Is it a Cyclops number? “Nobody” knows!












1












$begingroup$


Task:



Given an integer input, figure out whether or not it is a Cyclops Number.



What is a Cyclops number, you may ask? Well, it's a number whose binary representation only has one 0 in the center!



Test Cases:



Input | Output

--------------

1     | falsy

5     | truthy

12    | falsy    

27    | truthy

85    | falsy

101   | falsy

119   | truthy


Input: An integer or equivalent types. (int, long, decimal, etc.)
Output:




  • Truthy or falsy (e.g. true or false, 0 or 1)..


Challenge Rules:




  • Input that is less than 0 is assumed to be falsy.


  • If the length of the binary representation of the number is even, then the number cannot be a Cyclops number.



General Rules:




  • This is code-golf, so the shortest answers in bytes wins!.


  • Default loopholes are forbidden.


  • Standard rules apply for your answer with default I/O rules.




This is my first Programming Puzzles & Code Golf challenge, so any feedback on how I should improve would be much appreciated!






code-golf number decision-problem binary







share|improve this question









New contributor




Tau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    You've essentially made two separate challenges, one a decision problem and one about outputting the next number of a sequence. This will not do what you want, which is invite more answers, but instead put off users who now need to consider three options about what to program before posting. I'd recommend removing the option, and in the future you can try posting to our sandbox first where hopefully you will get helpful feedback before posting. Good luck!
    $endgroup$
    – FryAmTheEggman
    2 hours ago










  • $begingroup$
    @SriotchilismO'Zaic thanks for the feedback! I have edited the post.
    $endgroup$
    – Tau
    1 hour ago






  • 1




    $begingroup$
    I'd also recommend removing the hard coding restriction. By default solutions would need to handle all cyclops numbers in theory, and there are clearly infinitely many of them, so they couldn't all be hardcoded.
    $endgroup$
    – FryAmTheEggman
    1 hour ago






  • 1




    $begingroup$
    Note: This is A129868
    $endgroup$
    – tsh
    1 hour ago






  • 2




    $begingroup$
    Should 119 be true?
    $endgroup$
    – Quintec
    1 hour ago
















1












$begingroup$


Task:



Given an integer input, figure out whether or not it is a Cyclops Number.



What is a Cyclops number, you may ask? Well, it's a number whose binary representation only has one 0 in the center!



Test Cases:



Input | Output

--------------

1     | falsy

5     | truthy

12    | falsy    

27    | truthy

85    | falsy

101   | falsy

119   | truthy


Input: An integer or equivalent types. (int, long, decimal, etc.)
Output:




  • Truthy or falsy (e.g. true or false, 0 or 1)..


Challenge Rules:




  • Input that is less than 0 is assumed to be falsy.


  • If the length of the binary representation of the number is even, then the number cannot be a Cyclops number.



General Rules:




  • This is code-golf, so the shortest answers in bytes wins!.


  • Default loopholes are forbidden.


  • Standard rules apply for your answer with default I/O rules.




This is my first Programming Puzzles & Code Golf challenge, so any feedback on how I should improve would be much appreciated!






code-golf number decision-problem binary







share|improve this question









New contributor




Tau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    You've essentially made two separate challenges, one a decision problem and one about outputting the next number of a sequence. This will not do what you want, which is invite more answers, but instead put off users who now need to consider three options about what to program before posting. I'd recommend removing the option, and in the future you can try posting to our sandbox first where hopefully you will get helpful feedback before posting. Good luck!
    $endgroup$
    – FryAmTheEggman
    2 hours ago










  • $begingroup$
    @SriotchilismO'Zaic thanks for the feedback! I have edited the post.
    $endgroup$
    – Tau
    1 hour ago






  • 1




    $begingroup$
    I'd also recommend removing the hard coding restriction. By default solutions would need to handle all cyclops numbers in theory, and there are clearly infinitely many of them, so they couldn't all be hardcoded.
    $endgroup$
    – FryAmTheEggman
    1 hour ago






  • 1




    $begingroup$
    Note: This is A129868
    $endgroup$
    – tsh
    1 hour ago






  • 2




    $begingroup$
    Should 119 be true?
    $endgroup$
    – Quintec
    1 hour ago














1












1








1





$begingroup$


Task:



Given an integer input, figure out whether or not it is a Cyclops Number.



What is a Cyclops number, you may ask? Well, it's a number whose binary representation only has one 0 in the center!



Test Cases:



Input | Output

--------------

1     | falsy

5     | truthy

12    | falsy    

27    | truthy

85    | falsy

101   | falsy

119   | truthy


Input: An integer or equivalent types. (int, long, decimal, etc.)
Output:




  • Truthy or falsy (e.g. true or false, 0 or 1)..


Challenge Rules:




  • Input that is less than 0 is assumed to be falsy.


  • If the length of the binary representation of the number is even, then the number cannot be a Cyclops number.



General Rules:




  • This is code-golf, so the shortest answers in bytes wins!.


  • Default loopholes are forbidden.


  • Standard rules apply for your answer with default I/O rules.




This is my first Programming Puzzles & Code Golf challenge, so any feedback on how I should improve would be much appreciated!






code-golf number decision-problem binary







share|improve this question









New contributor




Tau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Task:



Given an integer input, figure out whether or not it is a Cyclops Number.



What is a Cyclops number, you may ask? Well, it's a number whose binary representation only has one 0 in the center!



Test Cases:



Input | Output

--------------

1     | falsy

5     | truthy

12    | falsy    

27    | truthy

85    | falsy

101   | falsy

119   | truthy


Input: An integer or equivalent types. (int, long, decimal, etc.)
Output:




  • Truthy or falsy (e.g. true or false, 0 or 1)..


Challenge Rules:




  • Input that is less than 0 is assumed to be falsy.


  • If the length of the binary representation of the number is even, then the number cannot be a Cyclops number.



General Rules:




  • This is code-golf, so the shortest answers in bytes wins!.


  • Default loopholes are forbidden.


  • Standard rules apply for your answer with default I/O rules.




This is my first Programming Puzzles & Code Golf challenge, so any feedback on how I should improve would be much appreciated!






code-golf number decision-problem binary




code-golf number decision-problem binary






share|improve this question









New contributor




Tau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Tau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 1 hour ago







Tau













New contributor




Tau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 hours ago









TauTau

1165




1165




New contributor




Tau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Tau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Tau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    You've essentially made two separate challenges, one a decision problem and one about outputting the next number of a sequence. This will not do what you want, which is invite more answers, but instead put off users who now need to consider three options about what to program before posting. I'd recommend removing the option, and in the future you can try posting to our sandbox first where hopefully you will get helpful feedback before posting. Good luck!
    $endgroup$
    – FryAmTheEggman
    2 hours ago










  • $begingroup$
    @SriotchilismO'Zaic thanks for the feedback! I have edited the post.
    $endgroup$
    – Tau
    1 hour ago






  • 1




    $begingroup$
    I'd also recommend removing the hard coding restriction. By default solutions would need to handle all cyclops numbers in theory, and there are clearly infinitely many of them, so they couldn't all be hardcoded.
    $endgroup$
    – FryAmTheEggman
    1 hour ago






  • 1




    $begingroup$
    Note: This is A129868
    $endgroup$
    – tsh
    1 hour ago






  • 2




    $begingroup$
    Should 119 be true?
    $endgroup$
    – Quintec
    1 hour ago














  • 1




    $begingroup$
    You've essentially made two separate challenges, one a decision problem and one about outputting the next number of a sequence. This will not do what you want, which is invite more answers, but instead put off users who now need to consider three options about what to program before posting. I'd recommend removing the option, and in the future you can try posting to our sandbox first where hopefully you will get helpful feedback before posting. Good luck!
    $endgroup$
    – FryAmTheEggman
    2 hours ago










  • $begingroup$
    @SriotchilismO'Zaic thanks for the feedback! I have edited the post.
    $endgroup$
    – Tau
    1 hour ago






  • 1




    $begingroup$
    I'd also recommend removing the hard coding restriction. By default solutions would need to handle all cyclops numbers in theory, and there are clearly infinitely many of them, so they couldn't all be hardcoded.
    $endgroup$
    – FryAmTheEggman
    1 hour ago






  • 1




    $begingroup$
    Note: This is A129868
    $endgroup$
    – tsh
    1 hour ago






  • 2




    $begingroup$
    Should 119 be true?
    $endgroup$
    – Quintec
    1 hour ago








1




1




$begingroup$
You've essentially made two separate challenges, one a decision problem and one about outputting the next number of a sequence. This will not do what you want, which is invite more answers, but instead put off users who now need to consider three options about what to program before posting. I'd recommend removing the option, and in the future you can try posting to our sandbox first where hopefully you will get helpful feedback before posting. Good luck!
$endgroup$
– FryAmTheEggman
2 hours ago




$begingroup$
You've essentially made two separate challenges, one a decision problem and one about outputting the next number of a sequence. This will not do what you want, which is invite more answers, but instead put off users who now need to consider three options about what to program before posting. I'd recommend removing the option, and in the future you can try posting to our sandbox first where hopefully you will get helpful feedback before posting. Good luck!
$endgroup$
– FryAmTheEggman
2 hours ago












$begingroup$
@SriotchilismO'Zaic thanks for the feedback! I have edited the post.
$endgroup$
– Tau
1 hour ago




$begingroup$
@SriotchilismO'Zaic thanks for the feedback! I have edited the post.
$endgroup$
– Tau
1 hour ago




1




1




$begingroup$
I'd also recommend removing the hard coding restriction. By default solutions would need to handle all cyclops numbers in theory, and there are clearly infinitely many of them, so they couldn't all be hardcoded.
$endgroup$
– FryAmTheEggman
1 hour ago




$begingroup$
I'd also recommend removing the hard coding restriction. By default solutions would need to handle all cyclops numbers in theory, and there are clearly infinitely many of them, so they couldn't all be hardcoded.
$endgroup$
– FryAmTheEggman
1 hour ago




1




1




$begingroup$
Note: This is A129868
$endgroup$
– tsh
1 hour ago




$begingroup$
Note: This is A129868
$endgroup$
– tsh
1 hour ago




2




2




$begingroup$
Should 119 be true?
$endgroup$
– Quintec
1 hour ago




$begingroup$
Should 119 be true?
$endgroup$
– Quintec
1 hour ago










4 Answers
4






active

oldest

votes


















2












$begingroup$


Japt, 25 19 10 bytes



¢èT ¶1©¢ês


That's more like it...



Checks if there is only one 0 in the binary representation and that the binary representation is a palindrome.



Try it online!






share|improve this answer











$endgroup$













  • $begingroup$
    found an even shorter way, based on your solution :)
    $endgroup$
    – Embodiment of Ignorance
    1 hour ago





















2












$begingroup$

Japt, 8 bytes



¤øT ©¢ês


Based off of Quintec's solution.



Try it Online!



Old regex-based solution, 15 bytes



¤f/^(1*)01$/ l


Returns 1 for true, 0 for false.



Try it Online!






share|improve this answer











$endgroup$













  • $begingroup$
    Well played, I should really learn regular expressions sometime. :) +1
    $endgroup$
    – Quintec
    1 hour ago










  • $begingroup$
    @Quintec Regex is awesome :)
    $endgroup$
    – Embodiment of Ignorance
    1 hour ago










  • $begingroup$
    Update: found shorter way :)
    $endgroup$
    – Quintec
    1 hour ago



















0












$begingroup$


Perl 6, 23 bytes





{.base(2)~~/^(1*)0$0$/}


Try it online!



Regex based solution






share|improve this answer









$endgroup$





















    0












    $begingroup$


    JavaScript (Node.js), 32 bytes





    f=(p,q)=>p&1?f(p/2,q+q|2):!(p^q)


    Try it online!



    31 bytes version





    f=(p,q)=>p&1?f(p>>1,q+q|2):p==q


    This one will report falsy for 0.





    JavaScript (Node.js), 35 bytes





    p=>p.toString(2).match(/^(1*)01$/)


    Try it online!






    share|improve this answer











    $endgroup$













      Your Answer





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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$


      Japt, 25 19 10 bytes



      ¢èT ¶1©¢ês


      That's more like it...



      Checks if there is only one 0 in the binary representation and that the binary representation is a palindrome.



      Try it online!






      share|improve this answer











      $endgroup$













      • $begingroup$
        found an even shorter way, based on your solution :)
        $endgroup$
        – Embodiment of Ignorance
        1 hour ago


















      2












      $begingroup$


      Japt, 25 19 10 bytes



      ¢èT ¶1©¢ês


      That's more like it...



      Checks if there is only one 0 in the binary representation and that the binary representation is a palindrome.



      Try it online!






      share|improve this answer











      $endgroup$













      • $begingroup$
        found an even shorter way, based on your solution :)
        $endgroup$
        – Embodiment of Ignorance
        1 hour ago
















      2












      2








      2





      $begingroup$


      Japt, 25 19 10 bytes



      ¢èT ¶1©¢ês


      That's more like it...



      Checks if there is only one 0 in the binary representation and that the binary representation is a palindrome.



      Try it online!






      share|improve this answer











      $endgroup$




      Japt, 25 19 10 bytes



      ¢èT ¶1©¢ês


      That's more like it...



      Checks if there is only one 0 in the binary representation and that the binary representation is a palindrome.



      Try it online!







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 1 hour ago

























      answered 1 hour ago









      QuintecQuintec

      1,9781726




      1,9781726












      • $begingroup$
        found an even shorter way, based on your solution :)
        $endgroup$
        – Embodiment of Ignorance
        1 hour ago




















      • $begingroup$
        found an even shorter way, based on your solution :)
        $endgroup$
        – Embodiment of Ignorance
        1 hour ago


















      $begingroup$
      found an even shorter way, based on your solution :)
      $endgroup$
      – Embodiment of Ignorance
      1 hour ago






      $begingroup$
      found an even shorter way, based on your solution :)
      $endgroup$
      – Embodiment of Ignorance
      1 hour ago













      2












      $begingroup$

      Japt, 8 bytes



      ¤øT ©¢ês


      Based off of Quintec's solution.



      Try it Online!



      Old regex-based solution, 15 bytes



      ¤f/^(1*)01$/ l


      Returns 1 for true, 0 for false.



      Try it Online!






      share|improve this answer











      $endgroup$













      • $begingroup$
        Well played, I should really learn regular expressions sometime. :) +1
        $endgroup$
        – Quintec
        1 hour ago










      • $begingroup$
        @Quintec Regex is awesome :)
        $endgroup$
        – Embodiment of Ignorance
        1 hour ago










      • $begingroup$
        Update: found shorter way :)
        $endgroup$
        – Quintec
        1 hour ago
















      2












      $begingroup$

      Japt, 8 bytes



      ¤øT ©¢ês


      Based off of Quintec's solution.



      Try it Online!



      Old regex-based solution, 15 bytes



      ¤f/^(1*)01$/ l


      Returns 1 for true, 0 for false.



      Try it Online!






      share|improve this answer











      $endgroup$













      • $begingroup$
        Well played, I should really learn regular expressions sometime. :) +1
        $endgroup$
        – Quintec
        1 hour ago










      • $begingroup$
        @Quintec Regex is awesome :)
        $endgroup$
        – Embodiment of Ignorance
        1 hour ago










      • $begingroup$
        Update: found shorter way :)
        $endgroup$
        – Quintec
        1 hour ago














      2












      2








      2





      $begingroup$

      Japt, 8 bytes



      ¤øT ©¢ês


      Based off of Quintec's solution.



      Try it Online!



      Old regex-based solution, 15 bytes



      ¤f/^(1*)01$/ l


      Returns 1 for true, 0 for false.



      Try it Online!






      share|improve this answer











      $endgroup$



      Japt, 8 bytes



      ¤øT ©¢ês


      Based off of Quintec's solution.



      Try it Online!



      Old regex-based solution, 15 bytes



      ¤f/^(1*)01$/ l


      Returns 1 for true, 0 for false.



      Try it Online!







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 1 hour ago

























      answered 1 hour ago









      Embodiment of IgnoranceEmbodiment of Ignorance

      1,516123




      1,516123












      • $begingroup$
        Well played, I should really learn regular expressions sometime. :) +1
        $endgroup$
        – Quintec
        1 hour ago










      • $begingroup$
        @Quintec Regex is awesome :)
        $endgroup$
        – Embodiment of Ignorance
        1 hour ago










      • $begingroup$
        Update: found shorter way :)
        $endgroup$
        – Quintec
        1 hour ago


















      • $begingroup$
        Well played, I should really learn regular expressions sometime. :) +1
        $endgroup$
        – Quintec
        1 hour ago










      • $begingroup$
        @Quintec Regex is awesome :)
        $endgroup$
        – Embodiment of Ignorance
        1 hour ago










      • $begingroup$
        Update: found shorter way :)
        $endgroup$
        – Quintec
        1 hour ago
















      $begingroup$
      Well played, I should really learn regular expressions sometime. :) +1
      $endgroup$
      – Quintec
      1 hour ago




      $begingroup$
      Well played, I should really learn regular expressions sometime. :) +1
      $endgroup$
      – Quintec
      1 hour ago












      $begingroup$
      @Quintec Regex is awesome :)
      $endgroup$
      – Embodiment of Ignorance
      1 hour ago




      $begingroup$
      @Quintec Regex is awesome :)
      $endgroup$
      – Embodiment of Ignorance
      1 hour ago












      $begingroup$
      Update: found shorter way :)
      $endgroup$
      – Quintec
      1 hour ago




      $begingroup$
      Update: found shorter way :)
      $endgroup$
      – Quintec
      1 hour ago











      0












      $begingroup$


      Perl 6, 23 bytes





      {.base(2)~~/^(1*)0$0$/}


      Try it online!



      Regex based solution






      share|improve this answer









      $endgroup$


















        0












        $begingroup$


        Perl 6, 23 bytes





        {.base(2)~~/^(1*)0$0$/}


        Try it online!



        Regex based solution






        share|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$


          Perl 6, 23 bytes





          {.base(2)~~/^(1*)0$0$/}


          Try it online!



          Regex based solution






          share|improve this answer









          $endgroup$




          Perl 6, 23 bytes





          {.base(2)~~/^(1*)0$0$/}


          Try it online!



          Regex based solution







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 1 hour ago









          Jo KingJo King

          24.4k357126




          24.4k357126























              0












              $begingroup$


              JavaScript (Node.js), 32 bytes





              f=(p,q)=>p&1?f(p/2,q+q|2):!(p^q)


              Try it online!



              31 bytes version





              f=(p,q)=>p&1?f(p>>1,q+q|2):p==q


              This one will report falsy for 0.





              JavaScript (Node.js), 35 bytes





              p=>p.toString(2).match(/^(1*)01$/)


              Try it online!






              share|improve this answer











              $endgroup$


















                0












                $begingroup$


                JavaScript (Node.js), 32 bytes





                f=(p,q)=>p&1?f(p/2,q+q|2):!(p^q)


                Try it online!



                31 bytes version





                f=(p,q)=>p&1?f(p>>1,q+q|2):p==q


                This one will report falsy for 0.





                JavaScript (Node.js), 35 bytes





                p=>p.toString(2).match(/^(1*)01$/)


                Try it online!






                share|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$


                  JavaScript (Node.js), 32 bytes





                  f=(p,q)=>p&1?f(p/2,q+q|2):!(p^q)


                  Try it online!



                  31 bytes version





                  f=(p,q)=>p&1?f(p>>1,q+q|2):p==q


                  This one will report falsy for 0.





                  JavaScript (Node.js), 35 bytes





                  p=>p.toString(2).match(/^(1*)01$/)


                  Try it online!






                  share|improve this answer











                  $endgroup$




                  JavaScript (Node.js), 32 bytes





                  f=(p,q)=>p&1?f(p/2,q+q|2):!(p^q)


                  Try it online!



                  31 bytes version





                  f=(p,q)=>p&1?f(p>>1,q+q|2):p==q


                  This one will report falsy for 0.





                  JavaScript (Node.js), 35 bytes





                  p=>p.toString(2).match(/^(1*)01$/)


                  Try it online!







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                  share|improve this answer



                  share|improve this answer








                  edited 48 mins ago

























                  answered 1 hour ago









                  tshtsh

                  9,32511652




                  9,32511652






















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