Am I a Rude Number?












3












$begingroup$


For a while now, I've been running into a problem when counting on my fingers, specifically, that I can only count to ten. My solution to that problem has been to count in binary on my fingers, putting up my thumb for one, my forefinger for two, both thumb and forefinger for three, etc. However, we run into a bit of a problem when we get to the number four. Specifically, it requires us to put up our middle finger, which results in a rather unfortunate gesture, which is not typically accepted in society. This type of number is a rude number. We come to the next rude number at 36, when we raise the thumb on our second hand and the middle finger of our first hand. The definition of a rude number is any number that, under this system of counting, results in us putting up only the middle finger of any hand. Once we pass 1023 (the maximum number reachable on one hand), assume we continue with a third hand, with additional hands added as required.



Your Task:



Write a program or function that receives an input and outputs a truthy/falsy value based on whether the input is a rude number.



Input:



An integer between 0 and 10^9 (inclusive).



Output:



A truthy/falsy value that indicates whether the input is a rude number.



Test Cases:



Input:    Output:
0 ---> falsy
3 ---> falsy
4 ---> truthy
25 ---> falsy
36 ---> truthy
127 ---> falsy
131 ---> truthy


Scoring:



This is code-golf, so the lowest score in bytes wins.










share|improve this question











$endgroup$








  • 2




    $begingroup$
    assume we continue with a third hand, When it comes to being rude, teamwork makes the dream work.
    $endgroup$
    – Veskah
    1 hour ago






  • 1




    $begingroup$
    @Veskah turns out that for the bounds of the question, you only need 3 people to make any given number. Sure beats the old kind of counting on fingers.
    $endgroup$
    – Gryphon
    1 hour ago










  • $begingroup$
    Also, I don't have the time at the moment, but if anyone could figure out an equation for this sequence, that'd be great.
    $endgroup$
    – Gryphon
    1 hour ago










  • $begingroup$
    You can reach 1023 on one hand? O_o
    $endgroup$
    – ASCII-only
    15 mins ago










  • $begingroup$
    BTW, pretty sure there is no short equation for this sequence, but you could (kinda) make a separate equation for each hand
    $endgroup$
    – ASCII-only
    3 mins ago
















3












$begingroup$


For a while now, I've been running into a problem when counting on my fingers, specifically, that I can only count to ten. My solution to that problem has been to count in binary on my fingers, putting up my thumb for one, my forefinger for two, both thumb and forefinger for three, etc. However, we run into a bit of a problem when we get to the number four. Specifically, it requires us to put up our middle finger, which results in a rather unfortunate gesture, which is not typically accepted in society. This type of number is a rude number. We come to the next rude number at 36, when we raise the thumb on our second hand and the middle finger of our first hand. The definition of a rude number is any number that, under this system of counting, results in us putting up only the middle finger of any hand. Once we pass 1023 (the maximum number reachable on one hand), assume we continue with a third hand, with additional hands added as required.



Your Task:



Write a program or function that receives an input and outputs a truthy/falsy value based on whether the input is a rude number.



Input:



An integer between 0 and 10^9 (inclusive).



Output:



A truthy/falsy value that indicates whether the input is a rude number.



Test Cases:



Input:    Output:
0 ---> falsy
3 ---> falsy
4 ---> truthy
25 ---> falsy
36 ---> truthy
127 ---> falsy
131 ---> truthy


Scoring:



This is code-golf, so the lowest score in bytes wins.










share|improve this question











$endgroup$








  • 2




    $begingroup$
    assume we continue with a third hand, When it comes to being rude, teamwork makes the dream work.
    $endgroup$
    – Veskah
    1 hour ago






  • 1




    $begingroup$
    @Veskah turns out that for the bounds of the question, you only need 3 people to make any given number. Sure beats the old kind of counting on fingers.
    $endgroup$
    – Gryphon
    1 hour ago










  • $begingroup$
    Also, I don't have the time at the moment, but if anyone could figure out an equation for this sequence, that'd be great.
    $endgroup$
    – Gryphon
    1 hour ago










  • $begingroup$
    You can reach 1023 on one hand? O_o
    $endgroup$
    – ASCII-only
    15 mins ago










  • $begingroup$
    BTW, pretty sure there is no short equation for this sequence, but you could (kinda) make a separate equation for each hand
    $endgroup$
    – ASCII-only
    3 mins ago














3












3








3





$begingroup$


For a while now, I've been running into a problem when counting on my fingers, specifically, that I can only count to ten. My solution to that problem has been to count in binary on my fingers, putting up my thumb for one, my forefinger for two, both thumb and forefinger for three, etc. However, we run into a bit of a problem when we get to the number four. Specifically, it requires us to put up our middle finger, which results in a rather unfortunate gesture, which is not typically accepted in society. This type of number is a rude number. We come to the next rude number at 36, when we raise the thumb on our second hand and the middle finger of our first hand. The definition of a rude number is any number that, under this system of counting, results in us putting up only the middle finger of any hand. Once we pass 1023 (the maximum number reachable on one hand), assume we continue with a third hand, with additional hands added as required.



Your Task:



Write a program or function that receives an input and outputs a truthy/falsy value based on whether the input is a rude number.



Input:



An integer between 0 and 10^9 (inclusive).



Output:



A truthy/falsy value that indicates whether the input is a rude number.



Test Cases:



Input:    Output:
0 ---> falsy
3 ---> falsy
4 ---> truthy
25 ---> falsy
36 ---> truthy
127 ---> falsy
131 ---> truthy


Scoring:



This is code-golf, so the lowest score in bytes wins.










share|improve this question











$endgroup$




For a while now, I've been running into a problem when counting on my fingers, specifically, that I can only count to ten. My solution to that problem has been to count in binary on my fingers, putting up my thumb for one, my forefinger for two, both thumb and forefinger for three, etc. However, we run into a bit of a problem when we get to the number four. Specifically, it requires us to put up our middle finger, which results in a rather unfortunate gesture, which is not typically accepted in society. This type of number is a rude number. We come to the next rude number at 36, when we raise the thumb on our second hand and the middle finger of our first hand. The definition of a rude number is any number that, under this system of counting, results in us putting up only the middle finger of any hand. Once we pass 1023 (the maximum number reachable on one hand), assume we continue with a third hand, with additional hands added as required.



Your Task:



Write a program or function that receives an input and outputs a truthy/falsy value based on whether the input is a rude number.



Input:



An integer between 0 and 10^9 (inclusive).



Output:



A truthy/falsy value that indicates whether the input is a rude number.



Test Cases:



Input:    Output:
0 ---> falsy
3 ---> falsy
4 ---> truthy
25 ---> falsy
36 ---> truthy
127 ---> falsy
131 ---> truthy


Scoring:



This is code-golf, so the lowest score in bytes wins.







code-golf number decision-problem






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 42 mins ago







Gryphon

















asked 1 hour ago









GryphonGryphon

3,1891963




3,1891963








  • 2




    $begingroup$
    assume we continue with a third hand, When it comes to being rude, teamwork makes the dream work.
    $endgroup$
    – Veskah
    1 hour ago






  • 1




    $begingroup$
    @Veskah turns out that for the bounds of the question, you only need 3 people to make any given number. Sure beats the old kind of counting on fingers.
    $endgroup$
    – Gryphon
    1 hour ago










  • $begingroup$
    Also, I don't have the time at the moment, but if anyone could figure out an equation for this sequence, that'd be great.
    $endgroup$
    – Gryphon
    1 hour ago










  • $begingroup$
    You can reach 1023 on one hand? O_o
    $endgroup$
    – ASCII-only
    15 mins ago










  • $begingroup$
    BTW, pretty sure there is no short equation for this sequence, but you could (kinda) make a separate equation for each hand
    $endgroup$
    – ASCII-only
    3 mins ago














  • 2




    $begingroup$
    assume we continue with a third hand, When it comes to being rude, teamwork makes the dream work.
    $endgroup$
    – Veskah
    1 hour ago






  • 1




    $begingroup$
    @Veskah turns out that for the bounds of the question, you only need 3 people to make any given number. Sure beats the old kind of counting on fingers.
    $endgroup$
    – Gryphon
    1 hour ago










  • $begingroup$
    Also, I don't have the time at the moment, but if anyone could figure out an equation for this sequence, that'd be great.
    $endgroup$
    – Gryphon
    1 hour ago










  • $begingroup$
    You can reach 1023 on one hand? O_o
    $endgroup$
    – ASCII-only
    15 mins ago










  • $begingroup$
    BTW, pretty sure there is no short equation for this sequence, but you could (kinda) make a separate equation for each hand
    $endgroup$
    – ASCII-only
    3 mins ago








2




2




$begingroup$
assume we continue with a third hand, When it comes to being rude, teamwork makes the dream work.
$endgroup$
– Veskah
1 hour ago




$begingroup$
assume we continue with a third hand, When it comes to being rude, teamwork makes the dream work.
$endgroup$
– Veskah
1 hour ago




1




1




$begingroup$
@Veskah turns out that for the bounds of the question, you only need 3 people to make any given number. Sure beats the old kind of counting on fingers.
$endgroup$
– Gryphon
1 hour ago




$begingroup$
@Veskah turns out that for the bounds of the question, you only need 3 people to make any given number. Sure beats the old kind of counting on fingers.
$endgroup$
– Gryphon
1 hour ago












$begingroup$
Also, I don't have the time at the moment, but if anyone could figure out an equation for this sequence, that'd be great.
$endgroup$
– Gryphon
1 hour ago




$begingroup$
Also, I don't have the time at the moment, but if anyone could figure out an equation for this sequence, that'd be great.
$endgroup$
– Gryphon
1 hour ago












$begingroup$
You can reach 1023 on one hand? O_o
$endgroup$
– ASCII-only
15 mins ago




$begingroup$
You can reach 1023 on one hand? O_o
$endgroup$
– ASCII-only
15 mins ago












$begingroup$
BTW, pretty sure there is no short equation for this sequence, but you could (kinda) make a separate equation for each hand
$endgroup$
– ASCII-only
3 mins ago




$begingroup$
BTW, pretty sure there is no short equation for this sequence, but you could (kinda) make a separate equation for each hand
$endgroup$
– ASCII-only
3 mins ago










5 Answers
5






active

oldest

votes


















3












$begingroup$


JavaScript (SpiderMonkey), 27 bytes





x=>/4/.test(x.toString(32))


Try it online!



This is a trivial solution, you just want to convert to base 32 and check if there is a 4 in it.






share|improve this answer









$endgroup$





















    2












    $begingroup$

    Ruby, 36 19 bytes





    ->n{n.to_s(32)[?4]}


    Try it online!



    Saved 17 bytes with @tsh's method.






    share|improve this answer











    $endgroup$













    • $begingroup$
      This returns true for 2207, which has a binary representation of 100010011111
      $endgroup$
      – Embodiment of Ignorance
      27 mins ago










    • $begingroup$
      @EmbodimentofIgnorance That is the correct result, is it not? The second hand is 00100.
      $endgroup$
      – Doorknob
      21 mins ago










    • $begingroup$
      I don't speak Ruby. But why not ->n{n.to_s(32)=~/4/}?
      $endgroup$
      – tsh
      18 mins ago












    • $begingroup$
      @tsh because I'm not as clever as you :)
      $endgroup$
      – Doorknob
      14 mins ago










    • $begingroup$
      Forgive me if I'm not understanding the question, but isn't the first hand of 2207 10001, the second 00111, and the third 11? None of them have their middle finger only up
      $endgroup$
      – Embodiment of Ignorance
      9 mins ago



















    2












    $begingroup$


    Perl 6, 16 bytes





    {.base(32)~~/4/}


    Try it online!



    Checks if there is a 4 in the base 32 representation of the number. Returns either Nil as false or a Match containing a 4.



    You can prove this by the fact that $2^5 = 32$ so each digit is the state of each hand.






    share|improve this answer









    $endgroup$





















      1












      $begingroup$

      Regex (ECMAScript), 37 bytes



      ^((?=(x+)(2{31}x*))3)*(x{32})*x{4}$



      Try it online!



      ^
      (
      (?=(x+)(2{31}x*)) # 2 = floor(tail / 32); 3 = tool to make tail = 2
      3 # tail = 2
      )* # Loop the above as many times as necessary to make
      # the below match
      (x{32})*x{4}$ # Assert that tail % 32 == 4





      share|improve this answer











      $endgroup$





















        0












        $begingroup$


        Japt, 5 bytes



        sH ø4


        Try it online!



        Explanation



              // Implicit input
        sH // To a base-H (=32) string
        ø // Contains
        4 // 4 (JavaScript interprets this as a string)





        share|improve this answer









        $endgroup$













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          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$


          JavaScript (SpiderMonkey), 27 bytes





          x=>/4/.test(x.toString(32))


          Try it online!



          This is a trivial solution, you just want to convert to base 32 and check if there is a 4 in it.






          share|improve this answer









          $endgroup$


















            3












            $begingroup$


            JavaScript (SpiderMonkey), 27 bytes





            x=>/4/.test(x.toString(32))


            Try it online!



            This is a trivial solution, you just want to convert to base 32 and check if there is a 4 in it.






            share|improve this answer









            $endgroup$
















              3












              3








              3





              $begingroup$


              JavaScript (SpiderMonkey), 27 bytes





              x=>/4/.test(x.toString(32))


              Try it online!



              This is a trivial solution, you just want to convert to base 32 and check if there is a 4 in it.






              share|improve this answer









              $endgroup$




              JavaScript (SpiderMonkey), 27 bytes





              x=>/4/.test(x.toString(32))


              Try it online!



              This is a trivial solution, you just want to convert to base 32 and check if there is a 4 in it.







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered 20 mins ago









              tshtsh

              9,15511650




              9,15511650























                  2












                  $begingroup$

                  Ruby, 36 19 bytes





                  ->n{n.to_s(32)[?4]}


                  Try it online!



                  Saved 17 bytes with @tsh's method.






                  share|improve this answer











                  $endgroup$













                  • $begingroup$
                    This returns true for 2207, which has a binary representation of 100010011111
                    $endgroup$
                    – Embodiment of Ignorance
                    27 mins ago










                  • $begingroup$
                    @EmbodimentofIgnorance That is the correct result, is it not? The second hand is 00100.
                    $endgroup$
                    – Doorknob
                    21 mins ago










                  • $begingroup$
                    I don't speak Ruby. But why not ->n{n.to_s(32)=~/4/}?
                    $endgroup$
                    – tsh
                    18 mins ago












                  • $begingroup$
                    @tsh because I'm not as clever as you :)
                    $endgroup$
                    – Doorknob
                    14 mins ago










                  • $begingroup$
                    Forgive me if I'm not understanding the question, but isn't the first hand of 2207 10001, the second 00111, and the third 11? None of them have their middle finger only up
                    $endgroup$
                    – Embodiment of Ignorance
                    9 mins ago
















                  2












                  $begingroup$

                  Ruby, 36 19 bytes





                  ->n{n.to_s(32)[?4]}


                  Try it online!



                  Saved 17 bytes with @tsh's method.






                  share|improve this answer











                  $endgroup$













                  • $begingroup$
                    This returns true for 2207, which has a binary representation of 100010011111
                    $endgroup$
                    – Embodiment of Ignorance
                    27 mins ago










                  • $begingroup$
                    @EmbodimentofIgnorance That is the correct result, is it not? The second hand is 00100.
                    $endgroup$
                    – Doorknob
                    21 mins ago










                  • $begingroup$
                    I don't speak Ruby. But why not ->n{n.to_s(32)=~/4/}?
                    $endgroup$
                    – tsh
                    18 mins ago












                  • $begingroup$
                    @tsh because I'm not as clever as you :)
                    $endgroup$
                    – Doorknob
                    14 mins ago










                  • $begingroup$
                    Forgive me if I'm not understanding the question, but isn't the first hand of 2207 10001, the second 00111, and the third 11? None of them have their middle finger only up
                    $endgroup$
                    – Embodiment of Ignorance
                    9 mins ago














                  2












                  2








                  2





                  $begingroup$

                  Ruby, 36 19 bytes





                  ->n{n.to_s(32)[?4]}


                  Try it online!



                  Saved 17 bytes with @tsh's method.






                  share|improve this answer











                  $endgroup$



                  Ruby, 36 19 bytes





                  ->n{n.to_s(32)[?4]}


                  Try it online!



                  Saved 17 bytes with @tsh's method.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 15 mins ago

























                  answered 1 hour ago









                  DoorknobDoorknob

                  54.9k17115352




                  54.9k17115352












                  • $begingroup$
                    This returns true for 2207, which has a binary representation of 100010011111
                    $endgroup$
                    – Embodiment of Ignorance
                    27 mins ago










                  • $begingroup$
                    @EmbodimentofIgnorance That is the correct result, is it not? The second hand is 00100.
                    $endgroup$
                    – Doorknob
                    21 mins ago










                  • $begingroup$
                    I don't speak Ruby. But why not ->n{n.to_s(32)=~/4/}?
                    $endgroup$
                    – tsh
                    18 mins ago












                  • $begingroup$
                    @tsh because I'm not as clever as you :)
                    $endgroup$
                    – Doorknob
                    14 mins ago










                  • $begingroup$
                    Forgive me if I'm not understanding the question, but isn't the first hand of 2207 10001, the second 00111, and the third 11? None of them have their middle finger only up
                    $endgroup$
                    – Embodiment of Ignorance
                    9 mins ago


















                  • $begingroup$
                    This returns true for 2207, which has a binary representation of 100010011111
                    $endgroup$
                    – Embodiment of Ignorance
                    27 mins ago










                  • $begingroup$
                    @EmbodimentofIgnorance That is the correct result, is it not? The second hand is 00100.
                    $endgroup$
                    – Doorknob
                    21 mins ago










                  • $begingroup$
                    I don't speak Ruby. But why not ->n{n.to_s(32)=~/4/}?
                    $endgroup$
                    – tsh
                    18 mins ago












                  • $begingroup$
                    @tsh because I'm not as clever as you :)
                    $endgroup$
                    – Doorknob
                    14 mins ago










                  • $begingroup$
                    Forgive me if I'm not understanding the question, but isn't the first hand of 2207 10001, the second 00111, and the third 11? None of them have their middle finger only up
                    $endgroup$
                    – Embodiment of Ignorance
                    9 mins ago
















                  $begingroup$
                  This returns true for 2207, which has a binary representation of 100010011111
                  $endgroup$
                  – Embodiment of Ignorance
                  27 mins ago




                  $begingroup$
                  This returns true for 2207, which has a binary representation of 100010011111
                  $endgroup$
                  – Embodiment of Ignorance
                  27 mins ago












                  $begingroup$
                  @EmbodimentofIgnorance That is the correct result, is it not? The second hand is 00100.
                  $endgroup$
                  – Doorknob
                  21 mins ago




                  $begingroup$
                  @EmbodimentofIgnorance That is the correct result, is it not? The second hand is 00100.
                  $endgroup$
                  – Doorknob
                  21 mins ago












                  $begingroup$
                  I don't speak Ruby. But why not ->n{n.to_s(32)=~/4/}?
                  $endgroup$
                  – tsh
                  18 mins ago






                  $begingroup$
                  I don't speak Ruby. But why not ->n{n.to_s(32)=~/4/}?
                  $endgroup$
                  – tsh
                  18 mins ago














                  $begingroup$
                  @tsh because I'm not as clever as you :)
                  $endgroup$
                  – Doorknob
                  14 mins ago




                  $begingroup$
                  @tsh because I'm not as clever as you :)
                  $endgroup$
                  – Doorknob
                  14 mins ago












                  $begingroup$
                  Forgive me if I'm not understanding the question, but isn't the first hand of 2207 10001, the second 00111, and the third 11? None of them have their middle finger only up
                  $endgroup$
                  – Embodiment of Ignorance
                  9 mins ago




                  $begingroup$
                  Forgive me if I'm not understanding the question, but isn't the first hand of 2207 10001, the second 00111, and the third 11? None of them have their middle finger only up
                  $endgroup$
                  – Embodiment of Ignorance
                  9 mins ago











                  2












                  $begingroup$


                  Perl 6, 16 bytes





                  {.base(32)~~/4/}


                  Try it online!



                  Checks if there is a 4 in the base 32 representation of the number. Returns either Nil as false or a Match containing a 4.



                  You can prove this by the fact that $2^5 = 32$ so each digit is the state of each hand.






                  share|improve this answer









                  $endgroup$


















                    2












                    $begingroup$


                    Perl 6, 16 bytes





                    {.base(32)~~/4/}


                    Try it online!



                    Checks if there is a 4 in the base 32 representation of the number. Returns either Nil as false or a Match containing a 4.



                    You can prove this by the fact that $2^5 = 32$ so each digit is the state of each hand.






                    share|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$


                      Perl 6, 16 bytes





                      {.base(32)~~/4/}


                      Try it online!



                      Checks if there is a 4 in the base 32 representation of the number. Returns either Nil as false or a Match containing a 4.



                      You can prove this by the fact that $2^5 = 32$ so each digit is the state of each hand.






                      share|improve this answer









                      $endgroup$




                      Perl 6, 16 bytes





                      {.base(32)~~/4/}


                      Try it online!



                      Checks if there is a 4 in the base 32 representation of the number. Returns either Nil as false or a Match containing a 4.



                      You can prove this by the fact that $2^5 = 32$ so each digit is the state of each hand.







                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered 11 mins ago









                      Jo KingJo King

                      23.7k257123




                      23.7k257123























                          1












                          $begingroup$

                          Regex (ECMAScript), 37 bytes



                          ^((?=(x+)(2{31}x*))3)*(x{32})*x{4}$



                          Try it online!



                          ^
                          (
                          (?=(x+)(2{31}x*)) # 2 = floor(tail / 32); 3 = tool to make tail = 2
                          3 # tail = 2
                          )* # Loop the above as many times as necessary to make
                          # the below match
                          (x{32})*x{4}$ # Assert that tail % 32 == 4





                          share|improve this answer











                          $endgroup$


















                            1












                            $begingroup$

                            Regex (ECMAScript), 37 bytes



                            ^((?=(x+)(2{31}x*))3)*(x{32})*x{4}$



                            Try it online!



                            ^
                            (
                            (?=(x+)(2{31}x*)) # 2 = floor(tail / 32); 3 = tool to make tail = 2
                            3 # tail = 2
                            )* # Loop the above as many times as necessary to make
                            # the below match
                            (x{32})*x{4}$ # Assert that tail % 32 == 4





                            share|improve this answer











                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              Regex (ECMAScript), 37 bytes



                              ^((?=(x+)(2{31}x*))3)*(x{32})*x{4}$



                              Try it online!



                              ^
                              (
                              (?=(x+)(2{31}x*)) # 2 = floor(tail / 32); 3 = tool to make tail = 2
                              3 # tail = 2
                              )* # Loop the above as many times as necessary to make
                              # the below match
                              (x{32})*x{4}$ # Assert that tail % 32 == 4





                              share|improve this answer











                              $endgroup$



                              Regex (ECMAScript), 37 bytes



                              ^((?=(x+)(2{31}x*))3)*(x{32})*x{4}$



                              Try it online!



                              ^
                              (
                              (?=(x+)(2{31}x*)) # 2 = floor(tail / 32); 3 = tool to make tail = 2
                              3 # tail = 2
                              )* # Loop the above as many times as necessary to make
                              # the below match
                              (x{32})*x{4}$ # Assert that tail % 32 == 4






                              share|improve this answer














                              share|improve this answer



                              share|improve this answer








                              edited 14 mins ago

























                              answered 21 mins ago









                              DeadcodeDeadcode

                              1,7241419




                              1,7241419























                                  0












                                  $begingroup$


                                  Japt, 5 bytes



                                  sH ø4


                                  Try it online!



                                  Explanation



                                        // Implicit input
                                  sH // To a base-H (=32) string
                                  ø // Contains
                                  4 // 4 (JavaScript interprets this as a string)





                                  share|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$


                                    Japt, 5 bytes



                                    sH ø4


                                    Try it online!



                                    Explanation



                                          // Implicit input
                                    sH // To a base-H (=32) string
                                    ø // Contains
                                    4 // 4 (JavaScript interprets this as a string)





                                    share|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$


                                      Japt, 5 bytes



                                      sH ø4


                                      Try it online!



                                      Explanation



                                            // Implicit input
                                      sH // To a base-H (=32) string
                                      ø // Contains
                                      4 // 4 (JavaScript interprets this as a string)





                                      share|improve this answer









                                      $endgroup$




                                      Japt, 5 bytes



                                      sH ø4


                                      Try it online!



                                      Explanation



                                            // Implicit input
                                      sH // To a base-H (=32) string
                                      ø // Contains
                                      4 // 4 (JavaScript interprets this as a string)






                                      share|improve this answer












                                      share|improve this answer



                                      share|improve this answer










                                      answered 17 mins ago









                                      ASCII-onlyASCII-only

                                      3,4601236




                                      3,4601236






























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