Question

Introduction

Try to convert words to two different unicode fonts.

Challenge

Your task is to transform your input string into the and unicode characters.

All uppercase words should become lowercase words.

• For Example: WORD ->

All lowercase words should become words

• For example: other words ->

All mixed case words should remain unchanged

• For example: Mixed Case Words -> Mixed Case Words

Periods and spaces should remain unchanged.

Words are separated by spaces or periods

• For example (highlighted things are words):

Hello. This is a word. S.O. are these

• Inputs: A string containing letters, spaces, and periods ([A-Za-z .]+)
• Output: The formatted string

As this is a golfing challenge, the lowest number of bytes wins

Example Input and Output

Input:

This is an example STRING that c.o.U.L.d. be INPUTTED. It can CONTAIN multiple sentences.

Output:

This ..... . It .

Reference

Math Sans Bold: (characters 120302 through 120327)

Math Sans: (characters 120250 through 120275)

Answer 1

QuadR, 45 43 bytes

-2 thanks to ngn.

\w+
⎕UCS a+(2>≢b)×120153+84×⊃b←∪96>a←⎕UCS⍵M

Since TIO scrambles Unicode output from QuadR, here is a screenshot of using QuadR as an APL library in an interactive session:

---

\w+ replace words with the result of applying the following code to them:

⍵M the found word
⎕UCS the Universal Character Set code points of that
a← store that in a
96> 0 or 1 for whether 96 is greater than each of those
∪ take just the unique; [0] or [1] or [0,1] or [1,0]
b← store that in b
⊃ pick the first from that
84× multiply 84 with that
120153+ add 120153 to that
(…)× multiply the following with that:
 ≢b the tally (length) of b (1 if single-case, 2 if mixed-case)
 2> 0 or 1 for whether two is greater than that (1 if single-case, 0 if mixed-case)
a+ the original code points added to that
⎕UCS convert the resulting code points back to characters

Answer 2

APL (Dyalog Unicode), 63 57 53 bytes

-6 thanks to Erik the Outgolfer. -4 thanks to ngn.

Anonymous tacit prefix function.

'\w+'⎕R{⎕UCS a+(2>≢b)×120153+84×⊃b←∪96>a←⎕UCS⍵.Match}

Since TIO scrambles Unicode output from Dyalog APL, here is a screenshot of the code in action:

---

'\w+'⎕R PCRE Replace words with the result of applying the following…

{...} anonymous lambda:

 ⍵.Match the found word

 ⎕UCS the Universal Character Set code points of that

 a← store that in a

 96> 0 or 1 for whether 96 is greater than each of those

 ∪ take just the unique; [0] or [1] or [0,1] or [1,0]

 b← store that in b

 ⊃ pick the first from that

 84× multiply 84 with that

 120153+ add 120153 to that

 (…)× multiply the following with that:

  ≢b the tally (length) of b (1 if single-case, 2 if mixed-case)

  2> 0 or 1 for whether two is greater than that (1 if single-case, 0 if mixed-case)

 a+ the original code points added to that

 ⎕UCS convert the resulting code points back to characters

Answer 3

Clean, 268 265 232 224 bytes

As a neat bonus, this works with strings containing any character. Including nulls.

import StdLib,StdInt,StdBool,Text.Unicode,Text.Unicode.UChar
u=isUpper
l=isAlpha
$c|l c=fromInt(toInt c+120153+if(u c)84 0)=c
?[h,s:t]=[if(u h<>isLower s)($c)c\\c<-[h,s:t]]
?[h]=[$h]
@s=[y\\x<-groupBy(\a b=l a&&l b)s,y<- ?x]

Try it online!

Defines the function @, taking a UString and returning a UString

Answer 4

C, 292 characters, 448 bytes (in UTF-8)

char*t;s,i,k;p(l){for(l=s=*t/96,i=k=strlen(t);i--;)t[i]/96-s&&++l;for(l=l-s&&write(1,t,k);!l&++i<k;)write(1,s?""+t[i]*4-388:""+t[i]*4-260,4);}f(char*s){char b[strlen(s)];for(t=b;*s;++s)*s<47?(*t=0),p(t=b),putchar(*s):(*t++=*s);*t=0;p(t=b);}

Try it online!

Unrolled:

char*t;
s,i,k;

p(l)
{
    for (l=s=*t/96, i=k=strlen(t); i--;)
        t[i]/96-s && ++l;

    for (l=l-s&&write(1, t, k); !l&++i<k;)
        write(1, s ? ""+t[i]*4-388
                   : ""+t[i]*4-260, 4);
}

f(char*s)
{
    char b[strlen(s)];

    for (t=b; *s; ++s)
        *s<47 ? (*t=0), p(t=b), putchar(*s) : (*t++=*s);

    *t = 0;
    p(t=b);
}

Answer 5

Java 8, 221 219 203 201 bytes

s->{StringBuffer r=new StringBuffer();for(String x:s.split("(?<=[\\. ])|(?=[\\. ])"))x.codePoints().forEach(c->r.appendCodePoint(c+(x.matches("[A-Z]+")?120237:x.matches("[a-z]+")?120153:0)));return r;}

I have to use a StringBuffer instead of a regular String to use .appendCodePoint, unfortunately..

Explanation:

Try it online.

s->{                           // Method with String parameter and StringBuffer return-type
  StringBuffer r=new StringBuffer();
                               //  Resulting StringBuffer
  for(String x:s.split("(?<=[\\. ])|(?=[\\. ])"))
                               //  Split by space or dot, and keep them as separate items,
                               //  and loop over all those substrings
   x.codePoints().forEach(c->  //   Inner loop over the codepoints of that substring
      r.appendCodePoint(       //    Convert int to char, and append it to the result:
        c                      //     The next codepoint of the substring
        +(x.matches("[A-Z]+")? //     If the word is fully uppercase:
           120237              //      Add 120237 to convert it to Math Sans Bold
          :x.matches("[a-z]+")?//     Else-if the word is fully lowercase:
           120153              //      Add 120153 to convert it to Math Sans
          :                    //     Else (mixed case, or a dot/space)
           0)));               //      Leave the codepoint (and thus the character) as is
  return r;}                   //  Return the resulting StringBuffer

Answer 6

Haskell, 172 170 bytes

(s#w)r=[x|all(`elem`s)w,c<-w,(x,k)<-zip r s,c==k]
t[]=[]
t w=filter(>[])[['A'..'Z']#w$[''..],['a'..'z']#w$[''..],w]!!0
f s|(a,b:c)<-span(>'.')s=t a++b:f c|1>0=t s

Try it online!

Fairly straightforward. The # operator takes the set s of charcters (upper or lower case) the word w, and the math sans set r. It returns the word in the math sans font if all the characters in the word are in s or the empty list otherwise. The t function takes a word and tries all three possiblities (all upper, all lower, or mixed), returning the first one that isn't empty. The f function finds the first word by using span, transforming it with t and concatenating it with the separator (either . or space) and recurring on the rest of the string. The alternate case is for if span can't find a separator; we just transform the string.

Edit: Thanks to @Laikoni for taking off 2 bytes! I'm not used to the whole "operator that takes three arguments" thing

Answer 7

Jelly, 34 bytes

e€ØBŒg
ṁǵŒl,Œuiị“¡ẓƬ“¡ẓġ“’×Ç+OỌµ€

Try it online!

Full program.

Answer 8

Japt, 34 33 32 31 bytes

Includes an unprintable (charcode 153) after the last #.

rV="%b%A+%b"Èc+#x#í
rVv Èc+#x#

Try it

---

Explanation

                        :Implicit input of string U
r                       :Replace
   "%b%A+%b"            :/\b[A-Z]+\b/g
 V=                     :Assign ^that to variable V
            È           :Run each match through a function
             c          :Map over the codepoints of the current match
              +#x#í     :  Add 120237
\n                      :Assign the result of that replacement to variable U
rVv                     :Another replacement, this time with V lowercased to give us the RegEx /\b[a-z]+\b/g
    Èc+#x#              :And, again, map over the codepoints of each match, this time adding 120153 to each

---

Original 32 Byte Japt v2 Solution

r/\b(\A+|\a+)\b/Èc_+#x#+#T*(X¶u

Try it

r                                     :Replace
 /\b(\A+|\a+)\b/                      :...all matches of this RegEx (\A=[A-Z], \a=[a-z])
                È                     :Pass each match through a function, with X being the current match
                 c_                   :Pass the codepoints of X through a function
                   +                  :Add to the current codepoint
                    #x#               :120153 (there's an unprintable after the second #)
                        +#T           :Plus 84
                           *          :  Multiplied by
                            (X¶u      :  Is X equal to its uppercase self

Answer 9

Python 3, 173 122 120 bytes

lambda s:''.join(chr(ord(c)+120153*t.islower()+120237*t.isupper())for t in re.split(r'\b(\w+)\b',s)for c in t)
import re

-51 bytes from ShreevatsaR

-2 bytes from abccd

Try it online!

Splits on word boundaries (re.split(r'\b(\w+)\b',s)), then maps lowercase words to (+120153*t.islower()), and uppercase words to (+120237*t.isupper()), and leaves mixed-case words alone, then joins the words back up.

Ungolfed and un-lambda-ed:

def f(s):
    words = re.split(r'\b(\w+)\b', s)
    ret = ''
    for word in words:
        for char in word:
            if word.isupper():
                ret += chr(ord(c) + 120237)
            elif word.islower():
                ret += chr(ord(c) + 120153)
            else:
                ret += c
    return ret

Answer 10

Retina, 84 bytes

/\b[A-Z]+\b/_(`.
ĵ$&
)T`L`ۮ-܇
/\b[a-z]+\b/_(`.
ĵ$&
)T`l`ں-ۓ
T`ÿ-߿`퟿-

Try it online! Explanation: Retina is a .NET application and therefore works in UTF-16 internally. Unfortunately as the Math Sans characters aren't in the BMP I can't directly transliterate them because the number of code points differs. Worse, I can't use unpaired surrogates at all. Instead, I shift the appropriate words into characters in the range 0xFF-0x7FF which conveniently only take two bytes to encode, plus I also prefix them with the 0x135 character. Finally I map that range onto the a range that overlaps the unpaired surrogates, creating valid BMP pairs.

Answer 11

JavaScript (ES6), 99 114 113 bytes

s=>s.replace(/\b([A-Z]+|[a-z]+)\b/g,e=>e.replace(/./g,f=>String.fromCodePoint(f.charCodeAt()+120153+(f<'a')*84)))

(Thanks to @pfg for pointing out an important flaw in my first solution.)

-1 bytes thanks to @Neil.

Snippet:

let f = 

s=>s.replace(/\b([A-Z]+|[a-z]+)\b/g,e=>e.replace(/./g,f=>String.fromCodePoint(f.charCodeAt()+120153+(f<'a')*84)))

d.innerHTML=f('This is an example STRING that c.o.U.L.d. be INPUTTED. It can CONTAIN multiple sentences.');

<p id="d">

Answer 12

05AB1E, 33 bytes

€aγ€g£εÐ.u•1Ù„•*s.l•1Ùm•*+sÇ+çJ}J

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