

Husk, ²² 19 bytes

05AB1E, 7 6 bytes

How?

05AB1E, 17 16 bytes

How again?

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primes
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Circular primes

13 Answers 13

J, 47 chars (w/ formatted output)

J, 23 chars (boolean only)

Python, 106 chars

Mathematica 62 61

Mathematica, 75 64

Ruby, 110 109 chars

GolfScript, 47 characters

R, 139

Python3.3, 160C

APL, 53

Smalltalk, 195 chars

Japt, 23 bytes

Husk, 22 19 bytes

05AB1E, 7 6 bytes

How?

05AB1E, 17 16 bytes

How again?

Your Answer

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Not the answer you're looking for? Browse other questions tagged code-golfprimes or ask your own question.

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Husk, ²² 19 bytes

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primes
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