Boolean to Text
 Included page "member:arimb" does not exist (create it now) Guest Wikidot community, I have an idea to make a program that takes a string of 1's and 0's in sets of five and translates them into letters so that A is 11111, B is 11110, C is 11101, D is 11100, all the way to Z. Then output this string of letters as a string. Please paste the text as a reply. Goodluck! Attachment: Reply | Edit | Link

 Included page "member:anonymous" does not exist (create it now) Guest Sooo… Reverse binary? Attachment: Reply | Edit | Link

 Included page "member:anonymous" does not exist (create it now) Guest If it is inverted binary, then this should do it: ``````:Delvar B :Input "String?",str1 :if 5≠length(str1 :stop For(Z,1,5 :B+1(sub(str1,Z,1)=2^(Z-1 end`````` should work :) Attachment: Reply | Edit | Link

 Included page "member:anonymous" does not exist (create it now) Guest Correction: ``````:Delvar B :Input "String?",str1 :if 5≠length(str1 :stop :For(Z,1,5 :B+1(sub(str1,6-Z,1)=2^(Z-1 :end Ans→X //not sure if needed, replace the x's on the following lines with Ans if it's not :If X<5 or X=31 :Stop :Sub("ABCDEFGHIJKLMNOPQRSTUVWXYZ",X,1`````` Attachment: Reply | Edit | Link

 Included page "member:anonymous" does not exist (create it now) Guest could you please explain what this does? Because I do not think we are thinking of the same thing Attachment: Reply | Edit | Link

 Included page "member:arimb" does not exist (create it now) Guest Oops…i forgot to login. The above post is mine Attachment: Reply | Edit | Link

 Included page "member:anonymous" does not exist (create it now) Guest It looks like what you're doing is inverted binary (instead of 1 being on and 0 being off, you have 1 represent off and 0 be on). The code I wrote allows you to input a 5 character string of 1's and 0's and it inverts it to give the intended answer (as you said, 11111 is A). Attachment: Reply | Edit | Link

 Included page "member:anonymous" does not exist (create it now) Guest Another correction (I really should get into the habit of testing my code more since I haven't written much recently >.> ``````:If X<5 or X=31 :Stop :Sub("ABCDEFGHIJKLMNOPQRSTUVWXYZ",X,1`````` should be ``````:If X<4 :Stop :output(2,1,Sub("ABCDEFGHIJKLMNOPQRSTUVWXYZ",32-X,1`````` Attachment: Reply | Edit | Link

 Included page "member:arimb" does not exist (create it now) Guest clarification: what i meant would look something like this ``````[input] [divide string into segments of 5] [compare segments of 5 to 11111 (A) all the way to 00110 (Z)] A:11111 B:11110 C:11101 ...11100,11011,11010,11001,11000,10111,10110,10101,10100,10011,10010,10001,10000,01111,01110,01101,01100,01011,01010,01001,01000,00111,00110 [add new letter to existing string] [output string once done]`````` Attachment: Reply | Edit | Link

 Moderator Guest ``````.2length(Str1→B For(A,1,B 0 For(C,-4,0 2Ans+expr(sub(Str1,5A+C,1 End Str1+sub("ZYXWVUTSRQPONMLKJIHGFEDCBA",Ans-5,1→Str1 End`````` I am sure there is a faster and smaller way, but I think that works (I didn't test it). Attachment: Reply | Edit | Link

 Included page "member:arimb" does not exist (create it now) Guest Xeda Elnara's code works, but it adds the answer to the end of the input. how could that be fixed? Attachment: Reply | Edit | Link

 Included page "member:anonymous" does not exist (create it now) Guest ``````" → Str0 .2length(Str1→B For(A,1,B 0 For(C,-4,0 2Ans+expr(sub(Str1,5A+C,1 End Str0+sub("ZYXWVUTSRQPONMLKJIHGFEDCBA",Ans-5,1→Str0 End`````` It stores it to Str0. Attachment: Reply | Edit | Link

 Moderator Guest Oh, sorry, I misread something and thought that you wanted it at the end of the string, sorry. You could add this to the end of my code to remedy that: ``sub(Ans,5B+1,int(B`` My code assumes that the input might not be a multiple of 5 digits. Attachment: Reply | Edit | Link